# Betti Numbers of Hypergraphs

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Betti Numbers of HypergraphsEric Emtander aa Department of Mathematics , Stockholm University , Stockholm, SwedenPublished online: 05 May 2009.

To cite this article: Eric Emtander (2009) Betti Numbers of Hypergraphs, Communications in Algebra, 37:5, 1545-1571, DOI:10.1080/00927870802098158

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• Communications in Algebra, 37: 15451571, 2009Copyright Taylor & Francis Group, LLCISSN: 0092-7872 print/1532-4125 onlineDOI: 10.1080/00927870802098158

BETTI NUMBERS OF HYPERGRAPHS

Eric EmtanderDepartment of Mathematics, Stockholm University, Stockholm, Sweden

In this article, we study some algebraic properties of hypergraphs, in particular theirBetti numbers. We define some different types of complete hypergraphs, which to thebest of our knowledge are not previously considered in the literature. Also, in a naturalway, we define a product on hypergraphs, which in a sense is dual to the join operationon simplicial complexes. For such product, we give a general formula for the Bettinumbers, which specializes neatly in case of linear resolutions.

Key Words: Betti numbers; Complete hypergraphs; Hypergraphs.

2000 Mathematics Subject Classification: 05C65; 13D02; 13A99.

1. INTRODUCTION

Let be a finite set and = E1 Es a finite collection of nonemptysubsets of . The pair = is called a hypergraph. The elements of arecalled the vertices, and the elements of are called the edges of the hypergraph. Ifwe want to specify what hypergraph we consider, we may write and forthe vertices and edges, respectively.

The hypergraphs that we will consider can all be seen as naturalgeneralizations of the ordinary complete graph Kn, on n vertices. Our main tools arefamiliar concepts in combinatorial algebra, such as Hochsters formula, the MayerVietoris sequence, and Knneths tensor formula.

A hypergraph is called simple if: (1) Ei 2 for all i = 1 s and (2) Ej Eiimplies i = j. If the cardinality of is n, we often just use the set n = 1 2 ninstead of .

We frequently identify a vertex vi of with a variable xi of a polynomialring kx1 xn over some field k, or with its corresponding characteristic vectorvvi = 0 0 1 0 0 in n, consisting of only zeros except in the ithposition were there is a 1. Hence we choose to consider 0 to be a natural number.This also allows us to identify a subset V of n with its characteristic vectorvV = iV vvi. We use bold letters to denote vectors and if w = w1 wn isa squarefree vector in n (i.e., a vector in which 0 wi 1 for i = 1 n), then

Received January 11, 2008; Revised March 27, 2008. Communicated by R. Villarreal.Address correspondence to Eric Emtander, Department of Mathematics, Stockholm University,

Stockholm S-10691, Sweden; E-mail: erice@math.su.se

1545

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we define its norm w by w = ni=1 wi. In this way, the cardinality V of V equalsthe norm of the characteristic vector vV.

Throughout the article, we denote by R the polynomial ring kx1 xn oversome field k, where n is the number of vertices of a hypergraph considered at themoment. We recall that the ring R is in a natural way both - and n-graded.Employing the ideas above, we may think of an edge Ei of a hypergraph as amonomial xEi = jEi xj in R. We use this notion to associate an ideal I Rto a hypergraph . The edge ideal, I, of a hypergraph is the ideal xEi Ei R, generated by the edges of .

The edge ideal was first introduced by Villarreal (1990), in the case ofsimple graphs. Since then, edge ideals have been studied widely, see for instanceFaridi (2002, 2005), H and Van Tuyl (2006, 2007), Morey et al. (2007), andZheng (2004). In H and Van Tuyl (2006), the authors give some nice recursiveformulas for computing Betti numbers. Furthermore, their techniques illustrate bothsome obstacles that occur when you try to generalize graph theoretical results tohypergraph theoretical, as well as ways of getting around such obstacles.

Another way of using hypergraphs to reveal connections between commutativealgebra and combinatorics was introduced by Faridi (2002). There, Faridi considers,the set of facets of a simplicial complex as a hypergraph. In this way a simplicialcomplex may be thought of as a higher dimensional graph. See Faridi (2002, 2005)and Zheng (2004) for details and examples.

Recall that an (abstract) simplicial complex on vertex set n is a collection

of subsets of n with the property that F G F G . The elements of

are called the faces of the complex and the maximal (under inclusion) faces arecalled facets. The dimension dim F of a face F in is defined to be F 1, and thedimension of is defined as dim = maxdim F F . The r-skeleton of is thecollection of faces of dimension at most r. Note that the empty set is the unique1 dimensional face of every complex that is not the void complex which has nofaces. The dimension of the void complex may be defined as .

The dimension dimR M of a R-module M , is by definition the Krll dimensionof R/AnnM .

Given a simplicial complex , we denote by its reduced chain complex,and by Hn k = Zn/Bn its nth reduced homology group with coefficientsin the field k. In general, we could use an arbitrary abelian group instead of k, butwe will only consider the case when the coefficients lie in a field. For convenience,we define the homology of the void complex to be zero.

If X and Y are two sets, we denote their disjoint union by X Y . Thus, supposewe have the two sets n and m. They both contain the number 1, but in n m,these two 1s are considered as distinct objects.

Let and be simplicial complexes on the disjoint vertex sets x1 xn andy1 ym, respectively. We define the join of and to be the simplicialcomplex on vertex set x1 xn y1 ym having faces xi1 xir yj1 yjs ,where xi1 xir and yj1 yjs are faces of and , respectively.

If n we denote by n the full simplex on n vertices. That is, the simplicialcomplex on n vertices in which every subset of n is a face. According to this, wemay think of the empty complex as a simplex on zero vertices.

Given a simplicial complex on n and a subset V n, we denote by Vthe simplicial complex on vertex set V , with faces F F V. We call this the

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restriction of to V . If j = j1 jn is a squarefree vector in n, by j we meanthe restriction to the set V n whose characteristic vector is j.

Now, let be a simplicial complex on n. The StanleyReisner ring R/I of

is the quotient of the ring R = kx1 xn by the StanleyReisner ideal

I = xF F

generated by the nonfaces of .Let n

k denote the set of all k-subsets (that is, subsets of cardinality k) of

n. If n < k, we interpret this as being empty. Furthermore, we let nk denote thecardinality of n

k, so nk = 0 if n < k.

In Section 2, we recall some basics that we will use throughout the article,while Section 3 is where the main results are found. In Theorems 3.1 and 3.5,respectively, we compute the Betti numbers of the d-complete and the d-completemultipartite hypergraphs, respectively. These results are very natural generalizationsof their graph theoretical counterparts. By considering the independence complexes,the ideas behind the proofs becomes transparent. In Section 3.4, we give a naturaldefinition of a product on hypergraphs. This in turn lets us compute the Bettinumbers of the da1 at-complete hypergraph. All these hypergraphs are in oneway or the other a natural generalization of the ordinary complete graph Kn. Inthe final section, Section 3.6, we define a class of hypergraphs that actually containall the previously considered ones. We show that the hypergraph algebra, R/I,corresponding to such hypergraph, has linear resolution.

