Tesi di Laurea — Diplomarbeit - UniTrentomatematita.science.unitn.it/4d/downloads/tesi.pdfTesi di Laurea — Diplomarbeit The concept of dimension: a theoretical and an informal

Tesi di Laurea — Diplomarbeit

The concept of dimension:a theoretical and an informal approach

Ester Dalvit

Relatore BetreuerProf. Dr. Domenico Luminati Prof. Dr. Jurgen Hausen

Anno Accademico 2006–2007

Contents

Introduction vii

Introduzione xi

Einleitung xv

I A survey on dimension 1

1 Dimension of manifolds 31.1 Basic concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Tangent space . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Sard’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Some results in Algebraic Topology . . . . . . . . . . . . . . . 151.5 Invariance of domain . . . . . . . . . . . . . . . . . . . . . . . 16

2 Dimension of affine varieties 212.1 Basic concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.2 Affine tangent spaces . . . . . . . . . . . . . . . . . . . . . . . 272.3 Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.4 Regular and singular points . . . . . . . . . . . . . . . . . . . 362.5 Transcendence degree . . . . . . . . . . . . . . . . . . . . . . . 382.6 Morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.7 Krull dimension . . . . . . . . . . . . . . . . . . . . . . . . . . 452.8 Zariski dimension . . . . . . . . . . . . . . . . . . . . . . . . . 472.9 Krull dimension for rings . . . . . . . . . . . . . . . . . . . . . 49

3 Dimension of metric spaces 533.1 Three concepts of dimension . . . . . . . . . . . . . . . . . . . 533.2 Some basic theory . . . . . . . . . . . . . . . . . . . . . . . . . 573.3 The large inductive dimension . . . . . . . . . . . . . . . . . . 603.4 Sum theorem for Ind . . . . . . . . . . . . . . . . . . . . . . . 66

3.5 Subspace and decomposition theorems for Ind . . . . . . . . . 693.6 Product theorem for Ind . . . . . . . . . . . . . . . . . . . . . 703.7 Covering dimension . . . . . . . . . . . . . . . . . . . . . . . . 713.8 Coincidence theorems . . . . . . . . . . . . . . . . . . . . . . . 803.9 The dimension of the Euclidean space . . . . . . . . . . . . . . 85

II Informal learning of mathematics 89

4 An exhibition about the 4D space 914.1 Some basic definitions . . . . . . . . . . . . . . . . . . . . . . 924.2 Contents of the exhibition . . . . . . . . . . . . . . . . . . . . 94

4.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 944.2.2 Models . . . . . . . . . . . . . . . . . . . . . . . . . . . 944.2.3 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . 974.2.4 Slices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 974.2.5 Dice . . . . . . . . . . . . . . . . . . . . . . . . . . . . 984.2.6 Flip a cube . . . . . . . . . . . . . . . . . . . . . . . . 984.2.7 Polytopes . . . . . . . . . . . . . . . . . . . . . . . . . 98

4.3 Flip a cube . . . . . . . . . . . . . . . . . . . . . . . . . . . . 994.3.1 Animations and images . . . . . . . . . . . . . . . . . . 1014.3.2 Flip a square . . . . . . . . . . . . . . . . . . . . . . . 1034.3.3 Flip a cube . . . . . . . . . . . . . . . . . . . . . . . . 106

4.4 Other animations . . . . . . . . . . . . . . . . . . . . . . . . . 1104.4.1 Folding up a (hyper)cube . . . . . . . . . . . . . . . . . 1104.4.2 Projections . . . . . . . . . . . . . . . . . . . . . . . . 1124.4.3 Hypersphere . . . . . . . . . . . . . . . . . . . . . . . . 117

4.5 Future work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

Acknowledgments 121

Bibliography 123

List of Figures

1.1 Charts and a chart transformation. . . . . . . . . . . . . . . . 4

4.1 The unfoldings of a cube and a hypercube. . . . . . . . . . . . 954.2 Parallel projections of a cube and a hypercube. . . . . . . . . 954.3 Schlegel diagram of a cube and a hypercube. . . . . . . . . . . 954.4 A model of the two tori of a hypercube (image by matematita). 964.5 The rotation axes of the cube in a kaleidoscope. . . . . . . . . 974.6 The cube present in the exhibition. . . . . . . . . . . . . . . . 1004.7 How to flip a square on the surface of a cube. . . . . . . . . . 1014.8 Flipping a square on a cube. . . . . . . . . . . . . . . . . . . . 1034.9 Folding up the cube to see the square. . . . . . . . . . . . . . 1044.10 Folding up the cube to see the square. . . . . . . . . . . . . . 1054.11 A square seems a trapezium in the Schlegel diagram. . . . . . 1064.12 A square seem a line segment in the Schlegel diagram. . . . . . 1064.13 Flipping a cube on a hypercube. . . . . . . . . . . . . . . . . . 1074.14 Changing the unfolding of the hypercube. . . . . . . . . . . . 1074.15 How to flip a cube on an (unfolded) hypercube. . . . . . . . . 1084.16 How to flip a cube on a hypercube (Schegel diagram). . . . . . 1094.17 Folding a cube and a hypercube. . . . . . . . . . . . . . . . . . 1104.18 How to fold a cube in the 3D space. . . . . . . . . . . . . . . . 1114.19 How to fold a hypercube in the 4D space. . . . . . . . . . . . . 1124.20 Projections of a hypercube. . . . . . . . . . . . . . . . . . . . 1134.21 Projections of a cube. . . . . . . . . . . . . . . . . . . . . . . . 1144.22 Projections of a hypercube. . . . . . . . . . . . . . . . . . . . 1154.23 A (deformed) sphere inside the solid tori of the hypersphere. . 1174.24 The sphere deformation in the hypersphere. . . . . . . . . . . 118

Introduction

In this work we analyze the concept of dimension from different viewpointsand we describe the preparation of an exhibition about the four dimensionalEuclidean space.

The concept of dimension plays a central role in many fields of mathe-matics, especially in geometry, being one of the first tools used to classifyobjects. In general, it is important to find some properties that will be usedto define and characterize the different classes of classes. The dimension isoften one of the first properties that are investigasted.

The most natural class of spaces one can refer to is the class of Euclideanspaces. When one gives a definition of dimension for a class of spaceswhich contains the Euclidean spaces, it is then important to verify that thisdefinition coincides with the usual one on Euclidean spaces.

The definitions of dimension are often given independently in the differentfields of mathematics. It is important to verify that they coincide on spacesthat can be considered as subjects of different fields. Nevertheless, even inthe cases where the coincidence of different definitions is a well known fact,the proofs are often difficult to be found in the literature.

Even if in mathematics one defines and studies spaces with an arbitrarynumber of dimensions — the dimension can be a natural or real numberor even infinite —, most people have only an intuitive notion of dimension,which derives from the experience. It is then natural to think of objects ashaving length, width, and height and also distinguish the different natureof linear, plane and solid objects. The relevant dimension of the thread isits length, while to measure a sheet one needs length and height, and for adie also a third dimension is to be considered. In some sense, we considerthe thread as one dimensional and the sheet as two dimensional, even ifboth objects are three dimensional. This is a first “intuitive and informalformalization” of the concept of dimension and is already clear to children.We can exploit this notion to give an idea of the properties of four dimensionalobjects and, reflecting on similarities and differences between linear, planeand solid objects, it is possible to “jump” to the four dimensions.

viii Introduction

The concept of n-dimensional space (n ≥ 4) is not easy to be illustrated,especially because people often try to find representations through drawings,but it can be very hard to find efficacious representations of objects in ahigher dimension and also to interpret and understand them.

Nevertheless, imagination and fantasy of many authors have beenfascinated by the four dimensional space and the literature has shown interesttowards the concept of dimension since the late 19th century. One of the firstworks about the concept of dimension and perhaps the most famous, whichhas inspired many other literary works, is Flatland: A Romance of ManyDimensions, published by Edwin A. Abbott in 1884. The author uses thenovel to criticize the society of his time, but this work is very interesting alsofrom the mathematical point of view. The protagonist is a square, livingin a two dimensional world, which gets in touch with a sphere and makes ajourney in the three dimensional world. In this manner the author presentsthe three dimensional objects from the point of view of the square, which isat first sceptical about the existence of a third dimension. This is perhapsthe first time that the general public is invited to reflect on the concept ofdimension and on the possibility that other universes with a different numberof dimensions can exist or at least can be imagined.

In the same period Charles H. Hinton published two “scientific romances”,as he called them, What is the Fourth Dimension? and A Plane World, wherehe presents some ideas about the concept of dimension. He also invented thewords ana and kata (from the Greek “up” and “down”), to indicate the twoopposite directions in the fourth dimension, the equivalents of left and right,forwards and backwards, and up and down. Vinn and vout are other wordswith the same meaning of ana and kata, coined by the science fiction authorRudy Rucker in his 2002 book Spaceland, a story in which a man meets a fourdimensional being. Henry More, an English philosopher who lived in the 17thcentury, had already given a name to the fourth spatial dimension, callingit spissitude. He thought that the spiritual beings are four dimensional andthey can reduce their extension in the first three dimensions to become largerin the fourth dimension.

Many other short stories and novels dealing with the concept of di-mensions have been written. Ian Stewart indagates various concepts ofdimensions in his 2001 book Flatterland. The Planiverse is a novel byAlexander K. Dewdney, written in 1984, which describes a two dimensionalworld, its physics, biology and chemistry. The novel Sphereland was writtenin 1965 by Dionys Burger as a sequel of Flatland and indagates the sphericalgeometry, to arrive to the concept that our universe is the hypersurface of ahypersphere. Robert A. Heinlein’s novel “–And He Built a Crooked House–”was written in 1941 and talks about an adventure in a hypercubic house.

Introduction ix

The science fiction, both in films and books, often uses the concept of“hyperspace” to give an atmosphere of mistery and suggestion. Exploitingthe people’s curiosity for these topics makes perhaps possible to draw theattention of the general public to an exhibition on four dimensional objectsand different ways of representing and visualizing them. Suggestive ideaslike hyperspace are powerful tools to make the visitor be interested inthe exhibition and many related topics can be presented and explained,for example projections, symmetries, rotations, polyhedra and polytopes,sections, knots and links. This is the aim of an exhibition which is beingprepared by the interuniversitary research centre for the communication andinformal learning of mathematics matematita. A part of the work for thisthesis was done within the project of the exhibition.

Many ideas of this work were inspired by the work of Thomas Banchoff,in particular the animations about the hypercube and the hypersphere.Banchoff uses electronic media to support his mathematical researches and asa powerful tool in mathematical educational. He developed many web-basedprojects1 to use computer graphics for education and informal learning ofmathematics. Also his book Beyond the Third Dimension and the web pages2

with the same title, which contain a virtual exhibition about mathemticalobjects, recall and explain the concept of dimension.

The present work is divided into two parts: in the first one, consisting ofthree chapters, we present various definitions of dimension, giving attentionto different fields of mathematics.

In the first chapter we consider the differential topology and in particularwe recall some basic theory about manifolds and the Sard’s Theorem. Withthese tools it can be proved that spheres and Euclidean spaces with differentdimensions can not be homeomorphic. Using homology theory, these andother results can be obtained more directly. In particular it can be shownthat, if n 6= m, then the Euclidean spaces with dimension n and m can not belocally homeomorphic (Theorem on Invariance of Domain). Using the latterresults, it can be shown that the dimension of topological manifolds is welldefined.

In chapter 2 we pass to consider affine algebraic varieties. Here weinvestigate the dimension of an algebraic set defined as dimension of the affinetangent space and the correlation between this concept and derivatives. Then

1Some of Thomas Banchoff’s projects are listed and linked in the web page

http://www.math.brown.edu/TFBCON2003/mathematics/. Other interesting web pages

about some four dimensional objects by Thomas Banchoff and Davide Cervone can be

found at http://www.math.union.edu/~dpvc/math/4D/.2http://alem3d.obidos.org/en/, realized by Banchoff and Cervone.

x Introduction

we recall the concepts of regular and singular points and show the connectionof this concept with the dimension. Finally the dimension as transcendencedegree and the Krull dimension are presented. It is proved that all thesedefinitions of dimension coincide. To conclude the chapter we outline theproof of the connections between this characterization of the dimension andthe stalks of holomorphic functions.

In chapter 3 three different definitions of dimension are presented formetric spaces. It is proved that the different definitions coincide on alarge class of metric spaces, and that the n-dimensional Euclidean spacehas dimension n also in the sense of metric space.

The second part of this work contains a chapter, where we describe thework done for the preparation of an exhibition about some four dimensionalobjects, which has taken place in Genova during the Festival della Scienza,from the 25th October to the 6th November 2007. We designed and developedsome interactive animations and animated images to illustrate some modelsof 4D objects and to explain some topics, like projections and rotations. Theanimations have been organized in some web pages, which can be found athttp://matematita.science.unitn.it/4d/ and, together with the sourcecode, in the CD–ROM included to this thesis.

Introduzione

In questo lavoro verra analizzato da vari punti di vista il concetto didimensione e sara descritta la preparazione di una mostra sullo spazioeuclideo quadridimensionale.

Il concetto di dimensione ha un ruolo centrale in molti campi dellamatematica, a cominciare dalla geometria, dove e uno dei primi strumentiutilizzati per classificare gli oggetti. In generale si rivela utile cercare quelleproprieta che servono per definire e caratterizzare le diverse classi di spazi;la dimensione e spesso una delle prime proprieta che si considerano.

La classe di spazi piu semplice e naturale da considerare e formata daglispazi euclidei. Quando viene definita la dimensione per una classe di spazidi cui fanno parte anche gli spazi euclidei, e importante verificare che essacoincida con la solita dimensione di spazio euclideo, cioe che Rn abbiadimensione n in ogni senso.

La dimensione viene spesso introdotta in modo indipendente nei diversicampi della matematica e si pone allora il problema di verificare che questedefinizioni coincidano sugli spazi che possono essere considerati oggetto distudio di diverse discipline. Spesso, anche nei casi in cui l’equivalenzadelle diverse definizioni e nota, le dimostrazioni non si trovano facilmentein letteratura.

Anche se in matematica si definiscono e si studiano spazi di dimensionearbitraria — la dimensione puo essere un qualsiasi numero naturale o realeo anche infinita —, la maggior parte delle persone ha soltanto un’ideaintuitiva di dimensione, idea legata alla propria esperienza. E percio naturaleconsiderare oggetti che abbiano una lunghezza, una larghezza e un’altezza eanche distinguere tra oggetti di natura diversa: oggetti lineari, piani o solidi.Per un filo la dimensione rilevante e la lunghezza, per un foglio di carta lo eanche la larghezza, mentre per un dado bisogna considerare anche una terzadimensione, l’altezza. In qualche senso, il filo e 1-dimensionale e il foglio2-dimensionale, anche se in realta entrambi hanno tre dimensioni. Questanozione intuitiva e informale di dimensione ci appartiene fin da bambini e sudi essa si puo far leva per far intuire quali possano essere le proprieta di un

xii Introduzione

oggetto a quattro dimensioni. Si puo riflettere su somiglianze e differenze traoggetti a una, due, tre dimensioni e quindi per analogia compiere il “salto”nello spazio a quattro dimensioni.

Il concetto di spazio n-dimensionale (n ≥ 4) non e semplice da spiegare,in particolare perche le persone tendono a cercare rappresentazioni grafiche,ma non e facile ne trovare immagini efficaci di oggetti multidimensionali neinterpretarle quando ci vengono proposte.

Nonostante questo, lo spazio quadridimensionale ha esercitato ed esercitaun grande fascino sull’immaginazione e sulla fantasia e la letteraturasi e interessata al concetto di dimensione gia a partire dalla fine deldiciannovesimo secolo. Una delle prime opere sul concetto di dimensione,e forse la piu famosa, che ha ispirato molti altri racconti e romanzi, eFlatland: A Romance of Many Dimensions (Flatlandia: racconto fantasticoa piu dimensioni), pubblicato da Edwin A. Abbott nel 1884. L’autore siserve del racconto per criticare la societa del suo tempo, ma il suo lavoro emolto interessante anche dal punto di vista matematico. Il protagonistae un quadrato che vive in un mondo a due dimensioni e che incontrauna sfera, con la quale compie un viaggio nel mondo tridimensionale. Inquesto modo l’autore presenta gli oggetti tridimensionali dal punto di vistadel quadrato, all’inizio molto scettico riguardo all’esistenza di una terzadimensione. Questa e forse la prima volta che il grande pubblico e invitatoa riflettere sul concetto di dimensione e sulla possibilita che universi con undiverso numero di dimensioni possano esistere o perlomeno essere immaginati.

Nello stesso periodo Charles H. Hinton pubblica i due “racconti scientifici”— cosı chiamati da lui stesso — What is the Fourth Dimension? (Che cos’ela quarta dimensione?) e A Plane World (Un mondo piano), in cui presentaalcune idee sul concetto di dimensione. Egli inventa anche le parole ana e kata(dal greco “su” e “giu”), per indicare le due direzioni opposte nella quartadimensione, in aggiunta a destra-sinistra, avanti-indietro e su-giu. Vinn evout sono altre parole con lo stesso significato di ana e kata, coniate dalloscrittore di fantascienza Rudy Rucker nel suo libro Spaceland, pubblicato nel2002, che racconta un incontro fantascientifico tra un uomo e un essere aquattro dimensioni. Gia un filosofo inglese del diciassettesimo secolo, HenryMore, aveva dato un nome alla quarta dimensione spaziale, chiamandolaspissitude. Secondo il suo pensiero, le quattro dimensioni sono proprie deglispiriti, che possono diminuire la propria estensione nelle tre dimensioni peracquistare maggior estensione nella quarta.

Sono stati scritti molti altri racconti e romanzi che hanno a che farecon il concetto di dimensione. Ian Stewart parla di diversi concetti didimensione nel libro Flatterland, pubblicato nel 2001. The Planiverse e unracconto di Alexander K. Dewdney, scritto nel 1984, che descrive un mondo

Introduzione xiii

bidimensionale, la sua fisica, biologia e chimica. Il racconto Sphereland,pubblicato nel 1965 da Dionys Burger, e un seguito di Flatland ed esamina lageometria sferica, per arrivare all’idea che il nostro universo e l’ipersuperficiedi un’ipersfera. Il racconto di Robert A. Heinlein intitolato “—And HeBuilt a Crooked House—” (La casa ipercubica) e stato scritto nel 1941 eracconta un’avventura fantascientifica in una casa che si richiude nella quartadimensione, assumendo la forma di un ipercubo.

La fantascienza, sia nei libri che nei film, usa spesso il concetto di“iperspazio” per creare un’atmosfera di mistero e suggestione. Sfruttando lacuriosita del pubblico per questi argomenti suggestivi e forse possibile richia-mare visitatori ad una mostra che racconti alcuni oggetti quadridimensionalie vari modi di rappresentarli e visualizzarli. Si possono allora presentare esviluppare alcuni argomenti riguardanti il concetto di dimensione, come adesempio proiezioni, simmetrie, rotazioni, poliedri e politopi regolari, sezionie nodi. Questo e lo scopo di una mostra che il Centro interuniversitario perla comunicazione e l’apprendimento informale della matematica matematitasta preparando e nel cui ambito e stata realizzata una parte del lavoro ditesi.

Molte delle idee presenti in questo lavoro, in particolare le animazioni suipercubi e ipersfere, sono state ispirate dal lavoro di Thomas Banchoff, unofra i primi ad usare la computer graphics sia nella ricerca matematica chenell’insegnamento. Egli ha sviluppato molti progetti sul web3, sfruttandola grafica tridimensionale per l’insegnamento informale della matematica.Anche il suo libro Beyond the Third Dimension (Oltre la terza dimensione)e le pagine web4 con lo stesso titolo, che contengono una mostra virtualedi oggetti matematici, richiamano e cercano di spiegare il concetto didimensione.

Il lavoro presente e diviso in due parti: nella prima, che contiene trecapitoli, vengono presentate varie definizioni di dimensione, considerandodiversi campi della matematica.

Nel primo capitolo viene presa in esame la topologia differenziale e inparticolare si richiama la teoria di base delle varieta differenziabili e il teoremadi Sard. Con questi strumenti si puo dimostrare che sfere e spazi euclidei didimensioni diverse non possono essere omeomorfi. Con la teoria omologicaquesti e altri risultati vengono ottenuti piu direttamente; in particolare si

3Alcuni links ai progetti di Banchoff si possono trovare nella pagina web

http://www.math.brown.edu/TFBCON2003/mathematics/. Altre pagine riguardanti

oggetti quadridimensionali, scritti da Thomas Banchoff e Davide Cervone, si possono

visionare all’indirizzo http://www.math.union.edu/~dpvc/math/4D/.4http://alem3d.obidos.org/en/ realizzato da Banchoff e Cervone.

xiv Introduzione

prova che, se n 6= m, due spazi euclidei di dimensione n e m rispettivamentenon possono essere nemmeno localmente omeomorfi (teorema di invarianzadel dominio). Questi risultati assicurano che la dimensione delle varietatopologiche e ben definita.

Nel capitolo 2 vengono considerate le varieta algebriche affini. Dapprimaviene esaminata la dimensione degli insiemi algebrici, definita come dimen-sione dello spazio tangente affine, e la correlazione di questo concetto con lederivazioni. Quindi viene richiamata la nozione di punti regolari e singolari,spiegando come essa abbia a che fare con la dimensione. Infine vengonopresentate la dimensione come grado di trascendenza e la dimensione di Krulle viene dimostrato che tutte queste definizioni di dimensione coincidono. Perconcludere il capitolo, viene illustrato uno schema di dimostrazione dellarelazione tra questa caratterizzazione di dimensione e i fasci di funzioniolomorfe.

Nel capitolo 3 vengono introdotte tre diverse definizioni di dimensione pergli spazi metrici. Si dimostra che esse coincidono su un’ampia classe di spazimetrici e che lo spazio euclideo Rn ha dimensione n anche secondo questedefinizioni.

La seconda parte di questo lavoro contiene un capitolo che descrivela preparazione di una mostra su alcuni oggetti quadridimensionali, chee stata allestita a Genova in occasione del Festival della Scienza, dal 25ottobre al 6 novembre 2007. Sono state progettate e sviluppate alcuneanimazioni interattive e immagini animate per illustrare alcuni modelli dioggetti quadridimensionali e per spiegare qualche argomento correlato, comeproiezioni e rotazioni. Le animazioni sono state organizzate in pagine web,disponibili all’indirizzo http://matematita.science.unitn.it/4d/ e nelCD–ROM allegato alla tesi, dove si trova anche il codice sorgente delleanimazioni.

Einleitung

In dieser Arbeit wird der Dimensionalbegriff aus verschiedenen Gesicht-spunkte analysiert und die Vorbereitung einer Ausstellung uber den vierdi-mensionalen euklidischen Raum wird beschrieben.

Der Begriff der Dimension spielt eine zentrale Rolle in vielen Beriche derMathematik, insbesondere in der Geometrie und ist ein der ersten Mittelzur Klassifikation von Objekte. Es ist namlich wichtig, die Eigenschaftenzu untersuchen, die man braucht, um die verschiedene Klassen von Raumezu definieren und kennzeichnen; die Dimension ist oft eine der erstenbetrachteten Eigenschaften.

Die einfachste und naturlichste Klasse von Raume, die man betrachtenkann, besteht aus den euklidischen Raume. Wenn die Dimension fur eineKlasse definiert wird, die auch die euklidische Raume enthalt, ist es dannwichtig zu prufen, dass die neue Definition mit der ubliche fur euklidischeRaume ubereinstimmt, d.h. Rn soll in jedem Sinn die Dimension n besitzen.

Die Dimension wird oft unabhangig in jedem Bereich eingefuhrt und dannentsteht das Problem, ob die Definitionen uber die Raume ubereinstimmen,die als Objekte verschiedener Bereiche betrachtet werden konnen. In vielenFallen hat diese Frage eine positive Antwort, aber der Beweis ist oft nichteinfach zu finden in der Literatur.

Obwohl in der Mathematik Raume beliebiger Dimension betrachtetwerden — die Dimension kann eine beliebige naturliche oder reelle Zahl sein,oder auch unendlich —, meistens haben Leute nur einen intuitiven Begriffder Dimension, der aus der praktischen Erfahrung stammt. Es ist namlichnaturlich, die Lange, die Breite und die Hohe der Gegenstande zu betrachten,und auch unter Sachen verschiedener Natur zu unterscheiden: lineare, flacheoder raumliche Sachen. Die bedeutende Dimension eines Fadens ist dieLange, fur ein Blatt ist auch die Breite wichtig, und fur ein Wurfel ist aucheine dritte Dimension zu betrachten, die Hohe. Irgendwie betrachten wir denFaden als eindimensional und das Blatt als zweidimensional, obwohl beideGegenstande drei Dimensionen besitzen. Diesen intuitiven und informellenBegriff der Dimension besitzen wir schon als Kinder, und er kann benutzt

xvi Einleitung

werden, um eine Idee zu geben, welche Eigenschaften ein vierdimensionalesObjekt besitzt. Es ist moglich, uber Ahnlichkeiten und Unterschiedeunter ein-, zwei- und dreidimensionale Gegenstande zu uberlegen und durchAnalogie in den vierdimensionalen Raum zu “springen”.

Es ist nicht einfach, den Begriff des n-dimensionalen Raumes informell zuerklaren, insbesondere weil Leute grafische Darstellungen suchen, aber es istnicht trivial weder wirksame Darstellungen zu finden noch sie interpretierenund verstehen.

Jedenfalls ubt der vierdimensionale Raum einen starken Einfluss aufdie Fantasie und die Imagination vieler Schriftsteller und die Literaturhat sich schon seit dem Ende des 19. Jahrhunderts fur dieses Themainteressiert. Eine der ersten literarischen Werke uber den Begriff derDimension, und vielleicht die beruhmteste, die viele Erzahlungen undRomane inspiriert hat, ist Flatland: A Romance of Many Dimensions, vonEdwin A. Abbott in 1884 veroffentlicht. Der Autor benutzt die Geschichte,um die Gesellschaft seiner Zeit zu kritisieren, aber dieses Werk ist auchaus dem mathematischen Gesichtspunkt interessant: der Protagonist ist einViereck, das in einer zweidimensionalen Welt lebt und eine Sphare trifft,mit der es eine Reise in der dreidimensionalen Welt macht. Der Autorbeschreibt die dreidimensionale Objekte aus der Sichtspunkt des Quadrats,das am Anfang skeptisch uber die Existenz einer dritten Dimension ist. Diebreite Offentlichkeit ist vielleicht zum ersten Mal eingeladen, uber den Begriffder Dimension zu uberlegen und uber die Moglichkeit der Existenz odermindestens der Vorstellung eines Universums mit einer anderen Anzahl vonDimensionen.

In der selben Zeit verofflicht Charles H. Hinton zwei sogennante “scientificromances” What is the Fourth Dimension? und A Plane World, wo ereinige Ideen uber den Begriff der Dimension vorstellt. Er erfindet dieWorter ana und kata (aus der alten griechischen Sprache “auf” und “unter”),um die gegenseitige Richtungen in der vierte Dimension zu kennzeichnen,analog zu links–rechts, vorwarts–ruckwarts, oben–unten. Vinn und vout sindgleichbedeutende Worter mit ana und kata, die von dem Science-Fiction-Schriftsteller Rudy Rucker gepragt wurden. In seinem Roman Spaceland,veroffentlicht in 2002, erzahlt Rucker, wie ein Mann ein vierdimensionalesLebewesen trifft. Schon der Philosoph Henry More, der im 17. Jahrhundertlebte, hatte einen Name zu der vierten Dimension gegeben: spissitude. Erdachte, die vier Dimensionen seien von den Geisten besitzt, die die erste dreiDimensionen kleiner machen konnten, um in die vierte sich zu erweitern.

Viele weitere Geschichte und Romane wurden geschrieben, die vomBegriff der Dimension handeln. Ian Stewart erzahlt von verschiedenenVorstellungen der Dimension im Buch Flatterland, veroffentlicht in 2001.

Einleitung xvii

The Planiverse ist eine Geschichte von Alexander K. Dewdney, geschriebenin 1984, die eine zweidimensionale Welt, seine Physik, Biologie und Chemiebeschreibt. , scritto nel 1984, che descrive un mondo bidimensionale, lasua fisica, biologia e chimica. Die Erzahlung Sphereland, veroffentlicht in1965 von Dionys Burger, ist eine Fortsetzung von Flatland und betrachtetdie spharische Geometrie, um zur Idee zu kommen, dass unsere Welt dieHyperflache einer vierdimensionalen Sphare ist. Die Geschichte “—And HeBuilt a Crooked House—”, von Robert A. Heinlein in 1941 veroffentlicht,erzahlt ein Science-Fiction-Abenteuer in einem Haus, das in der vierteDimension zusammenfaltet und die Form eines Hyperkubus annimmt.

Die Science-Fiction, sowohl in Bucher als auch in Filme, benutzt oftden Begriff von “Hyperrraum”, um eine geheimnisvolle und eindrucksvolleStimmung zu schaffen. Es ist vielleicht moglich, die Kuriositat des Pub-likums auszunutzen, um Besucher zu einer Ausstellung anzulocken, dieeinige vierdimensionale Gegenstande und deren verschiedene Darstellungeneinfuhrt. Man kann dann einige Themen vorstellen, die mit dem Begriffder Dimension verbunden sind, wie zum Beispiel Projektionen, Symmetrien,Rotationen, regelmassige Polyeder und Polytope, Schnitte und Knoten.Das ist das Ziel einer Ausstellung, die wird gerade vorbereitet von demUniversitatszentrum fur die Kommunikation und das informelles Lehrender Mathematik matematita. Im Rahmen dieses Projektes wurde Teil dervorliegenden Arbeit realisiert.

Viele in dieser Arbeit enthalten Ideen, insbesondere die Computeran-imationen uber Hyperkuben und Hyperspharen, wurden von der Arbeitvon Thomas Banchoff inspiriert, einer der ersten Mathematikern, der dieComputergrafik sowohl in der Forschung als auch im Lehren benutzt. Erhat mehrere Web-Projekte5 entwickelt, und benutzt die 3D Grafik fur dasinformelle Lehren der Mathematik. Sein Buch Beyond the Third Dimensionund die Webseiten 6 mit dem selben Titel, die eine virtuelle Ausstellunguber mathematische Objekte enthalten, erinnern und erklaren den Begriffder Dimension.

Die vorliegende Arbeit ist in zwei Teile geteilt: in dem ersten, der dreiKapitel enthalt, werden Definitionen von Dimension vorgestellt, die vonverschiedenen Bereiche der Mathematik untersucht werden.

