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Invariant Theory
Professor Gus Lehrer
Semester 2, 2010Notes by Clinton Boys
Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Chapter 1. Tensor products and related constructions . . . . . . . . . . 101.1. Construction of the tensor product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2. Functoriality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.3. The tensor algebra and quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.4. Graded algebras and Poincaré series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
Chapter 2. Elementary invariant theory of finite groups . . . . . . . . 192.1. Molien’s theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2. Elementary facts about finite group invariants . . . . . . . . . . . . . . . . . . . 242.3. Invariants of finite groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
Chapter 3. The fundamental theorems of invariant theory forGL(V ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
3.1. The first fundamental theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.2. Conclusion of the proof of the first fundamental theorem . . . . . . . . 363.3. The second fundamental theorem of invariant theory (type A) . . . 37
Chapter 4. The structure of the group algebra KSr . . . . . . . . . . . . 41
Chapter 5. Affine algebraic geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.1. Affine algebraic sets and ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.2. The Zariski topology and decomposition. . . . . . . . . . . . . . . . . . . . . . . . . 535.3. Integral dependence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555.4. Hilbert’s Nullstellensatz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 565.5. Affine algebraic varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575.6. Principal open subsets and rational functions . . . . . . . . . . . . . . . . . . . . 605.7. Group varieties. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 645.8. Birational morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 675.9. Density results for affine spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
Chapter 6. Invariant theory of the orthogonal group . . . . . . . . . . . 746.1. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746.2. The main lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 756.3. Second formulation of the first fundamental theorem for GLn . . . . 786.4. Some multilinear algebra and the orthogonal group . . . . . . . . . . . . . . 84
3
4 INVARIANT THEORY
6.5. The first fundamental theorem for the orthogonal group . . . . . . . . . 876.6. First formulation of O(V ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 916.7. The Brauer algebra and the first fundamental theorem . . . . . . . . . . 936.8. The second fundamental theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
Appendix A. Semisimple modules and double centralisertheory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .102
A1. Semisimple modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102A2. Double centraliser theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103A3. Semisimple rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .110
Preface
These notes were born out of a 26-lecture course on invariant theory given byProfessor Gus Lehrer at the University of Sydney in Semester 2, 2010. Mymethod was to take my laptop along to lectures, type away furiously, and thengo away and fastidiously correct and improve later. These notes are the resultof my own study and understanding of the course and have been significantlychanged and edited from their original form. I hope that this process hasresulted in a self-contained introduction to the subject which still retains theoriginal flavour which Professor Lehrer intended.The exercises at the ends of each chapter are essentially those things which thelecturer left as exercises in lectures. Where an exercise was simply to finish thelast line or two of a proof, I have often included this in the body of the notes.The numbering of sections is also somewhat erratic – results from Chapter 4and Appendix A are needed towards the end of Chapter 3.There will unavoidably remain some typographical errors in these notes, de-spite my best efforts to remove them all. Thanks to Jack, Jon and ProfessorLehrer for pointing some of them out to me.
5
Introduction
This course provides an introduction to invariant theory. The subject is clas-sically studied by introducing the so-called fundamental theorems of invarianttheory for different groups, and this is the approach we take here, proving thefirst and second fundamental theorems for the general linear group GL(V ) ofsome finite-dimensional K-vector space V , and also the first fundamental the-orem for the orthogonal group On(V ). Here we diverge from the traditionalapproach, and give an interesting and unique proof using basic concepts fromthe geometry of affine algebraic varieties. The course finishes by briefly dis-cussing an open problem in the explicit formulation of the second fundamentaltheorem for On(V ) using this approach.
Almost all of mathematics might naively be characterised as invariant theory.The characterisation depends on how general one wants to be when consideringinvariants. The following is a very general typical problem in invariant theory.
Problem 0.1. Suppose X is a space and ∼ an equivalence relation on X, andconsider the two following general problems.
(i) Find “invariant functions” f : X → A where A is some set(ii) Determine the structure of the set of all invariant functions.
Example 0.2. Consider the following examples.
(i) As an example of an equivalence relation on some set X, if G is a (semi)group of transformations of X, we can define an equivalence relationby x ∼ y if and only if y = gx (and x = hy) for some g, h ∈ G.
(ii) We give an example of invariant functions on a space. Suppose G isa group acting linearly on a vector space over a field K. Let S =K[V ] be the algebra of polynomial functions on V (i.e. polynomials incoordinates of V ). Then G acts on S by (gf)(v) = f(g−1v), and theset SG of invariant functions, i.e. functions f such that gf = f for allg ∈ G is an algebra.
Problem 0.3. A common problem in invariant theory, and one which we donot really tackle in this course, is to describe the algebra structure of SG,where S is the algebra of polynomial functions on some space. This is knownas geometric invariant theory.
6
INVARIANT THEORY 7
Example 0.4 (Knot invariants). Consider a braid and then join loose endstogether to create a link. Say two braids β1 and β2 are equivalent if they mapto the same link under this association. We look for functions
f : B∞ → A
where B∞ is the set of all braids. Such functions include the Alexander poly-nomial and the Jones polynomial.
Example 0.5. Suppose G acts linearly on a vector space W . Describe theendomorphisms of W which commute with G, i.e.
EndG(W ) = {ϕ : W → W | gϕ = ϕ for all g ∈ G}.We can formulate Fourier analysis in this way, as well as spherical functions (i.e.those invariant under the action of the orthogonal group). We can considerW = {f : R→ R | · · · } and let G = Z acting by translation. Then decomposeW = ⊕Wi and Pi : W → Wi ∈ EndG(W ) where Pi is the projection map.
Example 0.6. For a more specific example, let ζ = e2πi/n be a complexprimitive nth root of unity and let µn = 〈ζ〉 ⊂ C× be the cyclic group oforder n.
Then µn acts on C2 via
gζ
(ab
)=
(ζaζ−1b
).
Then S = C[x, y], where
x
(ab
)= a, y
(ab
)= b
are the coordinate functions. So
(gζx)
(ab
)= x
(ζ−1aζb
)= ζ−1a = ζ−1x
(ab
)so gζx = ζ−1x, and
(gζy)
(ab
)= y
(ζ−1aζb
)= ζb = ζy
(ab
), so gζy = ζy.
Let u = xy, v = xn and w = yn. Then u, v, w are invariant under our action.
It turns out that any f(x, y) ∈ SG, i.e. any polynomial invariant under ouraction, is a polynomial in u, v, w. Moreover, all relations between u, v, w followfrom un = vw. In particular, we have the following description of the orbitspace:
C2\µn ∼= {(u, v, w) | un = vw}.
Example 0.7 (Space of quadratic forms). Consider G = GL2(C) actingby linear substitution on the space V of quadratic forms, i.e. functions of theform
q(x, y) = ax2 + 2bxy + cy2.
8 INVARIANT THEORY
It is easy to show that this space is canonically isomorphic to
V ∼={(
a bb c
)= A ∈M2(C)
},
under the association q(x, y) = xtAx, where x =
(xy
). If
g =
(α βγ δ
)∈ GL2(C),
then (xy
)7→ g
(xy
),
x 7→ αx+ βy, y 7→ γx+ δy.
So we have
q
(g
(xy
))= xtgtAgx, i.e. g : A 7→ gtAg.
Lagrange in 1760 discovered that if g is unipotent, i.e. has the form
g =
(1 α0 1
),
then detA = ac− b2 is invariant. Gauss in 1801 showed ∆(gq) = (det g)2∆(q).It can be shown that, under SL2(C), ∆ is invariant.
Note that we can generalise this idea to several variables, i.e. let V be thespace of quadratic forms
q(x1, . . . , xr) =∑i
aiix2i +
∑i<j
2aijxixj = xtAx,
where A is a symmetric matrix in Mr(C).
Problem 0.8. Describe SG = C[V ]G (polynomial functions invariant underG).
Proposition 0.9. Under the action of G = SLn(C), q is equivalent to preciselyone of the following quadratic forms:
(i) Tr,δ = δx21 + x2
2 + . . .+ x2r, where δ ∈ C×,
(ii) Tp = x21 + x2
2 + . . .+ x2p, where p < r.
Before proving this we need the following lemma from linear algebra, a proofof which can be found in any standard linear algebra book (often the result isknown as Sylvester’s law of inertia).
Lemma 0.10. Let (·, ·) be a bilinear form on Cr. Then there exists a basisv1, v2, . . . , vr of Cr such that (vi, vj) = 0 for i 6= j and (vi, vi) = 1 for 1 ≤ i ≤ pand (vi, vi) = 0 for r ≥ i > p.
INVARIANT THEORY 9
Proof . (of Proposition 1.2) Follows easily from the lemma, using the fact that
GLr(C) =
δ 0 · · · 00 1 · · · 0
0 0. . . 0
0 0 · · · 1
SLr(C).
�
Corollary 0.11. K[V ]SLr(C) = C[∆] (and in particular is free), where ∆ :V → C : A 7→ detA. Moreover,
∆−1(δ) =
{{Tr,δ}, if δ 6= 0T0 t T1 t · · · t Tr−1, if δ = 0
Chapter 1
Tensor products and related constructions
1.1. Construction of the tensor product
Given two vector spaces V,W over a field K, there exists a new vector spaceV ⊗K W , called their tensor product, which can be defined in several ways.We first make an abstract definition and then in concrete terms describe itselements.
Definition 1.1. Let f : V ×W → U be a map between K-vector spaces. Wesay f is bilinear if it is linear in each variable, i.e. in both V andW . Explicitly,we have
f(λv + µv′, w) = λf(v, w) + µf(v′, w),
f(v, λw + µw′) = λf(v, w) + µf(v, w′)
for all v, v′ ∈ V , w,w′ ∈ W and λ, µ ∈ K. The tensor product V ⊗W comeswith a bilinear map V ×W → V ⊗W and has the universal property thatif f : V ×W → U is bilinear for any K-vector space U , then there exists aunique linear map f which makes the following diagram commute:
V ×W U
V ⊗W
f
ϕ f
We write v ⊗ w for ϕ(v, w).
We do not provide a construction of the tensor product (and only one suchconstruction must be carried out, since it will be unique). For a construction,and a proof of the following proposition, see Algebra by Serge Lang, [4].
Proposition 1.2 (Properties of the tensor product). Let V,W be vector spacesover a field K. Then
(i) V ⊗ W is spanned by elements of the form v ⊗ w, often called puretensors (so any element of V ⊗ W is a linear combination of pure
10
INVARIANT THEORY 11
tensors). These elements satisfy bilinearity:
(λv + µv′)⊗ w = λ(v ⊗ w) + µ(v′ ⊗ w),
v ⊗ (λw + µw′) = λ(v ⊗ w) + µ(v ⊗ w′).
(ii) If V has basis {vα}α∈A and W has basis {wβ}β∈B then V ⊗W has basis{vα ⊗ wβ | α, β ∈ A×B}.
(iii) If V,W are finite-dimensional then dimV ⊗W = dimV dimW .(iv) (V ⊗W )⊗ U ∼= V ⊗ (W ⊗ U) under the canonical isomorphism given
by(v ⊗ w)⊗ u 7→ v ⊗ (w ⊗ u).
Remark 1.3. The above construction works equally well for arbitrary modulesM,N over a ring A, that is, we can always form the tensor product M ⊗A N .For a more detailed construction for arbitrary modules, see Introduction toCommutative Algebra by Atiyah-Macdonald, [1].
1.2. Functoriality
As we mentioned above, the tensor product satisfies a very strong universalproperty. We now exploit this property to prove some important isomorphismswhich we will use throughout this course.
If α ∈ EndK(V ) and β ∈ EndK(W ) then we have a map
α⊗ β : V ⊗W → V ⊗Wdefined by α⊗ β(v ⊗w) = αv ⊗ βw and extending linearly. We need to checkthat α⊗ β is well-defined. In fact, we have
Lemma 1.4. If V and W are K-vector spaces, then
End(V ⊗W ) ∼= End(V )⊗ End(W )
and the isomorphism is canonical, i.e. does not depend on a choice of basis.
Proof . Consider the map
End(V )× End(W ) → End(V ⊗W )
(α, β) 7→ α⊗ βwhich is clearly bilinear. The universal property of the tensor product givesus a unique linear map
End(V )⊗ End(W ) → End(V ⊗W )
α⊗ β 7→ α⊗ β.We now show that this map is an isomorphism. Suppose {vi}i∈I and {wj}j∈Jare bases of V and W respectively. Then End(V ) has basis {αij} given by
αij : vp 7→ δpivj.
12 INVARIANT THEORY
(Note in matrix terms this is the matrix eij with a 1 in the (i, j)-entry andzeroes elsewhere). Similarly End(W ) has a basis {βkl}. Then if we define
αij ⊗ βkl : vp ⊗ wq 7→ δipδkqvk ⊗ wl,it is clear that αij ⊗ βkl ∈ End(V ⊗ W ) for all i, j ∈ I and all k, l ∈ J ,and that this endomorphism is the corresponding elementary basis element ofEnd(V ⊗W ), whence our isomorphism is established. �
We leave the proof of the following important result as an exercise. Noticethat once a map has been defined canonically, as below, it suffices to choosea particular basis in order to prove that it is an isomorphism. Note that thebasis elements on the right are those linear maps which map the whole spaceV onto one line in U – this seems restrictive, but in fact it is easy to show thatall linear maps are linear combinations of such maps.
Lemma 1.5. Let U and V be K-vector spaces and let
V ∗ = {ϕ : V → K | ϕ linear}be the corresponding space of linear functionals, i.e. the dual space of V . Thenthere is a canonical isomorphism
ξ : U ⊗ V ∗ → Hom(V, U)
u⊗ ϕ 7→ (ξu,ϕ : v 7→ ϕ(v) · u).
1.3. The tensor algebra and quotients
As a special case of the construction above, we now let V be a finite-dimensionalvector space over a field K, and form the tensor products V ⊗ V , V ⊗ V ⊗ V ,and so forth – these are the tensor powers of V .
Definition 1.6. We define the tensor algebra of V to be
T (V ) =⊕r≥0
T r(V ),
whereT r(V ) = V ⊗ V ⊗ · · · ⊗ V︸ ︷︷ ︸
r times,
the rth tensor power of V . Recall that a K-algebra is a K-vector spaceequipped with a multiplication which is compatible with the linear structure.
We use the convention that T 0(V ) = K. Noting that elements of T (V ) will beinfinite tuples with only finitely many nonzero terms, we define the multipli-cation in T (V ) as
v1 ⊗ v2 ⊗ · · · ⊗ vr · v′1 ⊗ v′2 ⊗ · · · ⊗ v′r := v1 ⊗ v2 ⊗ · · · ⊗ vr ⊗ v′1 ⊗ v′2 ⊗ · · · ⊗ v′rfor v1⊗v2⊗· · ·⊗vr and v′1⊗v′2⊗· · ·⊗v′r ∈ T r(V ) and extend pointwise to T (V ).This gives us a graded algebra structure since T r(V ) · T s(V )→ T r+s(V ).
INVARIANT THEORY 13
Remark 1.7. T (V ) is the free associative algebra on V , i.e. if A is anyassociative algebra and f : V → A is any linear map, there exists a uniquemap f : T (V ) → A which is a homomorphism of K-algebras, and such thatthe diagram
V A
T (V )
f
f
commutes. Note however that T (V ) is in general not commutative, i.e. that
v ⊗ w 6= w ⊗ v
in general. In fact it can be shown that v ⊗ w = w ⊗ v if and only if v, ware proportional. However we can take a quotient by an appropriate ideal andforce commutativity.
Definition 1.8. Let IS be the ideal of T (V ) generated by all elements of theform {v ⊗ w − w ⊗ v | v, w ∈ V }. Then
S(V ) = T (V )/IS
is called the symmetric algebra of V .
If V has basis {vi}i∈I , write vi for the image in S(V ) of vi – then by definitionof our ideal IS, we have vivj = vj vi. Then, in a canonical way,
S(V ) ∼= K[{vi}],
the algebra of polynomials in the vi with coefficients in K. Similarly, we mayform S(V ∗), where V ∗ is the dual space of V .
Remark 1.9. S(V ) is the free associative commutative algebra on V . (Theproof of this is not difficult given the above result about T (V ) – we have merelytaken the free associative algebra and imposed commutativity in a generalway). This means that, given a commutative (and associative) algebra A anda linear map f : V → A, there exists a unique K-algebra homomorphismf : S(V )→ A such that the diagram
V A
S(V )
f
f
commutes.
14 INVARIANT THEORY
This construction may be applied to form S(V ∗). Let F(V,K) be the K-algebra of all functions from V → K with pointwise addition and multiplica-tion, i.e. (f1 + f2)(v) = f1(v) + f2(v) and (f1f2)(v) = f1(v)f2(v). This is avery large commutative K-algebra. Since by definition
V ∗ = Hom(V,K),
we have a linear map V ∗ → F(V,K). Hence we have an algebra homomor-phism S(V ∗) → F(V,K). We can regard the elements of S(V ∗) as functionson V , and those particular functions are polynomial functions. For this reasonS(V ∗) is often called the algebra of polynomial functions on V .We write S(V ∗) = K[V ] and call it the coordinate ring of V .Remark 1.10. In this course, we insist all rings and algebras have identi-ties, and that all ring- and algebra- homomorphisms are homomorphisms ofrings and algebras with identity, so as to avoid situations as in the followingexamples.Example 1.11. Let A1 be the algebra
A1 =
{(0 a0 0
)| a ∈ K
}⊂M2(K).
Note any element of A1 has its square equal to zero, so A cannot have an iden-tity. Now let B = A1 ⊕M2(K) = {(a,m) | a ∈ A1,m ∈ M2(K)}. This againis an algebra without identity. However we have an injective homomorphismof algebras (but not of algebras with identity)
M2(K) ↪→ B.
Further, if we let C = M2(K)⊕M2(K), this is an algebra with identity, namely1C = (1, 1). But the inclusion mapM2(K) ↪→ C is an algebra homomorphism,but not an isomorphism of algebras with identity, since 1M2(K) 7→ (1, 0).By way of a second example, consider the algebra
A2 =
{(a aa a
)| a ∈ K
}⊂M2(K).
A calculation shows that this algebra has an identity, namely the elemente =
( 1/2 1/21/2 1/2
). Then the inclusion map A2 ↪→ M2(K) is an injective algebra
homomorphism, but not of algebras with identity, since e 67→ I.
We also can form the so-called exterior algebra.Definition 1.12. Let IE be the ideal of T (V ) generated by all elements of theform {v ⊗ w + w ⊗ v | v, w ∈ W}. Then
Λ(V ) = T (V )/IE
is called the exterior algebra of V .Let v1, . . . , vr ∈ V . We denote by v1 ∧ v2 ∧ · · · ∧ vr the image of v1 ⊗ · · · ⊗ vrunder the natural surjection η : T (V )→ Λ(V ).
INVARIANT THEORY 15
Remark 1.13. By definition of our ideal,
v1 ∧ v2 = −v2 ∧ v1.
Hence for σ ∈ Sr,
σ · (v1 ∧ · · · ∧ vr) = (sgn σ)(v1 ∧ · · · ∧ vr),
where the symmetric group acts on v1 ∧ · · · ∧ vr by permuting indices. Usingthis, it is easy to show that
Λr(V ) = η(T r(V ))
has basis{bi1 ∧ bi2 ∧ · · · ∧ bir | i1 < i2 < · · · < ir}.
Hence it follows that dim Λr(V ) =(nr
), where n = dimV . Hence
dim Λ(V ) =n∑r=0
(n
r
)= 2n.
These remarks all make sense in infinite dimensional vector spaces except forthe final remarks about dimension.
Remark 1.14. Λ(V ) is the free associative alternating algebra on V .
In summary, given a K-vector space V , we have constructed four graded al-gebras T (V ), S(V ), Λ(V ), K[V ] = S(V ∗). These constructions are in factfunctorial in a category theoretical sense.
1.4. Graded algebras and Poincaré series
Definition 1.15. Let G be a group. A graded G-module is a K-vector space
M =∞⊕r=0
Mr,
where, for each r ≥ 0, Mr is a KG-module. Our modules will be locally finite,which means dimKMr <∞ for all r ≥ 0.
Example 1.16. Let G ≤ GL(V ) where GL(V ) is the group of invertible lineartransformations of V . Then G acts on T (V ) via
g : V ⊗r → V ⊗r
v1 ⊗ · · · ⊗ vr 7→ gv1 ⊗ · · · ⊗ gvr,
i.e. EndV ⊗r ∼= (EndV )⊗r, g ⊗ · · · ⊗ g ← g.
This makes T r(V ) into aKG-module. Since g(v⊗w−w⊗v) = gv⊗gw−gw⊗gv,G acts on Sr(V ), similarly for Λr(V ).
16 INVARIANT THEORY
Remark 1.17. If G acts on V , it acts on V ∗ via
gϕ(v) = ϕ(g−1v), for all g ∈ G,ϕ ∈ V ∗, v ∈ V.This is called the dual or contragredient representation.
Definition 1.18. The Poincaré or Hilbert series of a locally finite gradedmodule is the formal power series
PM(t) =∞∑r=0
dimMrtr.
For g ∈ G (as above) we also define the equivariant Poincaré series
PM(g, t) =∞∑r=0
tr(g,Mr)tr.
In particular PM(1, t) = PM(t).
Definition 1.19. A graded module M is a graded algebra if we have a multi-plication M ⊗M →M (linear, associative) if
MrMs ⊂Mr+s.
Example 1.20. All of the examples from §1.3, i.e. T (V ), S(V ),Λ(V ), K[V ]are all graded algebras with GL(V )-module structure.
If M = ⊕Mr and N = ⊕Ns are graded modules, then so are M ⊕ N , where(M ⊕N)r = Mr ⊕Nr and M ⊗N , where
(M ⊗N)r =⊕i+j=r
Mi ⊗Nj.
Lemma 1.21. Let V1, V2 be K-vector spaces. Then
(i) S(V1 ⊕ V2) ∼= S(V1)⊗ S(V2).
(ii) Λ(V1 ⊕ V2) ∼= Λ(V1)⊗ Λ(V2).
Proof . Let {bi}, {cj} be bases of V1 and V2 respectively. Then
S(V1 ⊕ V2) ∼= K[bi, cj] ∼= K[bi]⊗K[cj] ∼= S(V1)⊗ S(V2).
Now Λp(V1 ⊕ V2) has basis
{bi1 ∧ bi2 ∧ · · · ∧ bik ∧ cj1 ∧ · · · ∧ cjl | i1 < · · · < ik, ji < · · · jl, k + l = p},and so
Λp(V1 ⊕ V2) =⊕k+l=p
Λk(V1)⊗ Λl(V2) = (Λ(V1)⊗ Λ(V2))p
by our remarks in Example 1.20, as claimed. �
Lemma 1.22. Let M,N be locally finite G-modules and let g ∈ G. Then
INVARIANT THEORY 17
(i) PM⊕N(g, t) = PM(g, t) + PN(g, t).
(ii) PM⊗N(g, t) = PM(g, t)PN(g, t).
Proof . The first statement is easy and follows from our remarks in Example1.20 and the fact that
tr(g,Mk ⊕Nk) = tr(g,Mk) + tr(g,Nk).