2. PRELIMINARIES

Here we recall some results and definitions which will be used throughout thearticle.

2.1. Hypergraphs and Independence Complexes

Our general reference concerning hypergraphs is Berge (1989). In this article,we will only consider simple hypergraphs, as defined in the introduction. Thus,hypergraph will always mean simple hypergraph.

Let be a hypergraph. A subhypergraph of is a hypergraph such that , and . If , the induced hypergraph on , , isthe subhypergraph with = and with consisting of the edges of that lies entirely in . A hypergraph is said to be d-uniform if Ei = d for everyedge Ei . Note that a 2-uniform hypergraph is just an ordinary simple graph.

Let = n be a hypergraph, and consider the edge ideal I R.Note that R/I is precisely the StanleyReisner ring of the simplicial complex

= F n E FE

This is called the independence complex of . Note that the edges in are preciselythe minimal nonfaces in .

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Let be an arbitrary simplicial complex on n. We then define the Alexanderdual simplicial complex to by

= F n n\F

Note that = .

2.2. Resolutions and Betti Numbers

To every finitely generated graded module M over the polynomial ring R =kx1 xn, we may associate a minimal (-)graded free resolution

0 jRjlj M

jRjl1j M

jRj0j M M 0

where l n and Rj is the R-module obtained by shifting the degrees of R byj. Thus, Rj is the graded R-module in which the grade i component Rji isRij .

The natural number ijM is called the ijth -graded Betti number of M .If M is multigraded, we may equally well consider the n-graded minimal freeresolution and Betti numbers of M . The difference lies just in the fact that we nowuse multigraded shifts Rj instead of -graded ones. The total ith Betti number isiM =

j ij . For further details on resolutions, graded rings and Betti numbers,

we refer the reader to Bruns and Herzog (1998, Sections 1.3 and 1.5).The projective dimension pdM of M is pdM = maxi ijM = 0.The Betti numbers of M occur as the dimensions of certain vector spaces over

k = R/m, where m is the unique maximal graded ideal in R. Accordingly, the Bettinumbers (and then of course the projective dimension) in general depend on thecharacteristic of k.

A minimal free resolution of M is said to be linear if for i > 0, ijM = 0,whenever j = i+ d 1 for some fixed natural number d 1. In this article, weonly consider resolutions of quotient rings R/I . Hence, the interesting parts of theresolutions are the degrees greater than zero. In the variuos formulas for Bettinumbers that we give, we thus assume that i > 0.

In connection to this we mention the EagonReiner theorem.

Theorem 2.1. Let be a simplicial complex and its Alexander dual complex. ThenR/I is CohenMacaulay if and only if R/I has linear minimal free resolution.

Proof. See Eagon and Reiner (1998, Theorem 3).

Since there is a 11 correspondence between StanleyReisner rings (orequivalently squarefree monomial ideals) and simplicial complexes, we get a 11correspondence between simple hypergraphs and StanleyReisner rings as well. Thisenables us to talk about resolutions, Betti numbers, and projective dimensions ofhypergraphs.

By a resolution, a Betti number, or the projective dimension of a hypergraph , we mean ditto of R/I. Thus ij = ijR/I and pd = pdR/I.

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One further result which we will use later on is the AuslanderBuchsbaumformula. If R is a finitely generated graded k-algebra for some field k and M = 0 afinitely generated graded R-module with pdM < , then the formula asserts that

pdM+ depthM = depthRFor a proof, see Bruns and Herzog (1998, Theorem 1.3.3).

2.3. Hochsters Formula

In topology one defines Betti numbers in a somewhat different manner.Hochsters formula provides a link between these and the Betti numbers definedabove. Hochsters formula will turn out to be a very useful tool of ours.

Theorem 2.2 (Hochsters Formula). Let R/I be the StanleyReisner ring of asimplicial complex . The nonzero Betti numbers of R/I are only in squarefree degreesj and may be expressed as

ijR/I = dimk Hji1j kHence the total ith Betti number may be expressed as

iR/I =Vn

dim HV i1V k

Proof. See Bruns and Herzog (1998, Theorem 5.5.1).

If one has n-graded Betti numbers, it is easy to obtain the -graded onesvia

ijR/I =

jnj =j

ijR/I

Thus,

ijR/I =VnV =j

dim HV i1V k

2.4. The MayerVietoris Sequence

Recall that if we have an exact sequence of complexes,1

0 L M N 0there is a long exact (reduced) homology sequence associated to it

HrN Hr1L Hr1M Hr1N

1That is, complexes of modules over some ring R.

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Later in this article, we will have great use of this homology sequence in the specialcase where it is associated to a simplicial complex as follows.

Suppose we have a simplicial complex N and two subcomplexes L and M , suchthat N = L M . This gives us an exact sequence of (reduced) chain complexes

0 L M LM N 0

The nontrivial maps here are defined by x xx and x y x + y.The long exact (reduced) homology sequence associated to this particular

sequence, is called the MayerVietoris sequence. The reason that we will have greatuse of the MayerVietoris sequence is that in the cases that we will consider, almostalways some of the considered chain complexes will turn out to be very easy tohandle. More about the MayerVietoris sequence can be found in Maunder (1996,Section 4.4).

2.5. Knneths Tensor Formula

If complexes L and M are given, then the tensor product LM may beconstructed and given the structure of a complex as well. The degree n component isdefined as LMn =

r+s=n Lr Ms. Now, suppose that we are considering chain

complexes corresponding to simplicial complexes L and M . It is a natural questionto ask if the (reduced) homology of the tensor product LM in some wayis related to the (reduced) homologies of L and M . The answer is given by Knnethstensor formula (Maunder, 1996, Theorem 10.1), which under suitable2 circumstancessays that

HnLM =

r+s=nrs0

HrL HsM

We will use of this formula in connection to the join operation. It is easy to verifythat the chaincomplex L M of the join of two simplicial complexes L and M ,is isomorphic to the tensor product LM1. This is the same as thecomplex LM if we just shift the degree by 1.

2.6. Some Results on Induced Hypergraphs

The formulas we have encountered so far actually yield a couple of easyresults.

Let be a d-uniform hypergraph. We say that two edges E and E are disjointif E E = . Then, by considering the Taylor resolution (see Bayer et al., 1998) ofR/I, one can prove the following results, which are essentially due to Jacques.

Proposition 2.3. Let be a d-uniform hypergraph. Then iid equals the numberof induced hypergraphs that consist of i disjoint edges.

2For example, when the coefficients of the homology groups are in a field k.

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Proof. For d = 2 this is Theorem 3.3.5 in Jacques (2004). The proof given thereholds also for d > 2.