Im ersten Kapitel wird die Differentialgeometrie betrachtet; insbesondere

5Einige Links zu den Banchoffs Projekte sind auf der Seite

http://www.math.brown.edu/TFBCON2003/mathematics/ zu finden. Weitere Webseiten

uber vierdimensionalen Objekte, von Thomas Banchoff und Davide Cervone, sind auf

http://www.math.union.edu/~dpvc/math/4D/ verfugbar.6http://alem3d.obidos.org/en/ realisiert von Banchoff und Cervone.

xviii Einleitung

werden die Grundlagen der Theorie fur differenzierbare Mannigfaltigkeitenund den Satz von Sard erinnert. Mit diesen Mittel kann man zeigen,dass Spharen und euklidische Raume unterschiedlicher Dimension nichthomoomorph sind. Mit der Homologie werden diese und weitere Ergeb-nisse unmittelbar erhalten. Insbesondere zeigt man, dass zwei euklidischeRaume, bzw. der Dimension n und m, wobei n 6= m, auch nicht lokalhomoomorph sein konnen. Diese Resultate sichern, dass die Dimension derMannigfaltigkeiten wohldefiniert ist.

Im Kapitel 2 werden algebraische affine Varietaten betrachtet. Zuerstwird die Dimension algebraischer Mengen analysiert, die definiert ist als Di-mension des affinen Tangentialraumes; dieser Begriff ist mit den Derivationenverbunden. Die Definition von regulare und singulare Punkte wird dann erin-nert und es wird erklart, was sie mit der Dimension zu tun hat. Dann werdendie Dimension als Transzendenzgrad und die Krulldimension vorgestellt undes wird gezeigt, dass alle Definitionen ubereinstimmen. Schliesslich wirdeine Beweisskizze der Korrelation zwischen diese Charakterisierungen derDimension und die Garbe holomorpher Funktionen vorgestellt.

Im Kapitel 3 werden drei Definitionen fur metrische Raume eingefuhrt.Es wird gezeigt, dass sie uber eine grosse Klasse von metrische Raumeubereinstimmen, und dass der euklidische Raum Rn auch in diesem Sinndie Dimension n besitzt.

Der zweite Teil dieser Arbeit beschreibt die Vorbereitung einer Ausstel-lung uber einige vierdimensionale Objekte, die in Genova eingerichtet wurde,im Rahmen des Festival della Scienza, vom 25. Oktober bis zum 6.November 2007. Einige Animation Bilder und interaktive Computeranima-tionen wurden geplant und entwickelt, um Modelle von vierdimensionalenObjekte zu erklaren und einige Themen wie Projektionen und Rotationenzu erlautern. Die Computeranimationen sind in Webseiten organisiert, dieauf http://matematita.science.unitn.it/4d/ und in der zu dieser Arbeitangelegten CD–ROM zu finden sind. In der CD–ROM ist auch den Quellcodeverfugbar.

Part I

A survey on dimension

Chapter 1

Dimension of manifolds

In this chapter we consider the dimension of manifolds, from the point ofview of the differential topology and the algebraic topology.

Recall that a topological manifold is a Hausdorff space, which is locallyhomeomorphic to some Rn and has a countable basis. The integer n is calledthe dimension of the manifold. However, the homeomorphism with Rn is alocal property, and one has to check that the n is the same for the wholemanifold.

It must also be proved that for each point there can not be twohomeomorphisms of some neighbourhoods to Euclidean spaces with differentdimension.

We will introduce the notion of regular points for a differentaible manifold,and prove the Sard’s Theorem 1.3.5. The latter result will be used to showthat spheres with different dimensions can not be homeomorphic.

Furthermore, using the homology theory, it can be proved that manifoldwith different dimensions are not locally homeomorphic. This is a conse-quence of the last result presented in this chapter, the Theorem on Invarianceof Domain 1.5.8, and proves that the dimension on manifolds is well defined.

1.1 Basic concepts

The objects considered in this chapter are topological manifolds, that meansHausdorff spaces, which are locally homeomorphic to some Rn and have acountable basis. The integer n is called the dimension of the manifold.

A chart (or coordinate system) of an n-dimensional manifold X is ahomeomorphism h : U → U ′ ⊆ Rn of an open set U ⊆ X (called chartdomain) onto an open set U ′ ⊆ Rn.

A collection of charts {hα}α∈A is an atlas for X if the union of the chart

4 Chapter 1. Dimension of manifolds

domains Uα covers the manifold,⋃

α∈A Uα = X.Given two charts hα : Uα → U ′

α ⊆ Rn and hβ : Uβ → U ′β ⊆ Rn, they are

both defined on the intersection of the domains Uα ∩ Uβ. By composition,we obtain a homeomorphism hαβ = hβ ◦ h−1

α : hα(Uα ∩Uβ) → hβ(Uα ∩Uβ) ofopen sets in Rn, which is called chart transformation.

X

Uα ∩ Uβ

Uα

hα

Rn

U ′α hα(Uα ∩ Uβ)

Uβ

hβ

Rn

hβ(Uα ∩ Uβ)

U ′βhαβ

Figure 1.1: Charts and a chart transformation.

An atlas of a manifold is called differentiable of class Ck if all its charttransformation are differentiable of class Ck.

A Ck-differentiable structure on a manifold is a maximal Ck-differentiableatlas. A Ck-differentiable manifold is a topological manifold together with aCk-differentiable structure.

A continuous map f : X → Y between Ck-differentiable manifolds is saidto be Ck-differentiable at a point p ∈ X if for some (and therefore for any)chart h : U → U ′ and k : V → V ′ with p ∈ U , f(p) ∈ V , the compositionk ◦ f ◦ h−1 is Ck-differentiable at the point h(p).

A Ck-diffeomorphism is an invertible Ck-differentiable map (that means,it is bijective and its inverse map is also Ck-differentiable).

In what follows we will suppose that all manifolds and maps are of classC∞ and we will refer to them simply as differentiable, or smooth.

Let X be a (n+ k)-dimensional differentiable manifold. A subset Y ⊆ Xis called an n-dimensional differentiable submanifold of X if for every point

1.2. Tangent space 5

p ∈ Y there exists a neighbourhood U ⊆ X and a chart around p

h : U → U ′ ⊆ Rn+k,

such that h(Y ∩U) = U ′∩Rn, where we consider Rn = Rn ×{0} ⊆ Rn ×Rk.The subset Y becomes a manifold, taking as an atlas all the maps h|Y ∩U ,where h is a chart of X. The number k is called the codimension of thesubmanifold.

A differentiable map f : X → Y between differentiable manifolds is anembedding if f(X) ⊆ Y is a differentiable submanifold, and f : X → f(X) isa diffeomorphism.

Note that every differentiable structure has a well defined dimension.In fact, there can not exist charts in the same differentiable structure withcodomains Rn and Rm respectively, if n 6= m, since there is no map Rn → Rm

which is invertible and differentiable.At this point, we could think that some manifold can have two differen-

tiable structures with different dimensions. We will see, as a consequence ofthe Theorem on Invariance of Domain 1.5.8, that this is never the case, andtherefore the dimension of every manifold is always well defined.

1.2 Tangent space

Given two manifolds X and Y , and a point p ∈ X with an open neighborhoodU ⊆ X, we say that two differentiable maps f, g : U → Y are equivalent ifthere is an open neighbourhood V ⊆ U of p such that f|V = g|V .

An equivalence class for this relation is called a germ of a map X → Y atp. We denote by f : (X, p) → Y or f : (X, p) → (Y, f(p)) the germ definedby the differentiable map f .

A function germ is a differentiable germ (X, p) → R. We denote by ε(p)the set of all function germs around p ∈ X. It is a real algebra. We willwrite εn for the set of germs (Rn, 0) → R.

Given a germ f : (X, p) → (Y, q), the mapping

f ∗ : ε(q) → ε(p)

φ 7→ φ ◦ f

is a contravariant functor between the algebras of all functions germs aroundthe points.

A derivation on ε(p) is a linear map δ : ε(p) → R which satisfies theproduct rule

δ(φ ◦ ψ) = δ(φ) ◦ ψ(p) + φ(p) ◦ δ(ψ).


Definition 1.2.1The tangent space TpX of a manifold X at a point p is the real vector spaceof the derivations of ε(p).

If δ is a derivation of ε(p), one can check that δ ◦ f ∗ is a derivation ofε(f(p)), hence we can define the mapping

Tpf : TpX → Tf(p)Y

δ 7→ δ ◦ f ∗,

called the differential of f at p, by setting

Tpf(δ)(φ) = δ ◦ f ∗(φ) := δ(φ ◦ f)

for each germ φ : (Y, q) → R.If h = (x1, . . . , xn) is a chart of X, among the derivations there are the

partial derivatives,∂

∂xi: εn → R

φ 7→ ∂

∂xiφ(0).

The following theorem holds (see [BJ82], Theorem (2.4)).

Theorem 1.2.2Let X and Y be manifolds of dimension n andm respectively. Introduce localcoordinates (x1, . . . , xn) around the point p ∈ X and (y1, . . . , ym) aroundq ∈ Y . Then the derivations ∂

∂xi, ∂

∂yjform vector space bases of TpX and

TqY respectively, and the tangential map of a germ f : (X, p) → (Y, q)with respect to these bases is given by the Jacobi matrix calculated in 0,Df0 : Rn → Rm.

Given a differentiable manifold of dimension n, the tangent space in eachof its points is a vector space of dimension n. Once again, we see that thedimension of the manifolds is well defined.

A differentiable germ f : (X, p) → (Y, q) is invertible if there is amapping f which defines the germ f and which maps a neighbourhood of pdiffeomorphically onto a neighbourhood of q.

For example, given an n-dimensional manifold X and a point p ∈ X,one can find a chart around p which defines an invertible germ h : (X, p) →(Rn, 0), and hence an isomorphism h∗ : εn → ε(p).

The following theorem (see [BJ82], Theorem (5.1)) holds.

1.2. Tangent space 7

Theorem 1.2.3 (Inverse Function Theorem)A differentiable germ is invertible if and only if its differential is anisomorphism.

The rank of a differentiable map f : X → Y at a point p ∈ X is

rkp f = rkTpf .

This is also the rank of the germ f : (X, p) → Y .The rank is lower semi-continuous, that means, if rkp f = r, then there is

a neighbourhood U of p such that rkq f ≥ r for all q ∈ U .A germ f : (X, p) → (Y, q) is said to have constant rank r if there is a

function f in the germ such that in some open neighbourhood of p the rankof f is constant.

Theorem 1.2.4 (Rank Theorem)Let f : (X, p) → (Y, q) be a germ of constant rank r, where X has dimensionn and Y has dimension m. Then, there are charts h around p and k aroundq such that the germ

k ◦ f ◦ h−1: (Rn, 0) → (Rm, 0)

is represented by the map

(x1, x2, . . . , xn) 7→ (x1, x2, . . . , xr, 0, . . . , 0).

For the proof, we refer to [BJ82], Theorem (5.4).

Definition 1.2.5Let f : X → Y be a C1 map.A point p ∈ X is regular or smooth if the differential Tpf is surjective.A point q ∈ Y is a regular value if all points of its preimage f−1(q) areregular.A point is critical or singular if it is non-regular.A differentiable map f : X → Y is called a submersion if rkp f = dimY forall p ∈ X, that is the same as to require that every point of X is regular, orthat every point of f(X) is a regular value.The map f is called an immersion if rkp f = dimX for all p ∈ X, that is thedifferential Tpf is injective at every point of X.

It is not hard to see that from the Rank Theorem, we can derive thefollowing theorem.


Theorem 1.2.6If f : X → Y is an injective immersion and f : X → f(X) a homeomorphism,where f(X) ⊆ Y has the subspace topology, then f is an embedding.

Proposition 1.2.7Let q be a regular value of the differentiable map f : X → Y , where dimX >dimY = n. Then f−1(q) is either empty or a differentiable submanifold ofX with codimension n.

For the proof see [Milnor78] §2, Lemma 1.

1.3 Sard’s Theorem

We will state and prove the Sard’s Theorem 1.3.5, which ensures that the setof critical values of a differentiable mapping of manifolds is “small” in somesense that we will see.

Definition 1.3.1A subset A ⊆ Rn has Lebesgue measure zero if for every ε > 0 it can becovered by countably many n-cubes of edge w with total n-volume

∑

i∈N wn <

ε.

Remark 1.3.2In the definition it does not matter if the cubes are open or closed, and wemay also take n-spheres instead of n-cubes.A countable union of sets of measure zero has again measure zero.

Proposition 1.3.3Let U be an open subset of Rn and A ⊆ U a subset of measure zero. Iff : U → Rm is differentiable, then f(A) has measure zero.

Proof. Since U is open, it is the union of a countable sequence of compactballs, and we may assume that A is contained in a compact ball, and that thecubes of a covering of A are also contained in a larger compact ball K ⊆ U .

By the mean value theorem, for [x, x+ h] ∈ K there is a constant c suchthat

| f(x+ h) − f(x) | ≤ c | h |If a cube W ⊆ K has edge length a, we have that | x − x′ | ≤ a

√n for

x ∈ W , and | f(x) − f(x′) | ≤ ca√n. Hence, f(W ) lies in a cube of volume

(2c√n)m |W | and, since the constant (2c

√n)m is independent of the cube,

the assertion follows.

1.3. Sard’s Theorem 9

This proposition allows us to speak of subsets with measure zero in adifferentiable manifold.

Definition 1.3.4A subset A of a differentiable manifold has measure zero if for every charth : U → Rn the image h(A ∩ U) ⊆ Rn has measure zero.

Theorem 1.3.5 (Sard’s Theorem)The set of critical values of a differentiable mapping of manifolds hasLebesgue measure zero.

In particular, by Proposition 1.2.7, if f : X → Rn is differentiable,then the preimage of a point is either empty or it is an n-codimensionalsubmanifold for almost all points of X.

Since a manifold has a countable basis for its topology, each atlas hasa subatlas with countable many charts. Applying Proposition 1.3.3 to thechart transformations, we obtain that A has measure zero if and only if forall charts hα of a fixed atlas, the image hα(A ∩ Uα) ⊆ Rn has measure zero.

Thanks to this result, in order to prove Sard’s Theorem, it is onlynecessary to consider subsets of Rn.

We also need another result, Fubini’s Theorem.

Theorem 1.3.6 (Fubini)Define

Rn−1t = {x ∈ Rn | xn = t},

let A be a compact subset of Rn and suppose At = A ∩ Rn−1t has measure

zero in Rn−1t

∼= Rn−1 for all t ∈ R. Then A has measure zero in Rn.

Proof. Since A is compact, we may assume that A ⊆ Rn−1 × I, where I =[0, 1], and that At has measure zero in Rn−1 ×{t}. Thus, for any ε > 0 thereis a countable covering of At, whose elements are open cubes W i

t ⊆ Rn−1t ,

such that∑

i∈N

|W it | < ε

Let Wt be the projection of ∪iWit on the first factor Rn−1 of Rn−1 × I. For

a fixed t, the function | xn − t | is continuous on A, and vanishes exactly onAt. Furthermore, outside of Wt × I, this function has a positive minumumvalue α, because A is compact and Wt is open.

Letting It be the interval (t− α, t+ α), we have

{x ∈ A | | xn − t | < α} ⊆Wt × It


The intervals It cover I, hence there is a finite subcovering {Ij}j=1,...,k withIj = Itj for some tj ∈ I. We may assume that their total volume

∑ | Ij |does not exceed 2, because we can chose the finite subcovering in such amanner that each point belongs to at most two open sets of the subcovering:if a point lies in more than two open sets, then there is one of them with asmallest initial point and one with a largest end point. We can cancel fromthe covering all other sets containing that point.

The sets {W itj× Ij | j = 1, . . . , k; i ∈ N} have total volume not exceeding

2ε and our set A is contained in their union. Hence, A has measure zero inRn.

It is not necessary that the subset A is compact, it is sufficient that itis a countable union of compact sets. Then, Fubini’s Theorem holds forclosed sets, open sets, images of open or closed sets under continuous maps,countable unions and finite intersections of open or closed sets.

Proof of Sard’s Theorem 1.3.5. We have to show the following:Let U ⊆ Rn be open, f : U → Rm differentiable and C ⊆ U the set of criticalpoints of f . Then f(C) ⊆ Rm has measure zero.

We proceed by induction on n. If n = 0, then Rn is a point, and f(C) isat most one point, and the theorem holds.

Suppose now the result is true for n − 1; we show it for n. Let Ci ⊆ Ube the set of points at which all partial derivatives of order up to i vanish.Then the sets Ci form a decreasing sequence of closed sets

C ⊇ C1 ⊇ C2 ⊇ . . . .

Note that C is a disjoint union

C = (C r C1) t (C1 r C2) t . . . t (Ck−1 r Ck) t Ck.

We will show that

(a) f(C r C1) has measure zero

(b) f(Ci r Ci+1) has measure zero for i ≥ 1

(c) there is a k such that f(Ck) has measure zero for all k ≥ k.

Then, C is a countable union of sets of measure zero, hence we are done.Note that all these sets are in the class to which Fubini’s Theorem is

applicable. Furthermore, it is sufficient to show that each point of the set D


we are considering (D = C r C1 or Ci r Ci+1 or Ck) has a neighbourhoodV such that f(V ∩ D) has measure zero, since each set D is covered bycountably many such neighbourhoods, because Rn is paracompact.

Now we prove (a).If m = 1, then C = C1 and the claim is trivial. Hence, assume m ≥ 2.

Let x ∈ C r C1, then some partial derivatives do not vanish at the point x.Assume ∂f1

∂x1(x) 6= 0. Then the map

h : U → Rn (x1, . . . , xn) 7→ (f1(x1), x2, . . . , xn)

is not singular at the point x. Therefore, the restriction of this map to aneighbourhood V of x is a chart h|V = h : V → V ′ and the transformed mapg = f ◦ h−1 : V ′ → Rm has the form

g(z1, . . . , zn) = (z1, g2(z), . . . , gm(z))

locally about h(x).

For a fixed t ∈ R, this map takes the hyperplane {z | z1 = t} into theplane {y | y1 = t}. Restrict g to

gt : ({t} × Rn−1) ∩ V ′ → {t} × Rm−1

Then a point in {t} × Rn−1 is critical for g if and only if it is critical for gt,because the Jacobi matrix of g is

1 0 . . . 0∗... Dgt

∗

By the induction hypothesis, the set of critical values of gt has measure zeroin {t} × Rn−1, hence the set of critical values of g has an intersection ofmeasure zero with each hyperplane {y | y1 = t}. Thus, by Fubini’s Theorem,also the set of critical values of g has measure zero in Rn, and (a) holds.

To prove (b), we proceed in the same way. For each point x ∈ Ck rCk+1

there is some derivative of order (k + 1), which does not vanish at the pointx. We may assume that

∂k+1f1

∂x1∂xν1. . . ∂xνk

(x) 6= 0


Define the function w : U → R, w = ∂kf1

∂xν1...∂xνk

. Then w(x) = 0 since x ∈ Ck,

and ∂w∂x1

(x) 6= 0 since x 6∈ Ck+1. The map

h : x 7→ (w(x), x2, . . . , xn)

defines a chart h : V → V ′ about x, and we have

h(Ck ∩ V ) ⊆ {0} × Rn−1 ⊆ Rn

Consider the transformed map g = f ◦ h−1 : V ′ → Rm. By the inductionhypothesis, its restriction

g0 : ({0} × Rn−1) ∩ V ′ → Rm

has a set of critical values of measure zero in Rm.Each point from h(Ck ∩V ) is critical for g0 because all partial derivatives

of g of order up to k vanish, and hence the same do the derivatives of g0.Thus, f(Ck ∩ V ) = g ◦ h(Ck ∩ V ) has measure zero.

To prove (c), suppose W ⊆ U be a cube with edges of length a and letk > n

m−1. We show that f(W ∩Ck) has measure zero. This will be sufficient,

since U is a countable union of cubes.For x ∈ Ck ∩W and x+ h ∈W , by the Taylor formula we have

f(x+ h) = f(x) + f (1)(x) + . . .+ f (k)(x)︸︷︷︸

0

+R(x, h),

with

|R(x, h) | ≤ f (k+1)(x1)

(k + 1)!| h | k+1 = c | h | k+1’

where c is a fixed constant depending only on f andW . DecomposeW into rn

cubes with edges of length ar

and suppose W1 is a cube of the decomposition,which contains a point x ∈ Ck. Then each point from W1 can be written asx+ h, where

| h | ≤ a

r

√n

since this is the length of the diagonal of the n-cube W1. From the remainderestimate of the Taylor formula, f(W1) lies in a cube with edges of length

2c | h | k+1 ≤ 2c(a

r

√n)k+1

=b

rk+1

where b is a constant dependng only on f and W and not on thedecomposition. Alle these cubes together have a total volume not exceeding

rn bm

rm(k+1)


and for m(k + 1) > n this volume converges to zero as r increases. Hence,chosing a decomposition into sufficiently small cubes, the total volume canbe made arbitrarily small.

Corollary 1.3.7The regular values of a differentiable map f : X → Y are dense in Y .

We prove in Corollary 1.3.10 that spheres with different dimensions cannot be homeomorphic. To do that, we use the Sard’s Theorem and thefollowing result.

Corollary 1.3.8Let f : X → Y be a differentiable map between manifolds of dimension nand m respectively, with n < m. Then f(X) has measure 0.

This follows directly from the fact that all points in f(X) are criticalvalues.

We recall the definitions of homotopy and smooth homotopy. Two mapsf, g : X → Y are called homotopic if there is a map F : X × I → Y suchthat

F (x, 0) = f(x) F (x, 1) = g(x)

for all x ∈ X. The map F is called a homotopy between f and g. Ifin addition the homotopy F is differentiable, then it is called a smoothhomotopy and f and g are said to be smoothly homotopic.

A mapping is called (smoothly) homotopically trivial if it is (smoothly)homotpic with a constant map.

Theorem 1.3.9Let n < m and

f : Sn → Sm ⊆ Rm+1

be a continuous function. Then f is homotopically trivial.

Proof. If f is differentiable, then it is not surjective by Corollary 1.3.8 andhence homotopically trivial.

We can extend f to an open subset U of Rn+1 with a continuous functionf : U → Sm. Then f|Sn = f .

Let K ⊆ U be a compact neighbourhood of Sn. Using an approximationtheorem (see for example [Hir76], chapter 2), we can approximate f on Kwith a polynomial function

g : K → Rm+1.


Then g|Sn : Sn → Rm+1 approximates f and since | f(x) | = 1, we mayassume 0 6∈ g(Sn).

Let g : Sn → Sm be defined through g(x) = g(x)| g(x) |

. It is well defined since

0 6∈ g(Sn) and it is differentiable, since g is a polynomial function. Definethe map

F : Sn × I → Sm

(x, t) 7→ (1 − t)f(x) + tg(x)

| (1 − t)f(x) + tg(x) | .

We claim that this is a homotopy between f and g. In fact, F (x, 0) =

f(x) and F (x, 1) = g(x)| g(x) |

= g(x) for each x ∈ Sn and we show that the

denominator is not 0. Since g|Sn approximates f , we can assume | f(x) −g(x) | < 1 for all x ∈ Sn. Furthermore, we know that | g(x) | 6= 0 and| f(x) | = 1 for all x ∈ Sn. By contradiction, suppose the denominator is 0.Then t 6= 0 and f(x) = t(f(x) − g(x)), hence

1 = | f(x) | = t | f(x) − g(x) | < t ≤ 1,

which is a contradiction. Hence | (1 − t)f(x) + tg(x) | 6= 0 and F is ahomotopy between f and g.

Since the map g : Sn → Sm is differentiable, it is not surjective, andhence homotopically trivial. Moreover, g is homotopic with f , hence f isalso homotopically trivial.

Corollary 1.3.10If n < m, then Sn and Sm are not homeomorphic.

Proof. Suppose by contradiction that f : Sn → Sm is a homeomorphism,then by Theorem 1.3.9, f is homotopically trivial and so is f−1 ◦ f = idSn .Hence, to find a contradiction, it suffices to show that the identity functionidSn is not homotopically trivial.

Let f : Sn → Sn be differentiable, and q a regular value. Then byProposition 1.2.7, f−1(q) is a submanifold of dimension 0, i.e. f−1(q) ={p1, . . . , pk}.The differential of f at a point pi is a map Tpi

f : TpiSn → TqS

n. Choose anorientation for TqS

n and define

ε(pi) =

{+1 if Tpi

Sn is consistently oriented with TqSn

−1 otherwise.

It can be proved that the number

dq(f) =k∑

i=1

ε(pi)

1.4. Some results in Algebraic Topology 15

does not depend on the choice of the regular value q. We call this numberthe degree of f , deg f . One can also prove that if f and g are smoothlyhomotopic, then deg f = deg g. Both proofs can be found in [Milnor78], §5.Furthermore, it is easy to see that deg id = 1 and deg c = 0 for each constantmap c. That means the identity map is not smoothly homotopically trivial.

To complete the proof, it can be shown that if idSn were homotopicallytrivial, then it would also be smoothly homotopically trivial.The proof usesapproximation on compact sets and is quite similar to the first part of theproof of Theorem 1.3.9.

1.4 Some results in Algebraic Topology

Many results stated so far in this chapter can be obtained more directly ifwe use the homology theory: that is what we do in the rest of the chapter.

We recall some results without giving the proofs, which can be found inthe book of Spanier [Spa66] or Greenberg [Gre81] for example.

Proposition 1.4.1A space X consisting in only one point is acyclic, that is H0(X) ∼= Z andHk(X) = (0) for k ≥ 1.

Proposition 1.4.2Let X be a topological space and X = ∪i∈IXi its decomposition in pathconnected components. Then for each k ≥ 0 holds:

Hk(X) =⊕

i∈I

Hk(Xi).

Proposition 1.4.3If X is a nonempty path connected space, then H0(X) ∼= Z.

Theorem 1.4.4 (Homotopy Theorem)Let f , g : X → Y be hotomopic maps between topological spaces. Then theinduced homomorphisms f∗, g∗ : Hk(X) → Hk(Y ) coincide for each k ≥ 0.

Theorem 1.4.5 (Mayer-Vietoris)Let U and V be open subsets of a topological space X such that X = U ∪V .Then, there is a long homology sequence

. . .ν−→ Hk+1(X)

∆−→ Hk(U ∩ V )µ−→ Hk(U) ⊕Hk(V )

ν−→ Hk(X)∆−→ . . .


Theorem 1.4.6 (Homology group of the sphere)For every n ≥ 1 holds:

Hk(Sn) =

{Z if k = 0 or k = n(0) otherwise.

Theorem 1.4.7Rn and Rm are homeomorphic if and only if n = m.

Proof. Assume there exists a homeomorpism f : Rn → Rm. Let p ∈ Rn

such that f(p) = 0 ∈ Rm. Then f|Rnr{p} : Rn r {p} → Rm r {0} is also ahomeomorphism. Since Rn r {p} has the same homotopy type as Sn−1, bythe Homotopy Theorem 1.4.4 we have

Hm−1(Sn−1) ∼= Hm−1(R

n r {p}) f∗−−−→∼=

Hm−1(Rm r {0}) ∼= Hm−1(S

m−1)

If m = 1, since H0(Sn−1) ∼= H0(S

0) ∼= Z ⊕ Z, it follows that n = 1.If m ≥ 2, since Hm−1(S

n−1) ∼= Hm−1(Sm−1) ∼= Z, by virtue of Theorem

1.4.6, it follows that n = m.

1.5 Invariance of domain

Definition 1.5.1A continuous function f : X → Y is called an embedding if X ∼= f(X), wheref(X) ⊆ Y has the topology induced by the topology of Y .Let Dr be a r-disc and f : Dr → Y an embedding. Then f(Dr) is called a(r-)ball in Y . Note that to define a ball we can also use Ir instead of Dr,since they are homeomorphic.Let f : Sr → Y be an embedding. Then f(Sr) is called a (r-)sphere in Y .

Proposition 1.5.2Let {U}j∈J be an open covering of a compact metric space X. Then there isa Lebesgue number ε > 0 such that for each subset A ⊆ X with diam A < εthere is an index j ∈ J such that A ⊆ Uj .

Lemma 1.5.3Let B ⊂ Sn be a r-ball (r < n); then Sn r B is acyclic, that is H0(X) ∼= Z

and Hk(X) = (0) for k ≥ 1.

Proof. Let B = f(Dr) ⊂ Sn, where f : Dr → Sn is an embedding. Weproceed by induction on the dimension r of the ball.

If r = 0, D0 is a point, hence Sn rB ∼= Sn r{point} ∼= Rn, that is acyclic.

1.5. Invariance of domain 17

Assume the result holds for r − 1. Let X = Sn r B, and z ∈ Zk(X) acycle for k ≥ 1, z = y−x for k = 0, where x and y are points of X. We wantto prove that z is a boundary, i.e. there is a chain c ∈ Sk+1(X) such that∂c = z. (Then H0(X) ∼= Z and for k ≥ 1 we have [z] = 0, also Hk(X) = (0)).

Since Dr ∼= Ir, we can assume that B = f(Ir), where f : Ir → B ⊂ Sn

is a homeomorphism. For each t ∈ I define Bt = f(Ir−1 × {t}); it is a(r − 1)-ball in Sn.

Let Xt = Sn r Bt; by the induction hypothesis there is a ct ∈ Sk+1(Xt)such that ∂ct = z.

The members of ct have compact images and Bt is compact, hence thereis an open neighbourhood Ut ⊆ Sn of Bt such that ct ∈ Sk+1(X r Ut). LetVt = f−1(Ut) ⊆ Ir. Since Ir−1 is compact and Ir−1 × {t} ⊆ Vt, there is anopen neighbourhood It ⊆ I of t, such that Ir−1 × It ⊆ Vt.

Let ε be a Lebesgue number for the open covering {It}t∈I of I and choosean integer m such that m ≥ 1

ε. For j = 1, . . . , m there is a tj ∈ I such that

Ij =[

j−1m, j

m

]⊆ Itj . Let Qj = f(Ir−1 × Ij) and cj = ctj for j = 1, . . . , m.

Then for each j: cj ∈ Sk−1(Sn rQj) and ∂cj = z.

We have B =⋃m

j=1Qj, and we may assume m ≥ 2. For each j letXj = Sn rQj .

The couple (X1, X2) is excissive for X1∪X2 = Snr(Q1∩Q2) ∼= SnrIr−1,hence by the Mayer-Vietoris Theorem the following sequence is exact:

Hk+1(X1 ∪X2) −→ Hk(X1 ∩X2)µ−→ Hk(X1) ⊕Hk(X2)

[z]X1∩X27→ ([z]X1

,−[z]X2)

By the induction hypothesis X1 ∪X2 is acyclic, that is Hk+1(X1 ∪X2) = (0),thus the kernel of µ is (0), that means µ is injective.

Since [z]X1= [∂c1]X1

= 0 and [z]X2= [∂c2]X2

= 0, there is a c12 ∈Sk+1(X1∩X2) = Sk+1(S

nr(Q1∪Q2)) such that ∂c12 = z, because [z]X1∩X2=

0.

Repeat this argument with X1 = Sn r (Q1 ∪ Q2) and X3 = Sn r Q3.Then there is a c123 ∈ Sk+1(S

n r (Q1 ∪Q2 ∪Q3)) with ∂c123 = z.After m − 1 times we get a c = c12...m ∈ Sk+1(S

n r (Q1 ∪ · · · ∪ Qm)) =Sk+1(S

n rB) such that ∂c = z.