For the second statement, consider
PM⊗N(g, t) =∑k
tr(g, (M ⊗N)k)tk
=∑k
tr (g,⊕i+j=kMi ⊗Nj) tk
=∑k
(∑i+j=k
tr(g,Mi ⊗Nj)
)tk
=∑k
(∑i+j=k
tr(g,Mi)tr(g,Nj)
)tk
=
(∑i
tr(g,Mi)ti
)(∑j
tr(g,Nj)tj
).
�
Example 1.23. We compute the Poincaré series of some easy spaces.
(i) Let V = Kn. Then by the geometric series formula
PT (V )(t) =∑k≥0
nktk =1
1− nt.
(ii) Note that Kn = K ⊕ · · · ⊕K. Hence Lemma 1.21 givesS[Kn] = S(K)⊗ · · · ⊗ S(K)
∼= K[x1]⊗ · · · ⊗K[xn]∼= K[x1, . . . , xn]
as we have seen before.(iii) Λ(Kn) = Λ(K)⊗ · · · ⊗ Λ(K). Now
Λ(K) = K1⊕Kxsince x ∧ x = 0, where x is any field element chosen to be the basiselement for the field K as a K-vector space. Hence
PΛ(K)(t) = 1 + t,
and so by our lemmaPΛ(Kn)(t) = (1 + t)n.
18 INVARIANT THEORY
So the binomial theorem tells us that dim Λr(Kn) =(nr
)as we have
seen above.
Remark 1.24. For every field K, K[t] is the algebra of polynomials in twith coefficients in K and K[[t]] the algebra of formal power series in t withcoefficients in K. Elements of K[[t]] are formal expressions
a0 + a1t+ a2t2 + . . .
which are multiplied and added formally as per the normal rules.
Exercises
Exercise 1.25. Write a proof of Lemma 1.5 for V,W finite-dimensional.
Exercise 1.26. Show how Lemma 1.5 implies Lemma 1.4.
Exercise 1.27. Write more elegant proofs of the two parts of Lemma 1.21,using the universal properties of the constructions.
Exercise 1.28. Show that for any two vector spaces V,W over a field K thereare canonical isomorphisms
(i) End(V )⊗ End(W )→ End(V ⊗W )(ii) V ⊗ V ∗ → End(V )(iii) (V ⊗W )∗ → V ∗ ⊗W ∗
(iv) End(V )→ (End(V ))∗
(v) (T r(V ))∗ → T r(V ∗)
Exercise 1.29. (i) In the isomorphism ξ : V ⊗ V ∗ → End(V ) in Exercise1.29(ii), compute tr(ξ(ϕ⊗ v)).
(ii) Show that in Exercise (iv), the identification arises from the non-degenerate form (α, β) 7→ tr(αβ).
Chapter 2
Elementary invariant theory of finite groups
In this chapter we discuss some basic results from the theory of finite groupinvariants. Throughout, we will assume V is a finite-dimensional vector spaceover a field K.
2.1. Molien’s theorem
Molien’s theorem gives elegant formulas for the Poincaré series of the spaceSG of G-invariants of S where S = S(V ) was constructed in Chapter 1. Wefirst need some more basic results about Poincaré series.
Proposition 2.1. Let g be a linear transformation of a vector space V = Kn.Then
PS(V )(g, t) =1
detV (1− gt), and PΛ(V )(g, t) = det(1 + gt).
Proof . Recall that PM(g, t) =∑
i tr(g,Mi)ti. It is important to recall that, in
order to be able to invert polynomials in the ring of formal power series werequire their constant terms to be nonzero. This is a special case of a generalresult, namely that f(x) a unit in A[[x]] if and only if f0 a unit in A, where Ais any commutative ring (this is an easy exercise – see [1], Exercise 1.2).
Assuming that K is algebraically closed, suppose that the eigenvalues of g onV are λ1, . . . , λn. Then there exists a basis v1, . . . , vn of V such that
gvi = λivi +∑j<i
µjivj, 1 ≤ i ≤ n.
Hence
g · (vm11 vm2
2 · · · vmnn ) = λm11 · · ·λmnn vm1
1 · · · vmnn + other monomials. (2.2)
Now 1 − gt is some matrix with entries in K[t] and so will have eigenvalues1− λit for i = 1, . . . , n, and so since monomials of degree m will span Sm(V ),
tr(g, Sm(V )) =∑
m1+...+mn=m
λm11 · · ·λmnn
19
20 INVARIANT THEORY
=n∏i=1
(1 + λit+ λ2i t
2 + . . .)∣∣∣tm
=
(1
1− λ1t
1
1− λ2t· · · 1
1− λnt
)∣∣∣∣tm
=
(1∏n
i=1(1− λit)
)∣∣∣∣tm
=
(1
detV (1− gt)
)∣∣∣∣tm,
(where by ·|tm we mean to take the coefficient of tm) and hence
PS(V )(g, t) =∑m
tr(g, Sm(V ))tm =1
detV (1− gt).
For the second formula, notice that
tr(g,Λp(V )) =∑
i1<i2<···<ip
λi1 · · ·λip
=
(n∏i=1
(1 + λit)
)∣∣∣∣∣tp
= (det(1 + qt))|tp
and so the desired formula follows. �
Remark 2.3. Note that if g has finite order then it is always diagonalisableand the “error term” in (2.2) in the above proof does not appear. To see this,put g in its Jordan form, so
g = su,
where s is semisimple (diagonalisable) and u is unipotent (i.e. all its eigenvaluesare 1) and su = us. Then
gm = smum = 1 if and only if sm = 1 and um = 1.
Now if g has finite order then both s and u must have finite order. Over a fieldof characteristic zero, the element
1 ∗ ∗ · · · ∗0 1 ∗ · · · ∗...
.... . . · · ·
......
... · · ·. . . ∗
0 · · · · · · · · · 1
has infinite order.
INVARIANT THEORY 21
Corollary 2.4. Let V be a K-vector space and g : V → V a linear transfor-mation. Then
PS(V )(g, t) · PΛ(V )(g,−t) = 1.
Proof . This follows immediately upon multiplying the two expressions for theseries in Proposition 2.1. �
Remark 2.5. The above corollary states that S(V ) and Λ(V ) are Koszul dualsof one another.
Definition 2.6. IfM is anyKG-module, we say thatm ∈M is semi-invariantif
gm ∈ K×m for all g ∈ G,where K× = K \ {0}.Lemma 2.7. If m ∈ M is nonzero and semi-invariant, there exists a grouphomomorphism χ : G→ K× such that
gm = χ(g)m for all g ∈ G.χ is known as a linear character of G.
Proof . Since g1(g2m) = (g1g2)m by definition of a group action, if we canwrite gm = χ(g)m for some function, the function will automatically be agroup homomorphism, since
χ(g1)χ(g2)m = g1(χ(g2)m) = (g1g2)m = χ(g1g2)m,
and similarly χ(1) = 1. �
Remark 2.8. In the special case when χ ≡ 1 (so that χ(g) = 1 for all g ∈ G),we say that m is invariant, or G-invariant.
Definition 2.9. Given a linear character χ : G→ K×, the set{m ∈M | gm = χ(g)m for all g ∈ G}
is a subspace of M , which we denote Mχ and call the subspace of χ-semi-invariants in M . When χ is the trivial charcter χ ≡ 1, then we write M1 =MG, the space of G-invariants in M .
Remark 2.10. If in addition to its linear structure, M also carries the struc-ture of a K-algebra, then
Mχ ·Mχ′ ⊆Mχ·χ′ ,
since if m ∈ m ∈ Mχ and m′ ∈ Mχ′ , then, assuming G acts as a group ofalgebra automorphisms,
g(mm′) = g(m)g(m′)
= χ(g)m · χ′(g)m′
= χ(g)χ′(g)mm′
= χχ′(g)mm′.
22 INVARIANT THEORY
In the special case when MχMG ⊆ Mχ for all χ, then Mχ is a module overMG for all χ. In particular, taking χ = 1 gives that MG is a subalgebra of M .
For the rest of this chapter, assume G is a finite group.
Recall that, by Maschke’s theorem, if V is a finite-dimensional KG-module,we can write
V = V1 ⊕ V2 ⊕ · · · ⊕ Vr (2.11)where the Vi are irreducible.
Assume for the moment that K is algebraically closed. Then the set of allnon-isomorphic irreducible KG-modules are in bijection with the irreduciblecharacters of G (this is a standard result from the representation theory offinite groups).
In the above decomposition, let
Vχ =⊕
{i|χ(Vi)=χ}
Vi,
which we call the χ-isotypic component of V . Then
V =⊕
χ∈ irr(G)
Vχ, (2.12)
where irr(G) is the set of irreducible characters of G. The decompositionin (2.11) is not canonical, but the decomposition (2.12) is. There exists acanonical projection pχ : V → Vχ, given by
pχ =χ(1)
|G|∑g∈G
χ(g−1)g.
This maps V → Vχ, since if V ′ is an irreducibleKG-module, then pχ commuteswith the G-action, i.e. gpχ = pχg for all g ∈ G. Therefore by Schur’s lemma,pχ is multiplication by a scalar. Computing
trpχ = χ(1)〈χ, χV ′〉we determine the value of this scalar, i.e.
pχ =
{idV ′ , if χV ′ = χ0, otherwise.
In particular this shows p2χ = pχ.
Lemma 2.13. Let M = ⊕iMi be a locally finite graded KG-module. Define
Mχ =⊕i
(Mi)χ.
ThenPMχ(t) = χ(1)〈χ, PM(·, t)〉,
INVARIANT THEORY 23
where PM(·, t) is the function G→ K[[t]] given by
g 7→ PM(g, t).
Note this is a class function, i.e. a function G → K× constant on conjugacyclasses of G, since
PM(xgx−1, t) = PM(g, t).
Proof . Using the notation above,
(Mi)χ = pχMi
=χ(1)
|G|∑g∈G
χ(g−1)gMi.
Since pχ is the projection Mi → (Mi)χ,
dim(Mi)χ = trMi(pχ),
since projections act with eigenvalue 1 on the subspace concerned and 0 onthe complement. Hence
dim(Mi)χ =χ(1)
|G|∑g∈G
χ(g−1)tr(g,Mi),
whence
PMχ(t) =∞∑i=1
dim(Mi)χti =
χ(1)
|G|∑g∈G
χ(g−1)PM(g, t)
as claimed. �
Corollary 2.14 (Molien’s theorem). Suppose V is a KG-module, and let
S = S(V ) =∞⊕i=0
Si(V ).
Then
PSG(t) =1
|G|∑g∈G
1
det(1− gt).
Proof . Note SG = (S)1, the isotypic component corresponding to the trivialcharacter 1. By Lemma 2.13,
PSG(t) =1
|G|∑g∈G
PS(g, t)
which gives the result by Proposition 2.1. �
24 INVARIANT THEORY
2.2. Elementary facts about finite group invariants
Lemma 2.15 (Lagrange interpolation). Let V = Kn (where K is any field).Let X = {v1, . . . , vs} be a finite set of points in V . Then, given a1, . . . , as ∈K, there exists a function f ∈ K[V ] = S(V ∗) such that f(vi) = ai for alli = 1, . . . , s.
Proof . It suffices to show there exist f1, . . . , fs ∈ K[V ] such that fi(vj) = δij,since then we can take
f =s∑i=1
aifi.
By symmetry it suffices to show there exists f1 with this property. We showthis by induction on s, for which the base case s = 1 is straightforward – wesimply take the constant function 1.
Assume now that s > 1, so that by inductive hypothesis there exists g1 suchthat g1(v1) = 1 and g1(v2) = g2(v2) = . . . = g1(vs−1) = 0. Take l ∈ V ∗ suchthat l(v1) 6= l(vs), and define
f1(v) =g1(v1)(l(v)− l(vs))
l(v1)− l(vs),
which clearly satisfies f1(v1) = 1 and f1(vj) = 0 for j 6= 1 as required. �
Remark 2.16 (Polynomial functions). F(V,K) is the algebra of all functionsV → K with pointwise operations. A map
ϑ : V ∗ → F(V,K)
extends uniquely by the universal property to an algebra homomorphism
p : S(V ∗) = K[V ]→ F(V,K).
In coordinates, if V has basis b1, . . . , bn and V ∗ has dual basis x1, . . . , xn,
f(x1, . . . , xn)(∑
αibi
)= f(α1, . . . , αn).
Note that if g ∈ GL(V ), then g acts on F(V,K) via gf(v) = f(g−1v).
Since g(V ∗) ⊂ V ∗, it follows that gS(V ∗) ⊂ S(V ∗) – “linear substitution”.
Lemma 2.17. If K is infinite, then p is injective.
Proof . Suppose f(x1, . . . , xn) ∈ ker p. Then f(α1, . . . , αn) = 0 for all α1, . . . , αn ∈Kn. Write
f =∑i
fi(x1, . . . , xn−1)xin,
where the sum over i is finite. Then for any α1, . . . , αn−1, the polynomial∑i
fi(α1, . . . , αn−1)xin
INVARIANT THEORY 25
vanishes everywhere, and so fi(α1, . . . , αn−1) = 0 for all α1, . . . , αn−1 and theresult follows by induction. �
In invariant theory, we are interested in K[V ]G = K[V ]1 as a commutativealgebra for certain G. If G is finite, write
AG =1
|G|∑g∈G
g,
an element of KG. Then it is easy to show that AG satisfies the followingproperties:
(i) for all g ∈ G, gAG = AG;(ii) A2
G = AG;(iii) if G acts on a vector space W then
AGW = WG,
(of course, AG = pχ where χ = 1 the trivial character);(iv) if g ∈ K[V ] = S(V ∗) then AGg ∈ K[V ]G;(v) if f1 ∈ K[V ]G and f2 ∈ K[V ] then AG(f1f2) = f1AG(f2).
2.3. Invariants of finite groups
Consider a finite group G ≤ GL(V ), acting on K[V ] = S(V ∗) as describedabove. We have a homogeneous decomposition
S = S0 ⊕ S1 ⊕ S2 ⊕ · · · ,
so since the group action preserves homogeneity
SG = (S0)G ⊕ (S1)G ⊕ (S2)G ⊕ · · · .
WriteSG+ = (S1)G ⊕ (S2)G ⊕ · · ·
(note that this is not an algebra in the sense that we have defined it since ithas no identity element – this is simply an ideal generated by invariants ofpositive nonzero degree). Let J be the ideal of S
J = 〈SG+〉 ={∑
i
′ Pifi | Pi ∈ S, fi ∈ SG+}.
Proposition 2.18. Let J be the ideal of S defined above. Then
(i) J has a finite set F1, . . . , Fr of homogeneous generators as an ideal of S.
(ii) This same set of generators F1, . . . , Fr generate SG as an algebra.
26 INVARIANT THEORY
Proof . K[x1, . . . , xn] is a Noetherian ring, i.e. we have the ascending chaincondition on ideals.1 The ascending chain condition is equivalent to every idealbeing finitely generated, and we may clearly extend any such finite generatingset of J until it consists entirely of homogeneous elements.
Now suppose P ∈ SG, and assume without loss of generality that P is ho-mogeneous. We proceed by induction on the degree of P , for which the casedegP = 0, i.e. P is constant, is immediate. Letting degP > 0, we must showthat P is a polynomial Q(F1, . . . , Fr) in the Fi, where the Fi generate J . NowP is invariant and non-constant and hence is in the ideal J , so is of the form
P =r∑i=1
PiFi, Pi ∈ S.
by our choice of generators Fi of J . Now, using the properties of the operatorAG from the previous section, we have
AGP = P =r∑i=1
AG(Pi)Fi.
Since the AG(Pi) have degree strictly lower than P for all i, by induction AGPiis a polynomial in the Fi, and so too is AGP = P . �
Corollary 2.19 (Noether’s theorem). Let K be any field of characteristic zero.Then K[V ]G is a finitely generated K-algebra.
Remark 2.20. This remains true for infinite reductive algebraic groups, butnot generally true (this is known as Hilbert’s fourth problem). Some unipotentgroups give counterexamples to the general case, including the well-knownNagata counterexample.
Proposition 2.21. Suppose that a finite group G acts on V = Kn. Then foru, v ∈ V , u and v are in the same G-orbit if and only if F (u) = F (v) for allF ∈ K[V ]G.
Proof . Suppose first that u, v are in the same G-orbit, say v = gu for someg ∈ G and F ∈ K[V ]G. Then
F (v) = F (gu) = (g−1F )(u) = F (u)
by invariance. Conversely, if u, v are not in the same G-orbit, we must findsome F ∈ K[V ]G such that F (v) 6= F (u). In this case, we know that the orbitGu = {gu | g ∈ G} is disjoint from Gv. By Lagrange interpolation (Lemma2.15), there exists f ∈ K[V ] such that f(gv) = 1 for all g ∈ G and f(gu) = 0
1The ascending chain condition on a commutative ring R states that if I1 ⊂ I2 ⊂ · · · isan ascending chain of ideals of R, then for some N , IN = IN+1 = · · · . It is obvious thatfields are Noetherian since their only ideals are 0 and R. Moreover, Hilbert’s basis theoremstates that R[x] is Noetherian if R is Noetherian, and so by an easy induction we see thatK[x1, . . . , xn] is a Noetherian ring and hence satisfies the ascending chain condition.
INVARIANT THEORY 27
for all g ∈ G (notice that it is important here that G be finite, otherwiseLagrange interpolation doesn’t work). Define F ∈ K[V ] by
F =∏g∈G
(gf).
Then F ∈ K[V ]g, and furthermore
F (v) =∏g∈G
gf(v) =∏g∈G
f(g−1v) = 1
by construction. Moreover,
F (u) =∏g∈G
gf(u) =∏g∈G
f(g−1u) = 0 6= F (v)
as required. �
Corollary 2.22. If SG = K[V ]G is generated as an algebra by F1, . . . , Fr ∈ S,then the map
ϕ : v 7→ (F1(v), . . . , Fr(v))
has image V \G, the space of orbits of G on V .
Proof . This follows since any two elements of SG take the same value onu, v ∈ V if and only if Fi(u) = Fi(v) for all i = 1, . . . , r. �
We recall some definitions and important results about field extensions andtranscendence degree. Suppose K ⊂ L is a field extension. Say that x ∈ L isalgebraic over K if x satisfies an equation
arxr + ar−1x
r−1 + . . .+ a1x+ a0 = 0
with ai ∈ K, not all zero. If x is not algebraic we say it is transcendental overK.
Say that the set S ⊂ L is algebraically independent if for any finite set ofmulti-indices {(m1, . . . ,mr) ∈M},∑
M
a(m1,...,mr)xm11 · · · xmrr = 0
with a(m1,...,mr) ∈ K implies a(m1,...,mr) = 0 for all m1, . . . ,mr.
The number of elements in any maximal algebraically independent subset,called a transcendence base, is the transcendence degree of L over K, writtentr degL/K, and it is an important fundamental result in field theory that thisis independent of the choice of transcendence base. The following result, oftenreferred to as Lüroth’s theorem, holds for finite and infinite field extensions,however we shall only need it in the finite case, when it states that any twotranscendence bases have the same number of elements.
Theorem 2.23 (Lüroth’s theorem). Any two maximal algebraically indepen-dent subsets of L have the same cardinal number.
28 INVARIANT THEORY
Proof . See van der Waerden [6] or Lang [4]. �
Example 2.24. Consider K(x1, . . . , xn) where xi are indeterminates. Thenthe set {x1, . . . , xn} is a maximal algebraically independent set, since everyelement is of the form
f(x1, . . . , xn)
g(x1, . . . , xn)= t
for some polynomials f, g, i.e. f(x1, . . . , xn) = tg(x1, . . . , xn). So the transcen-dence degree of K(x1, . . . , xn) is n.
Let us now put together everything from this chapter to state the followingfundamental result about the elementary invariant theory of finite groups.
Theorem 2.25. Let G be a finite group acting linearly on V = Kn as above.Then
(i) if dimV = n, there exist n algebraically independent homogeneous poly-nomials F1, . . . , Fn ∈ SG;
(ii) if degFi = di then |G| ≤ d1d2 · · · dn.
Proof . The transcendence degree of K(x1, . . . , xn) over K is n. Suppose Gacts on K(x1, . . . , xn) with fixed subfield K(x1, . . . , xn)G – then
deg(K(x1, . . . , xn) : K(x1, . . . , xn)G) = |G|by the Fundamental Theorem of Galois Theory.2 Furthermore, it is a factabout the transcendence degree that, given field extensions K ⊂ L ⊂M ,
tr degL/K + tr degM/L = tr degM/K.
Therefore, applying this to K ⊂ K(x1, . . . , xn)G ⊂ K(x1, . . . , xn), since everyelement of K(x1, . . . , xn)G is algebraic over K(x1, . . . , xn),
tr degK(x1, . . . , xn)G/K(x1, . . . , xn) = 0
and sotr degK(x1, . . . , xn)G/K = n.
We show thatK(x1, . . . , xn)G is equal to the field of fractions ofK[x1, . . . , xn]G.To see this, suppose P
Q∈ K(x1, . . . , xn)G. Then
gP
gQ=P
Q
2Since G is a subgroup of the Galois group AutKK(x1, . . . , xn) it corresponds underthe Galois correspondence to the subfield K(x1, . . . , xn) of elements fixed by G. Then theFundamental Theorem states that the degree of the field extension is equal to the relativeindex of the corresponding subgroups, i.e. [K(x1, . . . , xn) : K(x1, . . . , xn)G] = [1 : G] = |G|.It is also true that the representation of G on L as a vector space over LG is the regularrepresentation.
INVARIANT THEORY 29
for all g ∈ G. ButP
Q=P∏
g 6=1 gQ∏g∈G gQ
,
where the denominator is in K[x1, . . . , xn]G. Hence we can writeP
Q=P ′
Q′
with Q′ ∈ K[x1, . . . , xn]G. SogP ′
gQ′=gP ′
Q′=P ′
Q′
and so P ′ is invariant also, i.e. P ′ ∈ K[x1, . . . , xn]G, and so using the notationffracK[x1, . . . , xn]G for the field of fractions of K[x1, . . . , xn]G, we see P
Q∈
ffracK[x1, . . . , xn]G and so
K(x1, . . . , xn)G ⊆ ffracK[x1, . . . , xn]G,
whence equality holds since the other inclusion is obvious.
Now if F1, . . . , Fr is a set of homogeneous generators of K[x1, . . . , xn]G, whichexist by Proposition 2.18,
K(x1, . . . , xn)G = K(F1, . . . , Fr)
and hence {F1, . . . , Fr} contains an algebraically independent subset of cardi-nality n – for if that was not so, the field would be generated by fewer than nalgebraically independent elements, contradicting Lüroth’s theorem.
For the second claim, let F1, . . . , Fn be algebraically independent elements ofSG. Then K[F1, . . . , Fn] ⊂ SG is a subring. Hence
PK[F1,...,Fn](t) ≤ PSG(t)
where∑nit
i ≤∑mit
i if and only if mi, ni ∈ N and ni ≤ mi for all i. Now
K[F1, . . . , Fn] ∼= K[F1]⊗K[F2]⊗ · · · ⊗K[Fn]
and so
PK[F1,...,Fr](t) =n∏i=1
1
1− tdi≤ PSG(t) =
1
|G|∑g∈G
1
det(1− tg)(2.26)
by Molien’s theorem, Theorem 2.14. Now both series converge absolutely for|t| < 1, the rightmost term since
1
det(1− gt)=∏i
1
(1− λit)
where λi are the eigenvalues of gt. Multiply both sides of (6.5) by (1 − t)n.Then the left hand side becomes
n∏i=1
1− t1− tdi
=n∏i=1
1
1 + t+ t2 + . . .+ tdi−1
30 INVARIANT THEORY
and the right hand side1
|G|+∑g 6=1
1
|G|(1− t)n
det(1− tg).