Proposition 2.4. Let = n be a hypergraph and an inducedhypergraph. Then

ij ijProof. Since = for some n, we have

ij =VnV =j

dimk HV i1V k VV =j

dimk HV i1V k = ij

Corollary 2.5. Let = n be a hypergraph and an induced hypergraph.Then

i ipd pd

3. VARIOUS COMPLETE HYPERGRAPHS

Jacques (2004) obtains nice descriptions of the Betti numbers of some specialfamilies of graphs. We will generalize some of these to hypergraph analogues.

3.1. The d-Complete Hypergraph

The complete graph Kn on n vertices is a familiar object to all who haveencountered at least some graph theory. Since an ordinary simple graph is 2-uniform, it seems reasonable to consider d-uniform hypergraphs when seeking ahypergraph counterpart.

We make the following definition. The d-complete hypergraph Kdn on n verticesis the d-uniform hypergraph with Kdn = nd . We will now compute the Bettinumbers of Kdn .

Theorem 3.1. The -graded Betti numbers of the d-complete hypergraph Kdn on nvertices are independent of the characteristic of the field k and may be written as

ij(Kdn

) =(n

j

)(j 1d 1

)if j = i+ d 1

0 if j = i+ d 1Proof. Hochsters formula says

ij(Kdn

) = VnV =j

dimk HV i1(

(Kdn

)V k

)

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It follows from the definitions that Kdn is the d 2-skeleton of n. In the sameway, KdnV is the d 2-skeleton of nV V . Thus, the complexes KdnVcan only have nonzero homology in degrees less than or equal to d 2. But, byconsidering a minimal resolution of Kdn , it is also clear that ijK

dn = 0 if j < i+

d 1. This is simply because the generators of IKdn have degree d. Hence, wehave a linear resolution, and ijK

dn = 0 only if j = i+ d 1.

Now, consider KdnV for some V n. It is clear that every cycle inZd2Kdn is a linear combination of elementary cycles, by which we mean thederivatives of d 1-simplices in nV . Denote this generating set by V .

We note that we may actually extract a smaller generating set out of V .Namely, we claim that it is enough to consider the elements that contain a fixedvertex x V (by containing x we mean that some term in the cycle contains x).Denote this set by V x and consider an element x1 xd in V , that do notcontain x. This cycle is a linear combination of elements in V x, which may beseen by first forming the cone x x1 xd, and then taking the derivative of thed 1-skeleton of this cone. This proves our claim.

Furthermore, we may easily show that the images in the homology groupHd2Kdn k of the elements V x are linearly independent. Assume thatt

i=1 aii = 0, ai k = R/m (where m is the unique graded maximal ideal of R)and i V x. Every i contains a unique term which does not contain x. Thisis because i = i, where i is a d 1 simplex. Hence ai = 0 for every i =1 t.

Now we are done, since if V = j, the cardinality of V x clearly is j1d1 ,and the number of j-sets V are nj .

Due to Corollary 3.2, the above result also follows from Theorem 1 in Herzogand Khl (1984). Corollary 3.2 seems to be well known, but we did not manage tofind a previously published proof.

Since Kdn has a specially nice structure, it is easy to determine its Alexanderdual. As the minimal nonfaces of Kdn are all xi1 xid , xij n, the facets of

Kdn

are all xi1 xind , xij n. Whence Kdn Knd+1n .

Corollary 3.2. The ring R/IKdn is CohenMacaulay, and we have

iKdn =

(nj

)(j 1d 1

)pdKdn = n d 1

where j = i+ d 1.

Proof. The last two claims follow directly from the theorem. We know, by theEagonReiner theorem, that a StanleyReisner ring R/I of a simplicial complex

has a linear resolution precisely when the StanleyReisner ring R/I of theAlexander dual complex is CohenMacaulay. Since Kdn

Knd+1n , we aredone.

One should note that Kdn is in fact shellable. A shelling is easy to constructusing the lexicographic order on n-tuples.

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Corollary 3.3. The ring R/IKdn is CohenMacualay, and we have

dimKdn = n d 1

dimRR/IKdn = n dpdR/IKdn = d

Proof. The CohenMacaulayness is now clear, and the first equation followsfrom the definitions. The second equation follows from the first one sincedimR R/I = dim+ 1 for any simplicial complex (see Bruns and Herzog, 1998,Theorem 5.1.4). The second equation and the Cohen-Macaulayness together implythe third equation.

Jacques (2004) studies the graph algebra of Kn, which we denote K2n, and

obtains the formula

ijKn =

(n

j

)i if j = i+ 1

0 if j = i+ 1

Note that this is a special case of our formula for ijKdn ; just put d = 2, and use

the fact that j = i+ d 1.

3.2. The d-Complete Multipartite Hypergraph

Perhaps almost as familiar as the complete graph Kn, is the completemultipartite graph Kn1nt on a vertex set which is a disjoint union of t sets ni, withcardinality ni, respectively. Contrary to the situation of the complete graph, it is notclear how to generalize to hypergraphs. Again, it seems reasonable to look for ad-uniform hypergraph, but this can be done in several ways. In this article, we willconsider a few.

We define the d-complete multipartite hypergraph Kdn1nt on vertex set n1 n2 nt, to be the d-uniform hypergraph whose edge set consists of all d-edges except those of the form xi1 xid , where xij ni for all j = 1 d.

Lemma 3.4. The StanleyReisner ring R/IKdn1nt of the d-complete multipartitehypergraph has linear resolution, and ijK

dn1nt

= 0 only if j = i+ d 1.Proof. Contrary to case of the d-complete hypergraph, this time there may verywell exist d 1-faces x1 xd in Kdn1nt , since IKdn1nt is not generatedby all possible d-edges.

As in the proof of Theorem 3.1, ijKdn1nt

= 0 if j < i+ d 1. SupposeijK

dn1nt

= 0 and j > i+ d 1. Via Hochsters formula, we conclude thatthere must then exist a nonzero homology group HlK

dn1nt

V k, for some V n1 n2 nt and l d 1.

But a cycle in such a degree l has to be a sum of cycles, each of which liesentirely inside one of the simplices ni on vertices ni, respectively, which has nohomology at all. Thus, the cycle is a boundary, contrary to our assumptions.

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From now on, it will be understood that in a multipartite situation, i.e., when ahypergraph has some disjoint union n1 nt as vertex set, then ns denotesthe simplex on the ns vertices from the ns-component of . We now computethe Betti numbers of Kdn1nt .