Theorem 1.5.4Let n ≥ 1, 0 ≤ r ≤ n− 1 and S ⊂ Sn an r-sphere.Then for 0 ≤ r ≤ n− 2 holds:

Hk(Sn r S) ∼=

{Z for k = 0 and k = n− r − 1(0) otherwise.


For r = n− 1 holds:

Hk(Sn r S) ∼=

{Z ⊕ Z for k = 0(0) otherwise.

Proof. We proceed by induction on r.Let r = 0. S0 is the union of two distinct points, Sn r S0 = Sn r {p, q}.

Since Snr{p} ∼= Rn, it follows that SnrS0 ∼= Rnr{0} which is homotopicallyequivalent to Sn−1, since Sn−1 is a deformation retract of Rnr{0}. The claimfollows from Theorem 1.4.6.

Now, supposing the result is true for r − 1, we prove it for r. Let D+ ={z ∈ Sr | zr+1 ≥ 0} and D− = {z ∈ Sr | zr+1 ≤ 0}. Let f : Sr → S ⊆ Sn

be an embedding and B+ = f(D+), B− = f(D−) ⊆ S. Then S = B+ ∪ B−;furthermore T = B+ ∩ B− = f(D+ ∩D−) is a (r − 1)-sphere in Sn.

Let U = SnrB+, V = SnrB−; then U∪V = SnrT and U∩V = SnrS.

For k ≥ 1 we have the Mayer-Vietoris sequence

(0) = Hk+1(U) ⊕Hk+1(V ) −→ Hk+1(Sn r T )

∆−→∼=

Hk(Sn r S) −→ Hk(U) ⊕Hk(V ) = (0)

because U and V are acyclic by Lemma 1.5.3.By the induction hypothesis,

Hk(Sn r S) ∼= Hk+1(S

n r T ) ∼={

Z for k = n− r − 1(0) for k ≥ 1, k 6= n− r − 1

For k = 0 we have:

Z ⊕ Z ∼= H0(U) ⊕H0(V )ν−→ H0(U ∪ V ) = H0(S

n r T ) ∼= Z

([x], [y]) 7→ [x] + [y]

Then ker ν ∼= 〈[x0],−[x0]〉 ∼= Z for x0 ∈ Sn r S. Hence, the sequence

(0) = H1(U) ⊕H1(V ) −→ H1(Sn r T )

∆−→ H0(Sn r S)

µ−→ ker ν ∼= Z −→ (0)

is exact and splits because Z is a free group. Then H0(Sn r S) ∼= H1(S

n r

T ) ⊕ Z.

For r < n−1 the induction hypothesis implies H1(Sn rT ) = (0) because

n− r 6= 1. Then H0(Sn r S) ∼= Z.

For r = n− 1 it holds H1(Sn r T ) = Z. Then H0(S

n r S) ∼= Z ⊕ Z.

1.5. Invariance of domain 19

Corollary 1.5.5 (Jordan-Brouwer Separation Theorem)An (n − 1)-sphere S ⊂ Sn separates Sn into two components of which it isthe common boundary.

Proof. H0(Sn r S) ∼= Z ⊕ Z by Theorem 1.5.4, hence Sn r S has two path

connected components (this follows from Propositions 1.4.2 and 1.4.3). SinceSn rS is an open subset of Sn, it is locally path connected. Let U and V beits path components; they are also its components by Proposition 1.5.6.

U and V are open, hence their boundary is contained in S. We haveto show that S ⊆ U ∩ V . To prove that, let x ∈ S and N ⊆ Sn be aneighbourhood of x. Let A be a subset of S ∩ N such that S r A ∼= In−1.By Lemma 1.5.3, Sn r (S rA) is acyclic, hence it is path connected.

Let p ∈ U and q ∈ V ; then there is a path ω from p to q contained inSn r (S r A). Since p and q are in different path components of Sn r S,ω meets A. Hence A meets U and V . Therefore N meets U and V , andx ∈ U ∩ V because N was arbitrary.

Proposition 1.5.6Let A be a locally path connected space and A = B ∪ C its decompositionin path connected components. Then A is not connected.

Proof. We show that if a space is connected and locally path connected, thenit is also path connected. Then B and C are obviously connected but B ∪Cis not connected, otherwise it would be path connected.

Let D be connected and locally path connected, x a point of D. LetU = ∪{V | x ∈ V, V open and path connected}. It is obviously open andpath connected.

Let y ∈ ∂U , then since D is locally path connected, there is an open pathconnected set W which contains x. Since y ∈ ∂U , U ∪W is not empty, henceU ∪W is open and path connected, hence U ∪W ⊆ U , and y ∈ U , whichmeans that U is open and closed. Hence it must be U = D because D isconnected.

Remark 1.5.7The separation theorem holds also in Rn for n ≥ 2: an (n−1)-sphere S ⊂ Rn

separates the euclidean space into two components.

Let N = (0, . . . , 0, 1) ∈ Sn be the northpole and π : Sn r {N} → Rn thestereographic projection (which is an homeomorphism). Let f : Sn−1 → Rn

be an embedding and S = f(Sn−1).Then M = π−1(S) ⊆ Sn is an (n− 1)-sphere in Sn and thus by Theorem

1.5.4, we have H0(Sn rM) ∼= Z⊕Z. Hence, Sn rM = U1∪U2 has two path


components. We may assume N ∈ U2. Since Sn r M is open, also U1 andU2 are open in Sn.

The component U2 is path connected, and so is U2 r{N}. So V1 = π(U1)and V2 = π(U2 r {N}) are also path connected. Rn r S = V1∪V2 and thereis no path from V1 to V2.

Theorem 1.5.8 (Theorem on Invariance of Domain)If U and V are homeomorphic subsets of Sn and U is open in Sn, then V isopen in Sn.

Proof. Let f : U → V be a homeomorphism and let f(x) = y. Let Bbe a neighbourhood of x in U that is homeomorphic to In, with boundaryS homeomorphic to Sn−1. By Lemma 1.5.3, Sn r f(B) is connected, andby Theorem 1.5.5, Sn r f(S) has two components. Since Sn r f(B) andf(B) r f(S) are connected and Sn r f(S) = (Sn r f(B)) ∪ (f(B) r f(S)),they are the components of Sn r f(S). Hence f(B) r f(S) is an open subsetof Sn. Therefore, V is open in Sn, because y was an arbitrary point off(B) r f(S) ⊆ V .

The Theorem on Invariance of Domain assures that the dimension of amanifold is well defined.

If a point of the manifold has two neighbourhoods U , V , whichare homeomorphic to open subsets of Rn and Rm respectively, then theintersection U ∩ V is also homeomorphic to open subsets of Rn and Rm.Hence, by the Theorem on Invariance of Domain, they are both open in Rn

(supposing n ≥ m), that means it must be n = m, i.e. the dimension is welldefined.

Chapter 2

Dimension of affine varieties

There are some different definition of dimension over algebraic varieties, butat the end they all lead to the same result, therefore we could consider themas different characterizations of the dimension property, rather than differentdefinitions.

The most natural one, following the same approach as differentialtopology, is to consider the tangent space and define the dimension of thevariety as the dimension of the tangent vector space. There are at least twodifferent characterizations of the tangent space: one uses the gradient andthe Jacobi matrix, the other approach is based on the derivations on thealgebra of regular functions.

The tangent space is a local property of the varieties, as it is definedpointwise, and the definition of dimension for the whole variety is then theminimum of the dimensions of the tangent spaces in each point, seen as vectorspaces.

An important concept is then that of regular points, whose tangent spacehas constant dimension in a neighbourhood of the point. For algebraicvarieties over C, it turns out that regular points are smooth in the senseof differential topology, that is the variety is a smooth manifold in aneighbourhood of a regular point, and the dimension of the variety coincideswith the dimension of the manifold. In other words, the concept of dimensionof algebraic varieties and the concept of dimension of complex manifoldscoincide on objects belonging to both categories.

In addition to these concepts, there are two definition of dimension, whichare purely algebraic: one uses the transcendence degree of the field of rationalfunctions, and the other, called Krull dimension, considers the length of thepossible sequences of closed irreducible subsets of a variety, or, equivalently,of the prime ideals in the stalk in a point of the sheaf of regular functions.

22 Chapter 2. Dimension of affine varieties

2.1 Basic concepts

Throughout this chapter, we denote by K an algebraically closed field. Inthis section, we recall some basic definitions and results.

Let L ⊆ K[T1, . . . , Tn] be a set of polynomials in n variables; define thevanishing set (set of zeroes) of L as the subset

V (L) = {x ∈ Kn | f(x) = 0 for all f ∈ L} ⊆ Kn.

A set X ⊆ Kn is called an algebraic set if there are polynomials f1, . . . , fr ∈K[T1, . . . , Tn] such that

X = V (f1, . . . , fr) = {x ∈ Kn | fi(x) = 0 for all i = 1, . . . , r}.

The vanishing ideal of a subset X ⊆ Kn is the set of polynomials whichvanish on every point of X:

I(X) = {f ∈ K[T1, . . . , Tn] | f(x) = 0 for each x ∈ X}.

The algebraic sets of a Kn are the closed sets of a topology, called Zariskitopology. In the following, the concepts of open and closed set are alwaysintended with respect to the Zariski topology, if not explicitly statedotherwise. Note that if K = C, the Zariski topology is weaker as the usualtopology.

The coordinate ring of an algebraic set X ⊆ Kn is

K[X] = {f : X → K | f = F|X for a polynomial F ∈ K[T1, . . . , Tn]}.

This is an affine K-algebra, that is, a finitely generated K-algebra with nonilpotent elements.

If X ⊆ Kn is an algebraic set, the mapping

πX : K[T1, . . . , Tn] → K[X]F 7→ F|X

is a surjective morphism and its kernel is I(X). It follows that

K[X] ∼= K[T1, . . . , Tn]/I(X),

thus K[X] is a finitely generated K-algebra.The principal set of a function f ∈ K[X] is {x ∈ X | f(x) 6= 0} =

X r V (f).Let X be an algebraic set. For an ideal a ⊆ K[X] define the vanishing set inX

VX(a) = {x ∈ X | f(x) = 0 for all f ∈ a}.

2.1. Basic concepts 23

For a subset Y ⊆ X define the vanishing ideal in K[X]

IX(Y ) = {f ∈ K[X] | f(y) = 0 for all y ∈ Y }.

A non empty topological space X is said to be irreducible if it can notbe decomposed into two closed sets X1, X2 6= X. Other characterizations ofirreducibility are:

• every two non empty open subset of X have non empty intersection

• every non empty open set is dense in X.

Theorem 2.1.1A non empty algebraic set X ⊆ Kn is irreducible if and only if its vanishingideal I(X) ⊆ K[T1, . . . , Tn] is prime.

The proof can be found in [Milne05], corollary 1.15.

Theorem 2.1.2A non empty algebraic set X ⊆ Kn can be decomposed in an unique wayas X = X1 ∪ . . . ∪ Xn, where each Xi ⊆ X is closed and irreducible, andXi 6⊆ Xj if i 6= j. The sets Xi are called the irreducible components of X.

For the proof, see [Shaf77], chapter 1, § 3, Theorem 2.Let X ⊆ Kn be an algebraic set and U ⊆ X an open set. A function

f : U → K is said to be regular if for each x ∈ U there are g, h ∈ K[X] suchthat f = g

hin a neighbourhood of x.

The set of the regular functions on U forms the K-algebra of regular functionson U . It is denoted by OX(U).

A sheaf is a mapping F from the open sets of a topological space X tothe abelian groups (or rings, or K-algebras, ...)

F : {open subsets of X} → {abelian groups}, U 7→ F(U)

such that

(i) for every open sets V ⊆W ⊆ X there is a restriction homomorphism

ρWV : F(W ) → F(V ), f 7→ f|V

and if U ⊆ V ⊆ W are open subsets of X, these homomorphisms arecompatible,

ρWU = ρV

U ◦ ρWV


(ii) if U = ∪i∈IUi, where each Ui ⊆ X is open, then

(a) given f ∈ F(U) such that f|Ui= 0 ∈ F(Ui), for each i ∈ I, then

f = 0

(b) given fi ∈ F(Ui) for each i, such that fi|Ui∩Uj= fj|Ui∩Uj

for eachi, j, there is a f ∈ F(U) such that f|Ui

= fi for each i.

A mapping which satisfies only the first statement (i) is called a presheaf.For any algebraic set X ⊆ Kn, the mapping U 7→ OX(U), which maps theopen subset U of X into the K-algebra of regular functions on U , is a sheaf.

Given a commutative ring R with unity and a multiplicative system S ⊆R (that means 1 ∈ S and if a, b ∈ S then also ab ∈ S), we can consider theequivalence relation on R× S

(r, s) ∼ (r′, s′) ⇔ u(rs′ − r′s) = 0 for a u ∈ S

and write rs

for the equivalence class of (r, s). The quotient set is denoted byS−1R and is called fraction ring of R. In particular, for f ∈ R, we define thelocalization of R by f as Rf = S−1R, where S = {f r | r ∈ N}.

Theorem 2.1.3Let X ∈ Kn be an algebraic set and g ∈ K[X]. Then, the mapping

K[X]g → OX(Xg)fgr 7→

[

x 7→ f(x)g(x)r

]

is an isomorphism. In particular, taking g = 1, one has

OX(X) = K[X],

i.e. the K-algebra of regular functions is isomorphic to the coordinate ring.

Let X be a topological space and F a presheaf of abelian groups on it.Consider the disjoint union

F =⊔

x∈U⊆X

F(U)

where U runs through all the open neighbourhoods of x. On F define anequivalence relation as follows: for s ∈ F(U) and s′ ∈ F(U ′)

s ∼ s′ ⇔ there is a open neighbourhood V ⊆ U ∩ U ′ such that s|V = s′|V

2.1. Basic concepts 25

The stalk of F in a point x ∈ X is the quotient set

Fx = F/ ∼

For an element s ∈ F(U) and a x ∈ U , we denote by sx its residue class inFx and call it the germ of s in x.

The stalk Fx has the structure of abelian group: for sx, s′x ∈ Fx define

sx + s′x = (s|U∩U ′ + s′|U∩U ′)x.

In the same way, the stalks of a presheaf of rings (algebras, ...) are againrings (algebras, ...).

Let p be a prime ideal of a commutative ring R with unity. Then S = Rrp

is a multiplicative system and we define the localization of R at p as

Rp = S−1R.

Theorem 2.1.4Rp is a local ring, that is in Rp there is exactly one maximal ideal, which is

S−1p =

{r

s| r ∈ p, s ∈ S

}

.

Theorem 2.1.5Let X ⊆ Kn be an algebraic set and x ∈ X. Denote by mx = {f ∈K[X] | f(x) = 0} the maximal ideal corresponding to x. Then, the mapping

K[X]mx→ OX,x

fg

7→(

fg

)

x

from the localization of the coordinate ring onto the stalk (OX(X))x, is anisomorphism.

For the proof, see [Milne05], Proposition 3.6.A space with functions is a topological space X together with a sheaf OX

of K-valued functions.A continuous function ϕ : X → Y of spaces with functions is called a

morphism if for each open set V ⊆ Y , for each f ∈ OY (V ), the compositionf ◦ ϕ lies in OX(ϕ−1(V )).

Let ϕ : X → Y be a morphism of spaces with functions and V ⊆ Y anopen set. We define the comorphism

ϕ∗ : OY (V ) → OX(ϕ−1(V ))

f 7→ f ◦ ϕ.


Theorem 2.1.6Let X ∈ Kn and Y ∈ Km be algebraic sets and φ : X → Y be a function.Then the following statements are equivalent:

1. φ is a morphism of spaces with functions

2. φ is the restriction of a polynomial map Φ : Kn → Km.

Definition 2.1.7An affine K-variety is a space with functions which is isomorphic to analgebraic set in some Kn.

Definition 2.1.8Let F be a covariant (contravariant) functor of the category C in the categoryD. F is an equivalence if the following two conditions hold:

• for each object Y ∈ D there is an object X ∈ C such that F (X) = Y ;

• for every two objects X,X ′ ∈ C the mapping

Mor(X,X ′) → Mor(F (X), F (X ′))φ 7→ F (φ)

is a bijection. (The codomain is Mor(F (X ′), F (X)) in case of acontravariant functor.)

Theorem 2.1.9 (“Anti-equivalence” Theorem)Let X be an algebraic set. The functor which maps

X 7→ K[X] and ϕ 7→ ϕ∗

is a contravariant equivalence from the category of the algebraic sets to thecategory of the affine K-algebras. That is,

• it is a contravariant functor;

• for each affine algebra A there is an n ∈ N and an algebraic set X ⊆ Kn,such that A is isomorphic to K[X];

• if X ⊆ Kn and Y ⊆ Km are algebraic sets, then there is a bijection

{morphisms X → Y } → {homomorphisms K[Y ] → K[X]}ϕ 7→ ϕ∗.

Theorem 2.1.10Let X be an affine variety and f ∈ O(X). Then the principal set Xf = {x ∈X | f(x) 6= 0} with the subspace topology and the induced sheaf is an affinevariety, and O(Xf) = O(X)f .

2.2. Affine tangent spaces 27

2.2 Affine tangent spaces

Let f ∈ K[T1, . . . , Tn] be a polynomial and a ∈ Kn a point. Define

Lf,a =

n∑

i=1

∂f

∂Ti(a) (Ti − ai) = 〈gradf (a), T − a〉

The affine tangent space to an algebraic set X in a point a ∈ X is thespace

T affa X = V (Lf,a | f ∈ I(X))

We give T affa X the structure of vector space: for z = a + v, z′ = a + v′, and

a scalar c ∈ K, define

z + z′ = a + v + v′, cz = a+ cv,

that means also, a is the zero of T affa X.

Proposition 2.2.1If f1, . . . , fr generate I(X), then T aff

a X = V (Lfi,a | i = 1, . . . , r) and thedimension of the affine tangent space is linked with the rank of the Jacobimatrix by the relation

dim(T affa X) = n− rk

(∂fi

∂Tj(a)

)

1≤i≤r1≤j≤n

Proof. We have to prove that {Lf1,a, . . . , Lfr ,a} generate the same ideal in thepolynomial ring K[T1, . . . , Tn] as the set {Lf,a | f ∈ I(X)}, that means, wehave to write each Lf,a as linear combination of the Lfi,a. To do that, notethat for each f ∈ I(X) there are hi ∈ K[T1, . . . , Tn] such that f =

∑hifi,

because f1, . . . , fr generate I(X). Then

Lf,a = 〈gradf(a), T − a〉=∑

〈gradhi(a)fi(a), T − a〉 +

∑

〈hi(a)gradfi(a), T − a〉

=∑

〈hi(a)gradfi(a), T − a〉

=∑

hi(a)Lfi,a.

To prove the connection of the dimension with the Jacobi matrix, for each


point a ∈ Kn consider the matrix

Ma = (Lfi,a(ej + a))1≤i≤r1≤j≤n

=

(n∑

k=1

∂fi

∂Tj

(a)(Tk(ej + a) − ak)

)

i,j

=

(∂fi

∂Tj(a)

)

1≤i≤r1≤j≤n

If a ∈ X, then the dimension of T affa X is the dimension of the kernel of Ma,

and the proof is complete.

Remark 2.2.2Let X ⊆ Kn be an algebraic set. For each n ∈ N, the set

Ak = {x ∈ X | dim(T affx X) ≥ n}

is closed in X.In particular, the points whose tangent space has minimal dimension form anonempty open set in X.

Proof. Since

Ak = {x ∈ X | rk(Mx) ≤ n− k}= {x ∈ X | det(Nx) = 0 for all (n− k + 1)-minors Nx of Mx},

Ak is the intersection of X with an algebraic set in Kn, hence it is closed.The second assertion follows immediately.

Definition 2.2.3The Zariski dimension of an irreducible algebraic set X is defined throughthe dimension of the affine tangent space,

ZdimX = minx∈X

dimT affx X.

The Zariski dimension of an algebraic set is the maximum Zariski dimensionof its irreducible components.

In the next section, we extend the notion of Zariski dimension toirreducible affine varieties, see Definition 2.3.7.

2.3. Derivations 29

2.3 Derivations

Let R be a K-algebra with unity, and α : R → K a homomorphism. Anα-derivation is a K-linear map δ : R → K such that for each f, g ∈ R thefollowing product derivation rule holds:

δ(fg) = δ(f)α(g) + α(f)δ(g).

The set of α-derivations is a vector space on K; we denote it by Derα(R).With abuse of notation, we identify 1 · K ⊆ R with the field K and we

write α(α(g)) = α(g) and α(1) = 1.Let X ⊆ Kn be an algebraic set and x ∈ X a point. Let α be the

evaluation homomorphism, α : K[X] → K, f 7→ f(x). Its kernel is themaximal ideal mx ⊆ K[X]. The vector space of the α-derivations of K[X] isDerx(K[X]) and its elements are called the derivations of K[X] centered atx.

Let z = x + v ∈ T affx X. Assign to z the following derivation of the

polynomial ring K[T1, . . . , Tn] centered at x:

z 7→[ρz : f 7→ Lf,x(z) = 〈gradf (x), v〉

].

The usual properties of derivatives guarantee that ρz is actually a derivationand since z ∈ T aff

x X, this derivation vanishes on each element of I(X), andso it induces a derivation δz ∈ Derx(K[X]) with δz(f|X) = ρz(f). δz is calledthe directional derivative associated to z.Define Φ the canonical epimorphism

Φ : K[T1, . . . , Tn] → K[X], f 7→ f|X ,

then we have ρz = δz ◦ Φ.

Proposition 2.3.1Let R be a K-algebra, α : R → K a homomorphism, and δ ∈ Derα(R).Suppose m = kerα. Then

• for each k ∈ K, we have δ(k) = 0;

• for each f ∈ m2, δ(f) = 0.

Proof. We have δ(1) = 0, since

δ(1) = δ(1 · 1) = δ(1)α(1) + α(1)δ(1) = 2δ(1).

Hence, and since δ is K-linear, follows that δ(k · 1) = kδ(1) = 0.For the second part, suppose f ∈ m2; then it can be written as

∑higi with


suitable gi, hi ∈ m. Thus, by linearity and the product derivation rule, wehave

δ(f) = δ(∑

higi

)

=∑

(δ(gi)α(hi) + δ(hi)α(gi)) = 0,

since α(hi) = α(gi) = 0 for all i.

Theorem 2.3.2Let X ⊆ Kn be an algebraic set and a ∈ X. Then the mapping

T affa X → Dera(K[X]) z 7→ δz

is an isomorphism between vector spaces.

Proof. The mapping z 7→ δz is a homomorphism, since for z = a + v, z′ =a+ v′ ∈ T aff

a X, we have

• ρz+z′ : f → Lf,a(z + z′) =∑ ∂f

∂Ti(a)(vi + v′i) = Lf,a(z) + Lf,a(z

′).

• ρcz : f → Lf,a(cz) =∑ ∂f

∂Ti(a)(cvi) = cLf,a(z).

• ρe : f → Lf,a(e) =∑ ∂f

∂Ti(a), where e denotes the unity in T aff

a X.

Let Φ : K[T1, . . . , Tn] → K[X] be the canonical epimorphism defined byf 7→ f|X .

We first show that z 7→ δz is surjective: for that purpose let δ : K[X] → K

be a derivation centered at a, and look for some z such that δ = δz. Letδ′ = δ◦Φ; it is a derivation centered at a of the polynomial ring K[T1, . . . , Tn].Define v = (δ′(T1), . . . , δ

′(Tn)) and z = v + a. We claim that z ∈ T affa X. To

prove this, we must show that for each g ∈ I(X), the linear part Lg,a vanishesin z. We have

δ′(g) = δ(Φ(g)) = δ(0) = 0.

The Taylor expansion of g in a yields g = g(a) + Lg,a + h, with h ∈ m2,where m is the maximal ideal {f ∈ K[X] | f(a) = 0}. By proposition 2.3.1,applying δ′ to the Taylor expansion, we have

0 = δ′(g) = δ′(Lg,a) =n∑

i=1

∂g

∂Ti

(a)δ′(Ti) =n∑

i=1

∂g

∂Ti

(a)vi = Lg,a(z).

Thus, z ∈ T affa X is verified. The derivations δ′ = δ ◦ Φ and δz ◦ Φ coincide

by construction. Since Φ is surjective, it follows that δ = δz, hence z 7→ δz issurjective.

2.3. Derivations 31

To prove that z 7→ δz is injective, let z = v+a ∈ T affa X, with δz = 0. Then

ρz = δz ◦Φ is a derivation centered at a of the polynomial ring K[T1, . . . , Tn]and for each i

0 = ρz(Ti) =n∑

j=1

∂Ti

∂Tj

(a)vj =∂Ti

∂Ti

(a)vi = vi.

Hence, v = 0. Due to the definition of the structure of vector space on T affa X,

we have that z = a is the zero of T affa X.

Let X be an affine variety, x ∈ X and α : OX,x → K the evaluationhomomorphism

α : OX,x → K

fx 7→ f(x).

It is well defined, because f ∈ O(U) and f ′ ∈ O(U ′) are equivalent if andonly if f|V = f ′

|V for some open set V ⊆ U ∩ U ′ such that x ∈ V .

The K-vector space of the derivations on OX,x is defined as

Der(OX,x) = Derα(OX,x).

Thus, a derivation on the stalk OX,x is a K-linear mapping δ : OX,x → K

such that the product rule holds: δ(fg) = δ(f)g(x) + f(x)δ(g).We extend the concept of affine tangent space, which is so far defined

only for algebraic sets, also to affine variety.

Definition 2.3.3The affine tangent space to an affine variety X in a point x ∈ X is definedas

TX,x = Der(OX,x).

We will show in 2.3.6 that for every algebraic set X the affine tangentspace T aff

x X is isomorphic to the affine tangent space TX,x of X seen as anaffine variety.

Proposition 2.3.4Let R be a local K-algebra with maximal ideal m ⊆ R. Then there is adecomposition R = K ⊕ m and the mapping

αm : R = K ⊕ m → K

f = a⊕ f ′ 7→ a

is the unique homomorphism of algebras of R in K.


As in the case of the stalk of the sheaf OX , we define the vector space ofderivations on R:

Der(R) = Derαm(R).

Theorem 2.3.5Let R be a K-algebra, α : R → K a homomorphism of K-algebras andm = ker(α). Then there is a canonical isomorphism

ϕ : Derα(R) → Der(Rm)

δ 7→[

rs7→ δ(r)α(s)−α(r)δ(s)

α(s)2

]

.

Proof. We need show that

(1) ϕ is well defined;

(2) ϕ(δ) is a derivation of Rm;

(3) ϕ is an isomoprhism.

To prove (1), let r, r′ ∈ R and s, s′ ∈ Rrm such that rs

= r′

s′. By definition

of Rm there is a u ∈ Rr m such that u(rs′ − r′s) = 0. Then

0 = α(0) = α(u)α(rs′ − r′s)

and since u is not in the kernel of α, this holds if and only if

0 = α(rs′ − r′s) = α(r)α(s′) − α(r′)α(s). (2.3.1)

Furthermore, applying δ and the product rule to the equality u(rs′−r′s) = 0,we get

0 = δ(u(rs′ − r′s))= δ(u)α(rs′ − r′s) + α(u)δ(rs′ − r′s)= δ(u)(α(r)α(s′) − α(r′)α(s))+

+α(u) (δ(r)α(s′) + α(r)δ(s′) − δ(r′)α(s) − α(r′)δ(s)) ,

where the first summand is 0 by 2.3.1. Hence, since α(u) 6= 0, we obtain

δ(r)α(s′) − α(r′)δ(s) = δ(r′)α(s) − α(r)δ(s′). (2.3.2)

By 2.3.1, we have

α(r′) =α(r)α(s′)

α(s)and α(r) =

α(r′)α(s)

α(s′),

2.3. Derivations 33

hence, substituting these expressions in 2.3.2,

α(s′)

(

δ(r) − α(r)

α(s)δ(s)

)

= α(s)

(

δ(r′) − α(r′)

α(s′)δ(s′)

)

,

which gives

δ(r)α(s) − α(r)δ(s)

α(s)2=δ(r′)α(s′) − α(r′)δ(s′)

α(s′)2.

That means ϕ(δ)(

rs

)= ϕ(δ)

(r′

s′

)if r

s= r′

s′, hence ϕ does not depend on the

choice of the class representative, that is it is well defined.

Now we show (2). Clearly ϕ(δ) is K-linear and vanishes on K. To provethat the product rule holds, calculate

ϕ(δ)

(r

s

r′

s′

)

=δ(rr′)α(ss′) − δ(ss′)α(rr′)

α(ss′)2

=(δ(r)α(r′) + δ(r′)α(r))α(ss′) − (δ(s)α(s′) + δ(s′)α(s))α(rr′)

α(ss′)2

=δ(r)α(s) − δ(s)α(r)

α(s)2

α(r′)

α(s′)+δ(r′)α(s′) − δ(s′)α(r′)

α(s′)2

α(r)

α(s)

=α(r′)

α(s′)ϕ(δ)

(r

s

)

+α(r)

α(s)ϕ(δ)

(r′

s′

)

= α

(r′

s′

)

ϕ(δ)(r

s

)

+ α(r

s

)

ϕ(δ)

(r′

s′

)

.

Hence, ϕ(δ) is a derivation of Rm.

To prove (3), note that ϕ is a homomorphism of K-vector fields.To show that ϕ is injective, suppose ϕ(δ) = 0. Then for each r ∈ R

0 = ϕ(δ)(r

1

)

= δ(r),

hence δ = 0, and ϕ is injective.To show that ϕ is surjective, take a δ′ ∈ Der(Rm). Then, δ(r) = δ′

(r1

)defines

an α-derivation on R and

ϕ(δ)(r

s

)

=δ′( r

1)α(s) − α(r)δ′( s

1)

α(s)2

= δ′(r

1

)

α

(1

s

)

− α(r

1

)

δ′(s

1

)

α

(1

s2

)


Since

0 = δ′(s

s

)

= δ′(s

1

)

α

(1

s

)

+ δ′(

1

s

)

α(s

1

)

,

we have δ′(

1s

)= −δ′

(s1

)α(

1s2

). Hence

ϕ(δ)(r

s

)

= δ′(r

1

)

α

(1

s

)

+ α(r

1

)

δ′(

1

s

)

= δ′(r

s

)

,

which concludes the proof that ϕ is an isomorphism.

Proposition 2.3.6Let X be an algebraic set in Kn and x ∈ X a point.Then there are canonicalisomorphisms

T affx X ∼= Derx(O(X)) ∼= Der(OX,x) = TX,x.

Proof. By Theorem 2.3.2, T affx X ∼= Derx(K[X]) and by Theorem 2.1.3,

OX(X) = K[X]. By Theorem 2.3.5, Derx(O(X)) ∼= Der(OX(X)mx), and

since OX(X)mxis isomorphic to OX,x, the proof is complete.

Thanks to the previous result, we can now extend the definition of Zariskidimension given in Definition 2.2.3 to affine varieties.