For t ∈ R, 0 < t < 1, the left hand side will be less than or equal to the righthand side, and so this inequality will remain true as t → 1. Taking this limitof the two expression gives
n∏i=1
1
1 + t+ t2 + . . .+ tdi−1
t→1−−→n∏i=1
1
di=
1∏ni=1 di
,
and1
|G|+∑g 6=1
1
|G|(1− t)n
det(1− tg)
t→1−−→ 1
|G|,
and so the inequality translates ton∏i=1
di ≥ |G|
as claimed. �
Example 2.27. Suppose G is the symmetric group Sn acting on V by permu-tation of coordinates. Then if xi is the ith coordinate function,
F1 =∑
xi
F2 =∑
x2i
......
Fn =∑
xni .
It may be shown these are algebraically independent, and so
|Sn| ≤ n!.
Exercises
Exercise 2.28. The proof of Proposition 2.1.1 assumed thatK is algebraicallyclosed. Show that Proposition 2.1.1 holds for any field K. [Hint: Note thatthe statement is invariant under extensions of the field – if we enlarge the fieldand prove the result for a larger field, i.e. replacing K by its algebraic closureK, then it must hold for the subfield K since trace is invariant under fieldextensions].
Exercise 2.29. Write down a formula for f1 in Lemma 2.15.
INVARIANT THEORY 31
Exercise 2.30. Recall that the field extensionK ⊃ L isGalois if it is algebraic,and if for all x ∈ K the minimal polynomial mx(t) ∈ L[t] (which is irreducible)splits over K, i.e. mx(t) is a product of linear factors over K.Let K ⊃ L be fields of characteristic zero and let a, b be elements of K whichare algebraic over L. Show that L(a, b) = L(c) for some element c ∈ L(a, b).
Exercise 2.31. Deduce the following statement, known as the Theorem of thePrimitive Element.Let K ⊃ L be a finite extension of fields of characteristic zero. Then K = L(c)for some c ∈ K. The element c is called a primitive element of the extension.
Exercise 2.32. Let K ⊃ L be fields of characteristic zero. Assume that thereis an integer n ≥ 1 such that each element of K has degree at most n over L(i.e. satisfies a polynomial equation of degree ≤ n over L). Show that
(i) The degree [K : L] ≤ n, and(ii) K = L(α) for some α ∈ K.
[Hint: Note that if γ ∈ K has degree l, then [L(γ) : L] = l. Show that thisimplies, using Question 4, that if α ∈ K is such that [L(α) : L] is maximal,then L(α) = K. The hypothesis implies both that such α exists, and that Khas degree ≤ n].
Exercise 2.33. Let K be a field of characteristic zero and let G be a finitegroup of automorphisms of K, with fixed field KG. Show that K is a Galoisextension of KG, [K : KG] = |G| and G = Gal(K/KG).
Chapter 3
The fundamental theorems of invariant theory for GL(V )
Our goal in this chapter are proofs of the two fundamental theorems of invarianttheory for GL(V ). More precisely, we hope to describe the space of GL(V )-invariants in V ⊗r where V is a K-vector space of dimension n, and r ≥ 1.Often in more advanced treatments of the subject, the fundamental theoremsare proven for T rs (V ) = V ⊗r ⊗ (V ∗)⊗s (for example in Kraft and Procesi [3])– our proof can therefore be seen as the s = 0 case. Towards the end of thischapter we will occasionally need results from Appendix A and Chapter 4.
3.1. The first fundamental theorem
Let V be a vector space of dimension n over K, where K is a field of charac-teristic zero. Recall the definition of the symmetric group on r letters Sr asthe group of all permutations of the set {1, 2, . . . , r}. Suppose that the actionof the symmetric group Sr on T r(V ) = V ⊗r is defined as follows:
π · w1 ⊗ · · · ⊗ wr = wπ−11 ⊗ · · · ⊗ wπ−1r
for π ∈ Sr, where w1⊗ · · ·⊗wr ∈ V ⊗r. Note that this is indeed a well-definedgroup action, since if π, σ ∈ Sr, then
(πσ) · (v1 ⊗ · · · ⊗ vr) = v(πσ)−11 ⊗ · · · ⊗ v(πσ)−1r
= vσ−1(π−11) ⊗ · · · ⊗ vσ−1(π−1r),
whereas
π · (σ · (v1 ⊗ · · · ⊗ vr)) = π · (vσ−11 ⊗ · · · ⊗ vσ−1r)
= vσ−1(π−11) ⊗ · · · ⊗ vσ−1(π−1r)
as required.
Recall that we also defined an action of GL(V ) on V ⊗r in Example 1.16. Moreprecisely, if g ∈ GL(V ) and w1 ⊗ · · · ⊗ wr ∈ V ⊗r then
g(w1 ⊗ · · · ⊗ wr) = gw1 ⊗ · · · ⊗ gwr,i.e. g acts as g ⊗ · · · ⊗ g. We have the following important observation:
Proposition 3.1. If π ∈ Sr and g ∈ GLn then, as transformations of V ⊗r,we have
πg = gπ,
32
INVARIANT THEORY 33
i.e. we have an algebra homomorphism
µr : KSr → EndGL(V )V⊗r
Proof . We consider π as a linear transformation
V ⊗r → V ⊗r : v1 ⊗ · · · ⊗ vr 7→ π · v1 ⊗ · · · ⊗ vr.The statement follows easily since
π · (g · (w1 ⊗ · · · ⊗ wr)) = π · (gw1 ⊗ · · · ⊗ gwr)= gwπ−11 ⊗ · · · ⊗ gwπ−1r
= g · (wπ−11 ⊗ · · · ⊗ wπ−1r)
= g · (π · (w1 ⊗ · · · ⊗ wr).The algebra map
µr :∑σ∈Sr
aσσ 7→(v1 ⊗ · · · ⊗ vr 7→
∑σ∈Sr
aσσ · v1 ⊗ · · · ⊗ vr)
is hence a homomorphism by the above calculation. �
The first goal of this chapter is a proof of the following result.
Theorem 3.2 (First fundamental theorem of invariant theory for GL(V )).Let V be an n-dimensional vector space over K. Then the map µr definedabove is surjective, i.e.
EndGL(V )V⊗r = µr(KSr).
Definition 3.3. Recall the definition of the group algebra
KSr ={∑σ∈Sr
aσσ | aσ ∈ K},
i.e. the set of all formal sums of group elements with coefficients in K, withmultiplication and addition defined componentwise. For a subset H ⊆ Sr,define a(H), s(H) ∈ KSr as follows:
a(H) =∑h∈H
(sgn h)h, s(H) =∑h∈H
h;
respectively, the alternating and symmetric sums of H.
Remark 3.4. Notice that if H is a subgroup,
h · a(H) = h∑k∈H
(sgn k)k =∑k∈H
(sgn h−1hk)hk =∑p∈H
(sgn h−1)(sgn p)p
= (sgn h)a(H)
(making the substitution p = hk), and
h · s(H) = h∑k∈H
k =∑k∈H
hk =∑p∈H
p
34 INVARIANT THEORY
(making the same substitution) for all h ∈ H. Hence
a(H)2 = |H| a(H)
s(H)2 = |H| s(H).
Proposition 3.5 (Polarisation lemma). Suppose S is a spanning set of the K-vector space U , where K is a field of characteristic zero. Suppose in additionthat for each s, t ∈ S, there is an infinite set of λ ∈ K such that s + λt ∈ S.Then the ring of invariants (
U⊗r)Sr
is spanned by elements of the form s⊗ s⊗ · · · ⊗ s for s ∈ S.
Proof . The proof is a generalisation of the calculation
v ⊗ w + w ⊗ v = (v + w)⊗ (v + w)− v ⊗ v − w ⊗ w.
We first show that (U⊗r)Sr = s(Sr)U⊗r. For one inclusion, notice that if
x ∈ s(Sr)U⊗r,
x =∑σ∈Sr
σu, u ∈ U⊗r.
Then if π ∈ Sr,πx =
∑σ∈Sr
(πσ)u =∑ρ∈Sr
ρu = x,
making the substitution ρ = πσ. So x ∈ (U⊗r)Sr . For the other inclusion, letu ∈ (U⊗r)Sr . Then
u =∑σ∈Sr
1
r!u =
∑σ∈Sr
1
r!σu ∈ s(Sr)U
⊗r
as required.
Hence it suffices to show that s(Sr)u1 ⊗ · · · ⊗ ur ∈ X, where X is the spacespanned by all elements of the form {s ⊗ · · · ⊗ s | s ∈ S}. We prove this byinduction on r, for which the base case is trivial. Note that we can decompose
Sr = Sr−1 t (1 r)Sr−1 t · · · t (r − 1 r)Sr−1,
because (i r)Sr−1 = {π ∈ Sr | πr = i}. Hence
s(Sr) =(
1 + (1 r) + (2 r) + . . .+ (r − 1 r))s(Sr−1),
and so
s(Sr)u1 ⊗ u2 ⊗ · · · ⊗ ur =(1 + (1 r)+ . . . + (r − 1 r)
)(s(Sr−1)u1 ⊗ · · · ⊗ ur−1
)⊗ ur,
INVARIANT THEORY 35
and the rightmost factor is a linear combination of {s⊗ · · · ⊗ s⊗ t | s, t ∈ S}by our inductive hypothesis. Hence it suffices to show that(
1 + (1 r) + . . .+ (r − 1 r))s⊗ · · · ⊗ s⊗ t ∈ X.
Now(1 + (1 r) + . . .+ (r − 1 r)
)s⊗ · · · ⊗ s⊗ t
= (s⊗ · · · ⊗ s)⊗ t+ (1 r)(s⊗ · · · ⊗ s)⊗ t+ . . .+ (r − 1 r)(s⊗ · · · ⊗ s)⊗ t= s⊗ · · · ⊗ s⊗ t+ t⊗ s⊗ s⊗ · · · ⊗ s+ . . .+ s⊗ · · · ⊗ s⊗ t⊗ s
Letsi(s, t) =
∑u1 ⊗ u2 ⊗ · · · ⊗ ur,
where the sum is over all distinct rearrangements of the ui such that i of theuk’s are s and r− i of the uk’s are t. Now si(t, s) is a sum of
(ri
)terms, and we
are interested in sr−1(s, t). But si(t, s) ∈ X, for we can find infinitely manyλ ∈ K such that
qλ := (λs+ t)⊗ (λs+ t)⊗ · · · ⊗ (λs+ t) ∈ X,
where by a generalised binomial theorem-type calculation,
qλ =r∑i=0
λisi(s, t) (3.6)
(notice that this equation does not simplify further since we do not have com-mutativity of pure tensors in general by Remark 1.7). Let us choose r + 1distinct λi and arrange the r + 1 equations of the form (3.6) as a matrixequation,
qλ1
qλ2
...qλr+1
=
1 λ1 · · · λr11 λ2 · · · λr2...
.... . .
...1 λr+1 · · · λrr+1
s0(s, t)s1(s, t)
...sr(s, t)
.
For distinct λi, the matrix above is simply a Vandermonde determinant, andinverting will give us the si(s, t) in terms of the qλi and the λi, linear combi-nations of which must be in Σ since it is a K-subspace. Since sr−1(s, t) is justone of these si(s, t), in particular we will have si(s, t) ∈ Σ as required. �
Example 3.7. Suppose U = Matn(K) and S = GLn(K). We claim S satisfiesthe above additional hypothesis. Evidently
(s+ λt) = t(t−1s+ λI) ∈ S
unless λ is an eigenvalue of t−1s and hence is in S for infinitely many λ ∈ K,since all but finitely many elements of the field will not be eigenvalues (providedthe field is infinite).
36 INVARIANT THEORY
3.2. Conclusion of the proof of the first fundamental theorem
Recall we have a vector space V ∼= Kn of dimension n, where K is a field ofcharacteristic zero, and an action of GL(V ) on V ⊗r given by
g 7→ (g ⊗ g ⊗ · · · ⊗ g).
We have seen in Lemma 1.4 that End(V ⊗r) ∼= (End(V ))⊗r, where the isomor-phism is given by
α1 ⊗ · · · ⊗ αr 7→(v1 ⊗ · · · ⊗ vr 7→ α1v1 ⊗ · · · ⊗ αrvr
).
We are interested in EndGL(V )(V⊗r). Recall we have a map
µ : KSr → EndGL(V )(V⊗r),
and that the first fundamental theorem, whose proof we now conclude, statesthat this map is surjective. We first need a number of preliminary lemmas.
Lemma 3.8. If α ∈ End(V ⊗r) and π ∈ Sr, then π−1απ ∈ End(V ⊗r) is givenby
π−1(α1 ⊗ · · · ⊗ αr)π = απ1 ⊗ · · · ⊗ απr.
Proof . We compute(π−1(α1 ⊗ · · · ⊗ αr)π
)(v1 ⊗ · · · ⊗ vr) = π−1α1 ⊗ · · · ⊗ αr(vπ−11 ⊗ · · · ⊗ vπ−1r)
= π−1(α1vπ−11 ⊗ · · · ⊗ αrvπ−1r)
= απ1v1 ⊗ · · · ⊗ απrvr= απ1 ⊗ · · · ⊗ απr(v1 ⊗ · · · ⊗ vr)
as claimed. �
Let A = KSr, a semisimple ring (K-algebra). Then M = V ⊗r is a semisimpleA-module (a discussion of semisimple rings and modules is given in AppendixA). Let us compute EndA(V ⊗r).
Proposition 3.9. B = EndA(V ⊗r) is the linear span of
{g ⊗ · · · ⊗ g | g ∈ GL(V )}.
Proof . Notice that, by definition, α ∈ EndA(V ⊗r) = B if and only if πα = απfor all π ∈ Sr, i.e. if α commutes with the A-action (where A = Sr). This inturn happens if and only if π−1απ = α for all π ∈ Sr, which happens if andonly if α ∈ ((EndV )⊗r)Sr , where Sr acts via
π · (α1 ⊗ · · ·αr) = απ1 ⊗ · · · ⊗ απr, αi ∈ EndV,
by Lemma 3.8. Take U = EndV and S = GL(V ) in the Polarisation Lemma,Lemma 3.5. We must check S spans U and that there exist infinitely manyλ ∈ K such that s+ λt ∈ S.
INVARIANT THEORY 37
Supposing we have chosen a basis for V we have a basis for End(V ) {ξij}ni,j=1
where (ξij)kl = δikδjl, thinking of endomorphisms as matrices. Notice thenthat we can write
ξij = (id + ξij)− id,
i.e. as a linear combination of invertible transformations, whence GL(V ) spansEnd(V ).
Secondly, given two invertible matrices s, t ∈ GL(V ), we want to find infinitelymany field elements λ such that s + λt ∈ S, that is, if s, t are invertible wewant to show there are only finitely many λ such that s+ λt is not invertible.But
det(s+ λt) = det(s(I + λs−1t)
= det s det(I + λs−1t)
= det s det(λ(λ−1I + s−1t))
= λn det s det(s−1t− (−λ−1)I),
which is zero, i.e. s + λt is non-invertible, if and only if λ = 0 or −λ−1 is aneigenvalue of s−1t. But s−1t can only have finitely many eigenvalues, hencethe claim follows.
So by Lemma 3.5, (EndV ⊗r)Sr is spanned by the set
{s⊗ · · · ⊗ s | s ∈ S = GL(V )}as claimed. �
With these last two results and with material from Appendix A, we can nowconclude the proof of the First Fundamental Theorem of Invariant Theory.
Completion of the proof of Theorem 3.2. Clearly K ⊆ Z(KSr). By Maschke’sTheorem, V ⊗r is a semisimple KSr module and since it is a finite-dimensionalK-vector space, it is in particular a finitely-generated Z(KSr)-module. NowB = EndKSrV
⊗r ∼= GL(V ) by Proposition 3.9 and so by the Double Cen-traliser Theorem A.10,
µr : KSr → EndB(V ⊗r)
= EndGL(V )V⊗r
is surjective as required. �
3.3. The second fundamental theorem of invariant theory (type A)
The first fundamental theorem states that µ : KSr → EndGL(V )(V⊗r) is sur-
jective; the second fundamental theorem describes the kernel of µ. The proofof this theorem requires some results from Chapter 4 on the structure of thegroup ring KSr.
38 INVARIANT THEORY
Theorem 3.10 (Second fundamental theorem of invariant theory). Let r bean integer and let V be a K-vector space of dimension n.
(i) If r ≤ n then kerµ = 0.
(ii) Suppose r ≥ n+ 1, and regard Sn+1 as the subgroup S{1,2,...,n+1} of Sr.Recall that
a(Sn+1) =∑
σ∈Sn+1
(sgn σ)σ.
Thenkerµ = 〈a(Sn+1)〉,
where by 〈·〉 we mean the ideal of KSr.
Proof . First assume that n + 1 ≤ r, and let N = kerµ. This is a two-sidedideal of KSr, and hence by Remark 4.20, N is a sum of certain ideals I(λ),where
I(λ) =∑t∈T (λ)
KSre(t)
is defined in Theorem 4.18. For each integer k, define the ideal
Ik =∑λ∈P(r)l(λ)≥k
I(λ),
where l(λ) is the number of parts of λ – i.e. the sum is over partitions λ withat least k parts (as in Definition 4.1(i)). We write ak = a(Sk) ∈ KSr. Theassertion of the theorem is therefore that N = 〈an+1〉. We shall prove this inthree parts by showing that
〈an+1〉 ⊆ N ⊆ In+1 ⊆ 〈an+1〉.Claim 1: 〈an+1〉 ⊆ N . Since N = kerµ, we need only show that µ(an+1) ≡ 0.Since µ(an+1) : v1 ⊗ · · · ⊗ vr 7→ an+1 · v1 ⊗ · · · ⊗ vr, we must show thatan+1V
⊗r = 0. Now given an arbitrary element
v1 ⊗ · · · ⊗ vn+1 ⊗ · · · ⊗ vr,an+1 acts only on the first n+ 1 elements of this tensor product. Now
an+1v1 ⊗ · · · ⊗ vn+1
is Sn+1-alternating. Hence
an+1v1 ⊗ · · · ⊗ vn+1 ∈ Λn+1V.
But dimV = n and so Λn+1V = 0, whence an+1V⊗r = 0 since the element we
chose was an arbitrary basis element.
Claim 2: N ⊆ In+1. As mentioned above, N is a sum of certain I(λ). Weshow that if I(λ) 6⊆ In+1, then I(λ) 6⊆ N . Hence any I(λ) which appears as asummand in N also so appears in In+1, whence N ⊆ In+1.
INVARIANT THEORY 39
To say I(λ) 6⊆ In+1 simply means λ has n or fewer parts. Take such a partitionλ – we show that I(λ) 6⊆ N by finding a tableau t ∈ T (λ) and an elementv ∈ V ⊗r such that e(t)v 6= 0 – hence e(t) 6∈ kerµ = N and so I(λ) 6⊆ N .
Take any tableau t of shape λ and let b1, . . . , bn be a basis of V . For each i, letki be the row of t in which i lies. Now recall that t has at most n rows, and soki ≤ n for all i = 1, . . . , r. Define
vt := bk1 ⊗ bk2 ⊗ · · · ⊗ bkr .We shall show that e(t)vt 6= 0. First we make two simple observations.
First, if π ∈ R(t), the row group of t, then since i, πi are in the same row of tfor all i, we have
kπ−1i = ki,
and it follows that πvt = vt for all π ∈ R(t).
Secondly, if π ∈ C(t), and π 6= 1, then πvt 6= vt, say
πvt = bj1 ⊗ bj2 ⊗ · · · ⊗ bjr 6= vt,
since elements of the column group permute elements of the column, i.e. takesat least one entry of column i to a different row (providing the element is notthe identity).
It follows that the coefficient of vt in
e(t)vt = a(C(t))s(R(t))vt
is what we get from the trivial element of C(t) and all the elements of R(t),i.e. the coefficient is |R(t)| vt 6= 0, and hence I(λ) 6⊆ N .
Claim 3: In+1 ⊆ 〈an+1〉. We do this by showing that for any tableau t ∈ T (λ)where λ has at least n+ 1 rows, we have e(t) ∈ 〈an+1〉.Observe that we can find some π ∈ Sr, such that πt looks like
12...
c
. ..
C3C2
for some c ≥ n + 1 (since t has at least n + 1 rows), where the second, thirdand later columns of πt are C2, C3, . . .. Then
a(C(πt)) = a(Sc)a(C2)a(C3) · · ·and therefore
an+1
(a(C(πt))
)= an+1a(Sc)a(C2)a(C3) · · ·
= (n+ 1)! a(C(πt)),
40 INVARIANT THEORY
since a(H)2 = |H| a(H) (see Definition 4.3) and Sn+1 only affects the firstn+ 1 bits of Sc. Hence a(C(πt)) ∈ 〈an+1〉, which gives
e(πt) = a(C(πt))s(R(πt)) ∈ 〈an+1〉.But e(πt) = πe(t)π−1, and so e(t) ∈ 〈an+1〉.Having proved the second claim, we now take r ≤ n. But the same argumentas above will show that if κ ∈ kerµ, i.e. if κ acts as zero on V ⊗r, thenκ ∈ In+1 = 0. �
Remark 3.11. We have proved the first and second fundamental theorems forGL(V ) or SL(V ) in the first formulation. There is also a second formulation,which seeks to find GL(V )-invariant elements of (⊗rV )⊗ (⊗sV ∗), and a third,which seeks to find GL(V )-invariants in K[W ], where W = (⊕rV )⊕ (⊕sV ∗).We do not tackle these formulations in this course.
Exercises
Exercise 3.12. Check that the action of Sr on T r(V ) is indeed a left groupaction, i.e.
π1(π2(w1 ⊗ · · · ⊗ wr)) = (π1π2)w1 ⊗ · · · ⊗ wrfor π1, π2 ∈ Sr.
Chapter 4
The structure of the group algebra KSr
In this chapter we give an explicit decomposition of the group ring KSr intotwo-sided ideals, when K is a field of characteristic zero. The approach takenhere is slightly more general and less constructive than the normal approach– although we describe the collection of irreducible KSr-modules, we do notattempt to explicitly construct them here, as we do not need these so-calledSpecht modules for our purposes in Chapter 3. For this, we refer the reader toSagan [5]. First we need some elementary partition combinatorics.
Definition 4.1. Let r ≥ 0 be an integer.
(i) A partition λ of r is a sequence λ1 ≥ λ2 ≥ . . . ≥ λp > 0 such that∑pi=1 λi = r, and p is the number of parts of λ, often denoted l(λ).
(ii) A Young diagram is a diagram with p left-justified rows of boxes, suchthat row i contains λi boxes.
(iii) A tableau is a filling of the Young diagram of λ with the numbers{1, . . . , r}.
(iv) T (λ) is the set of all tableaux of shape λ and has size r!. Sr acts onT (λ) and is a copy of the regular representation. For t ∈ T (λ), t = (tij),πt = (πtij).