Theorem 3.5. The -graded Betti numbers of the d-complete multipartite hypergraphKdn1nt on vertex set n1 nt are independent of the characteristic of the field kand may be written as

ijKdn1nt

=

(N

j

)(j 1d 1

)

j1jttj1++jt=j

[ ts=1

(nsjs

)]

ts=1

js 1d 1

if j = i+ d 1

0 if j = i+ d 1

where N = ts=1 ns.Proof. In order to get the notations as clear as possible, we prove here only thecase where t = 2. It will be obvious that the same proof holds also when t > 2. Fort = 2, the formula in the theorem has the form

ijKdnm =

(n+m

j

)(j 1d 1

)

jj1=0

(n

j1

)(m

j j1

)[(j1 1d 1

)+

(j j1 1d 1

)]if j = i+ d 1

0 if j = i+ d 1

Our idea is to compare the terms HV i1KdnmV k occurring in Hochstersformula with the corresponding terms HV i1Kdn+mV k, which we encounteredwhen we computed ijK

dn+m.

We realize, simply because we have descriptions of the structures of theconsidered complexes, that

dimk HV i1KdnmV k dimk HV i1Kdn+mV k

for every set V n m. The possible difference lies in the fact that there mightvery well be faces F Kdnm such that F d. This would result in a nonzeroboundary group Bd2Kdnm in the chain complex of K

dnm.

It is an elementary fact that

dimk Hd2KdnmV k = dimk Zd2KdnmV dimk Bd2KdnmV

Since the cycle groups Zd2Kdn+mV and Zd2KdnmV clearly coincide, and

since Bd2Kdn+mV = 0, we only have to compute the dimension over k ofBd2KdnmV .

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If we write V = V1 V2, where V1 n and V2 m, it is clear that

Bd2KdnmV = Bd2KdnmV1 Bd2KdnmV2

This is because the potential d 1-faces of Kdnm lies either in Kdnmn or in

Kdnmm, which are disjoint.

Now, we have already proved how to compute dimk Bd2KdnmV, = 1 2.This was done when we computed the Betti numbers of Kdn . Thus,

dimk Bd2KdnmV1 =

(V1 1d 1

)dimk Bd2K

dnmV2 =

(V2 1d 1

)

If we put V1 = j1, the theorem follows as we simply sum over all possible V n m. Corollary 3.6. Given Kdn1nt with N =

ts=1 ns d, we have

iKdn1nt

=(Nj

)(j 1d 1

)

j1jttj1++jt=j

[ ts=1

(nsjs

)]

ts=1

(js 1d 1

)

pdKdn1nt = N d 1

where j = i+ d 1.Proof. The fact that pdKdn1nt N d 1 follows from directly from theformula. By putting j = N , we get(

N 1d 1

)

ts=1

(ns 1d 1

)

This expression is strictly greater than 0, which we may prove as follows. Considerthe set n1 n2 nt of N elements. Pick an arbitrary element, and removeit from the set. The first term above counts the number of ways of choosing d 1elements from the later set.

The sum in the above display counts the following: Start with the same setat before, and remove an arbitrary element xs from each one of the sets ns. Thenchoose d 1 elements from some ns\xs.

ts=1

ns1d1 is the total number of such

d 1-sets.Clearly, the difference of these two numbers is strictly greater than 0, just

consider a set of d 1-elements that do not lie entirely inside one set ns. As wehave assumed that N d, the claim follows. Example. Denote the vertex set of K323 by a b A BC. Then we have

K323 = abA abB abC aAB bAB aAC bAC aBC bBC

The Betti numbers are 0K323 = 1, 1K323 = 9, 2K323 = 13, 3K323 = 5.

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By construction, the edges in a hypergraph are the minimal nonfaces in

. This makes it easy to determine the facets in . As one easily realizes,they are the complements of the edges. Considering this, we get the followingexpression for the Alexander dual complex:

Kdn1nt =

ts=1

lsl=1

nsl+1ns n1++ns++ntd+l1n1 ns nt

where rns is the r-skeleton of ns , rn1 ns nt is the r-skeleton of n1 ns nt , means omit, and s = mind 1 ns.

Corollary 3.7. The ring R/IKdn1nt is CohenMacaulay, and we have

dimKdn1nt = N d 1

dimRR/IKdn1nt = N dpdR/IKdn1nt = d

Proof. The CohenMacaulayness follows from Lemma 3.4 and the EagonReiner theorem. By considering the above description of the Alexander dual, thefirst equation is clear and implies the second. The third equation follows sinceR/IKdn1nt

is CohenMacaulay.

Also in this case we have generalized a formula given by Jacques (2004). Bystudying the graph algebra of Knm, he obtains the formula

ijKnm =

j1j1=1

(n

j1

)(m

j j1

)if j = i+ 1

0 if j = i+ 1

A priori this looks quite different from our result. But, if one puts d = 2, andremember that

(nd

)is defined as 0 if n < d, our formula simplifies immediately to

this one.Contrary to when we considered Kdn, the structure of the Alexander dual

Kdn1nt is not transparent. One immediate question that appears is: When, if at

all, does the StanleyReisner ring of Kdn1nt both have linear resolution and theCohenMacaulay property? Since we already know that all considered resolutionsare linear, we only have to think about the CohenMacaulay property.

Lemma 3.8. Let N = ts=1 ns and ns d 1 for s = 1 t. Then KdN = Kdn1nt .Proof. Kdn1nt = KdN and Kdn1nt = KdN .

Proposition 3.9. The StanleyReisner ring R/IKdn1nt of a d-complete multipartitehypergraph on vertex set n1 n2 nt is CohenMacaulay precisely when ns d 1 for all s = 1 t.

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Proof. The AuslanderBuchsbaum formula tells us that

pdKdn1nt + depthRKdn1nt = N

where N = ts=1 ns. Since we already have computed the projective dimension, theabove formula says

depthRKdn1nt

= d 1

and it is clear that dimKdn1nt = maxni 1 d 2 i = 1 t. Thus, sincedepthRM dimR M holds for every finitely generated R-module M , R/IKdn1nt is CohenMacaulay precisely when ns d 1 for all s = 1 t. Furthermore,according to the lemma, we have Kdn1nt = KdN .

3.3. Hilbert Series

Let M be a -graded module (n-graded would work equally well). TheHilbert series HMt measures the dimensions over k = R/m of the graded piecesMi of M . More algebraically: Let M be such that every graded piece Mi has finitedimension over k. Then HMt is the formal power series

HMt =i

dimkMiti

The following is a well-known result. See for example Bruns and Herzog (1998,Theorem 4.1.13).

Lemma 3.10. Let R be the polynomial ring kx1 xn over a field k, and considera finitely generated -graded R-module M . Then

HMt =SMt

1 tn

where SMt =

ij1iijMtj .

If M is the StanleyReisner ring of a simplicial complex , one may rathereasily compute its Hilbert series. This is Corollary 1.15 in Miller and Sturmfels(2004). One gets

HR/It =1

1 tne

r=0fr1t

r1 tnr

where fr equals the number of r-faces of and e = dim+ 1.Note that this gives a nice connection between the geometric numbers fr

and the algebraic numbers ijR/I. In general though, it might be quite messyto handle the alternating sum of Betti numbers. But, if we consider a module Mwith linear resolution, the correspondence becomes much nicer.