Definition 2.3.7Let X be an irreducible affine variety. Its Zariski dimension is defined as

ZdimX = minx∈X

dimTX,x.

The Zariski dimension of an affine variety is the maximum Zariski dimensionof its irreducible components.

Definition 2.3.8Let X be an affine variety and x ∈ X be a point. Then the dual space(mx/m

2x)

∗is called Zariski tangent space of X at x.

Theorem 2.3.9Let R be a K-algebra, α : R → K a homomorphism of K-algebras andm = ker(α). Then the mapping

Derα(R) → (m/m2)∗

δ 7→ [uδ : f + m2 7→ δ(f)]

is an isomorphism of K-vector spaces.

2.3. Derivations 35

Proof. By Proposition 2.3.1, the map δ 7→ uδ is well defined. It is clearly alinear map.

To show that the mapping is injective, let δ ∈ Derα(R) such that uδ = 0,that means 0 = uδ(f + m2) = δ(f) for each f ∈ m. We have to show thatδ = 0, i.e. δ(f) = 0 for all f ∈ R. We have, by the linearity of δ,

δ(f) = δ(f − α(f) + α(f)) = δ(f − α(f)) + δ(α(f)).

Now, δ(α(f)) = 0 because α(f) ∈ K and since α is K-linear, α(f − α(f)) =α(f)−α(α(f)) = 0, that means f −α(f) ∈ kerα = m, thus δ(f −α(f)) = 0.Hence, δ(f) = 0 for all f ∈ R.

To show that the mapping is surjective, let u : m/m2 → K be a linearfunctional. Define the K-linear map

δ : R → K

f 7→ u(f − α(f) + m2).

Let us prove that it is an α-derivation: first note that δ vanishes on K andon m2.Furthermore, for each h ∈ R, we have

δ(h− α(h)) = u(h− α(h) − α(h− α(h)) + m2)

= u(h− α(h) + m2)

= δ(h).

Now, to obtain the product rule, consider

δ(fg) = δ(fg − α(fg))

= δ((f − α(f))α(g) + α(f)(g − α(g)) + (f − α(f))(g − α(g))

)

= δ((f − α(f))α(g)) + δ((g − α(g))α(f)) + δ((f − α(f))(g − α(g)))

= δ(f)α(g) + δ(g)α(f),

since (f − α(f))(g − α(g)) ∈ m2.

Now, to prove that δ 7→ uδ is surjective, note that for each f+m2 ∈ m/m2,by construction we have

uδ(f+m2) = δ(f) = u(f−α(f)+m

2) = u(f+m2)−u(α(f)+m

2) = u(f+m2).


By Theorem 2.1.5, the mapping

OX(X)mx→ OX,x

f

g7→(f

g

)

x

is an isomorphism, hence the stalk OX,x is a local ring, and its maximal idealis

mX,x =

{(f

g

)

x

| f ∈ mx, g ∈ OX(X) r mx

}

= {hx ∈ OX,x | h(x) = 0}.

Proposition 2.3.10Let X be an affine variety and x ∈ X be a point.Then there are canonicalisomorphisms

TX,x = Der(OX,x) ∼=(mX,x/m

2X,x

)∗

∼= Derx(OX(X)) ∼= (mx/m2x)

∗.

Proof. Immediate by Proposition 2.3.6 and Theorem 2.3.9.

2.4 Regular and singular points

We introduce the concept of regular points.

Definition 2.4.1A point x of an affine variety X is called smooth or regular if there is an openneighbourhood U of x such that dim(TX,u) = dim(TX,x) for each u ∈ U .The non-regular points are called singular.

Proposition 2.4.2Let X ⊆ Kn be an algebraic set, I(X) = 〈f1, . . . , fr〉 its vanishing ideal andx0 ∈ X. Then, the following statements are equivalent:

1. x0 is regular

2. x 7→ dim(TX,x) has a local minimum at x0

3. the rank of the Jacobi matrix(

∂fi

∂Tj(x))

has a local maximum at x0.

Proof. The equivalence of 2 and 3 is clear by virtue of Proposition 2.2.1 andRemark 2.2.2.It is clear that 1 implies 2.To show that 2 implies 1, let U be an open neighbourhood of x0, such that

2.4. Regular and singular points 37

dim(TX,x) = k is a local minimum in U of the mapping x 7→ dim(TX,x). ByRemark 2.2.2 the set Ak = {x ∈ X | dim(TX,x) ≥ k + 1} is closed in X.Hence, UrA is an open neighbourhood of x0 in X, and dim(TX,x) is constanton U r A.

Proposition 2.4.3A point x of an irreducible affine variety X is regular if and only ifdim(TX,x) = Zdim(X).

Proof. Let Zdim(X) = n. By Remark 2.2.2 the following set is open andnonempty:

V = {p ∈ X | dim(TX,p) = n} ⊆ X.

Suppose dim(TX,x) = n. Then x ∈ V , which is open, thus x is regular.Conversely, suppose x is regular, then there is an open set U ⊆ X such

that for each u ∈ U the dimensions dim(TX,x) = dim(TX,u) coincide. SinceX is irreducible, each two open subsets of X have nonempty intersection.Then, U ∩ V 6= ∅. Hence, there is a u ∈ U ∩ V , and, since dim(TX,u) = n,we can conclude dim(TX,x) = n.

Note that this does not hold in general if the affine variety is reducible.As an example consider the variety in C3 defined by X = V (z1z2, z1z3). Itsirreducible components are X1 = V (z1) and X2 = V (z2, z3). All points areregular except for the origin, but the dimension of the tangent space at theregular points of X1 is 2, while the dimension of the tangent space at theregular points of X2 is 1.

Theorem 2.4.4The set of regular points of an affine variety X is open and dense in X.

Proof. By definition, the set Xreg of regular points is open. We need onlyshow that it is dense in X.

Let X1, . . . , Xr be the irreducible components of X. We claim that eachset Ui = Xreg ∩Xi is dense in Xi. Consider

Vi = X r⋃

i6=j

Xj ⊆ Xi

It is not empty, because otherwise we could decompose X in r−1 irreduciblecomponents. Moreover, Vi is open in X.

Let V ′i ⊆ Xi be the set of points x ∈ Xi for which dim(TXi,x) is minimal

in Xi. V′i is not empty and open in Xi.

Let V ′′i = Vi ∩ V ′

i . Because of the irreducibility of Xi, since Vi and V ′i are

open in Xi, the set V ′′i is not empty; it is also open in X. Then OXi

(V ) =


OX(V ) for each open subset V ⊆ V ′′i . As a consequence, OXi,x

∼= OX,x forall x ∈ V ′′

i .Then for each point x of V ′′

i , dim(TXi,x) = dim(TX,x). By definition of V ′′i ,

on this set the function x 7→ dim(TX,x) is constant. Then V ′′i ⊆ Xreg ∩Xi =

Ui.As a consequence, Ui is not empty, and since it is open in Xi, which is

irreducible, Ui is dense in Xi.Finally,

X =

r⋃

i=1

Xi =

r⋃

i=1

Ui ⊆ Xreg

This proves that the set of regular points is dense in X.

2.5 Transcendence degree

Let K ⊆ L be a field extension. An element l ∈ L is said to be algebraic overK, if there is a non zero polynomial f ∈ K[T ] such that f(l) = 0. If l ∈ L isnot algebraic, it is said to be transcendent.A field extension K ⊆ L is algebraic if every element l ∈ L is algebraic overK.

A subset L ∈ L is said to be algebraically dependent over K if there exist apositive integer n, distinct elements l1, . . . , ln ∈ L and a non zero polynomialf ∈ K[T1, . . . , Tn] such that f(l1, . . . , ln) = 0. Otherwise L is said to bealgebraically independent over K.

A transcendence basis of K ⊆ L is an algebraically independent maximalset B ⊆ L. It is a well known fact that every two transcendence basis of thesame field extension have the same cardinality.

Every field extension K ⊆ L has a transcendence basis. The transcendencedegree of the extension is the cardinality of one of its basis B, trdegK(L) =|B | .

For any irreducible affine variety X, the ideal I(X) is prime, hence thecoordinate ring K[X] ∼= OX(X) is an integral domain, and we can speak ofthe quotient field of OX(X) and give the following definition.

Definition 2.5.1The field of rational functions K(X) of an irreducible affine variety X is thequotient field of the ring OX(X).

Definition 2.5.2The dimension of an irreducible affine variety X is defined as the transcen-dence degree of the field of rational functions,

dim(X) = trdegK(K(X)).

2.5. Transcendence degree 39

The dimension of an affine variety X with irreducible components X1, . . . , Xr

is the maximum of the dimensions of the irreducible components,

dim(X) = max1≤i≤r

dim(Xi).

For example, the affine variety Kn has dimension n, because its functionfield is K(T1, . . . , Tn) and a transcendence basis for the extension K ⊆K(T1, . . . , Tn) is {T1, . . . , Tn}.

Note that the transcendence degree of K ⊆ K(X) is always finite and sois the dimension, because K(X) = K(T1|X , . . . , Tn|X).

Proposition 2.5.3An affine variety X is finite if and only if dim(X) = 0.

Proof. Suppose X is finite and let X1, . . . , Xr be the irreducible componentsof X. Then each Xi consists of only one point (because it is irreducible).Hence, we need only show that each affine variety Y = {y} consisting ofexactly one point is zero dimensional. Since obviously O(Y ) = K, we haveK(Y ) = K, and trdegK(K(Y )) = 0.Conversely, if X has dimension zero, we can assume X is irreducible.Otherwise, we take one irreducible component with maximal dimension; weshow that it is zero dimenzional, hence the other components must also bezero dimensional. Since trdegK(K(X)) = 0, the extension K ⊆ K(X) isalgebraic, hence, because K is algebraically closed, K(X) = K. It followsthat K[X] = K.By the Anti-equivalence Theorem 2.1.9, each morphism Y → X correspondsbijectively to one homomorphism K[X] → K[Y ]. Now, taking a varietyY = {y} containing exactly one point, we have K[Y ] = K = K[X], hencethere is only one homomorphism K[X] → K[Y ]. Then the Anti-equivalenceTheorem ensures that there is only one morphism Y → X, that means Xconsists of exactly one point.

Theorem 2.5.4Let X be an affine variety and Y ⊆ X a closed subvariety. Then

dim(Y ) ≤ dim(X).

Proof. Without loss of generality we may suppose X and Y to be irreducible.We may also assume that X and Y are algebraic sets in some Kn. Then

OY (Y ) = K[T1|Y , . . . , Tn|Y ] K(Y ) = K(T1|Y , . . . , Tn|Y ).

Let d = dim(Y ). Then the extension K ⊆ K(Y ) has a transcendencebasis of the form {Ti1|Y , . . . , Tid|Y }, because we can choose d algebraically


independent elements among {T1, . . . , Tn} and these, restricted to Y , willform a basis of the field extension. It follows that if f ∈ K[Ti1 , . . . , Tid] withf(Ti1 , . . . , Tid)|Y = 0, then f = 0. Hence, the set

{Ti1|X , . . . , Tid|X} ⊆ K(X)

is algebraically independent over K, because for each f ∈ K[Ti1 , . . . , Tid] withf(Ti1 , . . . , Tid)|X = 0, we have f(Ti1 , . . . , Tid)|Y = 0, which implies f = 0.Hence, trdegK(K(X)) ≥ d.

Theorem 2.5.5Let X be an irreducible affine variety and Y ⊆ X a closed subvariety. Ifdim(Y ) = dim(X), then Y = X.

Proof. Without loss of generality let X and Y be algebraic sets in Kn, withX irreducible. Let d = dim(X) = dim(Y ). As in the proof of Theorem 2.5.4,choose a transcendence basis of the form {Ti1|Y , . . . , Tid|Y } for the extensionK ⊆ K(Y ). As in the previous proof, the set {Ti1|X , . . . , Tid|X} ⊆ K(X) isalgebraically independent, hence it is a transcendence basis for the extensionK ⊆ K(X).

By contradiction, assume now Y 6= X. Then IX(Y ) 6= 〈0〉 andthere is a function 0 6= f ∈ OX(X) such that f ∈ IX(Y ). The set{Ti1|X , . . . , Tid|X , f} ⊆ K(X) is algebraically dependent over K because itcontains d + 1 elements. Hence there are aj ∈ K[Ti1 , . . . , Tid ] (j = 1, . . . , kfor some k) and an equation on X

ak|Xfk + . . .+ a1|Xf + a0|X = 0.

Since X is irreducible, OX(X) has no proper divisor of zero. Since f 6= 0,we may assume a0|X 6= 0 (otherwise we may divide by the lowest powerof f). Since f ∈ IX(Y ), we have a0|X ∈ IX(Y ), that means a0|Y = 0.But a0|Y ∈ K[Ti1|Y , . . . , Tid|Y ] and since a0 6= 0, then {Ti1|Y , . . . , Tid|Y } arealgebraically dependent over K, which is a contradiction.

To prove the next results, we need to recall the Hilbert’s Nullstellensatz.

Recall that for any ideal a in a commutative ring R with unity we candefine the radical of a,

√a = {r ∈ R | rn ∈ a for some n ∈ N}.

This is again an ideal in R.

2.5. Transcendence degree 41

Theorem 2.5.6 (Hilbert’s Nullstellensatz)Let K be an algebraically closed field, and X ⊆ Kn an algebraic set. Then

IX(VX(a)) =√

a

for every ideal a ⊆ K[X].

The proof can be found in [ZS58], Chapter VII, § 3, Theorem 14.

Definition 2.5.7An affine variety X is purely k-dimensional if each of its irreduciblecomponents has dimension k.

Proposition 2.5.8Let X be an affine variety, f1, . . . , fr ∈ OX(X) pairwise not associated primeelements and f =

∏ri=1 fi. Then, VX(fi) (i = 1, . . . , r) are the irreducible

components of VX(f).

Proof. We may assume that X and Y are algebraic sets in Kn. By theCorollary 2.5.6 to the Hilbert’s Nullstellensatz, I(Yi) =

√

〈fi〉. Since each

fi is prime, then 〈fi〉 =√

〈fi〉 and it follows that each Yi is irreducible. Wehave to show that Yi 6⊆ Yj if i 6= j. Otherwise, we would have fj ∈ 〈fi〉for distinct i, j. But this contradicts the hypothesis that fi and fj are notassociated.

Theorem 2.5.9If f ∈ K[T1, . . . , Tn] is a non constant polynomial, then the variety V (f) ⊆ Kn

is purely (n− 1)-dimensional.

Proof. Since K[T1, . . . , Tn] is a unique factorization domain, we can writef = fµ1

1 · . . . · fµrr , where the fis are pairwise not associated. Since V (f) =

V (∏r

i=1 fi), we can apply Proposition 2.5.8, and we obtain that the V (fi)sare the irreducible components of V (f). We have to show that each V (fi)has dimension n− 1. Hence, we may assume that f is prime and prove thatV (f) has dimension n− 1.Let X = V (f) have dimension d. By Theorem 2.5.4, d ≤ n, and since f 6= 0,X is not the whole Kn. By Theorem 2.5.5, d < n. Hence, we need onlyshow that d ≥ n − 1. For that purpose, choose a j ∈ {1, . . . , n} such thatf depends on the variable Tj (such a j exists, because f is not constant).Consider the set

A = {T1|X , . . . , Tj−1|X, Tj+1|X, . . . , Tn|X}.

We claim that A is algebraically independent over K. Suppose it is not,then there is a polynomial g ∈ K[T1, . . . , Tn], not depending on Tj, such that


g|X = 0. Since g ∈ I(X) = 〈f〉, we have g = hf for some polynomial h. Thisis a contradiction to the fact that g does not depend on Tj .Hence, A ⊆ K(X) is algebraically independent over K and we have d =trdegK(K(X)) ≥ n− 1.

The following result considers the inverse situation.

Theorem 2.5.10If X ⊆ Kn is a purely (n − 1)-dimensional algebraic set, then the vanishingideal I(X) is generated by a polynomial f ∈ K[T1, . . . , Tn], I(X) = 〈f〉.

Proof. First, we consider X be irreducible. Then its vanishing ideal I(X) ⊆K[T1, . . . , Tn] is prime. By Theorem 2.5.5, X is a proper subset of Kn, henceI(X) 6= 〈0〉. Thus, there is an element 0 6= g ∈ I(X). Since I(X) is aprime ideal, there is a prime factor f ∈ I(X) of g. Then X ⊆ V (f) and byTheorem 2.5.9, dim(V (f)) = n − 1 = dim(X). By Theorem 2.5.5, we getV (f) = X. By the Corollary 2.5.6 to the Hilbert’s Nullstellensatz, we haveI(X) =

√

〈f〉. Since f is prime, I(X) =√

〈f〉 = 〈f〉.Now we are ready for the general case. Suppose X1, . . . , Xr are the

irreducible components of X. By the first part of the proof, we haveI(Xi) = 〈fi〉 for some prime elements f1, . . . , fr. Since Xi 6= Xj for i 6= j, itholds 〈fi〉 6= 〈fj〉. Hence the fi are pairwise not associated.

It is clear that 〈f1 · . . . · fr〉 ⊆ I(X). Conversely, let f ∈ I(X). Thenf ∈ I(Xi) = 〈fi〉 for each i, and each fi divides f . Since the fis are pairwisenot associated prime elements, we have that f1 · . . . ·fr divides f , that meansf ∈ 〈f1 · . . . · fr〉.Hence I(X) = 〈f1 · . . . · fr〉.

2.6 Morphisms

Definition 2.6.1Let ϕ : X → Y be a morphism between affine varieties.ϕ is said to be a closed embedding if the image ϕ(X) is closed in Y and ϕ isan isomorphism between X and ϕ(X).ϕ is called dominant or dominating if ϕ(X) is dense in Y .

Proposition 2.6.2ϕ is a closed embedding if and only if the comorphism ϕ∗ is surjective.ϕ is dominant if and only if the comorphism ϕ∗ is injective.

2.6. Morphisms 43

A dominant morphism ϕ : X → Y between irreducible affine varietiesdefines a monomorphism between the function fields

ϕ∗ : K(Y ) → K(X)g

h7→ ϕ∗(g)

ϕ∗(h).

Definition 2.6.3ϕ is said to be birational if ϕ∗ : K(Y ) → K(X) is an isomorphism.

Proposition 2.6.4If there is a dominant morphism ϕ : X → Y between affine varieties, thendim(X) ≥ dim(Y ).

Proof. First, we consider the case where both X and Y are irreducible. Sincethe comorphism is injective, we have a field extension K(Y ) ∼= ϕ∗(K(Y )) ⊆K(X). Hence

dim(Y ) = trdegK(K(Y )) ≤ trdegK(K(X)) = dim(X)

Consider the general case, let X1, . . .Xr be the irreducible components ofX. Then

Y = ϕ(X) = ϕ(X1) ∪ . . . ∪ ϕ(Xr).

Each ϕ(Xi) is irreducible, because it is the image of an irreducible spaceunder a continuous map. Thus, also the closure ϕ(Xi) are irreducible.Since the decomposition in irreducible components is unique, the componentsof Y are some of the ϕ(Xi). In particular there is a component Y0 = ϕ(Xj)with maximal dimension. As in the first part of the proof, we obtain

dim(X) ≥ dim(Xi) ≥ dim(Y0) = dim(Y ).

Remark 2.6.5If there is a birational morphism ϕ : X → Y between affine varieties, thendim(X) = dim(Y ), because the fields K(X) and K(Y ) are isomorphic.

The following results will be used in 2.8.4.

Theorem 2.6.6Let X and Y be irreducible affine varieties with function fields K(X) andK(Y ).

1. For any homomorphism ψ : K(Y ) → K(X), there is a principal setXf ⊆ X and a dominant morphism ϕ : Xf → Y such that ϕ∗ = ψ.


2. For any isomorphism ψ : K(Y ) → K(X), there are principal sets Xf ⊆X and Yg ⊆ Y and an isomorphism ϕ : Xf → Yg such that ϕ∗ = ψ.

Let ϕ : X → Y be a morphism between affine varieties, x ∈ X andϕ(x) = y. Note that ϕ∗(my) ⊆ mx and ϕ∗(m2

y) ⊆ m2x, hence ϕ∗(my/m

2y) ⊆

mx/m2x. Then, for δ ∈ TX,x, we have δ ◦ ϕ∗ ∈ TY,y =

(my/m

2y

)∗and we can

give the following definition.The differential of ϕ in x is the linear mapping

Txϕ : TX,x → TY,y

δ 7→ δ ◦ ϕ∗.

Let ϕ : X → Y and ψ : Y → Z be morphisms between affine varieties.Let x ∈ X and y = ϕ(x). Then it is not hard to see that

Tx(ψ ◦ ϕ) = Tyψ ◦ Txϕ

and it is obvious that Tx id = id. In particular, if ϕ is an isomorphism, thenalso its differential Txϕ is an isomorphism for all x ∈ X.

Given a morphism ϕ : X → Y between affine varieties, ϕ∗(O(X)) is asubring of O(X), hence O(X) has a structure of ϕ∗(O(X))-module.

Definition 2.6.7We say that a morphism ϕ : X → Y between affine varieties is finite if O(X)is finitely generated over ϕ∗(O(Y )).

Proposition 2.6.8If ϕ : X → Y is a finite dominant morphism between irreducible affinevarieties, then ϕ∗(K(Y )) ⊆ K(X) is an algebraic field extension of finitedegree, and dim(X) = dim(Y ).

Theorem 2.6.9Let ϕ : X → Y be a finite morphism of affine varieties and suppose A ⊆ X isa closed subset. Then ϕ(A) is closed in Y and the restriction ϕ|A : A→ ϕ(A)is a finite morphism.

The result 2.6.8 holds even if X and Y are not irreducible, in the followingsense.

Proposition 2.6.10If ϕ : X → Y is a finite dominant morphism between affine varieties, thendim(X) = dim(Y ).

2.7. Krull dimension 45

Proof. By Proposition 2.6.4, it is sufficient to show that dim(X) ≤ dim(Y ).Choose an irreducible component X ′ of X with maximal dimension, thatmeans dim(X) = dim(X ′). The image Y ′ = ϕ(X ′) is also irreducible, becauseϕ is continuous. By Theorem 2.6.9 the restriction ϕ|X′ : X ′ → Y ′ is afinite morphism of affine varieties. By Proposition 2.6.8 we have dim(X ′) =dim(Y ′). Applying Theorem 2.5.4, we get dim(Y ′) ≤ dim(Y ), because Y ′ isa closed subvariety of Y . Hence

dim(X) = dim(X ′) = dim(Y ′) ≤ dim(Y ).

2.7 Krull dimension

Definition 2.7.1The Krull dimension of a topological space X is the supremum over allintegers r ∈ N which allow a strictly ascending sequence

∅ 6= X0 ( . . . ( Xr ⊆ X

of irreducible closed subsets Xi.

Note that this definition is analogous to a characterization of thedimension of vector spaces: in this case maximal chains of vector subspacesare considered to define the dimension of the vector space.

Let ψ : R → S be a homomorphism of commutative rings with unity. Wesay that ψ is a finite homomorphism if S is a finitely generated module overψ(R).

Theorem 2.7.2 (Noether’s normalization Lemma)Let R be a finitely generated K-algebra. Then there is an integer d ∈ N anda finite monomorphism K[T1, . . . , Td] → R.

For the proof, we refer to [Shaf77], chapter 2, § 5.

Proposition 2.7.3 (Noether’s normalization Lemma, version 2)Let X be an affine variety. Then there is an integer d ∈ N and a finitesurjective morphism X → Kd.

Theorem 2.7.4The Krull dimension of an affine variety is equal to its dimension.


Proof. First, we prove that the Krull dimension of an affine variety X isbounded by dim(X). Let ∅ 6= X0 ( . . . ( Xr ⊆ X be an ascending sequenceof irreducible closed subsets. By Theorem 2.5.4 we have dim(Xi) ≤ dim(X)and by Theorem 2.5.5 dim(Xi) ≤ dim(Xi+1). Then r ≤ dim(X).

Conversely, we have to prove that dim(X) is bounded by the Krulldimension of X. If X = Kn, we have a sequence {0} ( K ( K2 ( . . . ( Kn.Then the Krull dimension of Kn is at least n = dim(Kn).

Let X be an arbitrary affine variety. By the Noether’s NormalisationLemma 2.7.3 there is a surjective finite morphism ϕ : X → Kn, where n =dim(X).

LetXn ⊆ X be a n-dimensional irreducible component ofX. By Theorem2.6.9, the morphism

ϕn = ϕ|Xn: Xn → Kn

is finite. Let Yn = ϕn(Xn). By Proposition 2.6.8 dim(Yn) = n. By Theorem2.5.5 we have Yn = Kn.

Consider Kn−1 ( Kn; the restriction of ϕn to ϕ−1n (Kn−1) is finite and

surjective. Then, ϕ−1n (Kn−1) has dimension n − 1, by Proposition 2.6.10.

In particular there is a (n − 1)-dimensional irreducible component Xn−1 ⊆ϕ−1

n (Kn−1). It is Xn−1 ( Xn and there is a finite surjective morphism

ϕn−1 = ϕn|Xn−1: Xn−1 → Kn−1.

In the same way we can find a (n − 2)-dimensional irreducible Xn−2 (

Xn−1, and so on. This proceeding leads to a strictly ascending sequenceX0 ( . . . ( Xn−1 ( Xn of irreducible closed sets Xi ⊆ X. Since the lengthof the sequence is n, it follows that the Krull dimension of X is at least n.

The following theorem extends the result 2.5.9.

Theorem 2.7.5 (Krull’s principal ideal Theorem)Let X be an n-dimensional irreducible affine variety and suppose f ∈ O(X)such that ∅ 6= VX(f) 6= X. Then, VX(f) is purely (n− 1)-dimensional.

For the proof, we refer to [ZS58].

Corollary 2.7.6Let X be an irreducible affine variety and suppose f1, . . . fr ∈ O(X) suchthat VX(f1, . . . , fr) 6= ∅. Then, each irreducible component of VX(f1, . . . , fr)has at least dimension dim(X) − r.

Proof. We use induction over r. The case r = 1 is the statement of Krull’sTheorem 2.7.5.For the induction step, consider the subvariety X ′ = VX(f1, . . . , fr−1). We

2.8. Zariski dimension 47

have VX(f1, . . . , fr) = VX′(fr|X′). For each irreducible component Y ofVX(f1, . . . , fr), choose an irreducible component Y ′ ⊆ X ′ such that Y ⊆ Y ′.By induction hypothesis we have dim(Y ′) ≥ dim(X) − r + 1. Hence, byKrull’s Theorem 2.7.5,

dim(Y ) = dim(VY ′(fr|Y ′)) ≥ dim(X) − r.

Corollary 2.7.7Let X be an irreducible affine variety of dimension n and let x be a point inX. Then there is a strictly ascending chain

{x} ( X1 ( . . . ( Xn,

where the Xi are irreducible and closed subsets of X.

Proof. We use induction over n = dim(X). The case n = 0 is trivial. Toprove the induction step, define Xn = X and choose a nonzero f ∈ O(Xn)with f(x) = 0. By Theorem 2.7.5 VX(f) has an irreducible component Xn−1

of dimension n − 1 with x ∈ Xn−1. Now apply the induction hypothesis onXn−1.

2.8 Zariski dimension

Recall that for an irreducible affine variety X, in 2.3.7 we have defined theZariski dimension as

ZdimX = minx∈X

dim(TX,x).

Definition 2.8.1A polynomial f ∈ K[T ] is said to be separable if each of its irreducible factorshas only simple roots in its splitting field.Let K ⊆ L be a field extension. An element l ∈ L is separable over K if it isalgebraic over K and its minimum polynomial fl ∈ K[T ] is separable.The field extension K ⊆ L is separable if it is algebraic and each elementl ∈ L is separable over K.

If l1, . . . , lk ∈ L are separable elements of a field extension K ⊆ L, thenK ⊆ K(l1, . . . , lk) is a separable field extension. If K ⊆ L and L ⊆ M areseparable extensions, also K ⊆ M is separable.If K ⊆ L is a simple extension, and L = K(l), then l is called a primitiveelement for the extension. Recall the following classical result of field theory.


Theorem 2.8.2 (Theorem of the primitive element)Each finite separable extension K ⊆ L is simple, i.e. L = K(l) for someprimitive element l ∈ L.

Proposition 2.8.3Let K be an algebraically closed field and K ⊆ L a finite field extension.Then there is a transcendence basis {a1, . . . , ad} ⊆ L of the extension, suchthat K(a1, . . . , ad) ⊆ L is a separable extension.

For the proof we refer to [ZS58].

Theorem 2.8.4Let X be an n-dimensional irreducible affine variety. Then, there is anirreducible polynomial p ∈ K[T1, . . . , Tn+1] such that the function field ofX is isomorphic to that of Y = V (p) ⊆ Kn+1. Furthermore, there exist twoisomorphic principal sets X0 ⊆ X and Y0 ⊆ Y .

Proof. By Proposition 2.8.3 there is a transcendence basis {f1, . . . , fn} ⊆K(X) for the field extension K ⊆ K(X), such that we can decompose K ⊆K(X) in a transcendent extension K ⊆ L and in a separable extension L ⊆K(X), where L = K(f1, . . . , fn).

Let h be a primitive element for the extension L ⊆ K(X), thenK(X) = L(h). Consider the minimal polynomial g ∈ L[S] of h over L.The isomorphism defined by

Φ : L[S] → K(T1, . . . , Tn)[Tn+1]fi 7→ Ti

S 7→ Tn+1

maps g to Φ(g) = pq, where p ∈ K[T1, . . . , Tn+1] and q ∈ K[T1, . . . , Tn] have

no common factor. Since g is irreducible, so is also p. Denote by Q(R) thequotient field of the integral domain R. We have

K(V (p)) = Q(K[T1, . . . , Tn+1]/〈p〉)∼= Q(K(T1, . . . , Tn)[Tn+1]/〈p/q〉)∼= Q(L[S]/〈g〉)∼= L[S]/〈g〉∼= L(h)

= K(X)

We can obtain the isomorphism of principal sets using Theorem 2.6.6.

2.9. Krull dimension for rings 49

Theorem 2.8.5The Zariski dimension of an irreducible affine variety coincides with itsdimension (defined as transcendence degree).

Proof. Let dim(X) = n be the dimension of the irreducible varietyX, definedas transcendence degree. We will show that there is a nonempty open subsetU ⊆ X such that dim(TX,u) = n for each u ∈ U . Then, since the set ofregular points Xreg is open and dense, the set Xreg ∩ U is not empty, hencethe dimension of TX,x at each x ∈ Xreg is n and so Zdim(X) = n.

Let p ∈ K[T1, . . . , Tn+1] be an irreducible polynomial as in Theorem 2.8.4,such that V (p) and X have isomorphic function fields and there are two

principal sets V ⊆ X and W ⊆ V (p) and an isomorphism ϕ : V∼=−→W .

Since p is irreducible, I(V (p)) = p and for each regular point y ∈ V (p)

TV (p),y = V (Lf,y | f ∈ I(V (p)) = p) = V (Lp,y).