(v) Given t ∈ T (λ), we have the corresponding row stabiliser R(t) andcolumn stabiliser C(t), both subgroups of Sr. For X ⊆ {1, . . . , r}, wewrite Stab(X) = {π | πx ∈ X for all x ∈ X}. Then
R(t) = Stab(row 1) ∩ · · · ∩ Stab(row p),
C(t) = Stab(column 1) ∩ · · · ∩ Stab(column q)
if t has q columns.
We denote the set of all partitions of r by P(r) and write λ ` r when λ ∈ P(r).
Proposition 4.2. If π ∈ Sr and X ⊆ {1, . . . , r}, then πStab(X)π−1 =Stab(πX). In particular
R(πt) = πR(t)π−1, C(πt) = πC(t)π−1.
Proof . σ ∈ StabπX if and only if σπx = πx for all x ∈ X. This in turn willoccur if and only if π−1σπx = x for all x ∈ X, i.e. if and only if π−1σπ ∈Stab(X), which is equivalent to σ ∈ πStab(X)π−1. �
41
42 INVARIANT THEORY
Definition 4.3. Let H ⊆ Sr and define
s(H) =∑h∈H
h
a(H) =∑h∈H
(sgn h)h
and note that s(H) and a(H) ∈ KSr. If H is a subgroup, then for x ∈ H,
xs(H) = s(H)
xa(H) = sgn xa(H)
s(H)2 = |H| s(H)
a(H)2 = |H| a(H).
For t ∈ T (λ), definee(t) = a(C(t))s(R(t)).
Lemma 4.4. For any π ∈ Sr, e(πt) = πe(t)π−1.
Proof . Since C(πt) = πC(t)π−1,
a(C(πt)) = πa(C(t))π−1,
and similarly for R(πt). Hence
e(π(t)) = a(C(πt))s(R(πt))
= πa(C(t))π−1πs(R(t))π−1
= πe(t)π−1
as claimed. �
Definition 4.5. Let λ, µ ` n. We say λ dominates µ and write λ D µ if theYoung diagram of λ is obtained from the Young diagram of µ by moving boxesupwards.
A more workable definition is the following. We see λD µ if, for all i,i∑
j=1
µj ≤i∑
j=1
λj.
Lemma 4.6 (Dominance lemma). Suppose λ, µ ∈ P(r), and that t, s aretableaux of shape λ, µ respectively. Suppose that for all i, the elements of rowi of s are all in distinct columns of t. Then λD µ.
Proof . Clearly we can sort the entries of each column of t so that the elementsof rows 1, 2, . . . , i of s all occur in the first i rows of t, since no column can
INVARIANT THEORY 43
contain more than one entry from the same row of s. Thusλ1 + . . .+ λi = number of elements in the first i rows of t
≥ number of elements of s in the first i rows of t= µ1 + . . .+ µi,
whence λD µ by definition. �
Lemma 4.7. Let t ∈ T (λ). Then g = pq ∈ R(t)C(t) (so p ∈ R(t), q ∈ C(t))if and only if for all pairs a, b in the same row of t, a and b are not in thesame column of gt.
Proof . Suppose first that g = pq is of the stated form, and let a, b be inthe same row of t ∈ T (λ). Then a, b are in the same row of pt. Hencefor any element q′ ∈ C(pt), a, b are not in the same column of q′pt. ButC(pt) = pC(t)p−1 and so we may take q′ = pqp−1. Hence a, b are not in thesame column of pqp−1pq = gt.Conversely, suppose that for all a, b in the same column of gt, a, b are indistinct rows of t. We must show g has the required form for some p ∈ R(t) andq ∈ C(t). Now by hypothesis, the entries of the first column of gt are in distinctrows of t, and so there exists p1 ∈ R(t) such that p1t has the same entries inits first column as gt (possibly in a different order). Repeating this argument,as many times as there are columns of t, we see that there is an elementp = p1p2 · · · ∈ R(t) such that gt and pt have the same entries in each column,up to ordering of columns. Hence there exists some q′ ∈ C(pt) = pC(t)p−1
such that gt = q′pt. Writing q′ = pqp−1 for some q ∈ C(t), we see thatgt = pqp−1pt = pqt and hence g = pq as claimed. �
These are the combinatorial tools we require for our decomposition of KSr.We now state several results of a group theoretic nature.
Lemma 4.8. If H,K are subgroups of Sr and (a b) ∈ H ∩K then
s(H)a(K) = 0.
Proof . Since (a b) ∈ H ∩K,s(H)a(K) = s(H)(a b)a(K) = −s(H)a(K) = 0
as claimed. �
Corollary 4.9. If a, b are in the same row of t and in the same column of sthen e(t)e(s) = 0.
Proof . We havee(t)e(s) = a(C(t))s(R(t))a(C(s))s(R(s)) = 0
since (a b) ∈ R(t) ∩ C(s). �
Proposition 4.10. Let λ, µ ∈ P(r) and let t ∈ T (λ) and s ∈ T (µ). Then
44 INVARIANT THEORY
(i) s(R(t))a(C(s)) 6= 0 implies µD λ;
(ii) If λ = µ and s(R(t))a(C(s)) 6= 0 then there exists some π ∈ C(s) suchthat t = πs.
Proof . By Corollary 4.9, there does not exist (ab) ∈ R(t) ∩ C(s), i.e. if a, bare in the same row of t, then a, b are in distinct columns of s, and by theDominance Lemma 4.6, µD λ.
Suppose now λ = µ. By the argument in the proof of the Dominance Lemma4.6, there exists π ∈ C(s) such that πs = t. �
Lemma 4.11. Suppose x ∈ KSr is such that qxp = (sgn q)x for all q ∈ C(t)and p ∈ R(t) for some tableau t ∈ T (λ). Then x = γe(t) for some γ ∈ K.
Proof . Writex =
∑g∈Sr
αgg, αg ∈ K,
which we can do since G forms aK-basis forKG. Then for (q, p) ∈ C(t)×R(t),
q−1xp−1 =∑g∈Sr
αgq−1qp−1
=∑g∈Sr
αqgpg
= (sgn q)∑g∈Sr
αgg,
and comparing coefficients of g ∈ Sr we must have
αqgp = (sgn q)αg for all (q, p) ∈ C(t)×R(t). (4.12)
In particular, taking g = 1, αqp = (sgn q)α1.
Sincee(t) =
∑q∈C(t)p∈R(t)
(sgn q)qp,
we are finished if we can show that αg = 0 if g 6= qp for some q, p.
If g is not of the form g = qp for some q ∈ C(t), p ∈ R(t), then g−1 is not ofthe form p−1q−1. By Lemma 4.7, there exist some a, b in the same row of tand in the same column of g−1t, i.e.
(a b) ∈ R(t) ∩ C(g−1t) = R(t) ∩ g−1C(t)g.
But since (a b) ∈ R(t) and 1 ∈ C(g−1t), we have
αg(a b) = αg
INVARIANT THEORY 45
by (4.12). But, also by (4.12), since 1 ∈ R(t) and g(a b)g−1 ∈ g−1C(t)g wehave
α(g(a b)g−1)g = −αg.Hence since αg(a b) = α(g(a b)g−1)g, we must have αg = 0. Hence
x =∑g=qp
(sgn q)qp = α1e(t),
and so the result follows upon setting γ = α1. Notice that we do not necessarilyknow that γ is nonzero. �
Corollary 4.13. For all t ∈ T (λ), e(t)2 = γe(t) for some γ ∈ K.
Proof . Clearly e(t)2 satisfies the condition of Lemma 4.11. �
Lemma 4.14. e(t)2 = γe(t) for some γ 6= 0.
Proof . We have seen that e(t)2 = γe(t) for some γ ∈ K. So it remains toprove that γ 6= 0.
Consider the linear map
τ : KSr → KSr
x 7→ xe(t).
We compute the trace of τ in two different ways. First, note that KSr hasa basis {g | g ∈ Sr}; write Sr = {g1 = 1, g2, . . . , gr!}. If e(t) =
∑αigi, then
since C(t) ∩ R(t) = 1, α1 = 1 (since qp = 1 is possible only when q = 1 andp = 1). Hence the coefficient of gj in gje(t) = τ(gj) is 1, and so trτ = r!.
For the second method of computation, take b1, b2, . . . , bf as a basis of KSre(t)and complete it to a basis {b1, . . . , bf , bf+1, . . . , br!} of KSr. Now for i =1, . . . , f , bie(t) = αie(t)
2 for some αi ∈ KSr, and so
bie(t) = αie(t)2
= αiγe(t)
= γαie(t)
= γbi
for some γ ∈ K, by Corollary 4.13. Moreover, for i > f , τbi ∈ KSre(t) and sothese basis elements do not contribute to the diagonal. Hence trτ = fγ = r!.
Hence γ 6= 0, and in particular γ = r!f6= 0. �
Lemma 4.15. Let λ ∈ P(r) and let t have shape λ. Then
e(t)KSre(t) = Ke(t).
46 INVARIANT THEORY
Proof . By Lemma 4.11, if x ∈ KSr is such that qxp = (sgn q)x for all q ∈ C(t)and p ∈ R(t) then x ∈ Ke(t). Hence to show e(t)KSre(t) ⊆ Ke(t) we mustshow qe(t)xe(t)p = sgn qe(t)xe(t) for all x ∈ Sr. This follows since
qe(t)xe(t)p = q(∑r,s
(sgn r)rs)x(∑t,u
(sgn t)tu)p
=(∑r,s
(sgn r)(qr)s)x(∑t,u
(sgn t)t(up))
= (sgn q)(∑r,s
(sgn r)rs)x(∑t,u
(sgn t)tu)
= (sgn q)e(t)xe(t),
and so the first inclusion holds. For the reverse inclusion, notice that, givenα ∈ KSr, we have
αe(t) =(αγ
)γe(t) =
(αγ
)e(t)2 =
(αγ
)e(t) · 1 · e(t) ∈ e(t)KSre(t)
by Lemma 4.14, whence equality holds. �
Remark 4.16. Before we embark on the proof of the following theorem, weneed to recall some easy results from representation theory. Specifically, weneed the fact that, given two Sr-modules V and W ,
〈χ(V ), χ(W )〉 = dim HomSr(V,W ).
In particular, then, V is an irreducible Sr-module if and only if
dim EndSr(V ) = 1,
i.e. if and only if every Sr-endomorphism of V is multiplication by some scalar.
Recall also that the number of irreducible representations of a finite group isequal to the number of conjugacy classes of the group. A good reference forsuch claims, which all follow readily from the Schur orthogonality relationsand Schur’s lemma, is Curtis and Reiner [2].
We now give a complete classification of irreducible KSr-modules, and de-scribe the decomposition of the group algebra in terms of these irreducibles.We will need the following definition.
Definition 4.17. Let λ ∈ P(r). Define an ideal
I(λ) =∑t∈T (λ)
KSre(t).
Theorem 4.18. Let r ≥ 0 and let λ ` r.
(i) For each tableau t ∈ T (λ), the Sr-module KSre(t) is irreducible.Moreover if s ∈ T (λ), then KSre(s) ∼= KSre(t).
INVARIANT THEORY 47
(ii) If t ∈ T (λ) and s ∈ T (µ) and KSre(t) ∼= KSre(s), then λ = µ.
(iii) If tλ ∈ T (λ) for all λ ∈ P(r) then the modules {KSre(tλ)} form acomplete set of simple KSr-modules.
(iv) For each partition λ ∈ P(r), I(λ) is a minimal two-sided ideal of KSr
and the group algebra can be decomposed
KSr =⊕λ∈P(r)
I(λ).
Proof . For the first claim, let ϕ ∈ EndSrKSre(t). We show that ϕ(t) = α idfor some α ∈ K – this suffices by complete reducibility, and by Remark 4.16.
Let x = ϕ(e(t)). Then ϕ(e(t)2) = ϕ(γe(t)), whence e(t)ϕ(e(t)) = γϕ(e(t)).Hence
ϕ(e(t)) =1
γe(t)ϕ(e(t)) ∈ e(t)KSre(t).
But e(t)KSre(t) = Ke(t) by Lemma 4.11. Hence ϕ(e(t)) = αe(t) for someα ∈ K, and so ϕ = α id since e(t) generates KSre(t).
For the second claim, suppose ϕ : KSre(s)→ KSre(t) is a homomorphism ofSr-modules which is not identically zero. Then
ϕ(e(s)) = xe(t) for some x ∈ KSr.
But then
ϕ(e(s)) = ϕ(e(s)2
γs
)=e(s)xe(t)
γs
for some γs such that e(s)2 = γse(s), which exists and is nonzero by Corollary4.13 and Lemma 4.14 respectively. Therefore ϕ(e(s)) ∈ e(s)KSre(t) and ϕ 6= 0implies e(s)KSre(t) 6= 0.
Hence there exists some π ∈ Sr such that e(s)πe(t) 6= 0, i.e. e(s)πe(t)π−1 6= 0,i.e.
e(s)e(πt) 6= 0.
We need this to imply that for all i, elements of row s are all in differentcolumns of πt, which follows from Corollary 4.9. Hence by Lemma 4.6, λD µ.By the same argument µD λ and so λ = µ.
For the third claim, notice that there are |P(r)| pairwise inequivalent irre-ducible representations by (i) and (ii). There are also |P(r)| conjugacy classesof Sr, which must equal the number of irreducible representations by Remark4.16.
For the final claim, we first want to show that I(λ is a two-sided ideal. In factwe show that for every π ∈ Sr, πI(λ) = I(λ)π, which evidently shows that
48 INVARIANT THEORY
the ideal is two-sided. Noting that we can write
I(λ) =∑t∈T (λ)
KSre(t) =∑π∈Sr
KSre(πt0)
for some fixed tableau t0 of shape λ, observe that for π ∈ Sr,
πI(λ) =∑σ∈Sr
KSrπe(σt0)
=∑σ∈Sr
KSrπe(σt0)π−1π
=∑σ∈Sr
KSr(πσt0)π
=∑ρ∈Sr
KSre(ρt0)π
= I(λ)π
as asserted. We claim that
dim(e(s)KSre(t)) =
{0, if sh s 6= sh t1, if sh s = sh t.
Notice that the case shs = sht is precisely Lemma 4.15. If shs 6= sht, we knowfrom (ii) that
dim HomSr(KSre(s), KSre(t)) = 0. (4.19)Let ϕ : KSre(s) → KSre(t) be an Sr-module homomorphism. Then, givenx ∈ KSr, by a similar argument to the above we can show that
ϕ(e(s)) =e(s)xe(t)
γs,
and so since ϕ ≡ 0 by (4.19), we have e(s)xe(t) = 0 for all x ∈ KSr (since wemay define such homomorphisms for all x ∈ KSr), i.e. dim e(s)KSre(t) = 0,and hence, since
I(λ)I(µ) =( ∑t∈T (λ)
KSre(t))( ∑
s∈T (µ)
KSre(s)),
if λ 6= µ, every term will be of the form KSre(t)KSre(s) = 0, whence we have
I(λ)I(µ) =
{I(λ), λ = µ0, λ 6= µ
Now it is clear that I(λ) is the isotypic component of KSr of type ρλ, where
ρλ = KSre(tλ)
for some fixed tableau tλ of shape λ. Hence KSr = ⊕λI(λ) since any space isthe direct sum of its isotypic components.
If I is a minimal two-sided ideal, then I(λ)I = I for some λ, since KSr =∑I(λ) and hence the identity element can be decomposed. �
INVARIANT THEORY 49
Remark 4.20. Notice that any two-sided ideal of KSr is a sum of certainI(λ)’s.
Exercises
Exercise 4.21. Show that D is a partial order on P(r), i.e. is antisymmetricand transitive. Show also that λ D µ = (1r) and (r) D λ for all λ ` r. Showalso that the given definition for the partial order on P(r) is equivalent to∑i
j=1 µj ≤∑i
j=1 λj for all i = 1, . . . , r.
Exercise 4.22. Show that if s, t ∈ T (λ), then KSre(s)∼−→ KSre(t), and
construct the isomorphism.
Exercise 4.23. Prove that
dim(e(s)KSre(t)) =
{0, if sh s 6= sh t1, if sh s = sh t.
Exercise 4.24. Let t be a tableau of shape λ ` r and let e(t) = s(R(t))a(C(t))as defined in lectures. Let dimV = n and suppose that λ has at most n rows.
(i) Show that e(t) acts non-trivially on V ⊗r for each integer r.(ii) Explain briefly how this fact is used to prove the second fundamental
theorem of invariant theory for GL(V ).
Chapter 5
Affine algebraic geometry
We give a brief overview of the basic facts from the theory of affine varietieswhich we will need in later chapters. Our main goal is a density theoremwhich is integral in our proof in Chapter 6 of the first fundamental theorem ofinvariant theory for the orthogonal group.
5.1. Affine algebraic sets and ideals
LetK be any field, and V = Kn. Recall the construction of S = K[x1, . . . , xn] =S(V ∗), and that in Chapter 1 we saw how S may be regarded as an algebra offunctions on V .
For any subset T ⊆ S, define
V (T ) = {a = (a1, . . . , an) ∈ Kn | f(a) = 0 for all f ∈ T},called the variety of T .
Clearly if T1 ⊆ T2 then V (T1) ⊇ V (T2), and if 〈T 〉 is the ideal generated byT , then each element of T is of the form f =
∑fiti where fi ∈ S, ti ∈ T and
so if a ∈ V (T ) then f(a) = 0 and so V (T ) ⊆ V (〈T 〉). The other inclusion istrivially satisfied, so we have the equality V (T ) = V (〈T 〉).Essentially, we have defined a map
V : subsets of S → subsets of Kn
T 7→ V (T ).
On the other hand, given some subset A of Kn, we can define a subset of S by
I(A) = {f ∈ S | f(a) = 0 for all a ∈ A}.This gives us a map in the other direction, i.e.
I : subsets of Kn → subsets of SA 7→ I(A).
Observe that I(A) is an ideal of S, since if f1, f2 ∈ I(A) then (f1 + f2)(a) = 0for all a ∈ A, i.e. f1 + f2 ∈ I(A), and (ff1)(a) = f(a)f1(a) for all a ∈ A andfor all f ∈ S, i.e. ff1 ∈ I(A).
Hence we have a correspondence between ideals of K[x1, . . . , xn] and subsetsof Kn, and we would like to discuss properties of the two maps V, I.
50
INVARIANT THEORY 51
Lemma 5.1. Let K be a field.
(i) For all A ⊆ Kn, I(A)E S.
(ii) A ⊆ A′ implies I(A) ⊇ I(A′)
(iii) T ⊆ T ′ implies V (T ) ⊇ V (T ′)
(iv) V (T ) = V (〈T 〉)
(v) IV I = I and V IV = V .
Proof . We have seen (i) through (iv) already in our discussion above. For theproof of (v), let A ⊆ Kn. Then V I(A) ⊇ A, for if f ∈ I(A), then f(a) = 0for all a ∈ A, whence a ∈ V I(A) for all a ∈ A. We apply the order-reversingproperty (ii) of I and see that IV I(A) ⊆ I(A). For the other inclusion, notethat for any ideal J of S, IV (J) ⊇ J , for if a ∈ V (J) then f(a) = 0 forall f ∈ J , hence f ∈ IV (J). Take J = I(A) in this inequality – this givesIV I(A) ⊇ I(A), the desired inclusion.
On the other hand, suppose that T ⊆ S – then is a ∈ V (T ), f(a) = 0 forall f ∈ T and so f ∈ I(V (T )) for all f ∈ T – applying the order-reversingproperty (iii) of V gives V (T ) ⊇ V (I(V (T )). From the proof of the first part,we know V (I(A)) ⊇ A for every subset A ⊆ Kn – taking A = V (T ) we getthe other inclusion and equality follows. �
Definition 5.2. Sets of the form V (I) for some I ⊆ S are called (affine) alge-braic subsets of Kn and the ideals I(A) for some A ⊆ Kn are called algebraicideals. By Lemma 5.1, we have a bijection between algebraic ideals of S andalgebraic subsets of Kn.
Remark 5.3. Note that not all ideals are algebraic. For example if I =〈x2〉EK[x], then
V (I) = {0}, IV I = 〈x〉 ) 〈x2〉.
Lemma 5.4. Let K be a field and let S = S(V ∗) = K[x1, . . . , xn].
(i) Let {Jλ}λ∈Λ be any collection of ideals of S. Then
V(∑
λ
Jλ
)=⋂λ
V (Jλ).
(ii) If J, J ′ E S then
V (JJ ′) = V (J ∩ J ′) = V (J) ∪ V (J ′).
(iii) If {Aα}α∈A is a collection of subsets of Kn, then
I(⋃α∈A
Aα
)=⋂α∈A
I(Aα).
52 INVARIANT THEORY
Proof . If a ∈ ∩λV (Jλ) then a ∈ V (Jλ) for all λ and hence a ∈ V(∑
λ Jλ
).
Conversely, if a ∈ V(∑
λ Jλ
)then f(a) = 0 for all f ∈ Jλ for each λ. Hence
a ∈ ∩λV (Jλ).
For the second claim, note that a ∈ V (JJ ′) if and only if for all f ∈ Jand f ′ ∈ J ′, f(a)f ′(a) = 0, because the elements ff ′ generate JJ ′. So fora ∈ V (JJ ′), if a 6∈ V (J), there is some f such that f(a) 6= 0, and so a ∈ V (J ′),i.e. a ∈ V (J) ∪ V (J ′). Conversely, note that JJ ′ ⊆ J because J, J ′ are ideals,from which we deduce V (JJ ′) ⊇ V (J) and by symmetry V (JJ ′) ⊇ V (J ′),whence V (JJ ′) ⊇ V (J) ∪ V (J ′).
Finally, notice that f ∈ I(∪αAα
)implies f(a) = 0 for all a ∈ Aα for all α,
and so f ∈ I(Aα) for all α, so f ∈ ∩αI(Aα). Conversely, if f ∈ ∩αI(Aα), thenf(Aα) = 0 for all α, whence f ∈ I
(∪αAα
). �
Remark 5.5. These proofs seem trivial, but they mask a hidden complexity.
(i) We have seen that JJ ′ ⊆ J ∩ J ′. However we may not have equality ingeneral, for example in K[x],
〈x〉〈x〉 = 〈x2〉 6= 〈x〉 ∩ 〈x〉.
(ii) It is generally false that I(A1 ∩ A2) = I(A1) + I(A2). Notice that if iff1 ∈ I(A1) and f2 ∈ I(A2), then (f1 + f2)(a) = 0 for all a ∈ A1 ∩ A2,so I(A1) + I(A2) ⊆ I(A1 ∩A2) is true in general. The reverse inclusionmay be strict.
Assume the result of Exercise 5.68, i.e. that (y − x2) and (y) arealgebraic ideals of K[x, y]. The relevant picture is
A2
A1
Take A1 = V (y − x2) and A2 = V (y). Then A1 ∩ A2 = (0, 0), and soI(A1 ∩A2) = 〈x, y〉, i.e. the ideal of all polynomials with zero constantterm. But I(A1) + I(A2) = (y − x2) + (y) = (x2, y) 6= I(A1 ∩ A2).
INVARIANT THEORY 53
The statement is however true if A1, A2 “intersect transversally” in anappropriate sense.
5.2. The Zariski topology and decomposition
The results of Lemma 5.4 are suggestive of the definition of a topology. In thissection we formalise this idea using the notion of the Zariski topology on anaffine variety.
Lemma 5.6. Let K and S be as above.
(i) Every affine algebraic set is the set of common zeroes of a finite set ofpolynomials in S.