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Lemma 3.11. Let be a simplicial complex such that R/I has a linear resolution.Then we have

ijR/I =e

r=01jirfr1

(n rj r

)

Proof. From Lemma 3.10, we get one expression for 1 tnHR/It, and from thediscussion right after that lemma we get another. Just identify the coefficient of tj

from the two expressions.

This lemma gives us an alternative way of computing the Betti numbers ofKdn and K

dn1nt

. All we need is the f -vector f1 f0 f1 fe1. In the casesconsidered, the f -vectors have nice and simple descriptions.

Let us begin by considering Kdn. Since this is the d 2-skeleton of n,we see that dimKdn = d 2. Thus e equals d 1 in this case. The number ofr 1-faces clearly is ( nr ), so the f -vector is given by(

1 n(n2

)

(n

d 1))

According to the above, recalling that j = i+ d 1, we get the formula

ijKdn =

d1r=0

1d1r(nr

)(n rj r

)

This is without a doubt correct, but looks completely different from our earlierexpression. We obviously have

d1r=0

1d1r(nr

)(n rj r

)=

(nj

)(j 1d 1

)

This identity may also be proved in a combinatorial way, using the Principle ofInclusionExclusion. We give the main ideas here. The trick is to identify somethingthat is counted by both sides of the identity. This something is described below.

1) Consider a set of n elements. First choose j elements of these, and thenchoose one of the j and colour it. Then colour d 1 further elements chosen fromthe remaining j 1 elements. This can be done in j( nj )( j1d1 ) ways.

We now claim that the following process counts the same thing.

2) Choose d 1 elements of the n-set and colour them. Choose j d + 1elements out of the remaining n d + 1 elements not previously choosen. Thenchoose one of the j elements choosen so far and colour it. This can be done inj(

nd1

)(nd+1jd+1

)ways. We realize that we have counted more coloured sets than in 1)

in this process, for example, those in which only d 1 element became coloured.In an attempt to adjust this we subtract j

(n

d2)(

nd+2jd+2

)from j

(n

d1)(

nd+1jd+1

). This

number is created using the same choice argument as before. Then we subtractthe number of coloured sets in which only d 1 elements were coloured. But we

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subtract too much, since we also subtract the number of sets in which only d 2elements is coloured. Thus, we have to add back.

Continuing this process, according to the Principle of InclusionExclusion,after a finite number of steps, we will stop, and the resulting number counts preciselythe same thing as 1). Finally, we just divide every term by j to obtain our identity.

The number described in 1) and 2), counts the number of ways of: Choosinga j-set of n to form a football team, say, and then determining in how manyways one can have d of the players on the field, one of which is to be chosen asgoalkeeper.

Note that the above arguments makes sense only if j d. However, accordingto our earlier investigations, this is quite natural.

We also obtain a different formula for the Betti numbers ijKdn1nt

. Just asbefore, we only need to compute the f -vector. This is sufficient since we know thatKdn1nt has linear resolution.

The structure of Kdn1nt is easy to understand, and it follows that

fr1Kdn1nt

=

(N

r

)if r d 1

ts=1

(nsr

)if r d

where N = ts=1 ns. Using the lemma, and remembering that j = i+ d 1, weobtain the following formula

ijKdn1nt

=d1r=0

1d1r(Nr

)(N rj r

)

er=d

1dr(N rj r

)[ ts=1

(nsr

)]

where e = maxns 1 d 2 s = 1 t is the dimension of Kdn1nt . Weimmediately note one thing. The first sum in the display actually is nothing butijK

dN =

(Nj

)(j1d1

), where N = ts=1 ns. Thus, we realize that that the second sum

gives us an alternative expression for the difference ijKdN ijKdn1nt . In other

words, we have an identity

er=d

1dr(N rj r

)[ ts=1

(nsr

)]=

j1jttj1++jt=j

[ ts=1

(nsjs

)]

ts=1

(js 1d 1

)

3.4. The Alexander Dual of a Join

It is known, and proved in for example Frberg (1988), that the join oftwo simplicial complexes and is CohenMacaulay precisely when both and are CohenMacaulay. In that case, remember that the EagonReiner theorem tellsus that the Alexander dual complex has linear resolution.

In this article, we consider several classes of hypergraphs with linearresolutions. Therefore, it would be nice to be able to describe the Alexander dual

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of a join since we then rather easily can construct more hypergraphs with linearresolutions. In this section, we derive a description of the Alexander dual of a join,and also give a formula for the Betti numbers.

Let and be simplicial complexes on n and m, respectively. Denotethe minimal nonfaces of () by fi, i = 1 r (gj , j = 1 s, respectively).Remember that according to the identifications that are made in the introduction,we may consider the fis (gjs) as squarefree monomials in kx1 xn(ky1 ym). Using this identification, we consider the StanleyReisner ideal I kx1 xn (I ky1 ym) of ( , respectively). It is well known (Miller andSturmfels, 2004, Theorem 1.7) that

I = fi i = 1 r =f

f

where for a subset V n, V is the ideal xi i V kx1 xn, and by f wemean n\f . It is easy to realize that it is enough to take the intersection where f isa facet of . If we consider the Alexander dual in the same way, we get

I = f i i = 1 r =

f f

where f i , i = 1 r is the set of minimal nonfaces of (analogously, we denoteby gj , j = 1 s, the minimal nonfaces of ). Note that this shows the algebraicversion of Alexander duality. The association is by the above equivalent to

I = fi i = 1 r r

i=1fi = I

If we consider I and I as ideals in kx1 xn y1 ym, it follows that

I = I + I = fi gj i = 1 r j = 1 s

Hence, the StanleyReisner ideal of is( ri=1

fi)( s

j=1gj

)= I I = II

So, we conclude that

I = f i gj i = 1 r j = 1 s

By considering the minimal nonfaces f i gj of , we realize that

= m n

Note that we could have reached these conclusions also by considering the minimalnonfaces of . The form of the StanleyReisner ideal of is particularlynice since the generators correspond to edges in certain hypergraphs.

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Suppose that hypergraphs = n and = m are given.We define the product of and to be the hypergraph on vertex set n m and with edges x1 xr y1 ys, where x1 xr is an edge in andy1 ys is an edge in . In other words, may be thought of as thecartesian product .

Using the above results, we may easily prove the following theorem.

Theorem 3.12. Let = n and = m be d- and d-uniformhypergraphs respectively. Then is a d + d-uniform hypergraph, and has linearresolution if and only if both and have linear resolutions.

Proof. The fact that is d + d-uniform is clear from the definition. If weput = and = in the results deduced just before the theorem, weget that

=

This is clear considering the minimal nonfaces of both sides of the equation. Bythe EagonReiner theorem has linear resolution precisely whenboth and are CohenMacaulay. This is, again by the EagonReinertheorem, the same thing as saying that both and have linear resolutions.