Note that Lp,y =∑n+1

i=1∂p∂Ti

(y)(Ti − yi) is a non constant polynomial in n+ 1variables, hence by Theorem 2.5.9, V (Lp,y) is purely n-dimensional.

Hence we have

dim(TV (p),y) = dim(V (Lp,y)) = n.

The set U = ϕ−1(W∩V (p)reg) is open and nonempty inX, and dim(TX,u) = nfor every u ∈ U and the proof is complete.

2.9 Krull dimension for rings

The Krull dimension can also be defined on rings.

Definition 2.9.1The Krull dimension of a commutative ring with unity R is the supremumover all r ∈ N which allow a strictly ascending sequence

p0 ( . . . ( pr ⊆ R

of prime ideals.

Proposition 2.9.2Let X be an irreducible affine variety, and x a point in X. Then

dim(X) = dim(O(X)) = dim(OX,x).


Proof. There is a bijective correspondence between irreducible closed subsetsofX and prime ideals in O(X). In particular ifX1 ⊆ X2 are irreducible closedsubsets of X, then IX(X1) ⊇ IX(X2).

Then the first equality is a direct consequence of Theorem 2.7.4, whichsays that the Krull dimension of X coincides with dim(X).

To prove the second equality, observe that the canonical homomorphismOX(X) → OX,x defines a bijection

{Prime ideals in mx} → {Prime ideals in OX,x}q 7→ (OX,x r mX,x)

−1q.

By Corollary 2.7.7 the maximal length of the sequences of prime ideals in mx

is the dimension of X.

Definition 2.9.3Let R be a local ring with maximal ideal m. We say that R is regular, ifits Krull dimension is equal to the dimension of m/m2 as vector space overR/m,

dim(R) = dimR/m(m/m2).

Theorem 2.9.4Let X be an irreducible affine variety, and x a point in X. Then x is regularif and only if the local ring OX,x is regular.

Proof. Suppose x is a regular point, then

dim(OX,x) = dim(X) = dim(TX,x) = dim(mX,x/m2X,x).

Conversely, if OX,x is a regular ring,

dim(TX,x) = dim(mX,x/m2X,x) = dim(OX,x) = dim(X).

An important result which can be obtained using the theory described inthis chapter is that on objects which can be studied both as differentiablemanifolds over Cn and affine varieties on the ground field C, the definitionsof dimension given in this and in the first chapter coincide. In particular,one can show that x is a regular point of an affine variety X if and only ifit has an open neighbourhood which is diffeomorphic to some Cn. We give asketch of the proof.

In this chapter we defined the sheaf of regular functions. In the sameway, taking holomorphic function instead of regular ones, one can define thesheaf of holomorphic functions of a differentiable manifold. We denote byOX,x the stalk of holomorphic functions at the point x ∈ X.The following result can be found in [Ser56], §2, Corollary 2.

2.9. Krull dimension for rings 51

Theorem 2.9.5OX,x has the same dimension as OX,x.

Furthermore, defining mX,x as the maximal ideal {hx ∈ OX,x | h(x) = 0}in OX,x, it can be shown that

mX,x

m2X,x

andmX,x

m2X,x

are isomorphic. Hence OX,x is a regular local ring if and only if OX,x is aregular local ring.

Then, by Theorem 2.9.4, we have that a point x ∈ X is regular exactlywhen OX,x is regular.

Suppose x is a smooth point, i.e. there is a biholomorphism from an openneighbourhood U of x, U → Cn, such that x 7→ 0. Hence OX,x

∼= OCn,0 and

the maximal ideal mx = (x1, . . . , xn) has length n, thus krdim OX,x = n.Conversely, suppose that x ∈ X ⊆ Cn, where X = V (I) and I is the

ideal generated by f1, . . . , fs. Suppose the rank of the Jacobi matrix is r,

rkDfx = rk(

∂fi

∂xj(x))

i,j= r. Then by the Inverse Function Theorem 1.2.3

there is a system of coordinates for which the following holds:

if k ≤ r fk = xk + gk(xr+1, . . . , xs)if k > r fk = gk(xr+1, . . . , xs)

In other words, defining the set X ′ = {gk(xr+1, . . . , xs) | k > r} and the map

F : X ′ → Cn

(xr+1, . . . , xs) 7→ −fk(xr+1, . . . , xs) k = 1, . . . , r

then X is the graph of F . X is biholomorphic to X ′ and the Jacobi matrixof F is defined by equations with rank 0. Hence we may suppose r = 0, thatmeans I ⊆ m2

Cn,x.

Then it suffices to show that if OX,x is regular, then I = (0), that meansX = Cn.

Since I ⊆ m2Cn,x

mX,x

m2X,x

∼=(

mCn,x

m2Cn,x

)

/I ∼=mCn,x

m2Cn,x

hence dimmX,x

m2X,x

= dimmCn,x

m2Cn,x

= n.


Thus it suffices to prove that if I 6= (0), then krdim OX,x < n. We do

this by contradiction: suppose X ( Cn and krdim OX,x = n. Then there isa sequence of prime ideals

p0 ( p1 ( . . . ( pn.

Denoting by π the projection

π : OCn,x � OX,x

and letting qi = π−1(pi) ⊆ OCn,x, we find a sequence of prime ideals in OCn,x

q0 ( q1 ( . . . ( qn

with (0) 6= I ⊆ q0. Hence the sequence of prime ideals

(0) ( q0 ( q1 ( . . . ( qn

has length n+1, and krdimOCn,x > n, contradiction to the fact that Cn hasdimension n.

Chapter 3

Dimension of metric spaces

In this chapter we study the dimension of topological spaces, in particularof metric spaces. Three different definitions of dimension are given, and itwill be proved in Theorems 3.8.2 and 3.8.5 that they coincide on a large classof topological spaces. In Theorem 3.9.3 it will be proved that this notion ofdimension coincides with the usual definition for Euclidean spaces, i.e. Rn

has dimension n. From this fact it follows that this dimension extends theconcept of dimension given in chapter 1 for manifolds.

3.1 Three concepts of dimension

We first recall some concepts concerning collections of subsets of a space.

Definition 3.1.1Let U be a collection of subsets of a topological space R and p a point of R.The order of U at p, denoted by ordp U is the number of sets of U thatcontain p, if it is finite, and ∞ otherwise.The order of U is ord U = sup {ordp U | p ∈ R}, if it is finite, and ∞otherwise.

Definition 3.1.2A collection U of subsets of a space R is said to be locally finite if for eachpoint p ∈ R there is a neighbourhood of p which intersect at most finitelymany elements of U .A collection U is called point-finite if each point is contained in at mostfinitely many elements of U , i.e. the order of U is finite.A covering of a space R is a collection U of subsets of R such that

⋃U = R.A subcovering V of U is a covering with V ⊆ U .A refinement V of U is a covering such that for each V ∈ V there is a U ∈ Uwith V ⊆ U . Then we write V < U .

54 Chapter 3. Dimension of metric spaces

Note that each subcovering is a refinement.

There are three “classical” definitions of dimension.

Definition 3.1.3The covering dimension of a topological space R (also called Cech-Lebesguedimension or topological dimension) is denoted by dimR.We say that dimR ≤ n if for every finite open covering of R there exists anopen refinement with order not exceeding n+ 1. Then we define dim ∅ = −1and for R 6= ∅ we set dimR = ∞ if dimR > n for each integer n and

dimR = min{n | dimR ≤ n}

otherwise.

Remark 3.1.4If dimR ≤ n and {U1, . . . , Uk} is an open covering of R, then there is anopen covering W = {W1, . . . ,Wk} such that

• ord W ≤ n+ 1 and

• Wi ⊆ Ui for all i.

In fact, by the definition, there is a refinement V = {Vγ}γ∈Γ of U of order atmost n+ 1. Defining

W1 =⋃

Vγ⊆U1

Vγ

W2 =⋃

Vγ⊆UiVγ 6⊆U1

Vγ

...

Wi =⋃

Vγ⊆UiVγ 6⊆U1∪...∪Ui−1

Vγ

...

we get the desired covering {W1, . . . ,Wk}.

Two other definitions of dimension are based on the idea of separatingtwo disjoint closed sets, using an open set, whose boundary has a smallerdimension.

We write ∂A for the boundary of a set A, i.e. ∂A = Ar◦

A.

3.1. Three concepts of dimension 55

Definition 3.1.5The small (or weak) inductive dimension (also called Urysohn-Mengerdimension) of a topological space R is denoted by indR and defined asfollows.For the empty space, set ind ∅ = −1.For every non empty space R and for every n ≥ 0, define indR ≤ n if forevery neighbourhood U of every point p there exists an open neighbourhoodV of p such that p ∈ V ⊆ U and ind ∂V ≤ n− 1.Define

indR = min{n | indR ≤ n}if this minimum exists, and indR = ∞ otherwise.

Note that a space has small inductive dimension equal to zero if and onlyif it is nonempty and has a basis for its topology consisting of open and closedsets. In fact, an open set is closed exactly when its boundary is empty.

It follows that if a space has small inductive dimension equal to zero andits topology is not trivial, then it is not connected.

Definition 3.1.6The large (or strong) inductive dimension of a topological space R is denotedby IndR and defined as follows.For the empty space, set Ind ∅ = −1.For every non empty space R and for every n ≥ 0, define IndR ≤ n if forevery closed and disjoint subsets F , G of R there exists an open subset U ⊆ Rsuch that F ⊆ U ⊆ R rG and Ind ∂U ≤ n− 1.Define

IndR = min{n | IndR ≤ n}if this minimum exists, and IndR = ∞ otherwise.

Note that the properties used to characterize the three dimension areinvariant for homeomorphisms, i.e. these dimensions are well defined ontopological spaces.

Now, let us look at some examples.

Example 3.1.7Every nonempty discrete set is zero dimensional.

To show this in the sense of the covering dimension, note that every opencovering contains the refinement each of whose elements consists of a singlepoint. This refinement has order 1, hence the covering dimension of the spaceis at most 0, but the space is not empty, so its dimension is 0.


For the inductive dimensions, note that every subset is open and closed,and thus all the boundaries are empty (their inductive dimension is −1), thatis all subsets have inductive dimension 0.

Example 3.1.8Consider the real line R. Clearly, dim R > 0, since each refinement of thecovering {(−∞, 1), (−1,∞)} has order at least 2. We show that dim R ≤ 1:let U be a finite open covering, say containing k open sets U1, . . . , Uk; eachof them is a union of disjoint open intervals

Ui =⋃

j∈Ji

(aji , b

ji ) i = 1, . . . , k.

Consider the set S0 containing all these intervals,

S0 = {(aji , b

ji ) | j ∈ Ji, i = 1, . . . , k}.

Order the rational points q1, q2, . . . , qn, . . . and recursively define the sets

Sn = Sn−1 r{(aj

i , bji ) | qn ∈ (aj

i , bji )}

j∈Ji,i=1,...,k∪ {(a, b), (a, b)},

where (a, b) and (a, b) are intervals such that

a = min{aji | qn ∈ (aj

i , bji )} (3.1.1)

b = max{bji | qn ∈ (aji , b

ji )}. (3.1.2)

Note that such minima and maxima exist, since each point of R lies in finitelymany (more precisely at most k) intervals.

Clearly S =⋂

n∈N Sn is an open refinement of U ; we show that S hasorder at most 2, by proving that each point p ∈ R is contained in at mosttwo intervals of S. If p = qn is rational, note that Sn, and hence S, containsat most two intervals which contain qn, because all the other intervals thatcontain qn are not in Sn. If p ∈ R r Q and p lies in three intervals, of whoseb is the minimum of the right endpoints, then there is a rational point q suchthat p < q < b, and q is in three intervals, which is a contradiction.

We have to show that S covers R. Since the initial intervals cover R, it ispossible that S does not cover R if and only if all intervals containing somepoint are not contained in Sn if n is large enough. Hence, it is sufficient toshow that when an interval (a, b) is in some Sn but not in Sn+1, all points in(a, b) are still contained in some other intervals. To prove this fact, note that(a, b) ∈ Sn is not contained in Sn+1 exactly when there are intervals (c, d)

3.2. Some basic theory 57

and (c, d) such that:

(c, d) ∩ (c, d) 6= ∅,c ≤ a,

d ≥ b.

But then (a, b) ⊆ (c, d) ∪ (c, d), hence every point of R is contained in someintervals of S.

The small inductive dimension of the real line is also equal to 1. Clearly,ind R > 0, because no open subset of R has empty boundary. To show thatind R ≤ 1, let p be a point and U a neighbourhood of p. Let r > 0 be thedistance between p and R r U . Then p ∈ Sr/2(p) ⊂ U and ∂Sr/2(p) consistsof two points, which are zero-dimensional in the sense of the small inductivedimension.

Also Ind R = 1 holds, but we do not prove this result here. It will be clearafter section 3.8, where we show that the large inductive dimension and thecovering dimension coincide for every metric space. A coincidence theoremholds also for the small inductive dimensions: this dimension coincides withthe other two in every metric separable spaces.

In the next sections we prove some result that hold in metric spaces. Someresults can be generalized for larger classes of spaces, such as normal spaces,but the concepts of dimension do not generally coincide in such spaces, thuswe will not deal with them. The results can be found in the book by Nagata[Nag83].

We will consider mostly the topological and the large inductive dimension,while about the small inductive dimension we will only prove a theorem thatstates the coincidence of the three definitions of dimension for separablemetric spaces.

3.2 Some basic theory

In this section we recall some facts about metrizable spaces and state theStone’s coincidence Theorem and the Nagata-Smirnov Metrization Theorem.

In a metric space R, we denote by Sr(p) the ball of center p and radius r.Recall that a space R is metrizable if it is homeomorphic to a metric

space, i.e. if it is possible to define a metric ρ on R such that {S1/n(p)}n∈N isa neighbourhood basis of each p ∈ R. Such a metric is said to be compatiblewith the topology of R.


Definition 3.2.1Let U be a collection of subsets of a topological space R.The star in a point p is

S(p,U) =⋃

{U | p ∈ U ∈ U} .

The star of a set A is

S(A,U) =⋃

{U | U ∈ U and A ∩ U 6= ∅} =⋃

a∈A

S(a,U).

We denote by U∆ the family of sets {S(p,U) | p ∈ R} and by U∗ the family{S(U,U) | U ∈ U}.We write U∆∆ for

(U∆)∆

and U∗∗ for (U∗)∗.A star refinement (resp. a delta refinement) of a covering U is a coveringsuch that V∗ (resp. V∆) refines U .

Proposition 3.2.2U < U∆ < U∗ < U∆∆.

Proof. We need only prove the last assertion, since the other ones are obvious.Let S(U,U) be an element of U∗ and p ∈ U be a point. For any U ′ ∈ U suchthat U ∩U ′ 6= ∅, since there is a point p′ ∈ U ∩U ′, we have U,U ′ ⊆ S(p′,U).Hence p ∈ S(p′,U) ∈ U∆, which implies U ′ ⊆ S(p′,U) ⊆ S(p,U∆). Then,since this holds for any U ′ ∈ U not disjoint from U , we have S(U,U) ⊆S(p,U∆) ∈ U∆∆, i.e. U∗ < U∆∆.

Proposition 3.2.3Let U be a locally finite collection. Then

⋃

U∈U

U =⋃

U∈U

U .

It follows that⋃F is closed if F is a locally finite closed collection.

Definition 3.2.4A topological space R is said to be

• T1, if for every two distinct points p, q there is an open set U such thatp ∈ U , q 6∈ U . Note that this is equivalent to require that every setconsisting of one single point is closed;

• T2 or Hausdorff, if for every two distinct points p, q ∈ R there are opendisjoint sets U and V such that p ∈ U and q ∈ V ;

3.2. Some basic theory 59

• regular, if it is T1 and for every closed F and every point p ∈ R r Fthere are disjoint open sets U and V such that F ⊆ U and q ∈ V ;

• normal, if it is T1 and for every disjoint closed sets F and G there aredisjoint open sets U and V such that F ⊆ U and G ⊆ V . It can beshown that the sets U and V can be chosen such that U ∩ V = ∅;

• totally normal, if it is T1 and every open covering has a star refinement;

• compact, if for every open covering there is a finite subcovering;

• paracompact, if every open covering has a locally finite open refinement;

• separable, if it contains a countable dense subset.

Proposition 3.2.5Totally normal ⇒ normal ⇒ regular ⇒ T2 ⇒ T1.

Proof. We need only prove the first implication, since the rest is trivial.Let F and G be disjoint closed subsets of a totally normal space R. LetU = {RrF,RrG}. This is an open covering, hence it has a star refinementV. Consider the open sets U = S(F,V) and V = S(G,V). Obviously F ⊆ Uand G ⊆ V . Seeking for a contradiction, suppose U ∩V 6= ∅. Then there areV1, V2 ∈ V such that

V1 ∈ S(F,V) hence V1 ∩ F 6= ∅V2 ∈ S(G,V) hence V2 ∩G 6= ∅

V1 ∩ V2 6= ∅

Let p ∈ V1 ∩ V2. Then V1 ∪ V2 ⊆ S(p,V). Thus, S(p,V) ∩ F 6= ∅, whichimplies S(p,V) 6⊆ R r F and in the same way S(p,V) ∩ G 6= ∅ impliesS(p,V) 6⊆ RrG. This is a contradiction, since S(p,V) ∈ V∆ < V∗, which isa refinement of U = {Rr F,RrG}.Remark 3.2.6Every metric space is totally normal, because the balls

{S1/n(p)

}

n∈Nform a

neighbourhood basis of each point p.

Definition 3.2.7A collection U of subsets of a space R is said to be:

• star finite (resp. star countable) if every member of U intersect onlyfinitely many (resp. countably many) members of U , i.e. for everyU ∈ U

| {V ∈ U | U ∩ V 6= ∅} | <∞ (resp. ≤ ℵ0);


• closure preserving, if for every subcollection V of U the following holds:

⋃

U∈V

U =⋃

U∈V

U ;

• discrete, if it is closure preserving and any two closures of its membersdo not intersect.

We state two important results of general topology. They will be usefulto prove many results in the next sections, where we consider metric spaces.Here we are not interested in the proofs, for which the reader can refer tothe book of Nagata, [Nag83], chapters V and VI.

Theorem 3.2.8 (A.H. Stone’s Coincidence Theorem)A T2 space is paracompact if and only if it is totally normal.

Corollary 3.2.9Every metric space is paracompact.

Definition 3.2.10A collection U of subsets of a space is said σ-locally finite if it can be writtenas a countable union of locally finite collections of subsets,

U =⋃

i∈N

Ui

where each Ui is locally finite.

Theorem 3.2.11 (Nagata-Smirnov Metrization Theorem)A regular space is metrizable if and only if it has a σ-locally finite open basis.

3.3 The large inductive dimension

In this and in the following sections we suppose that the spaces are metric.Recall that every metric space is regular, paracompact, and hence normal.

In this section we present some results concerning the large inductivedimension, which we will call dimension for simplicity. In the next threesections we will use the results obtained here to prove some main theorems.The Sum Theorem 3.4.5 links the dimension of a space with the dimension ofits closed coverings. The Subspace Theorem 3.5.2 ensures the monotonicityof the dimension: every subset of every space can have at most the same

3.3. The large inductive dimension 61

dimension as the whole space. The Decomposition Theorem 3.5.3 statesthat each n-dimensional space can be decomposed in n+ 1 zero dimensionalsubsets. Finally, the Product Theorem 3.6.2 considers the dimension of finiteproduct of spaces.

The following lemma states that a closed subset of a space has dimensionnot greater than the dimension of the whole space. In Section 3.5 we willprove that this result holds for every subset of the space.

Lemma 3.3.1Let F be a closed subset of a metric space R. Then IndF ≤ IndR.

Proof. We proceed by induction on IndR.If IndR = −1, then R = ∅, hence F = ∅.Assume the statement for the spaces with dimension ≤ n − 1. Let R be

a space of dimension ≤ n and G, H closed disjoint subsets of F . Then bythe definition of Ind, there is an open set U ⊆ R such that G ⊆ U ⊆ RrHand Ind ∂U ≤ n− 1.

Let V = U ∩ F . It is open in F and G ⊆ V ⊆ F r H . Since ∂FV isclosed in F and also in ∂U (because ∂FV = ∂U ∩ F ), ∂FV ⊆ ∂FU ⊆ ∂U .Remember that Ind ∂U ≤ n−1. Hence by induction we get Ind ∂FV ≤ n−1.Then IndF ≤ n.

If two spaces A and B have large inductive dimension n and mrespectively, in general it is not true that Ind(A ∪ B) is the maximum orthe sum of the dimensions of A and B.

For example, Ind Q = Ind(R r Q) = 0, but the real line R has dimension1. The fact that the sets of rational point and of irrational points are zerodimensional will be proved in Example 3.3.7.

However, the dimension can not “spring” arbitrarily in a union of twospaces, as next Lemma shows for the particular case where one of the spacesis zero dimensional.

Lemma 3.3.2Let A and B be subsets of a metric space R with IndA ≤ n, IndB ≤ 0 andA ∪ B = R. Then IndR ≤ n+ 1.

To prove the previous Lemma, we need the following result.

Lemma 3.3.3Let C and D be subsets of a metric space R with C ∩D = ∅ and C ∩D = ∅.Then there are disjoint open sets M and N in R such that C ⊆ M andD ⊆ N .


Proof. To each point p ∈ C assign an εp ∈ R+ such that Sεp(p) ∩D = ∅.

To each point q ∈ D assign an εq ∈ R+ such that Sεq(q) ∩ C = ∅.

DefineM =

⋃

p∈C

S 1

2εp

(p) and N =⋃

q∈D

S 1

2εq

(q).

Proof of Lemma 3.3.2. Let F and G be closed disjoint subsets of R. SinceR is normal, there are open sets V , W such that F ⊆ V , G ⊆ W andV ∩W = ∅.

Without loss of generality, B is not empty (else A = R and all is trivial),hence IndB = 0, that is for each two closed disjoint subsets of B there is anopen and closed set in B which separates them. Hence, there is U ⊆ B openand closed such that V ∩B ⊆ U ⊆ B rW .

We claim that F ∪U and G∪ (BrU) satisfy the assumptions of Lemma3.3.3, hence there are open disjoint subsets M and N of R such that F ∪U ⊆M and G ∪ (B r U) ⊆ N . Then, since U is open and closed and by theproperty of N , ∂M ⊆ A. By Lemma 3.3.1, Ind ∂M ≤ IndA ≤ n.

Now, M is open and F ⊆M ⊆ RrG, Ind ∂M ≤ n, hence IndR ≤ n+1.

We have to prove the claim, i.e. that

(F ∪ U) ∩G ∪ (B r U) = ∅ and F ∪ U ∩ (G ∪ (B r U)) = ∅

It is sufficient to show that the following intersections are empty.

• F ∩G = ∅ by assumption.

• F ∩ B r U ⊆ F ∩ B r (V ∩B) ⊆ V ∩ B r V which is contained in

V ∩B r V ∩B ⊆ (V r U) ∩ B = ∅, since V ∩ B ⊆ U .

• U ∩ (B r U) = (U ∩ B) r U , but ∂U ⊆ R r B, since U is open andclosed in B, hence U ∩ B = U , hence our intersection is empty.

• This is exactly the same situation as above: since U is open and closedin B, also B r U is so.

• U ∩ G ⊆ (U ∩W ∩ A) ∪ (U ∩W ∩ B). The first part is contained in

B rW ∩W ⊆ [(BrW )∪∂W ]∩W = [(BrW )∩W ]∪ (∂W ∩W ) = ∅.The second part is contained in U ∩(BrU), which is empty, as alreadyshown.


The following result about point-finite open coverings of a normal spacewill be useful in the proofs of further results about the large inductivedimension.

Proposition 3.3.4Let U = {Uα}α∈A be a point-finite open covering of a normal space R. Then

there exists an open covering V = {Vα}α∈A such that Vα ⊆ Uα for each α ∈ A.(Note that the two coverings have the same cardinality).

Proof. By Zermelo’s theorem, we can well-order U and write U = {Uα}α<τ .We find, for each α < τ , an open set Vα such that

Rr

[⋃

β<α

{Vβ} ∪⋃

γ>α

{Uγ}]

⊆ Vα ⊆ Vα ⊆ Uα (3.3.1)

To do this, we use transfinite induction.The set Rr

⋃

0<γ Uγ is closed and contained in U0, since U is a covering.Since R is normal, there exists an open set V0 such that

Rr⋃

γ>0

Uγ ⊆ V0 ⊆ V0 ⊆ U0

thus, 3.3.1 holds for α = 0.Now, assume we have defined the open set Vβ ,satisfying the condition, for

every β < α. We want to define Vα. Let U ′ = {Vβ}β<α ∪ {Uγ}γ≥α.We claim that U ′ is a covering of R. For this purpose, let p ∈ R. If

p ∈ Uγ for a γ ≥ α, then clearly p ∈ Uγ ∈ U ′. Else, if p 6∈ Uγ for every γ ≥ α,then, since U is point-finite, there is a maximal β such that p ∈ Uβ (of courseβ < α).If p ∈ Vβ′ for a β ′ < β, then p ∈ Vβ′ ∈ U ′ and we are done. Else, ifp 6∈ Vβ′ for each β ′ < β, then by 3.3.1 (where α = β)

p ∈ Rr

[⋃

β′<β

Vβ′ ∪⋃

β′′>β

Uβ′′

]

⊆ Vβ

Then p ∈ Vβ ∈ U ′ and our claim is proved.Since U ′ is a covering, the closed set Rr [

⋃

β<α Vβ ∪⋃

γ>α Uγ ] is containedin Uα. By the normality of R, there is an open set Vα satisfying 3.3.1.

We only have to prove that V = {Vα}α<τ is a covering (Vα open andVα ⊆ Uα is clear). We proceed in the same way as for U ′: let p ∈ R. Then,since U is point-finite, there is a maximal β < τ such that p ∈ Uβ. If p 6∈ Vβ′

for each β ′ < β, then

p ∈ Rr

[⋃

β′<β

Vβ′ ∪⋃

β′′>τ

Uβ′′

]

⊆ Vβ


which proves that V covers R.

Lemma 3.3.5Let R be a metric space with IndR ≤ n. Then there exists a σ-locally finiteopen basis V of R such that Ind ∂V ≤ n− 1 for each V ∈ V.

Proof. By the Nagata-Smirnov metrization Theorem 3.2.11, R has a σ-locallyfinite open basis

⋃∞i=1 Ui, where each Ui is a locally finite open covering.

Let Ui = {U iα}α∈Ai

. Then, by Proposition 3.3.4, there is an open coveringWi = {W i

α}α∈Aisuch that W i

α ⊆ U iα for each α ∈ Ai.

Since IndR ≤ n, there are open sets V iα (α ∈ Ai) such thatW i

α ⊆ V iα ⊆ U i

α

and Ind ∂V iα ≤ n− 1. Let Vi = {V i

α}α∈Ai. Then V =

⋃∞i=1 Vi is a basis with

the requested properties: by construction it is open with Ind ∂V ≤ n− 1 foreach V ∈ V. V is σ-locally finite, because the Vi’s are locally finite, sinceeach V i

α is contained in U iα. Finally, V is a basis, because each V i

α ∈ V iscontained in U i

α, which is an element of a basis; furthermore V iα contains W i

α

and the W iα’s form a covering.

We consider the inverse situation for n = 0.

Lemma 3.3.6If a metric space R has a σ-locally finite open basis each of whose elementshas empty boundary, then IndR ≤ 0.

Proof. Let V =⋃∞

i=1 Vi be a basis with the stated properties, where each Vi

is a locally finite open collection, and ∂V = ∅ for each V ∈ V. Let F and Gbe closed disjoints subsets of R. Then for each i the set

Ui = Rr⋃

{V | V ∈i⋃

j=1

Vj and V ∩ F = ∅}

is open and closed (since ∂V = ∅ and ∪ij=1Vj is a locally finite collection of

open and closed sets).F is contained in each Ui. Then for the sequence U1 ⊇ U2 ⊇ . . . ⊇ Uk ⊇Uk+1 ⊇ . . . ⊇ F we have

⋂∞i=1 Ui = F . In fact, since V is a basis, it follows

that for each point out of F there is an open neighbourhood of it outside F .In the same way we obtain a sequence of open and closed sets Wi for G suchthat W1 ⊇W2 ⊇ . . . ⊇Wk ⊇Wk+1 ⊇ . . . ⊇ G and

⋂∞i=1Wi = G.

Let

U =∞⋃

i=1

(Ui rWi).


It is open because each Ui rWi is open. The set U is also closed: in fact wecan write

U = (U1 rW1) t∞⊔

i=2

((Ui ∩Wi−1) rWi)

which is a disjoint union of open and closed sets. Now, if a point x ∈ R iscontained in F , then it is also in U , hence in exactly one of the open andclosed disjoint sets. Otherwise, if x 6∈ F , then there is a k such that x 6∈ Uk

for all k ≥ k, hence x ∈ R r Uk. If x is contained in none of the disjointsets whose union is U , then RrUk is an open neighbourhood of x that doesnot intersect U . Conversely, x is contained in one of the open and closedsets, say W , then (R r Uk) ∩W is an open neighbourhood of x that doesnot intersect U . Hence U is closed. Furthermore, F ⊆ U ⊆ R r G. Since∂U = ∅, we have IndR ≤ 0.

Example 3.3.7We show that both Q and R r Q are zero dimensional in the sense of thelarge inductive dimension. We need only show that the dimension is ≤ 0,because the sets are not empty.

Consider first the irrational numbers. Let q ∈ Q, then the neighbourhoodbasis {S1/n(q)}n∈N of q has rational endpoints, hence each S1/n(q)∩ (R r Q)is open in R r Q and has empty boundary.

⋃

n∈N,q∈Q

{S1/n(q)}

is a σ-locally finite open basis each of whose elements has empty boundary,hence by Lemma 3.3.6 we have Ind(R r Q) = 0.

Now we do the same for the set of rational numbers. For each n ∈ N findan irrational number an such that 1

n+1< an <

1n

and define

S ′n(q) = (q − an, q + an) ⊆ R.

The set of intervals {S ′n(q)}n∈N is clearly a neighbourhood basis of q, each of

whose elements has irrational endpoints. Hence, the sets S ′n(q)∩Q are open

and have empty boundary in Q. Thus,⋃

n∈N,q∈Q

{S ′n(q)}

is a σ-locally finite open basis each of whose elements has empty boundary,hence by Lemma 3.3.6 we have Ind Q = 0.

The following result will be generalized in Section 3.5 for spaces of finitedimension.


Lemma 3.3.8Let A be a subset of a metric space R with IndR ≤ 0. Then IndA ≤ 0.

Proof. This is a direct consequence of Lemmata 3.3.5 and 3.3.6: from thefirst lemma follows that there is a σ-locally finite open basis of R each ofwhose elements has empty boundary. Intersecting each element of the basiswith A, we get a basis for A with the same properties, hence IndA ≤ 0.