(ii) The intersection of an arbitrary collection of affine algebraic subsets ofV is algebraic.
(iii) The union of two affine algebraic subsets of V is algebraic.
Proof . If A is an affine algebraic set then A = V (I) for some ideal I of S =K[x1, . . . , xn]. But K[x1, . . . , xn] is Noetherian, and so every ideal is finitely-generated. In particular, the ideal I is finitely generated by say f1, . . . , fl,and
A = V (I) = V (〈f1, . . . , fl〉) = V ({f1, . . . , fl}) = V (f1) ∩ V (f1) ∩ · · · ∩ V (fl).
The second claim above follows from Lemma 5.4(i), and the third from Lemma5.4(ii). �
In the proof of the following corollary, we will need to invoke Hilbert’s basistheorem, which states that a polynomial ring over a Noetherian ring is itselfNoetherian. A proof of this result can be found in Atiyah-Macdonald [1]. Alsorecall the descending chain condition, which stipulates that any chain of acertain class of sets stabilises, i.e. given a chain C1 ⊃ C2 ⊃ · · · , there existssome N such that CN = CN+k for all k ≥ 1.
Corollary 5.7. We say that a subset A ⊆ V is closed if A is algebraic, i.e. ifA = V (I) for some ideal I.
(i) This definition defines a topology on V = Kn, called the Zariski topol-ogy.
(ii) The closed subsets of V satisfy the descending chain condition.
Proof . The first claim follows Lemma 5.6, since this is precisely the def-inition of a topology. For the second claim, consider an ascending chainI(C1) ⊆ I(C2) ⊆ · · · which will stabilise since S is Noetherian by Hilbert’sbasis theorem. If we apply V to this chain, we will get the descending chainC1 ⊇ C2 ⊇ · · · which must stabilise. �
54 INVARIANT THEORY
Definition 5.8. We say that a subset A ⊆ Kn is irreducible if A is not theunion of two proper closed subsets, i.e. if A = A1 ∪ A2 with Ai closed, thenA = A1 or A = A2. Given a subset A, if we have written A = A1 ∪ A2 ∪ · · · ∪Ar for some r, with Ai irreducible for each i = 1, . . . , r, then we call this adecomposition of A – if there are no repetitions, we say the decomposition isirredundant. The Ai are called the irreducible components of A.
Proposition 5.9. Any algebraic subset A of V has a decomposition A =A1 ∪A2 ∪ · · · ∪Ar for some r ∈ N, where the Ai are closed and irreducible. Ifthis decomposition is irredundant, it is unique up to the order of the Ai’s.
Proof . We first prove existence. Suppose that A is a closed subset which hasno such decomposition, i.e. as a finite union of irreducibles. Then in particular,A = C1∪C ′1 with C1, C
′1 closed and not equal to A. Without loss of generality
we may assume that C1 has no irreducible decomposition. Repeating thisargument with C1 we obtain A ⊇ C1 ⊇ C2 · · · , an infinite descending chain;impossible by Corollary 5.7.
For uniqueness, suppose A = A1 ∪A2 ∪ · · · ∪Al = A′1 ∪A′2 ∪ · · · ∪A′m are twosuch irredundant decompositions. Then
A1 = (A1 ∩ A1) ∪ (A1 ∩ A2) ∪ · · · ∪ (A1 ∩ A′m)
and since A1 is irreducible, A1 ∩ A′j = A1 for some j. So A1 ⊆ A′j ⊆ Ai bysymmetry. So A1 = A′j (when i = 1). So each Ai is an A′j and each A′j is anAi. �
Lemma 5.10. Suppose A is an algebraic subset of Kn. Then A is irreducibleif and only if I(A) is a prime ideal.
Proof . If A is irreducible, suppose f1, f2 ∈ K[x1, . . . , xn] are such that f1f2 ∈I(A). Now since f1f2 ∈ I(A), for each a ∈ A we have f1(a)f2(a) = 0, i.e.f1(a) = 0 or f2(a) = 0 (since K is a field, K[x1, . . . , xn] is an integral domain).Hence A = (A ∩ V (f1)) ∪ (A ∩ V (f2)), and by irreducibility, A ∩ V (f1) = A(without loss of generality – by irreducibility one of them must equal A). Hencef1 ∈ I(A) whence I(A) is prime.
Conversely, if A is not irreducible, then A = A1∪A2 with Ai closed and Ai 6= A.Hence I(A) $ I(Ai) for i = 1, 2. So there exists some f1 ∈ I(A1) \ I(A) andsome f2 ∈ I(A2) \ I(A). But f1f2 ∈ I(A1)I(A2) and by Lemma 5.4,
V (I(A1)I(A2)) = V (I(A1)) ∪ V (I(A2)) = A1 ∪ A2 = A,
so f1f2 ∈ I(A), whence I(A) is not prime. �
Remark 5.11. We have just shown that in the bijection between algebraicsubsets of Kn and algebraic ideals of K[x1, . . . , xn], irreducible subsets corre-spond to prime ideals. In fact, every prime ideal is algebraic, but we needsome more machinery to be able to see this.
INVARIANT THEORY 55
5.3. Integral dependence
We give a brief discussion of integral dependence, only mentioning those resultswhich we will need in our exposition. For a more detailed discussion, see [1].
Definition 5.12. Let R ⊆ S be two integral domains. If x ∈ S, the smallestsubring of S which contains x and R will be denoted R[x]. Elements maybe written (not uniquely) as r1 + r2x + . . . + rn+1x
n with ri ∈ R. Note thatit is convenient to think of this space as consisting of polynomials in x withcoefficients in R, however there may be non-trivial relations between elementsof R which complicate matters.
Proposition 5.13. Let R ⊆ S be as above and take s ∈ S. Then the followingare equivalent:
(i) R[x] is finitely-generated as an R-module, i.e.
R[x] = Rx1 +Rx2 + . . .+Rxl
for some x1, . . . , xl.(ii) R[x] ⊆ T ⊆ S, where T is a subring of S, finitely-generated as an
R-module(iii) x satisfies an equation
xn + r1xn−1 + r2x
n−2 + . . .+ rn = 0
with ri ∈ R for all i.
Proof . See Atiyah-Macdonald [1], pp59-60. �
Definition 5.14. If x satisfies any (and hence all) of the equivalent conditionsin Proposition 5.13, we say x is integral over R.
Example 5.15. The idea of integrality is a familiar one.
(i) Take R = Z, S = C. Then x =√
2 is integral, since it satisfiesx2 − 2 = 0. However
√3
2is not integral since it satisfies the non-monic
equation 4x2 − 3 = 0. Notice that
R[√3
2
]3 3
4
and hence contains 14and hence 1
4nfor all n, and so Z
[√3
2
]is dense in
Q ⊆ C.(ii) An element x ∈ Q is integral over Z if and only if x ∈ Z. A similar
result holds for any Euclidean domain – this is known as Gauss’ lemma.
Theorem 5.16. Let R ⊆ S be integral domains and let
T = {x ∈ S | x is integral over R}.Then
56 INVARIANT THEORY
(i) T is a subring of S
(ii) If x ∈ S is integral over T then x ∈ T .
Proof . Follows readily from Proposition 5.13. �
Definition 5.17. T is called the integral closure of R in S. We say that anintegral domain R is integrally closed if R is integrally closed in its field ofquotients, i.e. if R is equal to its integral closure in its field of quotients.
The main application to geometry which we will use is the following.
Theorem 5.18. Let R ⊆ S be integral domains and suppose that S is integralover R. Then R is a field if and only if S is a field.
Proof . See Atiyah-Macdonald [1], Proposition 5.7 on p61. �
5.4. Hilbert’s Nullstellensatz
Hilbert’s Nullstellensatz is the fundamental result which underlies algebraicgeometry. It can be seen as a generalisation of the fundamental theorem ofalgebra. We can think of the following form of the Nullstellensatz as statingthat every maximal ideal of K[x1, . . . , xn] has a zero.
Theorem 5.19 (Hilbert’s Nullstellensatz, weak form). Let K be algebraicallyclosed and let m be a maximal ideal of K[x1, . . . , xn]. Then V (m) 6= ∅, i.e.m = 〈(x1 − a1), . . . , (xn − an)〉 for some (a1, . . . , an) ∈ Kn.
Proof . Atiyah-Macdonald [1]. �
Corollary 5.20. Let K be an algebraically closed field.
(i) There is a bijection between the maximal ideals of S and the points ofKn (often Kn is thought of as Specm(S), the maximal spectrum of S).
(ii) If I $ S is a proper ideal of S then V (I) 6= ∅.
(iii) Given a finite set t1, . . . , tr of polynomials in S, either(a) there exist f1, . . . , fr ∈ S such that
∑fiti = 1, or
(b) there exists a ∈ Kn such that ti(a) = 0 for all i = 1, . . . , r.
Proof . The first claim follows by considering the bijectionϕ : Specm(S)→ Kn : m = 〈(x1 − a1), . . . , (xn − an)〉 7→ (a1, . . . , an).
The second claim follows since given any proper ideal we can find a maximalideal containing the ideal, and by the order-reversing property of V . Thefinal claim too is straightforward – in (a) the ti generate the entire ring S andhence 1 ∈ S can be so expressed. If not, then by (ii) they must have a commonzero. �
INVARIANT THEORY 57
Recall the definition of the radical of an ideal, i.e. given some ideal J of S,
rad(J) = {f ∈ S | f r ∈ J for some r}.
Theorem 5.21 (Hilbert’s Nullstellensatz, strong form). Let K be algebraicallyclosed and let J be any ideal of S = K[x1, . . . , xn]. Then
I(V (J)) = rad(J).
Example 5.22. Recall Example 5.3, where we saw the non-algebraic ideal〈x2〉, which has radical 〈x〉.
Corollary 5.23. The algebraic ideals of S are precisely those ideals I suchthat I = rad(I).
Proof . Recall that an algebraic ideal is one which is I(A) for some A ⊆ Kn.The result now follows from Lemma 5.1(v) and Theorem 5.21. �
Corollary 5.24. Every prime ideal is algebraic.
Proof . Recall the alternative characterisation of the radical of an ideal I asthe intersection of all the prime ideals which contain I. Using this formulationit is easy to see that any prime ideal p satisfies rad(p) = p and so p = I(V (p))is algebraic by Theorem 5.21. �
We call a K-algebra reduced if it has no non-zero nilpotent elements.
Corollary 5.25. For any affine algebraic set A, K[A] = S/I(A), the ring ofpolynomial functions on A, is a finitely-generated reduced K-algebra.
Proof . K[A] = K[x1, . . . , xn]/I(A) is generated by the images of x1, . . . , xn asa K-algebra. If f is a nilpotent element of A then f r ∈ I(A) for some r, i.e.f ∈ rad(I(A)) = I(A), whence f = 0 in the quotient K[A]. �
5.5. Affine algebraic varieties
We have seen that for any ideal I EK[x1, . . . , xn], V (I) is a topological spaceunder the Zariski topology. It can be shown that the Zariski topology satisfiesthe T1 separation axiom (i.e. all singleton sets are closed), but that it is notT2, or Hausdorff. In fact, it could not be more different from being Hausdorff.
Lemma 5.26. Let V be an algebraic subset of Kn. Then the following areequivalent:
(i) V is irreducible;(ii) Each non-empty open subset of V is dense in V .(iii) Each non-empty open subset is connected.(iv) Any two non-empty open subspaces of V intersect nontrivially.
58 INVARIANT THEORY
Proof . (i)⇒(ii) Let U be an open subset of V . Then C := V \U is closed andV = C ∪U . However V is irreducible and so if V is expressed as such a union,we must have either V = C (so U = ∅), or U = V , i.e. U is dense in V .
(ii)⇒(iii) Let U be a non-empty open subset of V . Suppose U1 and U2 arenon-empty open subsets of U (and therefore of V ) such that U1 ∩ U2 = ∅ andU = U1 ∪ U2, i.e. that U is disconnected. Then U1 ⊆ V \ U2 which is closed,so U1 ⊆ V \ U2, and so U1 ∩ U2 6= ∅, since similarly U2 ⊆ V \ U1.
(iii)⇒(iv) Let U1, U2 be non-empty open subspaces of V . If U1 ∩ U2 = ∅, thenU = U1 ∪ U2 is not connected which contradicts (iii).
(iv)⇒(i) Suppose that V = V1 ∪ V2 where Vi is closed, non-empty and doesnot equal V , for i = 1, 2. Then
(V \ V1) ∩ (V \ V2) = ∅,which contradicts (iv), whence V is irreducible. �
Lemma 5.27. If W ⊆ V is also algebraic, then W is irreducible if and onlyif its closure W is irreducible.
Proof . Suppose W is irreducible and W = W1 ∪W2 where Wi is closed in W .Then since W = (W ∩W1) ∪ (W ∩W2) and W is irreducible, we must haveWi ⊇ W for i = 1 or 2. Taking i = 1 without loss of generality, so W1 ⊇ W ,then W1 ⊇ W since W1 is closed, and so W is irreducible.
Conversely, if W is irreducible and W = C1 ∪ C2, where the Ci are closed inW , then Ci = Wi ∩W , where Wi is closed in V (by definition of the subspacetopology). So W1 ∪W2 ⊇ W and W = (W1 ∩W ) ∪ (W2 ∩W ), from which wededuce from the irreducibility of W that W1 ⊇ W ⊇ W or W2 ⊇ W ⊇ W ,from which it follows that W is irreducible. �
We call a topological space X Noetherian if its open sets satisfy the ascendingchain condition. Recall the following easy lemma from topology.
Lemma 5.28. A topological space is Noetherian if and only if every opensubspace is quasi-compact, i.e. every cover by open sets has a finite subcover.
Remark 5.29. The traditional definition of compactness holds for Hausdorffspaces only, i.e. a space is compact (by definition) if it is Hausdorff and quasi-compact.
Recall that for any algebraic subset V of Kn we define the coordinate ringK[V ] = K[x1, . . . , xn]/I(V ) – also called the ring of regular or polynomialfunctions on V . Note that this is the algebra of restrictions of polynomials toV .
We have seen thatK[V ] is a finitely-generatedK-algebra, is an integral domainif and only if V is irreducible by Corollary 5.24, is reduced (i.e. has no nilpotent
INVARIANT THEORY 59
elements) if K is algebraically closed, and that the points of V are in bijectionwith the maximal ideals of K[V ].
We now give the formal definition of an affine variety over a field K. Through-out this course we generally only consider the case where K is algebraicallyclosed.
Definition 5.30. A set Z together with a K-algebra K[Z] of K-valued func-tions is an affine K-variety if there exists a Zariski-closed subset V of Kn forsome n such that there exists a bijection ϕ : Z → V such that ϕ∗ : K[V ] →F(Z,K) (functions from Z to K) defined by
ϕ∗(f) = f ◦ ϕ
maps K[V ] isomorphically to K[Z].
Remark 5.31. In other words, an affine algebraic variety is a set of pointstogether with a K-algebra of functions on that set which is isomorphic (in thespecific sense above) to an affine algebraic set of Kn.
Definition 5.32. A morphism f : (V,K[V ])→ (W,K[W ]) is a map f : V →W such that f ∗ as defined above maps K[W ] to K[V ].
Remark 5.33. Note that f a morphism implies that f ∗ is an algebra homo-morphism, since, given g, h ∈ K[W ],
f ∗(gh)(x) = (gh) ◦ f(x)
= (gh)(f(x))
= g(f(x))h(f(x))
= (g ◦ f(x))(h ◦ f(x))
= f ∗(g)f ∗(h)(x),
and for α ∈ K
f ∗(αg)(x) = (αg) ◦ f(x) = αg(f(x)) = αf ∗g(x).
The following result states that the converse is true, i.e. that given an algebrahomomorphism, we can construct a corresponding morphism of varieties in anatural way.
Theorem 5.34. Let V,W be affine K-varieties. There is a natural bijectionfrom the set of morphisms f : V → W to the set of K-algebra homomorphismsK[W ]→ K[V ]. Any such f is the restriction to V of a polynomial map.
Proof . Suppose without loss of generality that V ⊆ Kn and W ⊆ Km arealgebraic subsets. To see the last assertion, suppose that f : V → W is amorphism. Since K[W ] is generated by the restrictions toW of the coordinatefunctions y1, . . . , ym on Km, we have that f ∗(yi) ∈ K[V ], the set of all restric-tions to V of polynomial functions on Kn. Hence f ∗(yi) is the restriction to
60 INVARIANT THEORY
V of such a polynomial fi(x1, . . . , xn) ∈ K[x1, . . . , xn]. So f is the restrictionto V of the map
(a1, . . . , an) 7→(f1(a1, . . . , an), . . . , fm(a1, . . . , am)
).
Given a morphism f : V → W , we have seen in Remark 5.33 that f ∗ :K[W ] → K[V ] is a K-algebra homomorphism. For the converse, given analgebra homomorphism α : K[W ] → K[V ] we want to construct f : V → Wsuch that f ∗ = α.Let α(yi) = fi ∈ K[V ]. Define f : V → Km by
f(v) = (f1(v), f2(v), . . . , fm(v)).
Clearly this satisfies f ∗ = α, so it remains only to prove that f(V ) ⊆ W . Tosee this, take some polynomial g ∈ I(W ). Then
g(f1(v), f2(v), . . . , fm(v)) = g(αy1(v), . . . , αym(v))
= (αg)(y1, . . . , ym)(v)
since α is a K-algebra homomorphism. Butα : K[y1, . . . , ym]/I(W )→ K[x1, . . . , xn]/I(V ),
so we deduce that αg ∈ I(V ), so αg(v) = 0. �
Remark 5.35. This theorem states that there is a contravariant equivalencebetween the category of affine algebraic K-varieties and their morphisms, andthe category of finitely-generated reduced K-algebras and their algebra homo-morphisms. The equivalence is contravariant because
(f1 ◦ f2)∗ = f ∗2 ◦ f ∗1 ,i.e.
V1f−→ V2
K[V1]f∗←− K[V2]
The map f ∗ is called a comorphism.A consequence of this is that a map f : X → Y is an isomorphism if and onlyif the map f ∗ : K[X] → K[Y ] is an isomorphism. It can be shown that themorphisms defined above are continuous maps in the Zariski topology.
5.6. Principal open subsets and rational functions
Let (V,K[V ]) be an affine algebraic variety, and assume V ⊆ Kn. ThenK[V ] = K[x1, . . . , xn]/I(V ).
Take f ∈ K[V ] and defineVf : {x ∈ V | f(x) 6= 0}.
This is obviously an open set, and is called a principal open subset of V .
INVARIANT THEORY 61
Lemma 5.36. (i) (Vf , K[Vf ]) is an affine variety, where K[Vf ] = K[V ]f ,the localisation of K[V ] at f , i.e.
K[V ]f = {g/fn | f ∈ K[V ], for some n}where f is nonzero.
(ii) The principal open subset form a basis for the Zariski topology on V .
Proof . First, define If EK[x1, . . . , xn+1] by
If = 〈I(V ), 1− f(x1, . . . , xn)xn+1〉.Then a = (a1, . . . , an+1) ∈ V (If ) if and only if both a1, . . . , an ∈ V , andan+1f(a1, . . . , an) = 1. Thus we have a bijection
V (If )→ Vf
(a1, . . . , an+1)ψ−→ (a1, . . . , an)(
a1, . . . , an,1
f(a1, . . . , an)
)ϕ←− 1.
Now
ϕ∗(xi) =
{xi, if 1 ≤ i ≤ n1f, i = n+ 1,
and so ϕ∗ : K[V (If )]→ K[V ]f , and similarly ψ∗ = (ϕ∗)−1.
For the second claim, let U be open in V . Then C = V \U is closed and henceis the set
C = {v ∈ V | f(v) = 0 for all f ∈ I(C)},where I(C) is an ideal of K[V ]. Then
U = {v ∈ V | f(v) 6= 0 for some f ∈ I(C)}
=⋃
f∈I(C)
Vf .
In fact the union is finite. �
Definition 5.37. Let V be an irreducible affine variety. The function fieldof V is the field of quotients K(V ) of the coordinate ring K[V ] (which is anintegral domain).
If g = f1/f2 ∈ K(V ), then g defines a function g : Vf2 → K, called a rationalfunction on V . It is important to remember that rational functions are notdefined everywhere, but only defined on open sets. However recall that allopen sets are dense.
Example 5.38. Sometimes such a function can be extended to the whole ofV , but the extension may not be a regular function.
(i) Let V = K2, f = x/y, defined on {(a, b) | b 6= 0}.
62 INVARIANT THEORY
(ii) Let V = V (y − x2) ⊆ K2, so K[V ] = K[x, y]/(y − x2)∼−→ K[T ], where
the map isϕ∗ : x 7→ T, y 7→ T 2.
(T, T 2)
T
It can be shown that this is an isomorphism.(iii) Let V = V (y2 − x3), an irreducible variety. Define a map
A1 = K → V
t 7→ (t2, t3).
This is a bijective morphism.
However the map is not an isomorphism. We have a corresponding map
K[V ]f∗−→ K[A1]
= K[x, y]/(y2 − x3)→ K[t],
and
f ∗(x) = x ◦ f : t 7→ t2
f ∗(y) = y ◦ f : t 7→ t3.
So f ∗(x) = t2, f ∗(y) = t3, so
im(f ∗) = K[t2, t3] $ K[t],
whence f ∗ is not an isomorphism. Written another way, the “inverse”of f should be (x, y) 7→ y/x = t3/t2 = t, but this is not defined atx = 0. This reflects the fact that in this case, V has a singular point at(0, 0).
INVARIANT THEORY 63
(iv) V = V (x1x4−x2x3) ⊆ K4. This is a cone over the twisted cubic curve.Write · for residue mod F = x1x4−x2x3, soK[V ] = K[x1, . . . , x4]/(F ).Consider the function
f =x1
x2
=x3
x4
.
Then f is defined on the set
{(x1, x2, x3, x4) | x2 6= 0 or x4 6= 0}.
Remark 5.39. A regular function means a function inK[V ], i.e. a polynomialfunction or morphism. A rational function is a function in K(V ).
Definition 5.40. Let V be an irreducible affine variety.
(i) For any point P ∈ V , the local ring OP (V ) is
OP (V ) = {f ∈ K(V ) | f is defined at P}.
(ii) For f ∈ K(V ), the pole set, or set of singular points of f is
{P ∈ V | f(P ) is not defined}.
We say that f is defined at P if there exists an expression f = f1/f2, f1, f2
regular functions, i.e. f1, f2 ∈ K[V ], and f2(P ) 6= 0; then f(P ) = f1(P )/f2(P ).
Example 5.41. Take V = Kn2 , which we can think of as the set Mn(K) ofn× n matrices over K. Then
Vdet = {A | detA 6= 0} = GLn(K).
This is an irreducible variety, since Kn2 is irreducible, Vdet is open in an irre-ducible variety, hence is dense in Kn2 and so its closure is irreducible, whenceit is itself irreducible.
Proposition 5.42. Let (V,K[V ]) be an irreducible affine variety over an al-gebraically closed field K. Then
(i) OP (V ) is a local ring, with unique maximal ideal {f ∈ K(V ) | f(P ) =0},
(ii) the pole set of any f ∈ K(V ) is a closed subvariety of V ,
(iii) any element f ∈ K(V ) which is defined everywhere is in K[V ],
(iv) if f ∈ K(V ) is such that there exists g ∈ K[V ] with f(x) = g(x) forall x ∈ U where U is open and dense in V , then f = g in K(V ); inparticular f ∈ K[V ] and f(v) = g(v) for all v ∈ V .