Note that the topological information in the above theorem says that

= m n

Now, let V = V1 V2 n m, and consider the exact sequence

0 V mV n V V 0

If V1 or V2 is empty, then V will not have any nonzero homology. This issimply because there are no nonfaces (consider the relations in the StanleyReisnerring). Our aim is to compute the Betti numbers via Hochsters formula and hence,it is enough to consider the sets V = V1 V2 n m for which V1 n and V2 m both are nonempty. But, in this case both mV and n V arecones and accordingly have no homology at all. Thus, if we consider the MayerVietoris sequence obtained from the above exact sequence, we get that the followingequation holds for every V n m, V n = , V m = :

Hr V k Hr1 V k

Using the results in Section 2.5, it follows that

Hr V k

r1+r2=r2r1r20

Hr1V k Hr2V k

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Thus, by Hochsters formula, we get

ij =V =j

V=V1V2

r1+r2=ji3

dimk Hr1V1 k dimk Hr2V2 k

Of course, we want to extend this to products of more than two hypergraphs. Thiswe do inductively.

Theorem 3.13. The ijth -graded Betti number of the product 1 t ofhypergraphs i, i = 1 t on vertex sets ni, respectively, is given by the followingexpression:

ij1 t =V =j

V=V1Vt

r1++rt=ji2t1

ri0

tl=1

dimk Hrl lVl k

Proof. We have already seen that the formula holds for t = 2. It follows easily byinduction that

dimk Hs1 tV k =

r1++rt=s2t1

tl=1

dimk Hrl lVl k

Now consider the following, which by the case t = 2 clearly holds.

ij1 t+1 =V =j

V=V1Vt

s+rt+1=ji3

dimk Hs1 tV k

dimk Hrt+1t+1V k

By putting the expression for dimk Hs1 tV k in the above formula, weeasily see that the two equations

r1 + + rt = s 2t 1s + rt+1 = j i 3

may be collected into the single equation

r1 + + rt+1 = j i 2t + 1 1

By induction, we are done.

The above formula for the Betti numbers becomes much nicer if weknow that each i has linear resolution. The effect of this is that the innersummation symbol becomes superfluous, this since we already know that in this casedimk Hrl lVl k can only be nonzero in one specific degree for each l. Thesepieces of information yield the degree in which dimk Hs1 tV k is nonzero.But, this degree is also expressed by the equation r1 + + rt = j i 2t 1.

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Thus, we may indeed remove the summation symbol, and we have the followingtheorem.

Theorem 3.14. Let hypergraphs i, i = 1 t, on vertex sets ni, respectively,be given. Assume that for i = 1 t, the hypergraph i is ai-uniform with linearresolution. Then the ijth -graded Betti number of the product 1 t is given by thefollowing expression:

ij1 t =V =j

V=V1Vt

tl=1

dimk Hal2lVl k

3.5. The da1 at-Complete Multipartite Hypergraph

As we mentioned before, there are many ways of generalizing the multipartitegraph Kn1nt to a hypergraph analogue. We have already discussed the d-completemultipartite hypergraph Kdn1nt , and will now move on to consider another class ofhypergraphs.

The edge set Kdn1nt of Kdn1nt

consists of all d-edges except those ofthe form xi1 xid , xij ns for some s = 1 t. In the case of the ordinarygraph Knm, this just tells us that we have all edges between the disjoint sets nand m of vertices. This one may think of as an edge being a choice of twovertices, prescribing a certain number of vertices in each one of the sets n and m,namely, one in each. This is the idea behind what we now define. The da1 at-complete multipartite hypergraph Kda1atn1nt is the d-uniform hypergraph on vertex setn1 n2 nt and edge set Kda1atn1nt consisting of all d-edges such thatprecisely as elements comes from ns, as , as 1,

ts=1 as = d.3

Proceeding in the same spirit as before, we begin our investigation by showingthat R/IKda1atn1nt has a linear resolution. First, let us simplify the notation a bit. Inwhat follows, a1 at = a, n1 nt = n, and d =

ts=1 as. Thus, da1 at =

da and Kdan = Kda1atn1nt .

Lemma 3.15. The StanleyReisner ring R/IKdan of the da-complete mutlipartitehypergraph has linear resolution, and ijK

dan = 0 only if j = i+ d 1.

Proof. This will be proved in greater generality in Section 3.6. However, we givea short proof here as well. This is since it contains some interesting information,which we will not get out of the more general proof in Section 3.6.

By considering the definition of the Alexander dual complex, we immediatelyget the following expression:

Kdan = n1a11n1 ntat1nt

3The d occurring in the superscript in the symbol Kda1at n1nt does not have any real purposehere. However, when we continue our work, the d will be useful since the notations will become moreunified.

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Thus, Kdan is CohenMacaulay since we know that each nsas1ns is CohenMacaulay. Now, our lemma follows by the EagonReiner theorem.

We now compute the Betti numbers of Kdan . As one easily realizes, either fromthe above lemma or directly from the definition, we have that

Kdan =t

s=1Kasns

Thus, we may apply the results from the previous section.

Theorem 3.16. The -graded Betti numbers of the da-complete multipartitehypergraph Kdan on vertex set n1 nt are independent of the characteristic ofthe field k and may be written as

ijKdan =

r1++rt=i+t1ri1

[ tl=1

rlrl+al1Kalnl

]if j = i+ d 1

0 if j = i+ d 1

Proof. We know that dimk Hrl KalnlVl k = 0 only when rl = al 2, and in this

case, we have

dimk Hrl KalnlVl k =

(jl 1al 1

)

where jl = Vl. Using this, the expression

V =j

V=V1Vt

r1++rt=ji2t1

ri0

tl=1

dimk Hrl KalnlVl k

obtained from Theorem 3.13 simplifies, via the formula in Theorem 3.14, to

V =j

V=V1Vt

tl=1

(jl 1al 1

)

This may in turn be written as

jl0

j=j1++jt

tl=1

(nljl

)(jl 1al 1

)

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Now, if some jl al 1, the corresponding term is zero. So, we may write jl =rl + al 1, where rl 1 for l = 1 t. The above expression then becomes

rl1

r1++rt=jd+t

tl=1

(nl

rl + al 1)(

rl + al 2al 1

)

Since we know that the resolution is linear, we have the equation j = i+ d 1.Using this in the last display, we get the formula in the theorem.

Corollary 3.17. The -graded Betti numbers of the da b-complete bipartitehypergraph Kdabnm may be written as

ijKdabnm =

r+s=i+1rs1

(n

r + a 1)(

r + a 2a 1

)(m

s + b 1)(

s + b 2b 1

)

Furthermore, note that by putting a = b = 1, we get

ijKd11nm =

p+q=jpq1

(np

)(mq

)

Now Kd11nm = Knm, so we have given another proof of Jacques formula forijKnm.