3.4 Sum theorem for the large inductive di-

mension

The aim of this section is to prove that if a space has a locally countableclosed covering, its dimension does not exceed the dimension of each closedset in the covering. We do this incrementally.

Lemma 3.4.1Let {Fi}i∈N be a countable closed covering of a metric space R such thatIndFi ≤ 0 for each i ∈ N. Then IndR ≤ 0.

Proof. Let G and H be closed disjoint sets in R. Since IndF1 ≤ 0, there isan open and closed subset U1 of F1 such that F1 ∩ H ⊆ U1 ⊆ F1 r G. Thesets H ∪ U1 and G ∪ (F1 r U1) are closed and disjoint. Since R is normal,there are open sets V1 and W1 such that

H ∪ U1 ⊆ V1 G ∪ (F1 r U1) ⊆W1 V1 ∩W1 = ∅ (3.4.1)

Now, since IndF2 ≤ 0, there is an open and closed subset U2 of F2 suchthat

F2 ∩H ⊆ F2 ∩ V1 ⊆ U2 ⊆ F2 rW1 ⊆ F2 rW1 ⊆ F2 rG

The sets V1 ∪ U2 and W1 ∪ (F2 r U2) are closed and disjoint. Then, bynormality of R, there are open sets V2 and W2 such that

V1 ∪ U2 ⊆ V2 W1 ∪ (F2 r U2) ⊆W2 V2 ∩W2 = ∅

Proceding with these arguments, we get three sequences {Ui}, {Vi} and {Wi}(i ∈ N), where each Ui is an open and closed subset of Fi and all the Vi’s andWi’s are open in R, and

Fi ∩H ⊆ Ui ⊆ Fi rG (3.4.2)

Vi−1 ∪ Ui ⊆ Vi Wi−1 ∪ (Fi r Ui) ⊆ Wi (3.4.3)

Vi ∩Wi = ∅ (3.4.4)

3.4. Sum theorem for Ind 67

Then, the sets

V =

∞⋃

i=1

Vi W =

∞⋃

i=1

Wi

are open and, by virtue of 3.4.1, H ⊆ V , G ⊆W . By 3.4.3 and 3.4.4, V andW are disjoint and by 3.4.1 and 3.4.3 Vi ∪Wi ⊇ Ui ∪ (Fi r Ui) = Fi.

Then, V ∪W ⊇ ⋃∞i=1 Fi = R.

Since V is closed (because R r V = W , which is open) and open in Rand H ⊆ V ⊆ R rG, we have IndR ≤ 0.

Lemma 3.4.2Let F = {Fγ}γ∈Γ be a locally finite closed covering of a metric space R suchthat IndFγ ≤ 0 for each γ ∈ Γ. Then IndR ≤ 0.

Proof. For each i ∈ N let Ui = {Uδ}δ∈∆ibe an open covering such that

mesh Ui = supU∈Ui

{diam U} <1

i

and such that for each Uδ ∈ Ui, the closure Uδ intersects finitely many Fγ’s.Such a covering exists, since the space is metric and F is locally finite. We cansuppose Ui is locally finite, since the space is paracompact and the requiredproperties hold also for refinements.

Each Uδ ∩ Fγ is closed in Fγ , hence by Lemma 3.3.1, Ind(Uδ ∩ Fγ) ≤IndFγ ≤ 0. The collection {Uδ ∩Fγ}γ∈Γ is a finite closed covering of Uδ, thusby Lemma 3.4.1, IndUδ ≤ 0.

By Proposition 3.3.4, there is an open covering Vi = {Vδ}δ∈∆isuch that

Vδ ⊆ Uδ for each δ ∈ ∆i. Since IndUδ = 0, there is a subset Wδ ⊆ Uδ whichis open and closed in Uδ and such that Vδ ⊆Wδ ⊆ Uδ.

Since Wδ is open and closed in Uδ, its boundary ∂Wδ is all contained inR r Uδ. On the other hand, Wδ ⊆ Uδ implies Wδ ⊆ Uδ, hence the boundary∂Wδ is all contained in Uδ; hence ∂Wδ is empty, i.e. Wδ is open and closedin R.

The collection Wi = {Wδ}δ∈∆iis locally finite, because each Wδ ∈ Wi is

contained in Uδ ∈ Ui, and Ui is locally finite. Then, W =⋃∞

i=1 Wi is σ-locallyfinite. Furthermore, each element in W is open and closed by the choice ofthe Wδ. Finally, W is a basis of R, because mesh Wi ≤ mesh Ui <

1i.

Then, by Lemma 3.3.6, IndR ≤ 0.

Lemma 3.4.3Let {Fi}i∈N be a countable closed covering of a metric space R such thatIndFi ≤ n for each i ∈ N. Then IndR ≤ n.


Lemma 3.4.4Let {Fγ}γ∈Γ be a locally finite closed covering of a metric space R such thatIndFγ ≤ n for each γ ∈ Γ. Then IndR ≤ n.

Proof. We prove the two lemmata in the same way. We proceed by inductionon n. If n = −1, the assertions are trivially verified.

Assume that the assertions hold for n − 1. Since IndFi ≤ n, by Lemma3.3.5 there is a σ-locally finite open basis Vi =

⋃∞k=1 Vi,k of Fi such that

Ind ∂FiV ≤ n− 1 for each V ∈ Vi. (3.4.5)

Since Vi is σ-locally finite, so is also V ′i = {∂Fi

V }V ∈Vi.

Let Hi =⋃

V ∈Vi{∂Fi

V }. Then⋃∞

i=1 V ′i is a σ-locally finite closed covering

of⋃∞

i=1Hi. By induction hypothesis and by 3.4.5, we have

Ind∞⋃

i=1

Hi ≤ n− 1. (3.4.6)

Vi restricted to Fi r Hi is an open basis of Fi r Hi (it is also σ-locallyfinite and ∂FirHi

V is empty, by definition of Hi). From Lemma 3.3.6 followsInd (Fi rHi) ≤ 0.

Let Gi = Fi r⋃∞

j=1Hj ⊆ Fi r Hi. By Lemma 3.3.8, IndGi ≤ 0. EachGi is closed in

⋃

iGi, hence by Lemma 3.4.1 (or Lemma 3.4.2 for the secondlemma)

Ind∞⋃

i=1

Gi ≤ 0 (3.4.7)

Finally we have R =∞⋃

i=1

Hi ∪∞⋃

i=1

Gi, hence by 3.4.6, 3.4.7 and Lemma 3.3.2,

IndR ≤ n.

The following is the main result.

Theorem 3.4.5 (Sum Theorem)Let F = {Fγ}γ∈Γ be a locally countable closed covering of a metric space Rsuch that for each γ ∈ Γ holds IndFγ ≤ n. Then IndR ≤ n.

Proof. Since F is locally countable, there is an open covering V each of whoseelements intersects at most countably many elements of F . Since R is regularand paracompact, there is a locally finite open covering U = {Uδ}δ∈∆ suchthat U is a refinement of V. Then each Uδ meets at most countably manyelements of F . Since IndFγ ≤ n and Fγ ∩ Uδ is a closed subset of Fγ, itfollows by 3.3.1, that Ind (Fγ ∩ Uδ) ≤ n. Since {Fγ ∩ Uδ}γ∈Γ covers Uδ, byLemma 3.4.3 we have IndUδ ≤ n and by Lemma 3.4.4, since U is a locallyfinite closed covering, IndR ≤ n.

3.5. Subspace and decomposition theorems for Ind 69

3.5 Subspace and decomposition theorems

for the large inductive dimension

In Section 3.3 we saw that if R is a metric space with IndR ≤ n, then thereexists a σ-locally finite open basis V of R such that Ind ∂V ≤ n− 1 for eachV ∈ V. The converse is also true, in fact the following theorem holds.

Theorem 3.5.1Let n ≥ 0. For each nonempty metric space R the following holds: IndR ≤ nif and only if there exists a σ-locally finite open basis V of R such thatInd ∂V ≤ n− 1 for each V ∈ V.

Proof. We need only prove the “if” part of the statement. For n = 0, wehave already proved the result in Lemma 3.3.6.Let n > 0. Let A =

⋃

V ∈V ∂V and B = R r A. {∂V }V ∈V is a σ-locallyfinite closed collection. Since Ind ∂V ≤ n − 1, by the Sum Theorem 3.4.5,IndA ≤ n− 1.

V restricted to B is an open basis of B and satisfies the assumptions ofLemma 3.3.6. Hence IndB ≤ 0. By Lemma 3.3.2, IndR = Ind (A ∪ B) ≤(n− 1) + 1 = n.

Theorem 3.5.2 (Subspace Theorem)For every subset A of a metric space R, IndA ≤ IndR.

Proof. We proceed by induction on the dimension n of the space: theinduction basis corresponds to an empty space and is trivially verified.Toprove the induction step, let A be a subset of an n dimensional space R andassume the result holds for every n−1 dimensional space. By Theorem 3.5.1,there is a σ-locally finite open basis V of R with Ind ∂V ≤ n − 1 for eachV ∈ V. Restrict V to A. Then V|A is a σ-locally finite open basis of A and∂AV|A ⊆ ∂V for each V ∈ V, hence by induction hypothesis Ind ∂AV|A ≤ n−1for each V|A ∈ V|A. By Theorem 3.5.1, we have IndA ≤ n.

The following result says that a space of dimension n can be decomposedin n+ 1 zero-dimensional subsets.

Theorem 3.5.3 (Decomposition Theorem)Let n ≥ 0 and let R be a metric space. Then IndR ≤ n if and only if

R =n+1⋃

i=1

Ai,

where each Ai is a subset of R such that IndAi ≤ 0.


Proof. The “if” part was already proved in Lemma 3.3.2. To prove the inverseimplication, as in the proof of Theorem 3.5.1, we decompose R = A1 ∪ B1

in two disjoint sets such that IndA1 ≤ 0 and IndB1 ≤ n − 1. Then wedecompose B1 in A2 ∪B2, where IndA2 ≤ 0 and IndB2 ≤ n− 2. Proceedingby induction in this way we get n + 1 zero-dimensional sets Ai which coverthe space.

Corollary 3.5.4For every two subsets A, B of a metric space R,

Ind (A ∪ B) ≤ IndA+ IndB + 1.

3.6 Product theorem for the large inductive

dimension

In this section we consider the dimension of the product of two spaces andshow that it is bounded by the sum of the dimensions of the spaces. We needthe following elementary result.

Remark 3.6.1For every subsets V , W of metric spaces R, S respectively, the boundary ofthe product subset in the product space is given by the “derivation” formula

∂R×S (V ×W ) = ∂R (V ) ×W ∪ V × ∂S (W ) .

Theorem 3.6.2 (Product Theorem)Let R, S be metric spaces, not both empty. Then

IndR× S ≤ IndR + IndS.

Proof. Let IndR = n, IndS = m. We proceed by induction on n +m.If one of the two spaces is empty, the assertion is trivially verified.Let n,m ≥ 0. By Theorem 3.5.1 there are σ-locally finite open basis V =⋃∞

i=1 Vi and W =⋃∞

i=1 Wi of R and S respectively, where each Vi and eachWi is locally finite, such that

Ind ∂RV ≤ n− 1 for each V ∈ V (3.6.1)

Ind ∂SW ≤ m− 1 for each W ∈ W (3.6.2)

The collection Vi × Wi = {V ×W | V ∈ Vi,W ∈ Wj} is locally finite andopen in R×S. Then

⋃∞i,j=1 (Vi ×Wj) is a σ-locally finite open basis of R×S.

3.7. Covering dimension 71

By 3.6.1, 3.6.2 and the induction hypothesis, we have: for each V ∈ V, foreach W ∈ W:

Ind(∂RV ×W ) ≤ n− 1 +m

Ind(V × ∂SW ) ≤ n+m− 1

By Remark 3.6.1 and the Sum Theorem 3.4.5, for each V ×W ∈ ⋃∞i,j=1 Vi ×

Wj :

Ind ∂R×S(V ×W ) = Ind((∂RV ×W ) ∪ (V × ∂SW )

)≤ n+m− 1

Then by Theorem 3.5.1, Ind(R× S) ≤ n +m.

3.7 Covering dimension

We turn to consider the covering dimension and prove some results whichwe will use to prove the Theorems 3.8.2 and 3.8.5 where the coincidence ofthe different notion of dimension on metric (and separable for ind) spaces isstated.

Remark 3.7.1For each closed subset F of a space R, we have dimF ≤ dimR.In fact, setting dimR ≤ n, we have to show dimF ≤ n. Let U = {U1, . . . , Uk}be an open covering of F , i.e. each Ui can be written as Ui = F ∩ Vi for anopen subset Vi of R. Then V = {V1, . . . , Vk, RrF} is an open covering of Rand since dimR ≤ n, V has an open refinement W of order at most n + 1.Then {W ∩ F |W ∈ W} is a refinement of U , it is open in F and has orderat most n + 1, hence dimF ≤ n.

Note that this result holds more generally for arbitrary subsets of a metricspace: in fact, it holds for the large inductive dimension, and in the nextsection we show that dim and Ind coincide.

The following result proves that in metric spaces it is the same to considerfinite or arbitrary coverings in the definition of the covering dimension. Infact, each covering has a locally finite refinement because every metric spaceis paracompact, and Theorem 3.7.2 shows that one can consider finite orlocally finite coverings.

Theorem 3.7.2For every metric space R, dimR ≤ n if and only if each open covering of Rhas a locally finite refinement V such that ord V ≤ n+ 1.


Proof. The “if” part is trivial, we need only show the inverse implication. LetU = {Uγ}γ∈Γ be an open covering. Since every metric space is paracompact,we can assume U to be locally finite (otherwise we can take a locally finiterefinement). That means, for each point there is a neighbourhood whichintersect finitely many elements of U , hence there is an open covering A suchthat each A ∈ A intersects finitely many elements of U .

Since the space is paracompact, we can find a locally finite openrefinement B of A, and we suppose it to be indexed by an ordinal τ ,B = {Bν}ν∈τ . We can assume that, defining Fν = Bν for each ν, F = {Fν}ν<τ

is still a refinement of A (otherwise shrink the elements of B as in Lemma3.3.4). Hence F is a locally finite closed covering and, since F refines A,

each Fν meets at most finitely many elements of U . (3.7.1)

Remark 3.7.1 ensures that

dimFν ≤ n for each ν. (3.7.2)

Without loss of generality we can assume F0 = ∅. Using transfinite inductionwe construct a sequence of open sets {Uν,γ}γ∈Γ and a sequence of closed sets{Aν,γ}γ∈Γ, ν < τ such that:

1. Aν,γ ⊆ Fν ;

2. Uν,γ = Uγ r⋃

β<ν

Aβ,γ ;

3.⋃

γ∈Γ

(Uν,γ r Aν,γ) = R;

4. ord {(Fν ∩ Uν,γ) r Aν,γ}γ∈Γ ≤ n + 1.

Setting A0,γ = ∅ and U0,γ = Uγ, the conditions are verified for ν = 0. Letα < τ and suppose we have already defined Aν,γ and Uν,γ for all ν < α.Define

Uα,γ = Uγ r⋃

ν<α

Aν,γ,

then the condition 2 is fulfilled. The collection {Aν,γ}ν<α is locally finite,since Aν,γ ⊆ Fν and F is locally finite. Hence,

⋃

ν<αAν,γ is closed, as alocally finite union of closed sets. Thus Uα,γ is open for each γ.

We claim that the collection {Uα,γ}γ∈Γ is a covering of R.Suppose by contradiction that there is a point x 6∈ Uα,γ for each γ and define


the set Mx = {γ ∈ Γ | x ∈ Uγ}. It is a finite set, since U is locally finite.Since U is a covering, then x is in some Uγ for some γ, thus x 6∈ Uα,γ =Uγ r

⋃

ν<αAν,γ, hence x ∈ Aν(γ),γ for some ν(γ) < α and for some γ.It follows that for each γ ∈Mx there is an index ν(γ) < τ such that ν(γ) < αand x ∈ Aν(γ),γ . Since Mx is finite, there is an index δ < τ such that δ < αand ν(γ) ≤ δ for each γ ∈Mx.Now, we distinguish three cases: if γ ∈Mx such that ν(γ) = δ, then x ∈ Aδ,γ.If γ ∈Mx such that ν(γ) < δ, then x ∈ ⋃ν<δ Aν,γ , hence x 6∈ Uδ,γ.Finally, if γ 6∈Mx, then x 6∈ Uγ .Hence in every case, x 6∈ Uδ,γ r Aδ,γ. This is a contradiction to condition 3,and it is proved that {Uα,γ}γ∈Γ is a covering of R.

Since Fα meets finitely many Uγ and Uα,γ ⊆ Uγ, then only finitely manyelements of {Fα∩Uα,γ}γ∈Γ are nonempty and this collection covers Fα. SincedimFα ≤ n, there is an open covering {Bγ}γ∈Γ of Fα such that Bγ ⊆ Fα∩Uα,γ

for each γ and ord {Bγ}γ∈Γ ≤ n+ 1. Define

Aα,γ = (Fα ∩ Uα,γ) r Bγ

It is a closed set and Aα,γ ⊆ Fα for all γ. This verifies condition 1.Since (Fα ∩ Uα,γ) r Aα,γ = Bγ , then ord {(Fα ∩ Uα,γ) r Aα,γ}γ∈Γ ≤ n + 1,which verifies 4.Since {Bγ}γ∈Γ covers Fα, then Fα ⊆ ⋃γ∈Γ(Uα,γ rAα,γ).Since {Uα,γ}γ∈Γ covers R and Aα,γ ⊆ Fα for each γ, then R r Fα ⊆⋃

γ∈Γ(Uα,γ rAα,γ), hence⋃

γ∈Γ(Uα,γ rAα,γ) = R, and condition 3 is verified.

We have now defined the open sets Uα,γ and the closed sets Aα,γ for each γsuch that the conditions 1, 2, 3 and 4 are verified. The transfinite sequencesare thus defined. Let

Vγ = Uγ r⋃

ν<τ

Aν,γ.

From condition 1 it follows that the closed collection {Aν,γ}ν<τ is locallyfinite, hence

⋃

ν<τ Aν,γ is closed, hence Vγ is open.

We claim that V = {Vγ}γ∈Γ is a covering of R.Suppose by contradiction that there is a point x 6∈ ⋃

γ∈Γ Vγ. Then x 6∈Uγ r

⋃

ν<τ Aν,γ for all γ.Define the set Mx = {γ ∈ Γ | x ∈ Uγ}. For each γ ∈ Mx there is a ν(γ) < τsuch that x ∈ Aν(γ),γ . Since Mx is a finite set, there is some δ < τ such thatν(γ) ≤ δ for all γ ∈Mx.Now we distinguish three cases: if γ ∈Mx and ν(γ) = δ, then x ∈ Aδ,γ.If γ ∈Mx and ν(γ) < δ, then x ∈ ⋃ν<δ Aν,γ, thus x 6∈ Uδ,γ = Uδ r

⋃

ν<δ Aν,γ.


If γ 6∈Mx, then x 6∈ Uγ.In every case, x 6∈ Uδ,γ r Aδ,γ, and this is a contradiction to condition 3.Hence V is an open covering of R such that Vγ ⊆ Uγ for each γ ∈ Γ.

Now we show that ord V ≤ n+ 1.Note that since by definition

Fν ∩ Vγ = Fν ∩ (Uγ r⋃

ν<τ

Aν,γ) = (Fν ∩ Uγ) r⋃

ν<τ

Aν,γ

and by 2

(Fν ∩ Uγ) rAν,γ = (Fν ∩ (Uγ r⋃

β<ν

Aβ,γ)) rAν,γ = (Fν ∩ Uγ) r⋃

β≤ν

Aβ,γ

then

Fν ∩ Vγ ⊆ (Fν ∩ Uν,γ) r Aν,γ.

Hence by condition 4, ord {Fν ∩ Vγ}γ∈Γ ≤ n + 1 for each γ. Since Fν coversR, then ord {Vγ}γ∈Γ ≤ n+ 1.

Hence V is an open refinement of U with order not exceeding n + 1 andthe proof is complete.

The following four lemmata consider this situation: take a locally finitecollection of open subsets of a space R and for each of its elements Uγ take aclosed subset Fγ contained in it. We find sets Vγ that separates each Uγ fromthe corresponding RrFγ , in such a way that the collection of the boundaries∂Vγ has order not exceeding the dimension of the space. Then we show thatinstead of the boundaries we can find open sets with the same property.

Finally, we consider countable many collections of this type and show thesame result holds for the boundaries.

We will use these results in the proof of the coincidence Theorem 3.8.2 inthe next section.

Definition 3.7.3Two collections A = {Aγ}γ∈Γ and B = {Bγ}γ∈Γ are said to be similar if foreach finite subset M ⊆ Γ

⋂

γ∈M

Aγ = ∅ ⇔⋂

γ∈M

Bγ = ∅.

For a collection of sets U = {Uγ}γ, we denote by U the collection {Uγ}γ.


Lemma 3.7.4Let R be a normal space. Let {Uγ}γ∈Γ be a locally finite open collection ofsubsets of R, and {Fγ}γ∈Γ a closed collection such that Fγ ⊆ Uγ for eachγ ∈ Γ. Then there exists an open collection G = {Gγ}γ∈Γ such that

Fγ ⊆ Gγ ⊆ Gγ ⊆ Uγ

and F and G are similar.

Proof. Well order Γ. We use transfinite induction to construct open sets{Gγ}γ∈Γ such that Fγ ⊆ Gγ ⊆ Gγ ⊆ Uγ and for each ν ∈ Γ,

{Kνγ}γ∈Γ is similar to {Fγ}γ∈Γ (3.7.3)

where

Kνγ =

{Gγ if γ ≤ νFγ if γ > ν

We can assume F0 = ∅, and the initial condition is fulfiled defining G0 = ∅.Suppose µ ∈ Γ such that Gγ is already defined for each γ < µ with theproperty 3.7.3 for each ν < µ. Let

Lγ =

{Gγ if γ < µFγ if γ ≥ µ

Then we show that {Lγ} is similar to {Fγ}. Suppose γ1, . . . , γr ∈ Γ.Without loss of generality γ1 < . . . < γj < µ ≤ γj+1 < . . . < γr. Then sinceLγi

= Kγjγi for all i,

r⋂

i=1

Lγi=

r⋂

i=1

Kγjγi

hence, since {Kγjγ }γ∈Γ is similar to {Fγ}γ∈Γ by induction hypothesis,

r⋂

i=1

Lγi=

r⋂

i=1

Kγjγi

= ∅ ⇔r⋂

i=1

Fγi= ∅

Since Lγ ⊆ Uγ for each γ, the collection {Lγ} is locally finite. Denote byΛ the set of finite subsets of Γ. For each λ ∈ Λ let Eλ =

⋂

γ∈λ Lγ . Each Eλ

is a finite intersection of closed sets. Thus, the collection {Eλ}λ∈Λ is closedand locally finite.

Let E =⋃{Eλ | EΛ ∩ Fµ = ∅} be the union of all finite intersections of

Lγ ’s which do not intersect Fµ. It is closed (because union of locally finiteclosed sets) and E ∩ Fµ = ∅.


Since the space is normal, we can separate the closed sets Fµ and (R r

Uµ) ∪E, i.e. we can find an open set Gµ such that

Fµ ⊆ Gµ ⊆ Gµ ⊆ Uµ rE.

In particular, Gµ ∩ E = ∅.Now the open sets Gγ are defined for γ ≤ µ. We need only show that

{Kµγ } is similar to {Lγ}, which is similar to {Fγ}. For this purpose, let

γ1, . . . , γr ∈ Γ.

If⋂r

i=1Kµγi

= ∅, then⋂r

i=1 Lγi= ∅ since Lγi

⊆ Kµγi

for each i.

Conversely, suppose⋂r

i=1 Lγi= ∅. We have to show that

⋂ri=1K

µγi

= ∅.Without loss of generality γ1 < . . . < γj ≤ µ < γj+1 < . . . < γr. If γj 6= µ,we are done, because then Kµ

γi= Lγi

for all i. Suppose γj = µ, then

∅ =

r⋂

i=1

Lγi= Lγ1

∩ . . . ∩ Lγj−1∩ Fµ ∩ Lγj+1

∩ . . . ∩ Lγr. (3.7.4)

If the set

r⋂

i=1

Kµγi

= Lγ1∩ . . . ∩ Lγj−1

∩Gµ ∩ Lγj+1∩ . . . ∩ Lγr

(3.7.5)

is not empty, there is a point p ∈ Gµ. Hence p 6∈ E, and by constructionof E, p 6∈ Lγ for all γ such that Lγ ∩ Fµ = ∅. Since by 3.7.5, p ∈ Lγi

fori = 1, . . . , j−1, j+1, . . . , r, we have Lγi

∩Fµ 6= ∅, hence⋂r

i=1 Lγi6= ∅, which

contradicts 3.7.4.

Lemma 3.7.5Let R be a metric space with dimR ≤ n. Let {Uγ}γ∈Γ be a locally finite opencollection of subsets of R, and {Fγ}γ∈Γ a closed collection such that Fγ ⊆ Uγ

for each γ ∈ Γ. Then there exists an open collection {Vγ}γ∈Γ such that

Fγ ⊆ Vγ ⊆ Vγ ⊆ Uγ

and ord {∂Vγ}γ∈Γ ≤ n.

Proof. For each γ define the open covering Wγ = {Uγ , R r Fγ}. Define thecovering

∧

γ∈Γ

Wγ =

{⋂

γ∈Γ

Wγ |Wγ ∈ Wγ for each γ ∈ Γ

}


Note that each element of ∧{Wγ} is a finite intersection of Uγ ’s intersectedwith all other RrFγ . In other words, denoting by Λ the set of finite subsetsof Γ,

∧

γ∈Γ

Wγ =

{⋂

γ∈λ

Uγ ∩⋂

γ 6∈λ

(Rr Fγ) | λ ∈ Λ

}

.

Since the collection {Uγ} is locally finite, then ∧{Wγ} is locally finite, too.Each of its elements is open, because the Uγ’s in the intersection are finitelymany, and ∩(R r Fγ) is open, since Fγ is locally finite. Hence ∧{Wγ} is alocally finite open covering.

By Theorem 3.7.2 there is a locally finite open covering N = {Nδ}δ∈∆

which refines ∧{Wγ}γ∈Γ and with ord N ≤ n + 1. Each Nδ meetsfinitely many Fγ’s, because N refines ∧{Wγ} and {Uγ} is locally finite. ByProposition 3.3.4, there is a closed covering {Kδ}δ∈∆ with Kδ ⊆ Nδ for eachδ.

To each Nδ and each Fγ intersecting Nδ assign an open set Nδ(Fγ) suchthat:

Kδ ⊆ Nδ(Fγ) ⊆ Nδ(Fγ) ⊆ Nδ and (3.7.6)

Nδ(Fγ) ⊆ Nδ(Fγ′) (or viceversa) for γ 6= γ′. (3.7.7)

This is possible because there are only finitely many Fγ ’s which intersect Nδ.The sets Nδ(Fγ) are related to the sets Fγ only for the cardinality.

LetVγ =

⋃

δ∈∆

Nδ(Fγ).

Note that in the union we can consider only the δ’s such that Nδ intersectsFγ , since the other Nδ(Fγ)’s are empty. For each γ, let Dγ be the set ofindices δ ∈ ∆ such that Fγ ∩Nδ 6= ∅, then we can write Vγ =

⋃

δ∈DγNδ(Fγ).

We claim thatFγ ⊆ Vγ ⊆ Vγ ⊆ Uγ .

To show the first inclusion, note that {Kδ}δ∈Dγis a covering of Fγ, hence

Fγ ⊆⋃

δ∈Dγ

Kδ ⊆⋃

δ∈Dγ

Nδ(Fγ) = Vγ .

To show the last inclusion, note that since each Nδ intersects finitely manyFγ ’s, say Fγ1

, . . . , Fγk, Nδ ⊆ Uγ1

∩ . . .∩Uγk∩⋂γ 6=γi

(RrFγ). Hence Nδ(Fγ) ⊆Nδ ⊆ Uγ . Since Vγ is a locally finite union of Nδ(Fγ)’s,

Vγ =⋃

δ∈Dγ

Nδ(Fγ) =⋃

δ∈Dγ

Nδ(Fγ) ⊆⋃

δ∈Dγ

Nδ


it follows that Vγ ⊆ Uγ .We now prove that {Vγ} has the required properties. Clearly, it is an

open collection, and we have just seen that Fγ ⊆ Vγ ⊆ Vγ ⊆ Uγ. It remainsto prove that ord {∂Vγ} ≤ n.

By contradiction, suppose that ord {∂Vγ}γ∈Γ > n, that means we canfind distinct γi’s (i = 1, . . . , n+ 1) such that:

n+1⋂

i=1

∂Vγ 6= ∅

Since each Vγ is a locally finite union of some Nδ(Fγ), there are some Nδi(Fγi

)(i = 1, . . . , n+ 1) such that:

n+1⋂

i=1

(

Nδi(Fγi

) rNδi(Fγi

))

6= ∅ (3.7.8)

If δi = δj for i 6= j, from 3.7.7 we obtain(

Nδi(Fγi

) rNδi(Fγi

))

∩(

Nδj(Fγj

) rNδj(Fγj

))

= ∅

which is a contradiction to 3.7.8. Hence the δi’s (i = 1, . . . , n + 1) must beall distinct.

Let

p ∈n+1⋂

i=1

(

Nδi(Fγi

) rNδi(Fγi

))

(3.7.9)

From 3.7.6 it follows that p 6∈ ⋃n+1i=1 Kδi

.Since {Kδ} is a covering of R, there is a δ with p ∈ Kδ. Thus, δ 6= δi for

all i = 1, . . . , n+ 1.By 3.7.6 and 3.7.9 it follows that p ∈ ⋂n+1

i=1 Nδi, hence

p ∈ Kδ ∩n+1⋂

i=1

Nδi⊆ Nδ ∩

n+1⋂

i=1

Nδi

which is a contradiction to ord N ≤ n+ 1. Then, ord {∂Vγ}γ∈Γ ≤ n.

Lemma 3.7.6Let R be a metric space with dimR ≤ n. Let {Uγ}γ∈Γ be a locally finite opencollection of subsets of R, and {Fγ}γ∈Γ a closed collection such that Fγ ⊆ Uγ

for each γ ∈ Γ. Then there exist open collections {Vγ}γ∈Γ and {Wγ}γ∈Γ suchthat

Fγ ⊆ Vγ ⊆ Vγ ⊆Wγ ⊆ Uγ

and ord{Wγ r Vγ

}

γ∈Γ≤ n.


Proof. By Lemma 3.7.5 there is an open collection V = {Vγ}γ∈Γ such thatFγ ⊆ Vγ ⊆ Vγ ⊆ Uγ and ord {∂Vγ}γ∈Γ ≤ n.

Since the space is normal and U is locally finite, by Lemma 3.7.4 thereis an open collection H = {Hγ}γ∈Γ with ∂Vγ ⊆ Hγ ⊆ Hγ ⊆ Uγ and {∂Vγ} issimilar to H.