64 INVARIANT THEORY
Proof . First, let M ⊆ OP (V ) be the set of functions vanishing at P . Then
M = ker(f 7→ f(P ))
and so M is a maximal ideal. If g/h ∈ OP (V ) \ M , then h(P ) 6= 0 andg(P ) 6= 0, whence (g/h)−1 = h/g ∈ OP (V ). Hence M consists of the non-invertible elements of OP (V ), so is the unique maximal ideal and OP (V ) islocal.
Now let P (f) be the pole set of f ∈ K(V ). Let
D(f) = {b ∈ K[V ] | fb ∈ K[V ]}
(the denominator ideal of f). Now f is not defined at P if and only if b(P ) = 0for all b ∈ D(f), i.e. if and only if P ∈ V (D(f)).
Consider the ideal D(f). Notice that by the above paragraph, f defined ev-erywhere implies that for all P ∈ V , b(P ) 6= 0 for some b ∈ D(f). By theNullstellensatz in its form in Proposition 5.20(iii), either D(f) 3 1, in whichcase D(f) = K[V ] and so f is regular, or there exists some P ∈ Kn such thatti(P ) = 0 for each ti which generates D(f), which contradicts the fact that fis everywhere defined, by our remarks above.
Without loss of generality, we may take U = Vh = {v ∈ V | h(v) 6= 0} for someh ∈ K[V ], since every open set is a union of such sets. Since f is defined onVh, f ∈ K[Vh] = K[V ]h, i.e. f(v) = g(v)/hi(v) for some i, on Vh. Therefore
f(v)hi(v) = g(v) for all v ∈ U,
i.e. fhi − g = 0 in K[V ] (density). Hence f = g/hi ∈ K[V ]. �
5.7. Group varieties
5.7.1. Product varieties. By way of a first example, note we have al-ready met a product variety in the form of
Am × An = Am+n.
In general, let (X,K[X]) and (Y,K[Y ]) be affine varieties, X ⊆ Km, Y ⊆ Kn.Then X × Y is an affine variety of Km+n, where
I(X × Y ) = (I(X)× I(Y )).
For K-algebras R, S, define the K-algebra R⊗ S as follows:
As a vector space, R⊗ S = R⊗K S, with componentwise multiplication, i.e.
r ⊗ s · r′ ⊗ s′ = rr′ ⊗ ss′.
Proposition 5.43. In general, K[X × Y ] = K[X]⊗K[Y ].
INVARIANT THEORY 65
Proof . Define φ : K[X]⊗K[Y ]→ K[X × Y ] by
φ(f ⊗ g)(x, y) = f(x)g(y).
We show this is an isomorphism. φ is surjective since its image contains thecoordinate functions on X × Y (e.g. xi = φ(xi ⊗ 1)). To show φ is injective,take {fi}, {gj} linearly independent subsets of K[X] and K[Y ] respectively.Then we must show that
φ(∑
cijfi ⊗ gj)
= 0 implies cij = 0 for all i, j.
But for all x ∈ X, y ∈ Y , ∑i,j
cijfi(x)⊗ gj(y) = 0,
and therefore∑
i cijfi(x) = 0 for all j, for all x ∈ X, whence linear indepen-dence of fi gives cij = 0 for all i, j. �
Thus, (X×Y,K[X]⊗K[Y ]) is an affine variety – it is called the product varietyof X and Y .
Remark 5.44. Notice that, in general, the topology on X × Y is not theproduct topology of X and Y – it is much finer, since the product in the planeof finite sets gives only lines; but in the plane, curves are closed in the Zariskitopology. It can also be shown that the product of two irreducible varieties isan irreducible variety.
Definition 5.45. G is an (affine) algebraic group, or a group variety, if G is agroup with the structure of an affine variety (G,K[G]) such that multiplicationµ : G×G→ G : (x, y) 7→ xy and inversion ι : G→ G : x 7→ x−1 are morphismsof varieties.
Example 5.46 (Examples of affine algebraic groups). The following four ex-amples are important.
(i) G = GLn(K).(ii) G = SLn(K).(iii) G = On(K) = {A ∈ GLn(K) | AtA = I}.(iv) G = SOn(K) = SLn(K) ∩On(K).
Theorem 5.47. Let G be an algebraic group. Then the irreducible componentsof G are cosets of a normal subgroup G0 of G of finite index in G.
Proof . We know G is an affine algebraic variety and so it has an irreducibledecomposition. Let
G = V1 ∪ V2 ∪ · · · ∪ Vrbe the irreducible decomposition of G. We show first that the irreduciblecomponents Vj are disjoint.
66 INVARIANT THEORY
Suppose the identity element 1 ∈ G is in V1. We show 1 6∈ Vi for all i = 2, . . . , r.If 1 is in two distinct Vi and x ∈ V1. Multiplication by x permutes the Vi. Hencex is in at least 2 of the Vi. Since this is true for any x ∈ V1, this contradictsirredundancy of the decomposition. Hence 1 ∈ V1 \ (V2 ∪ · · · ∪ Vr). We cannow translate by any element of G and repeat the argument to see that eachelement of G is in a unique Vi, i.e. the Vi are disjoint.The transformation x 7→ x−1 also permutes the Vi and fixes V1. It follows thatV1 is a subgroup of G. Finally, the map z 7→ yzy−1 for y ∈ G is a continuousmap which also fixes V1, whence V1 = G0 is a normal subgroup, and Vi = xiG
0
for some xi ∈ G. �
Corollary 5.48. G is irreducible if and only if G is connected as a topologicalspace. Any closed subgroup of G of finite index is open.
Example 5.49. Let us return to our examples above.
(i) We have seen then GLn(K) is irreducible.(ii) Further, SLn(K) is irreducible.(iii) SOn(K) is a normal subgroup of On(K) of index 2. Hence On(K) is
not irreducible.
Proposition 5.50. SOn(K) = {g ∈ On | det g = 1} is irreducible.
Proof . We proceed by induction on n, proving that SOn(C) is an irreducibleaffine variety. Notice that SOn(C) has the “ordinary topology”, namely itstopology as a subspace of Cn2 . SOn(C) will be irreducible if and only if it isconnected in the Zariski topology, which will follow if we can prove the strongerstatement that SOn(C) is connected in the ordinary topology.Let Sn = {(z1, . . . , zn) |
∑z2i = 1}, where n > 1. Then Sn is irreducible, which
we prove by showing that f =∑z2i −1 is an irreducible polynomial (and then
appeal to Assignment Q9(i)). So assume that f = (L0+L1+L2)(M0+M1+M2)where degLi = degMi = i. Wlog L2 = 0. L1 6= 0 so M2 = 0. Hencef = (L0 + L1)(M0 +M1) and since L0M0 = −1 we can take L0 = 1,M0 = −1and hence
(1 + L1)(−1 +M1) =∑
z2i − 1,
so L1 = M1 and L21 =
∑z2i which is impossible if n > 1.
Now by Gram-Schmidt orthogonalisation, observe that SOn acts transitivelyon Sn. Moreover, if
v0 =
10...0
then the stabiliser (SOn)v0 of v0 is
{g ∈ SOn | gv0 = v0} ∼= SOn−1,
INVARIANT THEORY 67
via the identification {(1 00 A
)| A ∈ SOn−1
}.
Hence the map
SOn → Sn
g 7→ gv0
is a surjective morphism whose fibres are precisely the cosets of SOn−1. Hence
SOn/SOn−1∼−→ Sn.
The map SOn → SOn/SOn−1 is closed (obvious for SOn(C), not obvious overa general field because we have not defined quotient varieties). If G(= SOn) =G0 t x1G
0 t · · · t xrG0 is the irreducible decomposition of G, then since wemay assume by induction that SOn−1 is irreducible, SOn−1 ⊆ G0. Hence
Sn = G/SOn−1 = G0/SOn−1 t x1G0/SOn−1 t · · · xrG0/SOn−1,
and by irreducibility of Sn, r = 0 and SOn is irreducible. �
Remark 5.51. We want to prove SOn(K) is irreducible. This means thatthe ideal in K[x11, . . . , xnn] which is generated by the relations AtA = 1 anddetA = 1, which are polynomial relations, is prime, which remains true inde-pendently of the field chosen.
5.8. Birational morphisms
Definition 5.52. Let V,W be irreducible affine varieties. A rational mapϕ : V → W is a morphism
ϕ : V 0 → W,
where V 0 is an open (and therefore dense) subvariety of V . We say thatV,W are birationally equivalent (or birational) if there exists ϕ : V → W andψ : W → V , both rational maps, which induce inverse isomorphisms fromV 0 → W 0 where V 0 and W 0 are open in V and W respectively.
Remark 5.53. Elements of K(V ) define rational maps V → K. These arenot defined on the whole of V , just on an open subset.
Example 5.54. Consider the following examples.
(i) GLn(K)∼−→ An2 , A 7→ A−1 for example.
(ii) (x, y) 7→ (xy, y) is known as a quadratic map. This is a birationalequivalence from A2 to A2. The map in the other direction is given by(a, b) 7→ (a
b, b), called the “blow-up” map.
68 INVARIANT THEORY
(iii) Let S be the space of skew-symmetric matrices (aij), i.e. matricesA = (aij) such that At = −A. Since the diagonal entries must be zeroand the the upper triangular entries completely determine the lowertriangular ones, we see that
S ∼= A12n(n−1).
Define C : S → GLn(K) by
C(S) = (1 + S)(1− S)−1,
called the Cayley transformation. Notice that this map is not definedeverywhere, since we require that (1−S) be invertible. The polynomialdet(1− S) does not vanish on S since for example
det(I −
(0 −2−2 0
))6= 0.
Proposition 5.55. Let S be defined as above and let S ∈ S.
(i) C(S) ∈ SOn(K).
(ii) C is a birational equivalence from A 12n(n−1) → SOn(K).
Proof . First, let A = C(S) = (1 + S)(1− S)−1. Then
At = ((1− S)t)−1(1 + S)t
= (1 + S)−1(1− S)
= A−1
since 1 + S and 1− S commute. Hence A ∈ On(K). Now
detA = det(1 + S) det((1− S)−1)
= det(1 + S)t det((1− S)−1)
= 1,
so A ∈ SOn(K).
Now let T = (1 + S)(1− S)−1, then T (1− S) = (1 + S) and so
S = (T − 1)(1 + T )−1.
So we have ψ : (SOn)det(1+T )∼−→ Sdet(1−S), where the subscript means that
function does not vanish. �
Remark 5.56. The main point is that SOn is birationally equivalent to anopen affine subvariety of AN for some N , and the polynomials which define thisbirational equivalence have coefficients 0 or 1, and the equivalence is definedover Z.
INVARIANT THEORY 69
5.9. Density results for affine spaces
Suppose K ⊂ L ⊂ K is a field extension of K which has characteristic zero,and that L is a subfield of the algebraic closure K.
Suppose we are given an affine K-variety V . We will say that such a variety isdefined over K if it is the set of common zeroes of polynomials inK[x1, . . . , xn],say f1, . . . , fr. Its K-points V (K) are defined as
V (K) := {(a1, . . . , an) ∈ Kn | fi(a) = 0, i = 1, . . . , r}.Similarly we have a notion of defined over L and L-points. Notice that everypoint on the variety is a K-point.
Lemma 5.57. Let V be defined over K. Then the K-points of V are exactlythose K-points fixed by the Galois group, i.e.
V (K) = {a ∈ V | γ(a) = a for all γ ∈ Gal(K/K)}.
Proof . Since V is defined over K, V is defined by the vanishing of a finite set ofpolynomials in K[x1, . . . , xn], say f1, . . . , fr. So a ∈ V if and only if fi(a) = 0for all i. Note that b ∈ K lies inK if and only if γ(b) = b for all γ ∈ Gal(K/K).Further, if a = (a1, . . . , an), γ(a) = (γ(a1), . . . , γ(an)) and so a ∈ Kn is in Kn
if and only if γ(ai) = ai for all i. Since polynomials with coefficients in K willnot change under the Gal(K/K)-action, the result follows. �
Example 5.58. Consider Q ⊆ C. Consider A1(Q). The real rational pointsare Zariski dense in C. Similarly, given any affine C-variety V , the Q-pointsof V are simply the rational points on V .
We aim to prove a substantial density result (Theorem 5.63). Before we provethis in full, we deduce the result for slightly less general situations – the firstbeing when our variety V is the whole of affine n-space.
Lemma 5.59. Let V = An = Kn and suppose K ⊂ L ⊂ K are fields of
characteristic zero. Then
V (K) = {a ∈ Kn | γ(a) = 0 for all γ ∈ Gal(K/K)}= {(a1, . . . , an) | ai ∈ K}
is Zariski dense in V (L).
Proof . That V (K) has the stated form follows from our discussion in the proofof Lemma 5.57. Let f ∈ L[An] = L[x1, . . . , xn] and suppose that f |V (K) = 0 –we need to show that f ≡ 0. We proceed by induction.
Consider the base case when n = 1. Then f ∈ L[x], i.e. f =∑r
i=1 aixi with
ai ∈ L. Since f vanishes on K, f has an infinite number of zeroes and so isidentically zero.
70 INVARIANT THEORY
In general, f ∈ L[x1, . . . , xn]. Let us write
f =r∑i=1
fixin
with fi ∈ L[x1, . . . , xn−1]. It is clear that, since f vanishes on V (K) = Kn, foreach a1, . . . , an−1 ∈ Kn−1,
r∑i=1
fi(a1, . . . , an−1)xin
vanishes at each xn = a ∈ K, so is zero, i.e. fi(a1, . . . , an−1) = 0 for alla1, . . . , an−1 ∈ Kn−1 and so fi = 0 ∈ L[x1, . . . , xn−1], whence we are done byinduction. �
We now prove the result for principal open sets.
Lemma 5.60. Let K ⊂ L ⊂ K, and let V be a principal open set of An,which is defined over K, i.e. V = Vf = {a ∈ An | f(a) 6= 0} for somef ∈ K[x1, . . . , xn]. Then V (K) is dense in V (L).
Proof . We must show that, given f ∈ K[V (L)], that if f vanishes on V (K)then f vanishes on V (L). Given such an f , by definition of K[V (L)] (seeLemma 5.36) we can write f = F (x1,...,xn)
gn, where V = (An)g and g ∈ K[x1, . . . , xn]
(where F ∈ L[x1, . . . , xn]).
If f vanishes on V (K), then F (x1, . . . , xn)g(x1, . . . , xn) = 0 on Kn. By Lemma5.59, Fg = 0 and since g 6= 0 and L[x1, . . . , xn] is an integral domain, F = 0,whence f = 0 as required. �
Remark 5.61. If we only stipulate that f be defined over L, our set may bereduced to a point. For example if
V = {(x, y) | A2 | y = πx},then V is not defined over any algebraic extension of Q. Hence V (Q) = {0}.
We now generalise our result to arbitrary open varieties V .
Corollary 5.62. If V is any open K-subvariety of An, defined over K, thenV (K) is dense in V (L).
Proof . Notice that V defined over K implies that An \ V is equal to the setof common zeroes of f1, . . . , fr where fi ∈ K[x1, . . . , xn]. Therefore V =Anf1∪ · · ·An
frand
V (K) = Anf1
(K) ∪ · · · ∪ Anfr(K)
is clearly dense. �
We are now ready to state the theorem in full generality.
INVARIANT THEORY 71
Theorem 5.63. Suppose K ⊂ L ⊂ K are fields of characteristic zero. LetX be an irreducible affine K-variety which is birationally equivalent to An forsome n via a K-morphism1 Let X ′ ⊂ X be a dense open K-subvariety of X.Then X ′(K) is dense in X ′(L).
Proof . The hypotheses imply that we have aK-isomorphism ϕ : X0 → V ⊂ An
where X0 is an open K-subvariety of X, since this is the definition of X beingbirationally equivalent to An, and that ϕ is defined over K. We may restrictϕ to
X ′ ∩X0ϕ−→ V0 ⊂ V ⊂ An,
where X ′ ∩X0 is open and nonempty (since X ′, X0 are dense) and V0 is open.ϕ is an isomorphism X ′∩X0
∼−→ V0 by definition of birational equivalence, andhence by Remark 5.35,
ϕ∗ : K[V0]→ K[X ′ ∩X0]
is an isomorphism. Let F ∈ L[X ′] and suppose F vanishes on X ′(K). Let F0
be the restriction of F to X ′∩X0. It suffices to show that F0 = 0 ∈ L[X ′∩X0],since F0 is a function on X ′ and X ′ ∩X0 is a non-empty dense subset thereof.Now F0 ∈ L[X ′ ∩ X0] and so (ϕ∗)−1F0 ∈ L[V0] and the fact that F vanisheson X ′(K) implies that (ϕ∗)−1F0 vanishes on V0(K). This implies by Corollary5.62 that (ϕ∗)−1F0 = 0 and so F0 = 0. �
Corollary 5.64. With the above notation, the variety SOn(K) is Zariski densein SOn(L).
Proof . SOn(K) is birationally equivalent to An(n−1)/2 via the Cayley transformC, which is defined over Z. �
Remark 5.65. It is convenient to first think through these results withK = Qand L = K = C. In Chapter 6, we will apply Theorem 5.63 to a morecomplicated situation where L is a field of rational functions in
(n2
)variables.
1Here, we are saying that there is an open subvariety of X which is also defined over Kand an isomorphism from that subvariety to an open subvariety of An which is defined overK. This morphism has coefficient functions which are polynomials, and we require thesepolynomials to have coefficients in K.
72 INVARIANT THEORY
Exercises
Exercise 5.66. Prove that V IV = V (i.e. prove the other half of Lemma5.1(v).
Exercise 5.67. Complete the proof of Lemma 5.4(ii).
Exercise 5.68. Show that 〈y − x2〉 and 〈y〉 are algebraic ideals of K[x, y].
Exercise 5.69. Prove Lemma 5.28.
Exercise 5.70. Check that the maximal ideals of K[V ], i.e. the maximalideals of S which contain I(V ), are in bijection with the points of V .
Exercise 5.71. Finish the proof of Theorem 5.34 by showing that the mapf ↔ f ∗ is indeed a bijection.
Exercise 5.72. Check that the map ϕ∗ in Example 5.68(ii) is indeed an iso-morphism, and that moreover it is the comorphism corresponding to the map
K → V
T 7→ (T, T 2).
This shows that the parabola is isomorphic as an algebraic variety to K1.
Exercise 5.73. Prove that F = x1x4 − x2x3 is an irreducible polynomial andtherefore V as defined in Example 5.68(iv) is an irreducible variety. [Hint:Notice that the function is linear over K(x1, x2, x3)].
Exercise 5.74. Verify in Definition 5.40 that if f1/f2 = g1/g2 then the valueof f(P ) is independent of the choice of f1, f2 or g1, g2.
Exercise 5.75. Prove that the product of two irreducible varieties is an irre-ducible variety (follows from R, S integral domains implies R ⊗ S an integraldomain).
Exercise 5.76. Show that G0 is the smallest closed (normal) subgroup offinite index in G.
Exercise 5.77. Show that if S ⊂ G is a closed subsemigroup then S is asubgroup.
Exercise 5.78. Prove that the variety SLn(K) = {A | detA = 1} is irre-ducible. [Hint: As a variety, GLn(K) is the product K×× SLn(K) where K×is thought of as non-zero scalar matrices].
Exercise 5.79. Show that if G is an affine algebraic group and H is an irre-ducible closed subgroup of G, then H ⊆ G0.
Exercise 5.80. (i) Show that the hypersurface defined by f(x1, . . . , xn) =0 is irreducible if and only if the polynomial f is irreducible.
INVARIANT THEORY 73
(ii) Show that the following polynomials are irreducible:(a) y − x2;(b) x2 + y2 − 1;(c) x1x3 − x2x4 ∈ K[x1, x2, x3, x4];(d) y2 − x2 − x3.
(iii) What is the pole set of the rational function f = 1−yx
on x2 + y2 = 1?(iv) Show that the function t = y
xon y2 = x2 + x3 is not regular.
Exercise 5.81. Show that the morphism φ : V → V ′ of affine varieties hasdense image if and only if the corresponding homomorphism φ∗ : C[V ′]→ C[V ]is injective.
Exercise 5.82. For any polynomials f1, . . . , fl ∈ K[x1, . . . , xl] let ∂(f1,...,fl)∂(x1,...,xl)
denote the usual Jacobian det( ∂fi∂xj
). Show that f1, . . . , fl are algebraically
independent if and only if ∂(f1,...,fl)∂(x1,...,xl)
6= 0.
Exercise 5.83. Prove the following assertion. Suppose A is an integral domainand A′ ⊂ A is a subalgebra such that A is finitely generated as an A′-module.If M ′ is any maximal ideal of A′ there exists a maximal ideal M of A suchthat M ′ = M ∩ A′.
Exercise 5.84. Deduce the following corollary of Exercise 5.83. Suppose φ :V → V ′ is a morphism of affine varieties such that C[V ] is finitely generatedas a module over φ∗(C[V ′]). Then φ(V ) is closed in V ′.
Exercise 5.85. Deduce the following consequence of Exercises 5.83 and 5.84.Let G be a finite group acting linearly on V = Cl. Let S = C[V ] etc as usual.Let {F1, . . . , Fl} be a set of algebraically independent homogeneous elementsof SG such that SG is integral over C[F1, . . . , Fl]. Then the map φ : V → Cl
given by v 7→ (F1(v), . . . , Fl(v)) is surjective.
Chapter 6
Invariant theory of the orthogonal group
6.1. Preliminaries
Let K be a field of characteristic zero, V = Kn, and assume that we have anon-degenerate symmetric bilinear form
(·, ·) : V × V → K,
which is equivalent to the form
((a1, . . . , an), (b1, . . . , bn)) =n∑i=1
aibi,
i.e. that we can find a basis such that our form is equal to the above formwhen expressed in terms of this basis. Recall that a form is non-degenerate if(a, b) = 0 for all b ∈ Kn implies a = 0, and symmetric if (a, b) = (b, a).
This form provides an isomorphism V → V ∗ : v 7→ ϕv, where
(v, w) = 〈ϕv, w〉 for all v, w ∈ V ,
where
〈·, ·〉 : V ∗ × V → K
(ϕ, v) 7→ ϕ(v) := 〈ϕ, v〉
is the natural pairing.
Since SOn is birationally equivalent to An(n−1)/2 via the Cayley transform, itfollows that SOn(Q) is Zariski dense in SOn(K) for any field L ⊃ Q (a specialcase of Theorem 5.63). Moreover, since
On = SOn t diag{−1, 1, . . . , 1}SOn,
it follows that On(Q) is dense in On(L) for all L ⊃ Q (again by Theorem 5.63).If U is any open subvariety of On, then U(Q) is dense in U(L).
For the rest of this chapter, we assume that K = C.
We begin by studying polynomial functions f : Mn(K) → K such thatf(XA) = f(A) for all X ∈ On(K) (i.e. invariant under left multiplicationby the orthogonal group).
74
INVARIANT THEORY 75
Note that if A ∈ Mn(K) is non-singular then P = AtA is a symmetric non-singular matrix and if AtA = BtB then (Bt)−1AtAB−1 = I, i.e.
(AB−1)t(AB−1) = I, i.e. AB−1 = X ∈ On(K).