Corollary 3.18. Given Kdan , we have

iKdan =

r1++rt=i+t1

ri1

[ tl=1

rlrl+al1Kalnl

]

pdKdan = N d 1

where j = i+ d 1.

Proof. The first assertion is clear. If we put i = N d 1 in the formula, we get

Nd1Kdan =

tl=1

nlal1Kalnl

which is nonzero. At the same time, we see that if i > N d 1, every term inthe sum is zero because some factor in every term is zero.

Example. Consider = K5113333 . If we denote the set of vertices of this hypergraphby a b c A BC d e f, we get

= aAdef aBdef aCdef bAdef bBdef bCdef cAdef cBdef cCdef

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The Betti numbers are 0 = 1, 1 = 9, 2 = 18, 3 = 15, 4 = 6,5 = 1.

Corollary 3.19. The ring R/I

K

dan

is CohenMacaulay, and we have

dimKdan = N d 1

dimRR/IKdan = N dpdR/I

Kdan

= d

Proof. The CohenMacaulayness follows, for example, from the theorem and theEagonReiner theorem. By considering the description of the Alexander dual givenin Lemma 3.15, the first equation is clear and imply the second. The third is aconsequence of the fact that R/I

Kdan

is CohenMacaulay.

Proposition 3.20. The StanleyReisner ring R/IKdan of the da1 at-completemultipartite hypergraph on vertex set n1 nt is CohenMacaulay preciselywhen as = ns for all s 1 t but possibly one. This single ai is such that itmaximizes the expression ai +

tj =ij=1 nj .

Proof. Let Is ns. It is necessary and sufficient that at least one set Ii satisfyIi < ai, for I1 It to be a face of Kdan . Thus, the dimension of Kdan is

max{ai 2+

tj =ij=1

nj i = 1 t}

so

dimRR/IKdan = max

{ai 1+

tj =ij=1

nj

}

We know that pdKdan = N d 1, so depthR/IKdan = d 1. Now, sinceby construction d = ts=1 as, we are done.

Note that this again, in a sense, collapses to an ordinary d-completehypergraph.

One special, and rather intuitive, way of generalizing the complete bipartitegraph, is to consider the d1 1-complete mulitpartite hypergraph Kd11n1nt .According to the above its ijth Betti number is given by

ijKd11n1nt

= r1++rt=i+t1

rs1

tl=1

(nlrl

)

Berge (1989) defines what he calls the d-partite complete hypergraph. In ourlanguage, this is just Kd11n1nd , so his definition is a special case of ours.

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3.6. The dI1 It-Complete Hypergraph

In this section, we define another class of complete hypergraphs that actuallycontains all of the previously defined classes of complete hypergraphs. We then showthat the hypergraphs in this new class have linear resolutions. In this way, one maythink that some of our previous results are superfluous. We argue that they are not.This is because the main part of the results so far is about calculating the Bettinumbers. The fact that we have had linear resolutions have mainly been used as acomputational aid.

In the case of the d-complete hypergraph, we considered all possible d-edges,and in the case of the da1 at-complete hypergraph, we considered thosein which precisely as elements came from the vertex set ns. We are going tokeep the vertex set n1 nt of N =

ts=1 ns vertices, but define another edge

set. For each s = 1 t, let Is be an interval s s in 0 ns. We definethe dI1 It-complete multipartite hypergraph to be the d-uniform hypergraphon vertex set n1 nt, and with edge set consisting of all d-edges I1a1 Itat. Here Isas is a subset of ns of cardinality as Is and d =

ts=1 as.

We immediately see why this generalizes previously considered hypergraphs. IfIs = 0 ns for all s = 1 t, we have the d-complete hypergraph KdN . IfIs = 0 minns d 1, we obtain the d-complete multipartite hypergraphKdn1ns . By letting Is consist of only one nonzero element for all s, we obtain theda1 at-complete hypergraph K

da1atn1nt

. So, we already know that some specialinstances of KdI1Itn1nt have linear resolutions.

One easily realizes that two different sets of intervals I1 It and J1 Jtsay, may yield the same hypergraph. Just consider the case where Is = as for alls,

ts=1 as = d, and Js = as for all s = 1, J1 = a1 a1 + 1. However, to obtain a

different hypergraph, we just need to change J2 say, to a2 1 a2.From now on, without loss of generality, we will assume that the sequence

of intervals I1 It in a hypergraph KdI1Itn1nt

satisfies the following property: IfIs = s s for s = 1 t, then

s +j =s

j d

s +j =s

j d

hold for every s. In other words, we assume that there is no redundancy in the sensethat every element as Is is part of an edge in the hypergraph.

It is clear that a set of intervals I1 It corresponding to a hypergraphKdI1Itn1nt can be constructed from at least one sequence a1 at, d =

ts=1 as,

corresponding to a da1 at-complete hypergraph, by successively changing theintervals by extending one of them (or possibly two of them depending on thesituation) in such a way that the inequalities above remains true in each step.The following example will clarify this idea.

Example. Suppose a1 + a2 + a3 = d, and consider KdI1I2I3n1n2n3 with I1 = a1 1 a1 I2 = a2 1 a2 + 1 I3 = a3 a3 + 1. These intervals can be constructed

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from the trivial intervals I1 = a1, I2 = a2 and I3 = a3 in the following way:

a1 a2 a3 a1 1 a1 a2 a2 + 1 a3 a1 1 a1 a2 1 a2 + 1 a3 a3 + 1

Theorem 3.21. The StanleyReisner ring R/I

K

dI1It n1nt

of the dI1 It-complete

multipartite hypergraph has linear resolution, and ijKdI1Itn1nt

= 0 only if j = i+d 1.

Proof. We start by noting that if t = 1, then KdI1Itn1nt = Kdn which we knowhas linear resolution, and nonzero homology only in degree d 2. If n < d, we havethe d-uniform hypergraph with empty edge set and this also has linear resolution.Since the set of intervals I1 It that corresponds to a hypergraph K

dI1Itn1nt

can be constructed (as above) from several sequences a1 at, a1 + + at = dwe may, without loss of generality, assume that It = at at + r for some positiveinteger r . Having already gone through the case where t = 1, let us assume thatall hypergraphs KdI1It1n1nt1 have linear resolutions and nonzero homology only indegree d 2.

Given KdI1Itn1nt as above, the expression KdatsI1It1n1nt1 , s 0 makes sense. It

means precisely what it says but there is one little problem, the intervals I1 It1may no longer be as small as possible. There may very well be an element aj Ij forsome j = 1 t 1, that cannot be used in a partition of d at s. So, we reallyshould use some other symbols I 1 I

t1 for the intervals, as they may depend on

s. We will however, for convenience, allow this abuse of notation in this proof.The next observation we make is that

KdatsI1It1n1nt1 Kdats+1I1It1n1nt1

Indeed, a face in the first complex cannot contain an edge from Kdats+1I1It1n1nt1 ,since it would then automatically also contain an edge from KdatsI1It1n1nt1 .