For each γ ∈ Γ define Wγ = Vγ ∪Hγ . Then

∂Vγ ⊆Wγ r Vγ ⊆ Hγ

Hence, since {∂Vγ} is similar to {Hγ}, it is also similar to {Wγ rVγ}. Hence,ord {Wγ r Vγ} ≤ n.

Further, the following inclusions hold:

Fγ ⊆ Vγ ⊆ Vγ ⊆Wγ ⊆ Uγ

Lemma 3.7.7Let R be a metric space with dimR ≤ n. Let {Uγ}γ∈Γi

, i ∈ N, be locallyfinite open collections of subsets of R, and {Fγ}γ∈Γi

closed collections suchthat Fγ ⊆ Uγ for each γ ∈ Γi, for each i ∈ N. Then there exist open subsetsVγ (γ ∈ ⋃i∈N Γi) such that


and ord {∂Vγ}γ∈�

i∈N Γi≤ n.

Proof. Suppose the sets of indices are disjoint. Let γ ∈ Γ1, then by Lemma3.7.6 there are open collections V1 = {V 1

γ }γ∈Γ1and W1 = {W 1

γ }γ∈Γ1such

that

Fγ ⊆ V 1γ ⊆ V 1

γ ⊆W 1γ ⊆ Uγ

and ord {W 1γ r V 1

γ }γ∈Γ1≤ n.

Now let γ ∈ Γ1 ∪ Γ2, then by Lemma 3.7.6 there are open collectionsV2 = {V 2

γ }γ∈Γ1∪Γ2and W2 = {W 2

γ }γ∈Γ1∪Γ2such that

V 1γ ⊆ V 2

γ ⊆ V 2γ ⊆ W 2

γ ⊆ W 1γ if γ ∈ Γ1

Fγ ⊆ V 2γ ⊆ V 2

γ ⊆W 2γ ⊆ Uγ if γ ∈ Γ2

and ord {W 2γ r V 2

γ }γ∈Γ1∪Γ2≤ n.


Proceeding in this way, we get a sequence of open collections Vi ={V i

γ}γ∈Γ1∪...Γiand Wi = {W i

γ}γ∈Γ1∪...Γisuch that

Fγ ⊆ V iγ ⊆W i

γ ⊆ Uγ (3.7.10)

ord {W iγ r V i

γ}γ∈Γ1∪...∪Γi≤ n (3.7.11)

V iγ ⊆ V i+1

γ (3.7.12)

W iγ ⊇ W i+1

γ (3.7.13)

For γ ∈ Γi define

Vγ =∞⋃

i=1

V iγ

Then, since for each integer j, Fγ ⊆ V jγ and V j

γ ⊆ Uγ , we have


Let γ ∈ Γi. Then

Vγ r Vγ =∞⋃

j=1

V jγ r

∞⋃

j=1

V jγ ⊆

∞⋂

j=1

(

V jγ r V j

γ

)

⊆∞⋂

j=1

(

W jγ r V j

γ

)

which, for each fixed integer k, is contained in W kγ r V k

γ . The collection

{W kγ r V k

γ }γ has order not exceeding n, thus it holds ord {∂Vγ}γ ≤ n.

3.8 Coincidence theorems

Now we are ready to prove that in metric spaces the dimensions dim and Indcoincide.

Lemma 3.8.1Let n ≥ 0. Let V be a σ-locally finite open basis of a metric space R, suchthat ord {∂V }V ∈V ≤ n.Then IndR ≤ n.

Proof. We proceed by induction on n: the case n = 0 was already provedin Lemma 3.3.6. Let n > 0 and suppose the result holds for n − 1. LetV = {Vγ}γ∈Γ. Then Vγ = {∂Vγ ∩ Vγ′}γ′∈Γ is a σ-locally finite open basis of∂Vγ with ord {∂∂Vγ

V | V ∈ Vγ} ≤ n − 1, because every point of ∂Vγ lies inat most other n− 1 boundaries, since the order of ∂V is at most n.By induction hypothesis, Ind ∂Vγ ≤ n− 1; thus, by Theorem 3.5.1, IndR ≤n.

3.8. Coincidence theorems 81

Theorem 3.8.2 (Coincidence Theorem I)For every metric space R, dimR = IndR.

Proof. First, let IndR ≤ n; we show dimR ≤ n.By the decomposition Theorem 3.5.3, we can write R =

⋃n+1i=1 Ai, where

IndAi ≤ 0 for each i. Let U = {Uj}j=1,...,k be an open covering of R. Weclaim that for each i there is an open covering {Vj}j=1,...,k of Ai such that

ord {Vj}j=1,...,k ≤ 1 Vj ⊆ Uj (3.8.1)

To see that such a covering exists, fix an index i and let Zj = Uj ∩Ai. Then{Zj}j covers Ai. By Proposition 3.3.4 there is a covering {Yj}j of Ai suchthat Yj ⊆ Yj ⊆ Zj . Since IndAi ≤ 0, we can separate each Yj from Ai r Zj

with an open and closed set Xj, that means Yj ⊆ Xj ⊆ Zj . The collection{Xj}j is a covering of Ai, because {Yj}j , is a covering and each Xj containsYj.Now set V i

j = Xj r⋃

k<j Xj for each j. The Vj ’s are open and closed, pairwisedisjoint and cover Ai, furthermore Vj ⊆ Uj , that means this is a covering withthe requested properties.

For each point p ∈ Vj choose ε(p) > 0 such that

Sε(p)(p) ∩ Ai ⊆ Vj

Sε(p)(p) ⊆ Uj

and defineWj =

⋃

p∈Vj

{S 1

2ε(p)(p)}.

SinceWj∩Ai = Vj, then Wi = {Wj}j=1,...,k is an open covering of Ai, Wj ⊆ Uj

for each j.We want show that ord Wi ≤ 1. By contradiction, suppose there is

a point p ∈ Wj1 ∩ Wj2 for distinct indices 0 ≤ j1, j2 ≤ k. Then by thedefinition of the Wj’s, there are x ∈ Vj1 and y ∈ Vj2 such that

p ∈ Sε(x)/2(x) Sε(x)(x) ⊆ Uj1 Sε(x)(x) ∩Ai ⊆ Vj1

p ∈ Sε(y)/2(y) Sε(y)(y) ⊆ Uj2 Sε(y)(y) ∩Ai ⊆ Vj2.

It follows that

d(x, y) ≤ d(x, p) + d(p, y) <ε(x)

2+ε(y)

2≤ max{ε(x), ε(y)},

that means x ∈ Sε(y)(y) or y ∈ Sε(x)(x). Since these conditions are symmetric,suppose the first holds. Then, x ∈ Vj1 ⊆ Ai implies x ∈ Sε(y)(y) ∩ Ai ⊆ Vj2.


Hence, x ∈ Vj1 ∩Vj2, which contradicts 3.8.1. Thus, we have shown ord Wi ≤1.

Define W =⋃n+1

i=1 Wi. This is an open refinement of U with order ≤ n+1,thus dimR ≤ n.

Now, to complete the proof, let dimR ≤ n; we have to show IndR ≤ n.Since R is paracompact, we can take a sequence Ui = {Uγ}γ∈Γi

, i ∈ N, oflocally finite open coverings such that limi→∞ mesh Ui = supU∈Ui

{diam U} =0, as in the proof of Lemma 3.4.2. Using Proposition 3.3.4 we can constructclosed coverings Fi = {Fγ}γ∈Γi

, i ∈ N, with Fγ ⊆ Uγ for all γ. By Lemma3.7.7 we can find open sets Vγ, γ ∈ ⋃∞

i=1 Γi such that Fγ ⊆ Vγ ⊆ Vγ ⊆ Uγ

and ord {∂Vγ}γ∈�

Γi≤ n. Then V = {Vγ}γ∈

�Γi

is a σ-locally finite opencollection. That it is a basis can be seen using the same argument as in theproof of Lemma 3.4.2. Hence by Lemma 3.8.1, IndR ≤ n.

The Coincidence Theorem 3.8.2 can fail outside the class of metric spaces.We report an example of a compact Hausdorff space S such that

dimS = 1 and indS = IndS = 2.

We describe briefly how to construct the space S, without proving theassertions. The description of this example and all the proofs can be foundin [Pea75], chapter 4, §3.4.

Example 3.8.3Let Y1 and Y2 be two spaces, each of which is homeomorphic to the long line.Identify the last element of Y1 with the last element of Y2, and call this pointy0. Define the set

S = {(y, t) ∈ Y × I | y = y0 or t ∈ C} ,

where C denotes the Cantor set. For i = 1, 2, let

Si = {(y, t) ∈ S | y ∈ Yi} .

Write each point of C in its ternary expansion, i.e.∑∞

i=1tn3n , where each tn

is 0 or 2. Let D ⊆ C consist of the points in C such that writing the ternaryexpansion, tn = 0 almost always or tn = 2 almost always. Then D is densein C. Define a map

ψ1 : C → I∞∑

i=1

tn3n

7→∞∑

i=1

t′n2n

where

t′n =

{0 if tn = 01 if tn = 2.

3.8. Coincidence theorems 83

This map is continuous, order-preserving and surjective. Let M1 = ψ1(D) r

{0, 1}; it is the set of dyadic rationals in the unit interval. In particular, M1

is countable and dense in the unit interval.The function

f : I → I

x 7→ ex − 1

e− 1

is surjective and strictly ascending. We claim that f maps each rational in(0, 1) on an irrational number. In fact, suppose x = p

qis a rational number

in (0, 1) such that f(x) = r, where r is rational. Then

ep = (r(e− 1) + 1)q

i.e. e is solution of a polynomial equation, which is impossible, since e istranscendent.Define

ψ2 : C → Ix 7→ f ◦ ψ1(x)

and let M2 = ψ2(D) r {0, 1}. Then M2 is countable, and it is a dense subsetof the open unit interval.Note that M1 and M2 are disjoint. The maps ψi : C → I, i = 1, 2 arecontinuous, order-preserving and surjective.We say that two points t, s ∈ D,

t =∞∑

i=1

tn3n

s =∞∑

i=1

sn

3n

are related if there exists a positive integer m such that sn = tn for alln < m, sm = 0, tm = 2 and sn = 2, tn = 0 for all n > m (or the same holds,exchanging s with t). It is easy to see that D is partitioned into related pairs.If s, t ∈ C, then ψ1(s) = ψ1(t) if and only if s and t are related points inD. Considering C as the unit interval where open intervals are deleted, thenD consists of the endpoints of the deleted intervals, and a pair of points isrelated exactly when they are the endpoints of some deleted interval.Let Y1 and Y2 have the interval topology, and define

gi : Yi × C → Si

setting

gi(y, t) =

{(y, t) if y < y0

(y0, ψi(t)) if y = y0


Let Si have the identification topology with respect to the surjection gi, i.e.we take the largest topology on Si such that gi is continuous.Since Si is the continuous image of the Hasdorff compact space Yi ×C, Si iscompact and it can be proved that it is also Hausdorff.It can be shown (Proposition 4.3.2 [Pea75]) that dimSi = IndSi = indSi = 1.The set S1 ∩ S2 = {y0} × I is closed both in S1 and S2 and has its usualtopology both as a subspace of S1 and of S2. Hence we can give S the weaktopology with respect to the covering {S1, S2}, i.e. the largest topology onS that induces the given topology on S1 and on S2.Then S is a compact Hausdorff space and S1 and S2 are closed in S. It canbe shown that, since S = S1 ∪ S2 and the Si are closed,

dimS = 1.

Furthermore, one can prove that

indS = IndS = 2.

Remark 3.8.4Every separable metric space R has a countable basis.In fact, let A ⊆ R be a countable dense subset and define

B ={B1/2n(x) | x ∈ A, n ∈ N

}.

It is easy to see that B is a countable basis of R.

Theorem 3.8.5 (Coincidence Theorem II)For every separable metric space R, dimR = IndR = indR.

Proof. We need only prove IndR ≤ indR, since the equivalence of Ind anddim was already proved in the preceding Theorem and ind ≤ Ind is clear fromthe definitions of these concepts of dimension. We proceed by induction onindR. The induction basis, that is the case indR = −1, is trivial.

Assume IndS ≤ indS for each separable space S with indS ≤ n− 1 andlet indR = n, where R is a metric separable space, hence by Remark 3.8.4,it has a countable open basis {Ui}i∈N. We define open sets Wij, i, j ∈ N inthe following way:

• if there exists an open set W such that Ui ⊆ W ⊆ Uj with ind ∂W ≤n−1 (and Ind ∂W ≤ n−1 by induction hypothesis), then set Wij = W ;

• otherwise Wij = ∅.

3.9. The dimension of the Euclidean space 85

Since indR = n, {Wij}i,j∈N is an open basis with Ind ∂Wij ≤ ind ∂Wij ≤n− 1. By Theorem 3.5.1 IndR ≤ n.

The hypothesis that the space is separable is necessary: an example of ametrizable space P such that

indP = 0 and dimP = IndP = 1

was given by Prabir Roy in [Roy62] and [Roy68], and it is described also in[Pea75], chapter 7 §4.

3.9 The dimension of the Euclidean space

In this section we show that the dimension of the Euclidean space Rn is n.To prove it, we introduce the notion of geometrical complex and prove theSperner’s Lemma, which we will use to prove that dim Rn ≥ n.

Definition 3.9.1The (geometrical) n-simplex with vertices p0, . . . , pn ∈ Rm is:

[p0, . . . , pn] =

{n∑

i=0

λipi |n∑

i=0

λi = 1, 0 ≤ λi ≤ 1

}

.

The k-face [pi0 , . . . , pik ] is the set of points of the simplex such that λi = 0for each i 6= ij , where j = 0, . . . , k.A finite set K of geometrical simplexes in Rm is a (geometrical) complex if

• every face of each simplex in K belongs to K and

• the intersection of two simplexes in K is either a face of both simplexesor it is empty.

A triangulation of a space X consists of a complex K together with ahomeomorphism h : K → X, i.e. it is a subdivision of the space intosimplices.

Lemma 3.9.2 (Sperner’s Lemma)Let {Si}i=0,...,n be the (n − 1)-faces of an n-simplex T n and {Fi}i=0,...,n acovering of T n such that Fi ⊆ T n r Si for each i = 0, . . . , n. Let K bea complex which triangulates T n. Then there is an n-simplex of K whichintersects every Fi.


Proof. To each vertex p of K assign an integer i(p) such that p ∈ Fi(p). Toeach r-simplex T = [p0, . . . , pr] of K assign the set i(T ) = {i(p0), . . . , i(pr)}.Let {Tj}j=1,...,k be the n-simplexes of K and s the number of Tj ’s with i(Tj) ={0, . . . , n}.

We have to show s > 0. To do that, it is enough to prove that s is odd.We proceed by induction on n. If n = 0, the claim is trivially true.Assume the claim for the (n − 1)-simplexes. Let sj be the number of

(n − 1)-dimensional faces T ′ of Tj such that i(T ′) = {1, . . . , n}. For each jit must be sj = 0, 1 or 2. For, if sj 6= 0, there is a face T ′ = [p1, . . . , pn] ofTj = [p0, . . . , pn] such that i(T ′) = {1, . . . , n}. Now, if i(p0) = 0, T ′ is theonly face with the requested property, hence sj = 1. Else, 1 ≤ i(p0) ≤ nimplies sj = 2.

Let t be the number of Tj’s such that sj = 2. Then

k∑

j=1

sj = s+ 2t (3.9.1)

because the term s comes from the Tj’s with sj = 1 and the term 2t fromthe Tj ’s with sj = 2.

Let u be the number of (n− 1)-simplexes T ′ of K such that they are faceof exactly one of the Tj ’s, i.e.

{i(T ′) = {1, . . . , n}T ′ ⊆ ⋃n

i=0 Si(3.9.2)

Let v be the number of (n− 1)-simplexes T ′′ of K such that they are face oftwo of the Tj ’s, i.e.

{i(T ′′) = {1, . . . , n}T ′′ 6⊆ ⋃n

i=0 Si(3.9.3)

Then,∑k

j=1 sj = u+ 2v.Combining this with 3.9.1, we get u+ 2v = s+ 2t.It is sufficient to show that u is odd. If T ′ is an (n−1)-simplex with 3.9.2,

T ′ ⊆ S0 (because if T ′ ⊆ Si for an i 6= 0, since Fi ⊆ T n r Si, then i 6∈ i(T ′),contradiction).

Therefore, u is the number of (n− 1)-simplexes T ′ of K contained in S0

and such that i(T ′) = {1, . . . , n}. Applying the induction hypothesis to thecovering {Fi ∩ S0}i=1,...,n of the (n− 1)-simplex S0, we get that u is odd.

Theorem 3.9.3dim Rn = n.

3.9. The dimension of the Euclidean space 87

Proof. The covering dimension of R is ≤ 1, as we saw in Example 3.1.8. TheProduct Theorem 3.6.2 together with the Concidence Theorem 3.8.2 ensuresthat dim Rn ≤ n.

To show that dim Rn ≥ n, we consider a simplex T n ⊂ Rn. By Remark3.7.1, we have dimT n ≤ dim Rn. Then, it is enough to prove that T n hasdimension at least n.

Let {Si}i=0,...,n be the (n − 1)-faces of T n. Consider the open covering{T n r Si}i=0,...,n of T n. By Remark 3.1.4, we can find an open covering V ={Vi}i=0,...,n of T n such that Vi ⊆ T n r Si for each i and ord V ≤ dimT n + 1.Then there is a closed covering {Fi}i=0,...,n of T n with Fi ⊆ Vi.

We have to prove that ∩ni=0Fi 6= ∅. Then, ∩n

i=0Vi 6= ∅, hence ord V ≥ n+1,hence dimT n ≥ n.

By contradiction, suppose ∩ni=0Fi = ∅, then since T n is compact, {T n r

Fi}i=0,...,n is an uniform covering of T n, that means, there is an ε > 0 suchthat {Sε(p) | p ∈ T n} is a refinement. Here ε is called a Lebesgue numberfor the space.

Now construct a triangulation K of T n, such that each of its simplexeshas diameter at most ε. Then each simplex of K is contained in T n r Fi forsome i, contradicting Lemma 3.9.2.

Corollary 3.9.4Each open subset of Rn has dimension n.

In fact, each open subset of Rn contains a n-simplex as subset.Hence, every n-dimensional manifold (where the dimension is the one

defined in chapter 1) has (covering) dimension n, because the dimensionof manifolds is given via local homeomorphisms with open subsets of theEuclidean spaces. In other words, the concept of dimension for metricspaces studied in this chapter extends the concept of dimension for manifolds,studied in chapter 1.

As a last note, we investigate how the definitions of topological dimensionsapply to algebraic sets with the Zariski topology.

Remark 3.9.5The covering dimension of an algebraic set counts its cardinality if it is finite,and otherwise it is not defined.

Proof. Let X be an irreducible algebraic set (or an irreducible componentof an algebraic set) and take n distinct points x1, . . . , xn ∈ X. Define the(Zariski-)open sets Ui = X r {x1, x2, . . . , xi−1, xi+1, . . . , xn}, for 1 ≤ i ≤ n.Then U = {Ui}1≤i≤n is a covering and each of its refinements contains at


least n elements. Remember that every finite intersection of (Zariski-)opensets is not empty. Therefore every refinement of U has order at least n, andthe covering dimension of X can be made arbitrarily large, except in thecase where X is a finite set of points, where the covering dimension is thecardinality of X.

Remark 3.9.6The inductive dimensions on algebraic sets coincide with the Krull dimension.

Proof. We proceed by induction on the Krull dimension of the algebraic setX. If X is empty, it has inductive dimensions and Krull dimension −1.Suppose indX ≤ krdimX for every algebraic set X with krdimX ≤ n− 1.Let X be an irreducible algebraic set such that krdim(X) = n and let Y bea (Zariski-)closed set in X. Hence krdimY ≤ n−1. Take a point p ∈ X rYin case we consider the small inductive dimension or a closed set Y ′ ⊆ Xdisjoint from Y in case consider the large inductive dimension. Consider theopen set U = X r Y . It contains p or Y ′. Note that ∂U = U rU = Y , sinceevery open set is dense in X. Hence ind ∂U ≤ n− 1 by induction hypothesisand it follows that indX ≤ krdimX.To show that the dimensions coincide, let Y be an irreducible closed set inX with krdim Y = n − 1 (it exists by Theorem 2.5.9), and let p ∈ X r Y(or Y ′ ⊆ X r Y ). Then for each open neighbourhood U of p disjoint from Y(or each open set U ⊇ Y ′ disjoint from Y ), Y is contained in the closed setX r U , which has dimension at least n − 1, since it contains the closed setY of dimension n− 1. Furthermore, ∂U = X rU since X is irreducible andU is dense in X. That means indX = n = krdimX.

Part II

Informal learning ofmathematics

Chapter 4

An exhibition about the 4Dspace

The interuniversitary research centre for the communication and informallearning of mathematics matematita1 is preparing an exhibition about thefour dimensional Euclidean space. The exhibition is planned to be readyin the next future. Nevertheless a part of it has already been set up and“tested” in Genova, within the “Festival della Scienza” (Science Festival,25th October–6th November 2007). The work we are presenting here is partof this project; we attended to the realization of one of the exhibits and ofthe related interactive animations. In this chapter we describe the exhibitionwhich was set up in Genova giving particular attention to the section weworked for.

The title of the exhibition is “Un tuffo nella quarta dimensione” (“A Tripinto the Fourth Dimension”). It is thought for non specialist visitors witha middle or high education level and for schools. In the exhibition some4D objects are described and represented through 3D models, and problemsabout their properties are posed.

A deep comprehension of the topics is not expected from the visitor, whois mainly invited to reflect on the concept of dimension and on some factsthat characterize the different “usual” dimensions (i.e. the lower dimensions1, 2 and 3). Dimensional analogy is then used to explain what happens“jumping” from the three dimensional to the four dimensional space andcomparisons with the the passage from the 2D space to the 3D space areoften made.

1http://www.matematita.it/

92 Chapter 4. An exhibition about the 4D space

4.1 Some basic definitions

We introduce the definitions of the figures and concepts that are describedinformally in the exhibition.

A hyperplane in an n-dimensional space is an affine subspace of codimen-sion 1. It is the set of zeroes of a linear equation a1x1 +a2x2 + . . .+anxn = a0

where ai ∈ R (0 ≤ i ≤ n).

A polytope is a compact convex region of an n-dimensional space enclosedby a finite number of hyperplanes, with nonempty interior. The last conditionmeans that we consider only n-dimensional polytopes in a space of dimensionn. Polytopes in dimension 0, 1, 2 and 3 are points, segments, polygons andpolyhedra respectively.

A face is the intersection of an n-dimensional polytope πn with a tangenthyperplane, i.e. a hyperplane that does not intersect the interior of thepolytope. Every face of πn is a polytope πk for some k ∈ {0, 1, . . . , n − 1}and is therefore called k-face. A 0-face is called vertex, a 1-face is called edge,and an (n− 1)-face is called facet.

If the mid-points of all the edges that emanate from a given vertex O ofa n-dimensional polytope πn lie on a hyperplane, then they are the verticesof a (n− 1)-dimensional polytope, called the vertex figure of πn at O.

A polytope πn (n > 2) is said to be regular if its facets are regular andthere is a regular vertex figure at every vertex. Then it can be shown thatthe facets are all equal and the vertex figures are all equal. The k-skeletonof an n-polytope (0 ≤ k ≤ n) consists of all faces of dimension up to k.

An n-cube, also called measure polytope, is the product of n intervals ofequal length. It is an n-dimensional regular polytope, each of whose facetsis parallel to exactly one other facet (opposite to it) and orthogonal to eachother facet. A 2-cube is called square, a 3-cube simply cube, and we call a4-cube hypercube.

Informally speaking, an unfolding of a polyhedron is obtained by cuttingits surface along some edges and flattening it into the plane, in such a waythat the plane figure (also called net) is connected. The unfolding is notunique. The procedure of folding up a net in dimension 3 allows to fold thefigure along the edges. In general one has to specify which edges of the netare to be joined, since there might be ambiguity, but for the polyhedra weconsider the folding procedure is unique.

The same can be made in higher dimensions: an unfolding of an n-polytope is a (n−1)-dimensional connected diagram consituted by the facetsof the polytope. It can be folded up along the (n − 2)-faces to obtain theoriginal polytope in dimension n.

4.1. Some basic definitions 93

For example, a net of a four dimensional polytope is formed by polyhedraconnected by their faces. To fold it up in the four dimensional space, onewould have to fold along the 2-faces.

A way to give a formal definition of the unfolding of a polytope is thefollowing. Define the adjacency graph of the polytope, that is a graph whosevertices are the facets of the polyhedron and an edge between two verticesis drawn if and only if the facets corresponding to the vertices are adjacent(that means, they have a common (n − 2)-face, if n is the dimension of thepolytope). For such a graph one can consider a spanning tree, that is a treecontained in the graph, which contains all the vertices of the graph. Everyspanning tree is an unfolding of the polytope.

We describe two procedures for projecting an n-dimensional figure F ∈ Rn

onto an hyperplane A. Parallel projection: given a line r not parellel to A,through each point of F draw a line parallel to r, to meet A in an “image”point. If r is orthogonal to A, then the projection is called orthogonal. One-point perspective projection: suppose a point p ∈ Rn (called projection center)is given such that each line passing through p and a point of F is not parallelto A. Then through each point of F draw a line that passes through p, tomeet A in an “image” point.

For a polyhedron consider a graph whose vertices and edges representrespectively the vertices and the edges of the polyhedron, two verticesbeing adjacent exactly when the corresponding vertices of the polytopeare endpoints of the same edge. It is not hard to see that every regularpolyhedron has a planar graph, since one can project the polyhedron ontothe sphere in which it is inscribed and then project the sphere onto theplane with a stereographic projection. Such a planar graph is called Schlegeldiagram of the polyhedron. A Schlegel diagram of a polyhedron can alsobe obtained through a face-centered one-point perspective projection of the1-skeleton of the polyhedron onto a hyperplane. Face-centered projectionmeans that the line through the projection center and the center of a facet isorthogonal to the hyperplane of projection and the projection center is closeenough to the face.

In an analogous way it is possible to define the Schlegel diagram of an n-polytope, and also in this case the Schlegel diagram can be obtained througha facet-centered one-point perspective projection.

The unit n-sphere is the set of points Sn = {x ∈ Rn+1 : ‖x‖ = 1}. Ageneric n-sphere is the set of points {x ∈ Rn+1 : ‖x− c‖ = r}, where c ∈ Rn+1

is the center and r > 0 is the radius of the sphere. We call hypersphere a4-sphere.


4.2 Contents of the exhibition

The main subject of the exhibition is the hypercube. This polytope has beenchosen for variuos reasons: it is one of the simplest regular polytopes, itsproperties can be compared with the analogous ones of the cube and finallyit has some characteristics that allow to present nontrivial problems andresults.

A little part of the exhibition is dedicated to the other five regularpolytopes and to the hypersphere.

The exhibition is subdivided into seven parts: introduction, models,symmetry, slices, dice, flip a cube, regular polytopes. Some interactiveanimations and short movies complete the exhibition, presenting the themesdeveloped during the visit. Our animations are endowed with writtenexplanations, while the other ones are not, some of them being onlyrecreational or intended to give suggestions.

The proposed visit does not touch all the seven parts, but only theintroduction, the models, one among the central ones and at the end thesection about polytopes. This choice is due to the special arrangement insidethe “Festival della Scienza”: there are many little exhibitions and the visitordoes not have much time to dedicate to each one, usually not more than30–45 minutes.

We describe in some detail each part of the exhibition. The section “flip acube”, which is the subject of our work, will be described in detail in section4.3.

4.2.1 Introduction

Some posters narrate in a catchy way an imaginary dialogue between twofriends, one of them has just come back from a tour in the 4D space. Somephenomena that happen in the 4D space are recalled in the story, withoutdescribing anything in detail. The posters are not intended to explainanything, but only to involve the visitor.

A large model representing a projection of a 120-cell (one of the regular4D polytopes) completes this introductory section.

4.2.2 Models

This section is the central part of the exhibition. Its aim is to describe some3D models of the hypercube, mainly by analogy with 2D representations ofthe cube. The representations presented with models and explications arethe following:

4.2. Contents of the exhibition 95

• models obtained unfolding the cube and the hypercube, figure 4.1

• parallel projections, figure 4.2

• perspective projections (Schlegel diagram), figure 4.3.

Figure 4.1: The unfoldings of a cube and a hypercube.

Figure 4.2: Parallel projections of a cube and a hypercube.

Figure 4.3: Schlegel diagram of a cube and a hypercube.

There is also a model of two polyhedral tori (figure 4.4): it representstwo complementary solid tori of the hypercube and suggests with differentcolours which faces are to be identified. Each torus of the hypercube is madeof four cubes in the 4D space, but our model deforms them to represent thelinked tori in the 3D space.


Figure 4.4: A model of the two tori of a hypercube (image by matematita).

Two problems are posed to help the visitor think about the differentrepresentations and their properties. To find the solution it can be useful toconsider the different models at disposal.

The first problem asks to find, among some different constructions madeof 8 cubes each, which ones can be obtained unfolding a hypercube. In otherwords, the visitor is invited to think about some necessary properties thatthe constructions of cubes must have to be folded into the hypersurface ofa hypercube. The answer can be found comparing this problem with theanalogous one for the cube. To complete this section, two posters show the11 unfoldings of the cube among all possible plane constructions of six squaresglued at some edges, and the 261 unfoldings of a hypercube.

The second problem invites the visitor to think about a path on thesurface of the cube, that touches four faces forming a plane torus, i.e. thefaces of the cube excluding two parallel ones. Then the analogous problem inthe hypercube is considered. In the polytope the problem is more interesting,because the torus one can identify is solid, made of four cubes, and itscomplementary is again a solid torus made of four cubes, linked with thefirst one. In this way one can have an idea of the structure of two linkedpolyhedral tori represented in the model of figure 4.4. This concept is howeverdifficult for the average visitor and it is not explained in the exhibition, butonly mentioned and recalled by some models showing the tori. A particularlyinterested visitor can obtain more explanations from the guides.

4.2. Contents of the exhibition 97

4.2.3 Symmetry

This section investigates the rotation group of the cube, giving attention tothe fact that it acts transitively on vertices, edges and faces. In other words,vertices, edges and faces are not distinguishable: in any way one fixes two ofthem, it is possible to take one into the other with rotations.

Thanks to some kaleidoscopes, the symmetry of the cube is highlightedand its rotation axes are pointed out. They can be seen also in figure 4.5:one axis of order 4 through the centre of two opposite faces, one of order 3through two opposite vertices and one of order 2 through the centres of twoopposite edges.

Figure 4.5: The rotation axes of the cube in a kaleidoscope.

The visitor should note that these axes are not easy to see in the twodimensional representations of the cube. Analogously, the three dimensionalrepresentations of the hypercube do not point out its symmetry.

The aim of this section is not to study the symmetry of the hypercubeor to talk about symmetry planes of 4D objects, but only to convince thevisitor that also the hypercube is “somehow” symmetric, even if that is noteasy to note simply looking at its 3D representations.