Hence A = XB, i.e. AtA = BtB if and only if A and B are in the sameOn(K)-orbit.
Remark 6.1. If P = Q2 for some Q, for some symmetric Q, then any A suchthat AtA = P is of the form A = XQ for some X ∈ On(K).
Lemma 6.2. Suppose P ∈ Mn(K) has distinct eigenvalues λ1, . . . , λn ∈ K.Then
(i) for each i = 1, . . . , n, the operator
Ei =∏j 6=i
P − λjλi − λj
is the projection to the λi-eigenspace of P ;
(ii) assuming that√λi ∈ K for all i,
Q =n∑i=1
√λiEi
is a square root of P , i.e. Q2 = P .
Proof . Since P has distinct eigenvalues,V = Kn = Kv1 ⊕Kv2 ⊕ · · · ⊕Kvn,
where Pvi = λivi. It follows that
Eivk =∏j 6=i
P − λjλi − λj
vk =
{0, k 6= ivi k = i.
Secondly, Q =∑√
λiEi acts as√λi on vi for all i, so Q2vi = λivi for all i, i.e.
Q2 = P . �
Remark 6.3. Notice that Q is a polynomial in P , so Q commutes with P andif P is symmetric then Q is symmetric.
6.2. The main lemma
Our main task is to prove the following theorem.
Theorem 6.4. Let K be a field of characteristic zero and let f : Mn(K)→ Kbe a polynomial function such that
f(XA) = f(A) for all X ∈ On(K). (6.5)Then there exists a polynomial function F : SMn(K) → K such that f(A) =F (AtA), where SMn(K) is the set of all symmetric matrices over K.
76 INVARIANT THEORY
Remark 6.6. In other words, every orthogonal invariant function is really afunction of the inner products between the columns of A.
Traditional proofs of the first fundamental theorem for On often refer to theCapelli identities from combinatorics, whose role in invariant theory has neverbeen completely clear. Our proof will only involve basic techniques from alge-braic geometry introduced in Chapter 5.
We now make some preliminary remarks before proving the theorem.
Remark 6.7. Let (pij)1≤i,j≤n be a set of indeterminates such that pij = pji.Then P = (pij) is a symmetric matrix whose entries are indeterminates –in fact it is the generic symmetric matrix. Note that P ∈ Mn(L) whereL = K(p11, . . . , pnn). P has distinct eigenvalues in L, the algebraic closure of Lbecause one specialisation of P is the matrix diag{1, 2, . . . , n} – the condition tohave distinct eigenvalues is an open condition, on the characteristic polynomial.This polynomial cannot possibly vanish everywhere because we have foundsome specific values for which it does not vanish. By Lemma 6.2, there existsQ ∈ SMn(L) such that Q2 = P .
Proof . (of Theorem 6.4) Observe first that if K ⊃ K is any field extension ofK, then f : Mn(K)→ K is On(K)-invariant, because
f : On(K)×Mn(K)→ K
defined by f(X,A) = f(XA)−f(A) vanishes onOn(K)×Mn(K), and Theorem5.63 gives that On(K)×Mn(K) is dense in On(K)×Mn(K), and so by densityf = 0 on On(K)×Mn(K).
Next, observe that σ ∈ Gal(L/L) (automorphism of L which fixes L pointwise),then σ fixes P and since Q2 = P , we have (σ(Q))2 = P = Q2. By our remarksabove,
σ(Q) = XQ for some X ∈ On(L).
Hencef(σ(Q)) = σ(f(Q)) = f(Q).
Hence f(Q) ∈ L = K(p11, . . . , pnn). Hence f(Q) is a rational function of P ,i.e.
f(Q) =γ1(p11, . . . , pnn)
γ2(p11, . . . , pnn),
where γi ∈ R = K[p11, . . . , pnn].
A specialisation of P is obtained from a ring homomorphism α : R → K.Suppose we have such an α which satisfies
(i) α(γ2(p11, . . . , pnn)) = γ2(α(p11), . . . , α(pnn)) 6= 0, i.e. the specialisationwe take avoids all the zeros of γ2 (this is an open condition)
(ii) α(P ) has distinct eigenvalues, α ∈ An(n−1)/2.
INVARIANT THEORY 77
These two conditions form an open condition on α.
If α satisfies these two properties, then for any A ∈ Mn(K) such that AtA =α(P ), we have
f(A) = ϕ(α(P )), where ϕ =γ1
γ2
,
because any such A satisfies
A = XQ, where Q2 = α(P ), X ∈ On(K),
i.e. f(A) = f(Q) = ϕ(α(P )). We are now precisely in the situation of Propo-sition 5.42(iii), i.e. we have two functions f, ϕ from an open subvariety U ofMn(K) (defined by the conditions on α) to K, f is regular, ϕ is rational, andf = ϕ on U . Hence we conclude that f ≡ ϕ and so ϕ is regular, whencef(A) = ϕ(AtA). �
Corollary 6.8. Suppose f : GLn(K)→ K is a polynomial function such thatf(XA) = f(A) for all X ∈ On(K). Then there exists a polynomial function
F : SMn(K)det6=0 → K
such that f(A) = F (AtA).
Proof . Since K[GLn] = K[Mn]det, the localisation, we know that
f(A) =h(A)
(detA)l,
for some h(A) ∈ K[Mn(K)] and we may assume that l is even (since otherwisewe can just multiply h(A) by detA). Then since l is even say l = 2m,
h(XA) = f(XA)(det(XA))2m
= f(A)(detX)2m(detA)2m
= f(A)(detA)2m = h(A)
for all x ∈ On(K), since det(X) = ±1. Applying Theorem 6.4 to h, we see thatthere exists a polynomial function G : SMn → K such that h(A) = G(AtA),i.e.
f(A)(detA)2m = G(AtA).
Hence we have
f(A) =G(AtA)
(detA)2m=
G(AtA)
(det(AtA))m=
G
detm(AtA),
whence the result follows upon setting F = Gdetm
which is clearly a mapSMn(K)det6=0 → K as claimed. �
78 INVARIANT THEORY
6.3. Second formulation of the first fundamental theorem for GLn
Let us now return to the situation where V = Kn is a linear space with noorthogonal form necessarily present. We wish to give an alternative formulationof the first fundamental theorem for GLn(V ) which we gave in Chapter 3. Ourformulation in Chapter 3 was a statement about algebras, and we wish totranslate this into a linear statement. In order to do this we must first set upsome algebraic machinery.
Recall that the action of G = GLn(K) on V ∗ is given by
〈gϕ, v〉 = 〈ϕ, g−1v〉for ϕ ∈ V ∗ (where 〈·, ·〉 is just a notational convenience: 〈ψ,w〉 simply meansψ(w)).
Definition 6.9. Let ϕ : V → W be a map between two vector spaces equippedwith a G-action for some group G. We say ϕ is G-equivariant if it respects theG-action, i.e. if ϕ(gv) = gϕ(v) for all g ∈ G and v ∈ V . Notice that such a ϕis often called a G-homomorphism.
We define a pairing which demonstrates that the space T r(V ) ⊗ T r(V ∗) isself-dual. Define a pairing
[·, ·] : (T r(V )⊗ T r(V ∗))× (T r(V )⊗ T r(V ∗))→ K
by
[v1⊗· · ·⊗vr⊗ϕ1⊗· · ·⊗ϕr, v′1⊗· · ·⊗v′r⊗ϕ′1⊗· · ·⊗ϕ′r] =r∏i=1
〈ϕi, v′i〉r∏j=1
〈ϕ′j, vj〉.
for v1, . . . , vr, v′1, . . . , v
′r ∈ V and ϕ1, . . . , ϕr, ϕ
′1, . . . , ϕ
′r ∈ V ∗. This pairing
allows for the natural identification T r(V )⊗ T r(V ∗) with its dual via
v ⊗ϕ↔ [·,v ⊗ϕ],
where we use v to mean an arbitrary element v1 ⊗ · · · ⊗ vr ∈ T r(V ), and ϕ tomean an arbitrary element ϕ1 ⊗ · · · ⊗ ϕr ∈ T r(V ∗).
Lemma 6.10. The above pairing is perfect, i.e. the map
Φ : (T r(V )⊗ T r(V ∗))→ (T r(V )⊗ T r(V ∗))∗
v ⊗ϕ 7→ [·,v ⊗ϕ]
is a GL(V )-equivariant isomorphism.
Proof . That the map is injective follows since our pairing [·, ·] is non-degenerate.By finite-dimensionality, Φ is an isomorphism of vector spaces. Let v⊗ϕ andw ⊗ψ ∈ T r(V )⊗ T r(V ∗). Then for all g ∈ GL(V ) we have
〈gΦ(v ⊗ϕ),w ⊗ψ〉 = 〈Φ(v ⊗ϕ), g−1(w ⊗ψ)〉= [g−1(w ⊗ψ),v ⊗ϕ]
INVARIANT THEORY 79
Expanding these out and using the definition of the GL(V )-action, we see
〈gΦ(v ⊗ϕ),w ⊗ψ〉 = [g−1(w ⊗ψ),v ⊗ϕ]
=r∏i=1
〈g−1ψi, vi〉r∏i=1
〈ϕi, g−1wi〉
=r∏i=1
〈ψi, gvi〉r∏i=1
〈gϕi, wi〉
= [w ⊗ψ, g(v ⊗ϕ)]
= 〈Φ(g(v ⊗ϕ)),w ⊗ψ〉,whence gΦ(v ⊗ ϕ) = Φ(g(v ⊗ ϕ)) for all v ⊗ ϕ ∈ T r(V ) ⊗ T r(V ∗), i.e. Φ isindeed GL(V )-equivariant. �
Definition 6.11. For each σ ∈ Sr, define an operator γσ ∈ (T r(V )⊗T r(V ∗))∗by
γσ : T r(V )⊗ T r(V ∗)→ K
v ⊗ϕ 7→r∏i=1
〈ϕi, vσ(i)〉.
It is a straightforward exercise to check that γσ is linear. We also define, foreach 1 ≤ i, j ≤ r,
γij(v1 ⊗ · · · ⊗ vr ⊗ ϕ1 ⊗ · · · ⊗ ϕr) = 〈ϕi, vj〉.
Theorem 6.12. Let V = Kn, G = GL(V ) where K has characteristic zero.Then
(i) (First fundamental theorem)
For any integers r, s,((T r(V )⊗ T s(V ∗))∗
)G=
{0, if r 6= sK{tσ | σ ∈ Sr}, if r = s,
where the tσ are the images of the γσ under the canonical isomorphismΦ.
(ii) (Second fundamental theorem)
If r ≤ n, the γσ are linearly independent.
If r ≥ n + 1 then all linear relations among the γσ are linear conse-quences of
det(γij) i∈Sj∈S∗
∏k 6∈Sl 6∈S∗
γkl,
where S, S∗ ⊂ {1, . . . , r}, |S| = |S∗| = n+ 1.
80 INVARIANT THEORY
Proof . Since we have the canonical isomorphisms
T r(V )⊗ T s(V ∗) ∼= T r(V )⊗ (T r(V ))∗ ∼= Hom(T s(V ), T r(V )),
we see that the ring of invariants(T r(V )⊗ T s(V ∗)
)G ∼= HomG(T s(V ), T r(V )).
Recall that Z(G) = Z(GL(V )), the center of G, is the set of scalar matrices,i.e. the set
Z(G) = {diag{λ, λ, . . . , λ} | λ ∈ K} .An element z = diag{λ, . . . , λ} ∈ Z(G) acts on T r(V ) by
z · (v1 ⊗ · · · ⊗ vr) = λv1 ⊗ · · · ⊗ λvr = λrv1 ⊗ · · · ⊗ vr.So if H ∈ HomG(T s(V ), T r(V )) then for v ∈ T s(V ), we have, for z =diag{λ, . . . , λ} ∈ Z(G),
λsH(v) = H(λv)
= H(z · v)
= z ·H(v)
= λrH(v)
and so λr = λs for all λ ∈ K (if H 6= 0) and so r = s. Therefore if r 6= s,HomG(T s(V ), T r(V )) = 0.
We now assume r = s, and consider the following diagram:
KSr T r(V )⊗ T r(V ∗)(T r(V )⊗ T r(V ∗)
)∗
T r(End(V ))
End(T r(V ))
λ ∼=Φ
µ∗
ξ∼=
∼= Ψ
ξ∗
µ
Now we have seen the map
µ : KSr → End(T r(V ))
σ 7→ (v1 ⊗ · · · ⊗ vr 7→ vσ−1(1) ⊗ · · · ⊗ vσ−1(r))
in Chapter 3. The isomorphism End(T r(V )) → T r(End(V )) is the canonicalone which we have seen before, namely
Ψ : T r(End(V ))∼−→ End(T r(V ))
α1 ⊗ · · · ⊗ αr 7→ (v1 ⊗ · · · ⊗ vr 7→ α1v1 ⊗ · · · ⊗ αrvr),
INVARIANT THEORY 81
and we define µ∗ by the relation µ = Ψ ◦ µ∗. If we recall the canonical map
ξ : V × V ∗ → End(V )
(v, ψ) 7→ (w 7→ ψ(w)v),
then the map ξ is defined by
ξ : T r(V )⊗ T r(V ∗)→ T r(End(V ))
v ⊗ϕ 7→ ξ(v1, ϕ1)⊗ · · · ⊗ ξ(vr, ϕr).
We also recall the map Φ from Lemma 6.10 which is induced by our pairing[·, ·], and finally we define λ so as the make the diagram commute. We alsowrite λ∗ for Φ ◦ λ.
We claim that
λ∗(σ) = γσ−1 , (6.13)
i.e. more precisely that for all v ⊗ϕ ∈ T r(V )⊗ T r(V ∗),
[v ⊗ϕ, λ(σ)] = γσ−1(v ⊗ϕ) =r∏i=1
〈ϕi, vσ−1(i)〉. (6.14)
Let b1, . . . , bn be a basis for V and let β1, . . . , βn be the corresponding dualbasis for V ∗. We make the two following simple observations:
(a) for all v ∈ V and ϕ ∈ V ∗, we can write
v =n∑i=1
〈βi, v〉bi, ϕ =n∑j=1
〈ϕ, bj〉βj;
(b) for all v ∈ V and ϕ ∈ V ∗,
〈ϕ, v〉 =n∑i=1
〈βi, v〉ϕ(bi)
=n∑i=1
〈βi, v〉n∑j=1
〈ϕ, bj〉βj(bi)
=n∑i=1
〈ϕ, bi〉〈βi, v〉.
We define, for σ ∈ Sr, the following element tσ ∈ T r(V )⊗ T r(V ∗):
tσ =∑
1≤i1,...,ir≤n
bi1 ⊗ · · · ⊗ bir ⊗ βiσ−1(1)⊗ · · · ⊗ βiσ−1(r)
.
82 INVARIANT THEORY
Then by (a) and (b) above we see that
[v ⊗ϕ, tσ] =∑
1≤i1,...,ir≤n
[v ⊗ϕ, (tσ)i1,...,ir ]
=∑
1≤i1,...,ir≤n
[v ⊗ϕ, bi1 ⊗ · · · ⊗ bir ⊗ βiσ−1(1)⊗ · · · ⊗ βiσ−1(r)
]
=∑
1≤i1,...,ir≤n
n∏j=1
〈ϕj, bij〉〈βiσ−1(j), vj〉
=n∏j=1
∑1≤i1,...,ir≤n
〈ϕj, bij〉〈βij , vσ(j)〉
=n∏j=1
〈ϕj, vσ(j)〉
= γσ(v ⊗ϕ) (6.15)
for all v ⊗ϕ ∈ T r(V )⊗ T r(V ∗). We now compute ξ(tσ):
ξ(tσ) =∑
1≤i1,...,ir≤n
ξ(bi1 , βiσ−1(1))⊗ · · · ⊗ ξ(bir , βiσ−1(r)
).
Applying ξ(tσ) to bk1 ⊗ · · · ⊗ bkr we obtain
ξ(tσ)(bk1 ⊗ · · · ⊗ bkr) =∑
1≤i1,...,ir≤n
〈βiσ−1(1), bk1〉bi1 ⊗ · · · ⊗ 〈βiσ−1(r)
, bkr〉bir ,
and the summands are all zero except when iσ−1(j) = kj for all j = 1, . . . , r,i.e. bkσ(j)
= bij , i.e. if b = bk1 ⊗ · · · ⊗ bkr ,
ξ(tσ)b = µ∗(σ−1)(b),
where µ∗ = Ψ ◦ µ, where Ψ is the canonical isomorphism mentioned above.Hence we have
ξ(tσ) = µ∗(σ−1). (6.16)Now since we defined λ as the map which made our diagram commute, we seefor all v ⊗ϕ ∈ T r(V )⊗ T r(V ∗),
[v ⊗ϕ, λ(σ)] = [v ⊗ϕ, ξ−1(µ∗(σ))] by commutativity= [v ⊗ϕ, tσ−1 ] by (6.16)= γσ−1(v ⊗ϕ) by (6.15),
which is precisely the claim in (6.14). Now since ξ and Ψ are isomorphisms,λ is surjective if and only if µ is surjective, but µ is surjective by the firstfundamental theorem (Theorem 3.2). Hence λ is surjective, and it followsfrom the above that its image λ(KSr) is the span of the set
{γσ | σ ∈ Sr}.
INVARIANT THEORY 83
Further, since ξ is a GL(V )-equivariant isomorphism,
im(λ) = (T r(V )⊗ T r(V ∗))GL(V )
and similarly im(λ∗) = (T r(V )⊗ T r(V ∗))∗ – this proves (i) of Theorem 6.12.
Notice that tσ is independent of the choice of basis (∑bi ⊗ βi ∈ V ⊗ V ∗
corresponds to the idV ∈ End(V )).
Further, notice that (T r(V )⊗ T r(V ∗))GL(V ) = span{tσ | σ ∈ Sr}.
By the second fundamental theorem’s first formulation (Theorem 3.10), ker(µ)is the two-sided ideal of KSr generated by
an+1 = a(Sn+1) =∑
σ∈Sn+1
(sgn σ)σ.
But ker(µ) = ker(λ) and so we seek a linear description of ker(λ). Since KSr
is an algebra, if we consider it as a linear space, the ideal generated by an+1
will be the linear span of all elements of the form
σ1an+1σ2, σ1, σ2 ∈ Sr,
since in the algebra we may pre- or post-multiply by permutations. Considerλ(an+1). Computing, we see
λ(an+1) =∑
σ∈Sn+1
(sgn σ)λ(σ)
=∑
σ∈Sn+1
(sgn σ)γσ,
since each σ appears in the sum we may sum over σ−1 rather than σ. Now if weimagine our permutation σ in the center of a diagram with the permutationsσ1 and σ2 above and below it, it is clear that we will have some set S ⊆{1, . . . , r} of cardinality n + 1 at the top, some other set S ′ ⊆ {1, . . . , r} ofthe same cardinality at the bottom, and it will be between these sets that ourpermutation now interweaves. The remaining r − (n+ 1) dots at the top andbottom of the diagram remain arbitrary. Considering now the correspondingelement to λ(σ) in the dual space under our canonical identification (since itis easier to compute things in the dual space), we see, if we let λ(σ) = t, thenfor v ⊗ϕ ∈ T r(V )⊗ T r(V ∗),
[v ⊗ϕ, t] =(∑σ∈Sr
sgn σ〈ϕi, vσ(i)〉) r∏j=n+2
γj,j
=(
det1≤i,j≤n+1
(γij(v ⊗ϕ))) r∏j=n+2
γj,j.
84 INVARIANT THEORY
SoΦ(λ(σ1an+1σ2)
)=(
deti∈Sj∈S′
(γij)) ∏k∈{1,...,r}\Sl∈{1,...,r}\S′
γkl
where S and S ′ are subsets of {1, . . . , r} of cardinality n+ 1, which depend on
σ1 and σ2. It is clear then that any linear relation in(
(T r(V ) ⊗ T r(V ∗))∗)G
is a consequence of this relation, and we may now use Φ−1 to translate theserelations into the original space T r(V )⊗ T r(V ∗). �
Remark 6.17. The first fundamental theorem as given in our previous chap-ter was a statement about an algebra homomorphism. This statement aboveis a purely linear statement, which is nonetheless equivalent to the previousformulation.
6.4. Some multilinear algebra and the orthogonal group
We develop some important algebraic identities concerning the orthogonalgroup. Recall that if W = Kn, we can define K[W ], the space of regularfunctions on W , and we have
K[W ] ⊂ F(W,K),
where F(W,K) is the set of all functions from W to K. GL(W ) acts on allfive spaces T r(W ), T r(W ∗), Sr(W ), Sr(W ∗) ∼= K[W ] and F(W,K). Moreover,the canonical surjection T r(W )→ Sr(W ) respects the GL(W )-action.
Also recall the definitions of the tensor algebra and the symmetric algebra,
T (W ∗) =⊕r
T r(W ∗) and S(W ∗) =⊕r
Sr(W ∗).
Lemma 6.18. Let W be defined as above. Then the composite
γ : T (W ∗)→ S(W ∗)∼−→ K[W ] ↪→ F(W,K)
is described explicitly on T r(W ∗) as follows:
γ(ϕ1 ⊗ · · · ⊗ ϕr) : w 7→∏i
〈ϕi, w〉,
and by linear extension to T (W ∗).
Proof . γ : T (W ∗) → F(W,K) as defined is a homomorphism of K-algebras,where T (W ∗) has the (tensor) algebra structure, and F(W,K) has the algebrastructure given by the pointwise product of functions. Hence it factors throughS(W ∗) and the result follows. �
INVARIANT THEORY 85
Let V be a K-vector space of dimension n and suppose V has an orthogonal,non-degenerate, symmetric, bilinear form (·, ·). This gives an explicit identifi-cation V → V ∗, namely
ϕ : V → V ∗ (6.19)v 7→ ϕv,
where ϕv(w) = (v, w). Notice the important notational consequence of this,i.e. that
〈ϕv, w〉 = (v, w). (6.20)
Definition 6.21. Let A ∈ End(V ). We define At, the transpose of A, by thecondition
(Atv, w) = (v, Aw) for all v, w ∈ V,Lemma 6.22. Using the above definition of transpose, (AB)t = BtAt for allA,B ∈ End(V ).
Proof . Simply notice that for all A,B ∈ End(V ), for all v, w,∈ V we have((AB)tv, w) = (v, (AB)w) = (v,A(Bw)) = (Atv,Bw) = (BtAtv, w),
and so (AB)t = BtAt. �
We now give a concrete algebraic definition of the orthogonal group O(V ) onV , using our orthogonal form.
Definition 6.23. The orthogonal group O(V ) on V is defined by the followingcondition. Let g ∈ GL(V ). Then g ∈ O(V ) if and only if (gv, gw) = (v, w) forall v, w ∈ V .
Remark 6.24. Notice that this definition does fit in with our usual definitionof the orthogonal group being defined by the condition AtA = I on matrices.Notice that (gv, gw) = (v, w) for all v, w ∈ V is in turn true if and only if(gtgv, w) = (v, w) for all v, w ∈ V , i.e. if and only if
gtg = idV .
Definition 6.25. Let g ∈ GL(V ). Then we define g∗ ∈ GL(V ) by
g∗(v) = (g−1)t(v), for v ∈ V .