By considering the minimal nonfaces in the complex, one realizes that

KdI1Itn1nt has the following expression:[

KdatI1It1n1nt1 nt L0

n1++nt1 at2nt M0

]

[

Kdatr+1I1It1n1nt1 nt

Lr1

n1++nt1 at+r3nt Mr1

]

[

KdatrI1It1n1nt1 nt

Lr

n1++nt1 at+r2nt Mr

]

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• BETTI NUMBERS OF HYPERGRAPHS 1569

Here the expression in each row corresponds to the faces that do not contain onetype of minimal nonface. We take the intersection of these to obtain the faces thatdo not contain any minimal nonface, in other words the whole complex.

Put A0 = L0, B0 = M0, 0 = A0 B0, and define recursively Ar = Lr r 1, Br = Mr r 1, r = Ar Br . We will now deduce explicit formulas forAr , Br , and Ar Br . Having done this, easy use of MayerVietoris together with theinduction hypothesis will give our result.

We start by considering Ar :

Ar = Lr r 1 = Lr Ar1 Br1 = Lr Lr1 Mr1 r 2= Lr r 2 = = Lr 0 = KdatrI1It1n1nt1 nt

The expression for Br , we will prove by induction.

Claim. Br = r1

s=0 KdatsI1It1n1nt1 at+s1nt n1++nt1 at2nt.

If we interpret the expression in brackets as when r = 0, it is clear that theformula holds for r = 0. Now

B1 = n1++nt1 at1nt A0 B0= KdatI1It1n1nt1 at1nt n1++nt1 at2nt

so the formula holds for 1 as well. Assume that the formula holds for r. Then

Br+1 = Mr+1 Ar Br

Let us investigate Mr+1 Ar and Mr+1 Br separately. Using the expression for Arthat we already have, we immediately get

Mr+1 Ar = KdatrI1It1n1nt1 at+r1nt

Furthermore,

Mr+1 Br = Mr+1 [( r1

s=0

KdatsI1It1n1nt1 at+s1nt

) n1++nt1 at2nt

]

=[( r1

s=0

KdatsI1It1n1nt1 at+s1nt

)] n1++nt1 at2nt

So,

Br+1 = Mr+1 Ar Mr+1 Br

=[ r

s=0

KdatsI1It1n1nt1 at+s1nt

] n1++nt1 at2nt

and we have proved the claim.

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• 1570 EMTANDER

Lastly, we consider Ar Br . Using the expressions that we have just deduced,we get

Ar Br =[

KdatrI1It1n1nt1 nt

][( r1

s=0

KdatsI1It1n1nt1 at+s1nt

) n1++nt1 at2nt

]=

(

KdatrI1It1n1nt1 at+r2nt

) (KdatrI1It1n1nt1 at2nt)

= KdatrI1It1n1nt1 at+r2nt

Now, since we have descriptions of Ar , Br , and Ar Br , it will be rather easy tofinish the proof.

Ar is a cone and hence has no homology at all. For Br , we write

Br =[ r1

s=0

KdatsI1It1n1nt1 at+s1nt

]

L

n1++nt1 at2nt M

M is a cone and has no homology at all, and L can only have homology indegree d 2. This follows easily by induction on r using MayerVietoris. The casewhere r = 1 follows directly from the results in Section 2.5. Furthermore, one easilysees that L M = KdatI1It1n1nt1 at2nt. It follows again from the result inSection 2.5 that this complex can only have homology in degree d 3. Thus, usingMayerVietoris on Br we conclude that the homology of Br can be nonzero only indegree d 2.

Using Knneths formula again, we immediately conclude that Ar Br onlyhas homology in degree d 3. Thus, the exact sequence

0 Ar Br ArBr Ar Br 0

gives, via MayerVietoris, that the homology of Ar Br can be nonzero only indegree d 2. Now we are almost done. We aim to use Hochsters formula toconclude that the resolution of KdI1Itn1nt is linear. Hence, we would like to concludethat the homology of every restriction KdI1Itn1nt V behaves precisely like that ofthe whole complex. In other words, that it only can exist in degree d 2. If V ns = for all s = 1 t and the induced hypergraph KdI1Itn1nt V has nonemptyedge set, then the restriction KdI1Itn1nt V has precisely the same form as theoriginal complex KdI1Itn1nt . Thus, by the above, it may only have homologyin degree d 2. Next assume that V ns = for at least one s. Then F

KdI1Itn1nt V if and only if ns F KdI1Itn1nt V . Hence, in this case KdI1Itn1nt Vis a cone and have no homology. The last case to consider is if V ns = for alls, but the edge set of KdI1Itn1nt V is empty. Also in this case, we have a cone, in fact,a simplex. This is easy since if KdI1Itn1nt V has no edges, then there are no minimalnonfaces in KdI1Itn1nt V . We have thus proved that if dimk HlK

dI1Itn1nt

V k =0, then l = d 2. Hence Hochsters formula gives that ijKdI1Itn1nt = 0 only ifj = i+ d 1, and we are done.

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• BETTI NUMBERS OF HYPERGRAPHS 1571

Example. Consider = K5I1I2I3333 , where I1 = 1 2 I2 = 1 I3 = 2 3. Thereare 36 5-edges in this hypergraph, and using a computer, one easily computes theBetti numbers. They are 0 = 1, 1 = 36, 2 = 90, 3 = 87, 4 =39, 5 = 7.

By considering the edges in KdI1Itn1nt and the description of the Alexander dualof Kdan , we obtain the following description of KdI1Itn1nt

:

KdI1Itn1nt =

a1++at=dasIs

n1a11n1 ntat1nt

We immediately get the following corollary.

Corollary 3.22. The ring R/I

K

dI1It n1nt

is CohenMacualay, and we have

dimKdI1Itn1nt = N d 1

dimRR/IKdI1It n1nt = N d

pdR/I

K

dI1It n1nt

= d

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J. Algebra 309(1):405425.Herzog, J., Khl, M. (1984). On the betti numbers of finite pure and linear resolutions.

Comm. Algebra 12:16271646.Jacques, S. (2004). Betti Numbers of Graph Ideals. PhD thesis, University of Sheffield.

arXiv:math/0410107.Maunder, C. R. F. (1996). Algebraic Topology. New York: Dover publications Inc.Miller, E., Sturmfels, B. (2004). Combinatorial Commutative Algebra. Springer.Morey, S., Reyes, E., Villarreal, R. H. (2007). Cohen-Macaulay, Shellable and Unmixed

Clutters with a Perfect Matching of Knig Type. arXiv:0708.3111v3.Villarreal, R. H. (1990). Cohen-Macaulay graphs. Manuscripta Math. 66:277293.Zheng, X. (2004). Resolutions of facet ideals. Comm. Algebra 32(6):23012324.

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