4.2.4 Slices

Some models present different slices of the cube. The visitor is invited to usethem to find out which polygons, among some given ones, can be obtainedas slices of a cube. This problem is in general not trivial, but the polygonsare chosen such that one needs only some evident properties to tell whetherthey can be slices of a cube.

This question is posed to prepare the visitor to an interactive animation,that shows the 3D slices of a hypercube along four hyperplanes.


4.2.5 Dice

How many ways are there to dispose the numbers on the faces of a cubic die?

The visitor has thirty different dice at disposal and is asked to tell whetherthey represent all the possible cases. One way to arrive to the solution is thefollowing. Fix a face, giving it the same number for all dice and consider theopposite face: there can be five possible numbers on it. Now fix a lateralface and consider the opposite one, for which there are three possibilities. Todispose the numbers on the last two faces, there are other two possibilities.Then, there are 30 ways to dispose the numbers on the faces of a die.

Using the same argument, the visitor can be convinced that there are 210different ways to dispose the numbers on the hyperfaces of a hypercube.

4.2.6 Flip a cube

This part is the subject of our work and is described in detail in the nextsection 4.3.

4.2.7 Polytopes

The visitor is helped to understand how many regular polyhedra there exist,by disposing regular polygons around a vertex and checking when they canbe folded up in the 3D space. The visitor should understand that there areat most five regular polyhedra: one can dispose three equilateral trianglesaround a vertex (obtaining the tetrahedron), four equilateral triangles(octahedron), three squares (cube) or three pentagons (dodecahedron).

Analogously, the visitor can try and dispose regular polyhedra aroundan edge, without filling the whole space around that edge. Six possibilitieswill be found: three, four or five tetrahedra around each edge (that gives thehypertetrahedron, the hyperoctahedron and the 600-cell respectively), threecubes (hypercube), three octahedra (24-cell) or three dodecahedra (120-cell).

Some models suggest a way to obtain a polytope from one in a lowerdimension: one considers the construction of the pyramid (to obtain atetrahedron from a triangle and a hypertetrahedron from a tetrahedron),of the prism (to obtain a cube from a square and a hypercube from acube) and of the bipyramid (to obtain an octahedron from a square and ahyperoctahedron from an octahedron). Here it is suggested that tetrahedron,cube and octahedron have a corresponding polytope in every dimension (andin fact they are the only regular polytopes that exist in every dimension, butthis is not described).

4.3. Flip a cube 99

4.3 Flip a cube

The phenomenon we want to explain is mathematically elementary: SO(3)preserves the orientation of R3, but one can act with SO(4) on a threedimensional object (that is, contained in R3 ⊆ R4) such that the imageis again in R3, but has the opposite orientation. In other words, it is possibleto realize a reflection of R3 as a rotation of R4.

The group of rotations SO(3) of R3 is a subgroup of the group of rotationsof R4 via the following mapping:

SO(3) ↪→ SO(4)

A 7→(A 00 1

)

Also the set O(3)− = O(3) r SO(3) of orthogonal 3 × 3 matrices withdeterminant −1 can be seen as a subset of SO(4):

ψ : O(3)− ↪→ SO(4)

A 7→(A 00 −1

)

Then, since SO(4) is connected, there is a rigid movement Mt which takesthe identity matrix in ψ(A), i.e. such that M0 = id and M1 = ψ(A). Inother words, a three dimensional object can be rotated in R4 and broughtback into R3 such that is is flipped. It is then possible to flip a cube, butone can not notice when it is flipped, because of its symmetry, described alsoin a section of the exhibition. Hence we need to introduce an orientationon the cube to note when it is flipped. To do that, we use different colours,arranged clockwise or counterclockwise around a vertex.

We chose to “flip” a cube, because it can be moved in the hypersurfaceof a hypercube, and cubes and hypercubes are the principal subject of thewhole exhibition, which the visitor should be familiar with, after visiting thefirst part of the exhibition. In order to explain the movement of the cubewe use the analogy, treating first the case of flipping a square in the threedimensional space.

Let us consider a square first. To orientate a square it suffices to paintits edges with at least three different colours, to distinguish two edges whichconcur in a vertex. In the animations and in the models for the exhibitionfour colours were used. The reason is purely aesthetic and “psychological”:most people who saw the models painted with three colours asked why twoedges were painted with the same colour. That way of colouring seems to beperceived as “innatural” in some sense by many people.


The visitor can observe and use a model of a square coloured in this way.The orientation does not change if one simply rotates the square maintainingit adherent to the plane of the table. However, moving the square in the 3Dspace, it can be brought back onto the plane with the opposite orientation.

This movement can also be performed keeping the edges of the squarealways adherent to the surface of a cube, having the same edges’ length asthe square. The visitor is asked to try and find this movement, using theobject shown in figure 4.6, a plexiglas cube with a metallic square slightlylarger than the faces of the cube. The square remains adherent to the surfaceof the cube thanks to some magnets placed in the cube. Figure 4.7 gives anidea of the movement that filps the square.

Figure 4.6: The cube present in the exhibition.

Since a cube can be represented in two dimensions, also the movement ofthe square on the surface of the cube can be drawn in two dimensions. Thisstep is necessary in order to understand the corresponding movement for acube in the hypercube. In fact, for a square on a cube we can see both the2D representation and the 3D object, while when we consider the hypercube,we can at best see one of its 3D representations and one can only try andimagine what happens in the 4D space.

The visitor can then use the computers where some interactive animationshelp him to be aware of the phenomenon. The movement of the squareis represented in two 2-dimensional models of the cube (its unfolding andits Schlegel diagram: see figure 4.8) and the movement of the cube inthe hypercube is represented in the corresponding models of the hypercube(figure 4.13). We describe these animations in section 4.3.2 and 4.3.3.

4.3. Flip a cube 101

Figure 4.7: How to flip a square on the surface of a cube.

4.3.1 Animations and images

To develop the animations and to create animated images we used freesoftware:

• JavaTMPlatform, Standard Edition 6 Development Kit (JDKTM)2,a development environment for building applications, applets, andcomponents using the Java programming language;

• JoGL3 (Java OpenGL), a wrapper library to use OpenGL in the Javaprogramming language;

• NetBeans IDE 5.54, a free, open-source Integrated Development Envi-ronment for software developers;

2http://java.sun.com/3https://jogl.dev.java.net/4http://www.netbeans.org/


• POV-Ray5 (Persistence of Vision Raytracer), a free tool (with sourcecode available) for creating three-dimensional graphics.

Nine interactive animations were developed in the Java programminglanguage. Four of them describe how it is possible both to flip a squareon a cube and a cube on a hypercube.

The other ones are about related topics, and are described in the nextsection 4.4. Two animations illustrate some two dimensional representationsof a cube and some three dimensional representations of a hypercube. Twoanimations show different projections of a hypercube. The last one describesanother phenomenon: how to reverse a sphere in a hypersphere.

Some common characteristics of the interactive animations are listed here.

• There is a control panel under the drawing. It allows the user to interactwith the animation in an immediate way.

• One or more sliders control the parameters (such as angles of rotationor frame of the animation) that manage the main actions.

• A button lets the animation play and stop (not available in someapplets).

• Other buttons control specific parameters for the applet, for exampleto choose the model to display.

• The left mouse button controls a trackball to rotate the object.

• The right mouse button controls the zoom, which changes the size ofthe figure.

Some animated images were realized using POV-Ray to support thedescription of the interactive animations. Some images show the analogousphenomenon as an animation, but in a lower dimension. Some images showexactly the same as an animation, and they are useful as short films tobe placed in the exhibition. They are intended not necessarily to explainsomething, but also to give some suggestions and to make the exhibition moreinteresting. The advantages of these images versus the interactive animationsare the better quality of the 3D graphic and the lower use of cpu.

The applets are organized into web pages. Each page contains both theinstructions to interact with the applet and a brief informal explanation ofwhat is going on from a mathematical point of view. The latter text containsalso links to the other pages.

5http://www.povray.org/


The web pages can be found in the CD–ROM included to this thesis,where also the source code and the stand alone applications are available.

4.3.2 Flip a square

The animations in figure 4.8 show the square moving on the surface of acube till it is flipped. The cube is represented by one of its unfoldings inthe first animation and by the Schlegel diagram in the second, i.e. a centralperspective projection where the centre of the projection is located near thecentre of the face that is parallel to the projection plane and far from it.Central projection means that the projection plane is parallel to two facesof the cube. In other words, the Schlegel diagram is a graph whose verticesrepresent the vertices of the cube and the edges, that represent the edges ofthe cube, meet only in the vertices.

Figure 4.8: Flipping a square on a cube.

In both animations it is possible to see the object as two dimensional oras a representation which gives the idea of 3D.

We report the text written for the web page containing the first animation.

Moving a square on a plane, it is not possible to flip it. Paint its edgesyellow, blue, red and green, disposing the four colours clockwise alongthe perimeter. Any movement of the square on the plane lets thecolours be disposed in the same way.

However, if the square is free to move in the 3D space, it is easyto flip it and bring it back onto the plane with the colours disposedcounterclockwise.It is also possible to flip the square without detaching its edges fromthe surface of a cube.If the cube is represented by one of its unfolding, it is then possible


to follow the movement of the square on a plane, even if it takesplace in the 3D space.

The square seems to be broken and in some moments the edgesare painted twice, but folding up the cube one can verify that thesegments always form a square.

In an analogous way, it is not possible to flip a cube in the usualspace, but that becomes possible in the 4D space and one can followthe movement on an unfolded hypercube.

It is possible to see the movement of the square also in another planerepresentation of the cube, its Schlegel diagram.

Figure 4.9: Folding up the cube to see the square.

The user should note that in the two dimensional representations someproperties of the 3D object are not preserved.

In particular, in the unfolded cube, the edges of the square do not seemto be always adjacent and the square seems to “open”. In fact the cubecan be seen as quotient of its unfoldings, where some of the edges must beidentified following the relations which permit to fold up the model in the3D space. Another property of this representation that can be noted by theuser is that the edges of the square are drawn twice when they coincide withthe cube edges that must be identified. The user can comprehend that these


“errors” are due to the representation we choose for the cube: the model cannot preserve all the properties of the cube because it is contained in a spacewith lower dimension as the cube itself. In this sense it is important that theuser can “close” the unfolding to obtain the cube and verify that the figurehe sees is always a square.

Figures 4.9 and 4.10 show two examples where the unfolded cube is foldedup to obtain the 3D object: in this manner the user can verify that thefigure drawn is always a square, even if it does not appear so in the 2Drepresentation.

Figure 4.10: Folding up the cube to see the square.

In the second animation, the default view mode is 2D, that means theprojection of the cube is fixed (Schlegel diagram), but one can switch to a3D view, activating the trackball to make possible to change the projectioncentre and see different projections. The shading helps interpret the objectsas three dimensional.

In the Schlegel diagram there are representation “errors” as well: thesquare becomes a trapezium and even a line segment when the point of viewis complanar with the square. Moreover, the edges length varies during themovement. Examples of these drawings can be see in figures 4.11 and 4.12,where the 2D view is shown together with the 3D view from the same pointof view and from another point, to make easier to comprehend how the 3Dfigure looks like.


Figure 4.11: A square seems a trapezium in the Schlegel diagram.

Figure 4.12: A square seem a line segment in the Schlegel diagram.

The text for the web page is almost the same as for the previousanimation, but the paragraph on the characteristics of the 2D model issubstituted by the following.

The square seems to become larger and smaller, deform into atrapezium and degenerate into a line segment, but looking atthe cube from different positions, it is possible to verify that thequadrilateral is in fact always a square.

The visitors are now prepared to see what happens in a higher dimension:they have just seen that a 2D model of a 3D object might cause somerepresentation “errors”, and can now accept that the same holds for 3Dmodels of 4D objects.

4.3.3 Flip a cube

The animations in figure 4.13 show how a cube can be flipped moving inthe hypercube. The usual two representations of the hypercube are used:the first animation considers some different unfoldings, while the second onereproduces a Schlegel diagram of the hypercube.

In the animation with the unfolded hypercube, we chose to let the userchange the unfolding, to see how some of the 2-faces are identified, in orderto verify that the cube faces are always adjacent even if the cube seems tobe “open” and sometimes some faces are drawn twice. These are exactly the


Figure 4.13: Flipping a cube on a hypercube.

same “errors” as for the square in the cube and can be seen in figure 4.15.Some of the infoldings of the hypercube are shown in figure 4.14, togetherwith the intermediate movements that take an unfolding in the other.

Figure 4.14: Changing the unfolding of the hypercube.

The text written for the web page containing this animation is thefollowing.


Figure 4.15: How to flip a cube on an (unfolded) hypercube.

Moving a cube in the usual 3D space, it is not possible to flip it.Paint three faces - concurrent in a vertex - yellow, blue and red,disposing the colours counterclockwise around the chosen vertex.Any movement of the cube in the space lets the colours be disposedin the same way.

As for a square, if we move the cube in the four dimensional space,we can flip it and bring it back to our space with the colours disposedin the opposite way.It is possible to flip the cube without detaching its faces from thehypersurface of a hypercube.If the hypercube is unfolded, it is then possible to follow themovement of the cube in our space, even if it takes place in the4D space.

The cube seems to be broken and in some moments the faces arepainted twice, but changing the unfolding of the hypercube, one canverify that the six faces are always adjacent.

It is possible to see the movement of the cube also in another 3D


representation of the hypercube, its Schlegel diagram.

To understand the relation between the two models of the hypercube,an animation shows how to fold up and unfold the hypercube.

Figure 4.16: How to flip a cube on a hypercube (Schegel diagram).

The text written for the web page containing the second animationis almost the same but the paragraph about the properties of the 3Dmodel is substituted by the following, which describes the “errors” in therepresentation of the cube that can be seen in the Schlegel diagram. Theycan be noted also looking at figure 4.16.

The cube seems to become larger and smaller, deform into atruncated pyramid and degenerate into a square. That happensbecause we are representing a 4D object in the 3D space: as someof the cubes in the hypercube, also the cube is deformed by theprojection from the 4D to the 3D space.

The same movement of the cube in an unfolded hypercube and in theSchlegel diagram is the subject of two animated images, some frames ofwhich can be seen in figure 4.15 and 4.16.


4.4 Other animations

Other interactive animations were developed to illustrate topics related tothe previous one.

4.4.1 Folding up a (hyper)cube

The animations in figure 4.17 should help the visitor to understand howthe two representations of the hypercube (the unfolding and the Schlegeldiagram) are linked: we can apply the same projection which gives theSchlegel diagram also to the hypercube while we fold and unfold it.

Figure 4.17: Folding a cube and a hypercube.

The first animation is a two dimensional projection of an unfolded cubethat folds up in the 3D space to construct the cube. In the default view theunfolding lies in a plane that is parallel to the projection plane, but the usercan rotate the object.

The same construction is the subject of an animated image, figure 4.18shows some of its frames.

The text written for the web page containing this applet is the following.

There are different ways to represent a cube in the plane, for exampleunfolding it (there are 11 different plane unfoldings) or projecting itto obtain the Schlegel diagram.If we observe the “most classical” 2D unfolding of a cube, the latincross, that folds up in the 3D space, we obtain a two dimensionalrepresentation of the passage from a model to the other.

First the four faces adjacent to the basis are folded up till they reacha vertical position, then the lid is shut.

4.4. Other animations 111

Figure 4.18: How to fold a cube in the 3D space.

In an analogous way, one can represent the 3D unfolding of ahypercube folding up in the 4D space, to obtain the Schlegel diagramof the hypercube.

The second animation is the analogon of the first one, in a higherdimension. It shows how a three dimensional unfolding of a hypercube canbe folded up in the four dimensional space to obtain the 4D object. The 4Dobject is projected into the 3D space.

In this animation two projections are applied to the hypercube: the firstone maps the 4D object into the 3D the space, and the second one projectsthe 3D object onto a 2D representation on the monitor.

The user can change the centre of the second projection, that is rotatethe 3D model, but the first projection is fixed, the centre being close to thecentre of a hyperface.

The text for the corresponding web page is the following:

There are different ways to represent a hypercube in the 3D space,for example unfolding it (there are 261 different solid unfoldings) or


Figure 4.19: How to fold a hypercube in the 4D space.

projecting it to obtain the Schlegel diagram.As for a cube, also the hypercube can be “constructed” folding upin the 4D space one of its three dimensional unfoldings.

Through a projection, we can represent this construction in the 3Dspace. First the six hyperfaces adjacent to the basis are folded uptill they reach a “vertical” position, then the lid is shut.

An animated image has the same subject, some frames of which are infigure 4.19.

4.4.2 Projections

The two animations in figure 4.20 show different projections of a hypercube.The first one allows the user to act on a slider, controlling the projection

centre: it is at first near the hypercube (giving a perspective projection), butcan be made far away from the object, since one obtains a parallel projection.

To understand what happens with the centre of projection, the user canlook at an image that describes the projection of a cube onto a plane. Some


Figure 4.20: Projections of a hypercube.

frames of this image are in figure 4.21. Here the projection centre and therays through the cube vertices are drawn, and define the cube shadow on theprojection plane.

The web page contains the animated image of figure 4.21 and the followingtext.

To obtain a two dimensional representation of the cube, it is sufficientto project it onto a plane. We can choose the plane to be parallelto a face of the cube (one-point perspective). The image obtainedthrough the projection changes according to the projection centre.If this is located near the centre of a face, we obtain the Schlegeldiagram of the cube, where

• the face closest to the plane seems to be the largerst one andto contain all the other ones;

• parallel edges that are also parallel to the plane are representedby parallel segments;

• parallel edges that are not parallel to the plane are representedby segments that converge to a point.

If we move the projection centre far away from the cube, till it reachesan “infinite distance”, the projection rays become parallel and weobtain an axonometric projection, where

• the faces parallel to the plane are projected onto equal squares;

• parallel edges are represented by parallel segments.

In an analogous way we can represent the hypercube by projectingit into the usual 3D space. We obtain different representationsaccording to the projection centre. When this is close to the centre


of a hyperface, we obtain the Schlegel diagram, while if it is infinitelyfar from the hypercube, we have a parallel-edges representation.

To discover more about this topic, you can look at other projections,obtained by rotating the hypercube in the 4D space or moving theprojection centre in different ways.

Figure 4.21: Projections of a cube.

Some frames of this animation can be seen in figure 4.22: the first frameshows the Schlegel diagram, a central projection (or one-point projection),while the last one shows an axonometric projection.

The edges of the hypercube are parallel to the coordinate axis in the 4Dspace. The edges that are parallel to each other are drawn with the samecolour. Since the space of projection is parallel to three of the coordinateaxes, the edges along these three directions are parallel also in the centralprojection, while the edges that are parallel to the fourth axis appear asconvergent to a point. The two cubes parallel to the 3D space of projectionare not equal in the projection: the one closest to the space of projection isrepresented by a smaller cube because it is far from the projection centre.In the axonometric projection the projection rays are parallel to each other.Every group of edges, which are parallel to an axis, is projected onto parallelsegments and the two cubes parallel to the 3D space of projection areprojected onto equal cubes.

The meaning of the coloration is explained in the text that accompaniesthe next animation. There we also mention what happens to parallel edges,but we are not expecting that every visitor comprehends this concept. In


Figure 4.22: Projections of a hypercube.

fact the second animation concerning the hypercube projections is probablymore difficult to use and understand for a visitor since more parameters canbe controlled: the position of the projection centre and the position of thehypercube with respect to the 3D projection space.

We recall a simple fact about rotations in order to describe how theanimation is constructed: we claim that the group SO(4) of rotations in R4

is generated by the rotations about the six distinct planes each of whose isdetermined by two coordinate axes.In fact, suppose R is an element of SO(4). Note that if R fixes the firstcoordinate axis, Re1 = e1, then proving our claim is equivalent to say thatSO(3) is generated by the rotations about the three coordinate axes. This isa well known fact. Let Re1 = v 6= e1. To prove our claim, it suffices to findsubsequent rotations about the coordinate planes that map v = (v1, v2, v3, v4)to e1.

Note that one can map (a, b) to (∗, 0), defining

(x, y) 7→

a√a2 + b2

b√a2 + b2

− b√a2 + b2

a√a2 + b2

(xy

)

This is a rotation, because the matrix has determinant 1. It follows that wecan map v to e1 through subsequent rotations, in the following way. Thereis a rotation Rθ1

e2,e3such that Rθ1

e2,e3v = (v′1, v2, v3, 0). A rotation Rθ2

e2,e4maps

(v′1, v2, v3, 0) to (v′′1 , v2, 0, 0). Finally a rotation Rθ3

e3,e4maps (v′′1 , v2, 0, 0) to


(w1, 0, 0, 0). Since the rotations preserve the norm, we have w1 = ±1. Ifw1 = −1, take Rθ3+π

e3,e4instead of Rθ3

e3,e4.

Then Rθ3

e3,e4Rθ2

e2,e4Rθ1

e2,e3v = e1 and the claim is proved.

From this fact follows that we can determine how the hypercube is rotatedwith respect to the 3D space fixing six angles. These angles are controlledby six sliders.

There is also the trackball, which rotates the 3D projection of thehypercube. It gives the idea of threedimensionality to the projection. Ifthe projection centre is on the fourth axis then rotating the projection isthe same as rotating the hypercube about the planes that contain the fourthaxis. But when the projection centre is in a general position, this does nothold anymore.

Four more sliders control the position of the 4D projection centre, fixingthree angles and the distance from the origin. The three angles define theline through the origin and the centre of projection, and the distance definesuniquely the position of the centre.

The text for the web page is the following.

The projection of a hypercube in the three dimensional space dependson the position of the projection centre, but also on the position ofthe hypercube with respect to the 3D projection space.

In the animation at first two hyperfaces are parallel to the 3D space,that has coordinates x, y and z. The edges are parallel to thecoordinate axes and are drawn red, green, blue (those parallel tothe axis x, y, z respectively) and white (those parallel to the fourhaxis).

The projection centre is close to the centre of a hyperface parallelto the 3D space. In this way we get the Schlegel diagram. The rededges are projected onto parallel segments, and the same holds forthe green ones and the blue ones, while the edges parallel to thefourth axis are projected onto segment converging to a point.

In the 4D space an object can rotate about a plane. If we rotatethe hypercube about the xy plane, acting on the top right slider,the red segments remain parallel, and also the green ones, sincethe corresponding edges are still parallel to the 3D space after therotation. On the other hand, the blue edges, that are no more parallelto the 3D space, are now projected onto segments converging to apoint.

When the projection centre is at infinite distance, we get axonometricprojections and the edges with the same colour are in any case


projected onto parallel segments.

4.4.3 Hypersphere

The last animation developed for the exhibition is about a different object:the hypersphere. We want to show how a sphere can be reversed in thehypersphere: in contrast to the movement of the cube, here we use topologicaldeformations, not only rigid movements. We want to make an inversion:bring the sphere into itself, in such a way that the inside is now outside andviceversa. Figure 4.23 shows a step in the movement of the sphere.

If we define the hypersphere S3 as a subset of C2,

S3 = {(x, y) ∈ C2 | | x | 2 + | y | 2 = 1}

then the set

S3 ∩{

(x, y) ∈ C2 | | x | 2 ≤ 1

2

}

defines a solid torus. The torus S3 ∩ { | x | 2 ≥ 12} defines a second torus,

which meets the first in the common boundary. Hence we can represent thehypersphere by two linked solid tori.

The meridians of a torus must be identified with the parallels of the otherand viceversa. The identification is helped by some meridians and parallelsdrawn with two different colours.

We chose this representation with two tori because it is analogous to oneof the representations of the hypercube presented in the exhibition (figure4.4).

Figure 4.23: A (deformed) sphere inside the solid tori of the hypersphere.

The inside surface of the sphere is painted with a different colour as theoutside.


We describe figure 4.24, which reports some frames of the animation. Atthe beginning the sphere is inside a torus, then is becomes larger till it touchesthe surface of the torus in one meridian. So it touches also the surface of thesecond torus, along a parallel.

Figure 4.24: The sphere deformation in the hypersphere.

Then we inflate the sphere, that remains in the first torus till it becomeshalf the torus, delimited by two disks whose boundaries are two oppositemeridians. Now we push the sphere into the second torus, till it becomesmaximal, i.e. has the same radius as the hypersphere. In this moment thesphere “cuts” each torus into two symmetric parts. One torus is cut alongtwo opposite meridians and the other along two opposite parallels.

We deflate the sphere in the first torus, exactly in the same way as weinflated it. The sphere is then entirely contained in the first torus and can


be moved inside it till it reaches the same position it had at the beginning.Now the inside of the sphere has become outside and viceversa: this can benoted observing the colour of the surface.

This is exactly the same as “inflating” a circumference on the surfaceof a sphere till it becomes maximal and then “deflating” it in the otherhemisphere. The interior has now taken the place of the exterior.

Colouring the sphere is nothing else as choosing a vector field that isnormal or transverse to the surface of the sphere. We chose a “complicated”,not smooth, movement for the sphere, because we wanted it to have thesurface on the surface of the tori for the longest time possible, in orderto visualize it in our representation of the hypersphere. However, thephenomenon we are showing here can be obtained also with a diffeotopy (thatis, a differentiable ambient isotopy). We describe a diffeotopy f : S3×I → S3

that takes the 2-sphere S ⊆ S3 into itself, exchanging its interior with theexterior. Note that such a diffeotopy exists: one can inflate the spheremaintaining it parallel to itself till it becomes maximal, then deflate it inthe other hemisphere of the hypersphere till it has the same radius as atthe beginning and finally apply a rotation of π about the plane parallelto the sphere and passing through the centre of the hypersphere. Thenft : S3 → S3 is a diffeomorphism for each t ∈ I such that f0 = id, f1(S) = Sand f1|S : S → S reverses the orientation. Furhermore, ft(S) is a smoothsphere for every t ∈ I (this is not true for the movement chosen in theanimation).

We claim that ft : S3 → S3 always maintains the orientation. Supposenot, then there is a t′ ∈ I such that dft′(x) = 0 for some x, which is acontradiction to the fact that ft is a diffeomorphism for every t ∈ I.

Choose a vector field n that is normal to the surface of the sphere att = 0. Then dft · n is a transverse field to S in every moment, thatmeans the coloration is defined in every moment by the normal component.Furthermore, df1 · n = −kn, where k is a positive function. In other words,the interior surface of the sphere has become the exterior and viceversa.

The text written to explain the animation is probably very difficult forthe visitor.

To describe the hypersphere, the 4D analogon to the sphere, wecan think of two solid tori that are glued together in such a waythat the meridians on the surface of one torus coincide with theparallels on the surface of the other one, and viceversa. In theanimation meridians and parallels are painted with different coloursto emphasize the identification.

A sphere, painted blue outside and pink inside, is in the hypersurface


of the hypersphere, inside one torus. We inflate the sphere till ittouches the surface of the torus in a meridian. Since the surfacesof the tori are identified, the sphere appears also on the other torus,along a parallel.

We inflate the sphere further and the part on the surface of the toripasses entirely inside the second torus, till the sphere is representedby two disks (in the first torus) which are adjacent to an annulus(inside the second torus).

Now we can deflate the sphere symmetrically and it returns insidethe torus where it was at the beginning, but painted pink outside.In other words, it is reversed: the inside surface is now outside andviceversa.

This text is surely the most difficult for the visitor. The main reason isthat the exhibition does not present the hypersphere and in this page only avery short explanation about the representation of the hypersphere is given.

4.5 Future work

For the future we are planning to explain the construction of the hypersphereas two linked solid tori in analogy with the polyhedral solid tori in thehypercube (see figure 4.4). We also plan to talk about the (topological)“similarity” between cube and sphere and translate this in a higher dimensionto give a better idea of a hypersphere.

Another idea is to present knots and links in three dimensions and to showthat they can be untied in four dimensions. It would be also interesting toinvestigate how a sphere can be knotted in four dimensions. However thisis probably too difficult to be presented in an exposition and explained togeneral public.

Acknowledgments

I want to thank Valter Cavecchia (CNR, Istituto di Fotonica e Nanotec-nologie - Trento) for the suggestions and the help about openGL and forsolving some troubles in my animations and applets.

Special thanks to Maria Dedo (Dipartimento di Matematica “FederigoEnriques” - Universita degli Studi di Milano) for the seminary about the4D taught to scholars in Salorno, where I saw her way of teaching that hasfascinated me.

I also thank my supervisor in this work, Domenico Luminati (Laboratoriodi Didattica e Comunicazione della Matematica del Dipartimento di Mate-matica - Universita di Trento) for the enthusiasm that he has in his workand that he has transmitted to me.

Ringrazio le persone che mi hanno stimolata ad aprire gli occhi durantequesti anni passati (non solo) all’universita: Laura, con cui ho condivisoesperienze meravigliose, Damiana che a Tubingen e diventata una sorella,la comix che mi ha permesso di non perdere il contatto con il mondoreale anche se studiavo matematica, il mio fratellone Bruno con cui dialogosoprattutto durante le camminate in montagna, i miei genitori che non mihanno ostacolata nelle mie scelte anche se per loro erano difficili da accettare,Giorgio che mi ha sorpresa con un regalo indimenticabile alla stazione diBolzano.

Grazie anche agli amici conosciuti all’universita, per i momenti trascorsiinsieme; mi dispiace di avervi spesso solo sfiorato, senza conoscervi per benecome meritavate. In ordine sparso: Viviana, Cipro, Thomas, Rosj, Maria,Maurizio, Sara, Trentista, Giorgio, Francesca, Marianna, Silvia, Marco,Marina, Francesca, Stefania, Bolo, Demelza, Ilaria, Davide, Mattia, Robi,Bis...

Grazie ancora a chi ha fatto parte delle squadre delle Facoltiadi, inparticolare quelle del 2007, per l’emozione irripetibile della doppia vittoria;al Nutria Team di dragonboat per la fatica e la prima vittoria a Borgo,

122 Acknowledgments

nella gara piu spettacolare; alle squadrette di calcio Piallaenne e Gardolo(femminile) e alla squadra seria del Lavis, che lotta per la promozione!

Non dimentico gli amici di Tubingen, in particolare Johannes, Claudia eAlex, i Mitbewohner Yang, Irina, Daniel e Max, e gli incoscienti che hannovissuto con me l’avventura della doppia laurea: Kirk, Giorgia, Bose, Gab,Pietro, Davide.

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[Hir76] M. W. Hirsch,Differential Topology,Springer, 1976.

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[Eng77] R. Engelking,General Topology,Polish Scientific Publishers, 1977.

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[Roy62] P. Roy,Failure of equivalence of dimension concepts for metric spaces,Bull. Amer. Math. Soc. 68, 1962 pag. 609-613.

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Ich erklare hiermit, dass ich die Arbeit stelbstandig und mit den angegebenenHilfsmitteln verfaßt habe.

Trento, den 19. Dezember 2007

Ester Dalvit

Documents

Tesi di Laurea — Diplomarbeit - UniTrentomatematita.science.unitn.it/4d/downloads/tesi.pdfTesi di Laurea — Diplomarbeit The concept of dimension: a theoretical and an informal