Remark 6.26. Note that (g1g2)∗ = g∗1g∗2, i.e. ∗ is a homomorphism, since
Lemma 6.27. Using notation as defined above,
(i) for all g ∈ GL(V ), if ϕ is the map defined in (6.19), we have a com-mutative diagram
V V ∗
V V ∗
ϕ
g∗ g
ϕ
86 INVARIANT THEORY
i.e. we have ϕg∗v = g ◦ ϕv for all v ∈ V ;(ii) we have a canonical map ξ∗ : V ∗ ⊗ V ∗ → End(V ) given by
ξ∗(ϕv ⊗ ϕw) : x 7→ 〈ϕw, x〉v = (w, x)v.
for all v, w ∈ V . If g ∈ GL(V ), then
ξ∗(gτ) = g∗ξ∗(τ)g−1
for any τ ∈ V ∗ ⊗ V ∗.(iii) Let σ : V ∗ ⊗ V ∗ → V ∗ ⊗ V ∗ be the interchange map – explicitly
σ(ϕv ⊗ ϕw) = ϕw ⊗ ϕv.
Then
ξ∗(στ) = (ξ∗(τ))t.
(iv) Let S = {A ∈ End(V ) | At = A}, the set of symmetric matrices. Thenthere is an isomorphism of GL(V )-modules
ξ∗ : S2(V ∗)→ S
where the action of g ∈ GL(V ) on S is
g : A 7→ g∗Ag−1.
Proof . In all cases the proofs are routine, and all proofs depend heavily on(6.20). For the first claim, notice that ϕg∗v = g ◦ ϕv (action of G on dualspace) since
〈ϕg∗v, w〉 = (g∗v, w) = ((g−1)tv, w)
= (v, g−1w)
= 〈ϕv, g−1w〉= 〈gϕv, w〉.
For the second claim, take τ = ϕv ⊗ ϕw:
ξ∗(gτ) = ξ∗(gϕv ⊗ ϕw)
= ξ∗(ϕg∗v ⊗ gϕw)
x 7→ 〈gϕw, x〉g∗v= 〈ϕw, g−1x〉g∗w= (w, g−1x)g∗v
= ξ∗(ϕv ⊗ ϕw)(g−1x)
= g∗ξ∗(ϕv ⊗ ϕw)g−1(x).
INVARIANT THEORY 87
For the third claim,
ξ∗(σ(ϕv ⊗ ϕw)) = ξ∗(ϕw ⊗ ϕv)x 7→ (x, v)w
(ξ∗(τ)t(x), y) = (x, ξ∗(τ)y)
= (x, (w, y)v)
= (x, v)(w, y).
The final claim follows from (ii) and (iii). �
Lemma 6.28. Let F be a homogeneous polynomial map from S2(V ∗)→ K ofdegree `. Then the corresponding linear map L : T 2`(V ∗) → K satisfies, forall g ∈ GL(V ),
LF◦g = LF ◦ g.Here, LF is defined by LF = F ◦ γ2 ◦ γ1 where
T 2`(V ∗)γ1−→ T `(S2(V ∗))
γ2−→ S`(S2(V ∗))→ K
Proof . This is clear because γ1, γ2 are canonical maps which commute withthe action of GL(V ), and polynomial functions correspond to linear functionson the tensor powers by a lemma above. �
6.5. The first fundamental theorem for the orthogonal group
In this section we aim to prove an orthogonal group analogue of Theorem 6.12– in fact it will become apparent in the proof in this section that we developedimportant techniques in that proof. Specifically, we shall prove the followingtheorem.
Theorem 6.29. Let V be a K-vector space of dimension n, equipped withan orthonormal form (·, ·). Let O(V ) be the orthogonal group on V . LetW = T r(V ). Then
(W ∗)O(V ) =
{0, if r is oddK{δσ | σ ∈ Sr}, if r = 2s is even
where
δσ(v1 ⊗ · · · ⊗ v2s) =s∏i=1
(vσ(2i−1), vσ(2i)).
Proof . We begin our proof by making several observations. Let ϕ : W → Kbe an O(V )-invariant linear map, i.e. such that
ϕ(g · v1 ⊗ · · · ⊗ vr) = ϕ(gv1 ⊗ · · · ⊗ gvr) = ϕ(v1 ⊗ · · · ⊗ vr)
88 INVARIANT THEORY
for all g ∈ O(V ), for all v1 ⊗ · · · ⊗ vr ∈ W . Given such a ϕ, define a map
fϕ : End(V )×W → K
(A,v) 7→ ϕ(Av), (6.30)
where by Av we mean A(v1⊗· · ·⊗vr) = Av1⊗· · ·⊗Avr. Now by construction,fϕ is linear in v, and is a polynomial of homogeneous degree r in A.
We may obtain from fϕ a map f ∗ϕ : End(V )→ W ∗, via
f ∗ϕ : A 7→ (v 7→ fϕ(A,v) = ϕ(Av)),
a polynomial map of degree r. Now for g ∈ O(V ), we have f ∗ϕ(gA) = f ∗ϕ(A),since for v ∈ W , we have
f ∗ϕ(gA)(v) = ϕ(g(Av)) = ϕ(Av) = f ∗ϕ(A)(v),
by O(V )-invariance of ϕ. So we now have the property of our original map fϕthat
fϕ(A,v) = f ∗ϕ(A)(v) = f ∗ϕ(gA)(v) = fϕ(gA,v)
for all v ∈ W and g ∈ O(V ). Hence for each v ∈ W we have a map fϕ,v :End(V )→ K which is O(V )-invariant – hence our main lemma, Theorem 6.4,there exists a polynomial function
Hv : S → K,
where S is the symmetric elements of End(V ) such that
fϕ,v(A) = Hv(AtA),
and defining
H : S ×W → K
(S,v) 7→ Hv(S),
we see that, for each A ∈ End(V ) and v ∈ W ,
fϕ(A,v) = fϕ,v(A) = Hv(AtA) = H(AtA,v). (6.31)
We observe some properties of the map H.
Take any g ∈ GL(V ) – then by the properties of the maps above we have
fϕ(A,v) = ϕ(Av) = ϕ(Ag−1 · gv) = fϕ(Ag−1, gv)
for all v ∈ W , and so, expressing this in terms of our map H,
H(AtA,v) = H((Ag−1)tAg−1, gv)
= H((g−1)tAtAg−1, gv)
= H(g∗(AtA)g−1, gv)
= H(g · AtA, g · v)
by the definition of the GL(V )-action in Lemma 6.27(iv), whence H is aGL(V )-invariant polynomial function on S ×W .
INVARIANT THEORY 89
Consider the following diagram of maps, where the bottommost isomorphismfollows from Lemma 6.27(iv):
S ×W K
S ⊗W
S2(V ∗)⊗W
H
∼=
H ′
Then by the universal property of the tensor product, the diagram commutes(so if ι : S ×W ↪→ S ⊗W is the inclusion, H ′ ◦ ι = H), and so we have aunique GL(V )-invariant polynomial function
H∗ : S2(V ∗)⊗W → K
with all the properties of H. We may assume H∗ is homogeneous of degree din S2(V ∗), i.e. H∗ may be identified with a GL(V )-invariant linear map
Sd(S2(V ∗))⊗W → K.
In the sequel we use S and S2(V ∗) interchangeably. Consider the chain ofGL(V )-equivariant maps
T 2d(V ∗)⊗W (1)−→ T d(End(V ))⊗W (2)−→ T d(S)⊗W (3)−→ Sd(S)⊗W H∗−→ K,
where
(1) is an isomorphism since V ∗ ⊗ V ∗ ∼= End(V ), canonically;(2) is a canonical surjection, since End(V ) ∼= T 2(V ∗), and S2(V ∗) is a
quotient of T 2(V ∗);(3) is a canonical surjection, since Sd(S) is a quotient of T d(S),
and call the resulting composite surjection
H : T 2d(V ∗)⊗W → K.
Diagrammatically,
T 2d(V ∗)⊗W T d(End(V ))⊗W
Sd(S)⊗W T d(S)⊗W
K
(ξ∗)⊗d ⊗ id
H∗
∼=
H
90 INVARIANT THEORY
where ξ∗ is the canonical isomorphism introduced in Lemma 6.27(ii).
Since all maps in the diagram areGL(V )-equivariant, H is aGL(V )-equivariantmap T 2d(V ∗)⊗W → K.
By the First Fundamental Theorem for GL(V ) (in its second formulation, i.e.Theorem 6.12(i)), we deduce that for H, and hence H∗, not to be the zeromap, then r must equal 2d, and H must be a K-linear combination of mapsγσ, where we recall
γσ(ϕ1 ⊗ · · · ⊗ ϕ2d, v1 ⊗ · · · ⊗ v2d) =2d∏i=1
〈ϕi, vσ(i)〉,
i.e. we must have, for some coefficients aσ ∈ K,
H =∑σ∈S2d
aσγσ.
Hence it only remains to see what H is, given that it takes the form above.
If σ ∈ S2d, write δσ for the map T d(End(V )) ⊗W → K corresponding to γσ– and write H for the map T d(End(V ))⊗W → K which corresponds to H∗.We claim that, for each σ ∈ S2d,
δσ(A1 ⊗ · · · ⊗ Ad ⊗ v1 ⊗ · · · ⊗ v2d) =d∏i=1
(vσ(2i−1), Aivσ(2i)) (6.32)
for A1 ⊗ Ad ∈ T d(End(V )) and v1 ⊗ · · · ⊗ v2d ∈ W .
To prove this claim, let us choose an orthonormal basis b1, . . . , bn of V , i.e. abasis such that (bi, bj) = δij for 1 ≤ i,≤ n. Recall our notation of ϕv for themap to which v ∈ V corresponds in the dual space V ∗, and our canonical mapξ∗ defined in Lemma 6.27(ii), which in this case gives
ξ∗(ϕbi ⊗ ϕbj) : v 7→ (bj, v)bi.
Write Eij = ξ∗(ϕbi ⊗ ϕbj) ∈ End(V ). Since the claim is linear in each Aj, itsuffices to take Aj = Ei2j−1,i2j .
In this case, we have
A1 = Ei1,i2 = ξ∗(ϕbi1 ⊗ ϕbi2 )
. . .
Ad = Ei2d−1,i2d = ξ∗(ϕbi2d−1⊗ ϕbi2d )
and so
A1⊗ · · · ⊗ Ad ⊗ v1 ⊗ · · · ⊗ v2d
= ((ξ∗)⊗d ⊗ id)(ϕbi1 ⊗ ϕbi2 ⊗ · · · ⊗ ϕbi2d ⊗ v1 ⊗ · · · ⊗ v2d
)= ((ξ∗)⊗d ⊗ id)(α),
INVARIANT THEORY 91
where
δσ(α) =2d∏j=1
(bij, vσj) (6.33)
But with this choice of Aj, the right-hand side of (6.32) reduces tod∏i=1
(vσ(2j−1), Ei2j−1,i2jvσ(2j)) =d∏j=1
(vσ(2j−1), (vσ(2j), bi2j)bi2j−1
)
=d∏j=1
(vσ(2j−1), bi2j−1)(vσ(2j), bi2j)
=2d∏j=1
(bij , vσ(j)),
which equals the right-hand side of (6.33). This proves the claim in (6.32).
Taking A = I, the identity map on V , we see that
ϕ(v) = fϕ(I,v)
= H(I tI,v)
= H(I tI ⊗ · · · ⊗ I tI ⊗ v)
= H[((ξ∗)⊗d ⊗ id)−1(I tI ⊗ · · · ⊗ I tI ⊗ v)]
= H(v),
say, under a canonical association. Then we have shown that
ϕ =∑σ∈S2d
aσδσ,
i.e. ϕ is a K-linear combination of the δσ as asserted, since we took ϕ to bean arbitrary element of (W ∗), invariant under the O(V )-action. �
6.6. First formulation of O(V )
Let V, (·, ·) be as above. We wish to find a presentation for the space EndO(V )V⊗r.
As always, we have the map
µ : KSr → EndO(V )V⊗r
σ 7→ (v1 ⊗ · · · ⊗ vr 7→ σ · v1 ⊗ · · · ⊗ vr).
But by comparing our two fundamental theorems we have discussed in thischapter, we can conclude that EndO(V )V
⊗r % µ(KSr), i.e. that there areextra elements in EndO(V )V
⊗r which are not in µ(KSr). Our goal in thissection is to describe these elements.
92 INVARIANT THEORY
Lemma 6.34. Let V be K-vector space with dimension n, with basis {bi}ni=1,and let {βi}ni=1 be the corresponding dual basis for V ∗. Then the element γdefined by
γ =n∑i=1
bi ⊗ βi ∈ V ⊗ V ∗
has the following properties:
(i) γ is independent of the choice of basis;(ii) γ is GL(V )-invariant; and(iii) γ corresponds to idV under the canonical map ξ : V ⊗ V ∗ → End(V )
If, further, V is an orthogonal space with orthogonal form (·, ·), and orthonor-mal basis {bi}ni=1 to which {βi}ni=1 is dual, then the element γ0 defined by
γ0 =n∑i=1
bi ⊗ βi
has the following properties:
(i) γ0 is independent of the choice of orthonormal basis;(ii) γ0 is O(V )-invariant; and(iii) γ0 corresponds to idV under the map ξ′ = ξ ◦ η, where η is defined by
η : V ⊗ V → V ⊗ V ∗
v ⊗ w 7→ v ⊗ ϕw.
Proof . Observe that since by definition of our canonical map
ξ(v ⊗ ϕ) : x 7→ 〈ϕ, x〉v,we have
ξ(γ)(x) =n∑i=1
ξ(bi ⊗ βi)x
=n∑i=1
〈βi, x〉bi
= x.
Hence ξ(γ) = idV , and we have
ξ(gt) = g−1ξ(t)g for all g ∈ GL(V ).
Since g−1idV g = idV for all g, we have gγ = γ for all g ∈ GL(V ).
For the second claim, observe that under the map
ξ′ : V ⊗ V → V ⊗ V ∗ → End(V )
γ0 7→n∑i=1
bi ⊗ ϕbi 7→ idV
INVARIANT THEORY 93
where {ϕbi} is another notation for the basis of V ∗ dual to {bi}.
If g ∈ O(V ), then
g : v ⊗ w 7→ gv ⊗ gw 7→ gv ⊗ ϕgw = gv ⊗ gϕw 7→ gξ(v ⊗ ϕw)g−1,
and γ0 is an O(V )-invariant element of V ⊗2 as claimed, since gϕv = ϕg∗v,where g∗ = (g−1)t = g if g ∈ O(V ). �
We have a linear version of the first fundamental theorem for O(V ), and wewant an algebra version, i.e. a version which applies to associative, non-commutative algebras, that is, we want to describe EndO(V )(V
⊗r) ) µr(KSr),as we have seen.
Define a map e : V ⊗ V → V ⊗ V by
e(v ⊗ w) = (v, w)γ0.
Corollary 6.35. e ∈ EndO(V )V⊗2.
Proof . If g ∈ O(V ) then
ge(v ⊗ w) = (v, w)gγ0 = e(v ⊗ w)
eg(v ⊗ w) = (gv, gw)gγ0 = e(v ⊗ w),
whence g · e = e · g, so e commutes with the O(V )-action and e ∈ EndO(V )V⊗2.�
This provides us with r − 1 elements e1, e2, . . . , er−1 ∈ EndO(V )V⊗r, where ei
acts on factors i, i+ 1 of V ⊗ V ⊗ · · · ⊗ V︸ ︷︷ ︸r times
.
6.7. The Brauer algebra and the first fundamental theorem
We shall study the subalgebra E of EndV ⊗r which is generated by µ(KSr)and e1. We have (among others) the following relations:
σi = µ(i, i+ 1)
e2i = nei.
since
e2(v ⊗ w) = (v, w)eγ0 = (v, w)e( n∑i=1
bi ⊗ bi)
= n(v, w)γ0.
We also have the relationeiei±1ei = ei,
94 INVARIANT THEORY
called a Temperley-Lieb relation. We give a proof of this relation by showingthat e1e2e1 = e1 – the general case is no more difficult.
v1 ⊗ v2 ⊗ v3e1−→ (v1, v2)
∑i
bi ⊗ bi ⊗ v3
e2−→ (v1, v2)∑i,j
bi ⊗ (bi, v3)bj ⊗ bj
= (v1, v2)∑j
v3 ⊗ bj ⊗ bj
e1−→ (v1, v2)∑j,k
(v3, bj)bk ⊗ bk ⊗ bj
= (v1, v2)∑k
bk ⊗ bk ⊗ v3
= e1(v1 ⊗ v2 ⊗ v3).
Proposition 6.36. The following relations hold in E. The ei satisfy
e2i = nei,
eiei±1ei = ei,
eiej = ejei if |i− j| ≥ 2,
collectively called the Temperley-Lieb relations. We also have
σ2i = 1
σiσj = σjσi if |i− j| ≥ 2
σiσi+1σi = σi+1σiσi+1,
which are the relations which define the symmetric group, and finally the mixedrelations
σiei = eiσi = ei
σi+1eiσi+1 = σiei+1σi
σiej = ejσi if |i− j| ≥ 2.
Remark 6.37. These relations do not fully define the endomorphism algebra.While they clearly define an algebra, this algebra, known as the Brauer algebrais not semisimple. However we will find a surjective homomorphism from theBrauer algebra onto our endomorphism ring.
We may describe the action of E by diagrams:
bb
bb
bb
rr
rr
rrrb
rbr
b · · ·bb
bb
rr
rr
bb
bb
Such diagrams can be concatenated:
INVARIANT THEORY 95
This is a diagrammatic way of representing the symmetric group, and its groupalgebra is simply the algebra of all formal linear combinations of such diagramswith concatenation as its multiplication.
Let
A (Brauer) diagram is any way of joining dots in the two rows of a diagram inpairs. The number of Brauer diagrams on r dots is
(2r − 1)(2r − 3) · · · 5 · 3 · 1 = (2r)!!.
The Brauer algebra Br(n;K) is defined as the K-linear span of the Brauerdiagrams with multiplication given by concatenation, remembering we delete
96 INVARIANT THEORY
interior circle and replace with n. We have seen
dimK Br(n) = (2r)!!
Proposition 6.38. Let Br(n) be the Brauer algebra defined above.
(i) Br(n) is defined by {s1, . . . , sr, f}.(ii) If fi = τifτi where τi = (1 i)(2 i+ 1), i.e.
then the relations in Proposition 6.36 are satisfied with si replacing σiand fi replacing ei.
(iii) Br(n) is generated by s1, . . . , sr−1, f1, . . . , fr−1 subject to the relationsin Proposition 6.36.
(iv) For s = 0, 1, . . . ,[r2
](integer part), let
then
Br(n) =
[ r2 ]⊕s=0
KSrasKSr,
where KSr = 〈s1, . . . , sr−1〉 is thought of as a subalgebra of Br(n).
Proof . (Sketch) Let’s prove (iv) first. Every diagram has a certain number ofthrough strings and a certain number of horizontal arcs.
KSrasKSr is the linear span of diagrams with r−2s through strings. (i) alsofollows by a similar argument. For (ii) we need to check the relations – forexample, we check that s2f1s2 = s1f2s1.
INVARIANT THEORY 97
Finally, we check (iii). To show there are no relations other than those fromProposition 6.36, we show that the dimension of the algebra which is generatedby 〈s1, . . . , sr−1, f1 | relations in Proposition 6.36〉 is what we expect. (iv) givesan inductive way to prove this. �
Corollary 6.39. There is a surjective homomorphism Br(n)→ E given by
si 7→ σi
fi 7→ ei.
Proof . This follows immediately because E satisfies more relations than Br(n).�
Consider the diagram below, which we call (∆):
Br(n)
T 2r(V ) End(V ⊗r) ∼= (End(V ))⊗r
ν µ
A
98 INVARIANT THEORY
Where we recall that ξ(v ⊗ w) ∈ End(V ),
ξ(v ⊗ w) : x 7→ (w, x)v,
A is defined by
A(v1 ⊗ · · · ⊗ vr ⊗ w1 ⊗ · · · ⊗ wr) = ξ(v1, w1)⊗ · · · ⊗ ξ(vr, wr),
and we define ν as follows. Take a diagram D ∈ Br(n), and define
ν(D) = tD ∈ V ⊗2r,
wheretD =
∑i1,...,i2r
ij=ik if (j, k) is an edge of D
bi1 ⊗ · · · ⊗ bi2r ,
given that the vertices in the top row of a Brauer diagram are labelled 1, 2, . . . , rand the bottom row r + 1, r + 2, . . . , 2r.
For example,
Sotid =
∑bi1 ⊗ · · · ⊗ bir ⊗ bi1 ⊗ · · · ⊗ bir .
Note that
[v1 ⊗ · · · ⊗ v2r, tid] = (v1, vr+1)(v2, vr+2) · · · (vr, v2r).
We aim to prove that the diagram above is commutative. We begin by makingthe following observation.
Lemma 6.40. Let D ∈ Br(n).
(i) [v1 ⊗ · · · ⊗ v2r, tD] =∏
(i,j) edge of D(vi, vj).(ii) tD ∈ (V ⊗2r)O(V ).(iii) imν = (V ⊗2r)O(V ).
Proof . (i) follows from the fact that
[v ⊗ w, γ0] = (v, w)
(this is a way of expressing Parseval’s identity). (ii) we already have observed,since taking the inner product with tD is a function which is invariant un-der O(V ) since the form is invariant under O(V ). (iii) follows from the firstfundamental theorem. �
INVARIANT THEORY 99
Next, observe that in the diagram (∆), each space is a module for Sr ×Sr ⊂S2r. The commutativity of the diagram is not immediately obvious, since E isnot immediately defined as a permutation (?). The actions of Sr ×Sr on therespective spaces are given by, for (σ, τ) ∈ Sr ×Sr:
(σ, τ) ·D 7→ σDτ−1, (D ∈ Br(n))
(σ, τ) · (v ⊗w) 7→ µ(σ)v ⊗ µ(τ)w, v,w ∈ T r(V )
(σ, τ) · A 7→ µ(σ)Aµ(τ)−1, A ∈ End(V ⊗r),
where σDτ−1 is the diagram given by
Lemma 6.41. Each of the maps in (∆) respects the action of Sr ⊗Sr.
Proof . We must prove three things. First, µ respects the Sr × Sr-actionbecause it is an algebra homomorphism. Secondly, A respects the Sr × Sr-action, since taking t = v ⊗w ∈ T 2r(V ) we see
A(t) = ξ(v1, w1)⊗ ξ(v2, w2)⊗ · · · ⊗ ξ(vr, wr)
: x 7→r∏i=1
(wi, xi)v1 ⊗ vr
A((σ, τ) · t) = ξ(vσ−11, wτ−11)⊗ · · · ⊗ ξ(vσ−1r, wτ−1r)
: x 7→∏i
(wτ−1i, xi)µ(σ)v
=∏i
(wi, xτi)µ(σ)v
= µ(σ)A(t)µ(σ−1)x.
Hence A((σ, τ) · t) = µ(σ)A(t)µ(τ)−1.
Finally we must show that ν commutes with (σ, τ), i.e. we need to check thatthe edges of σDτ−1 are the same as the indices in (σ, τ)tD which are equal.We have three cases.
Case (a): suppose j, k are in the top row of D.
(j, k) is moved by (σ, τ) to (σ−1j, σ−1k) in (σ, τ)D, and µ(σ) ⊗ µ(τ)−1tD hasequal indices with positions σ−1j, σ−1k.
100 INVARIANT THEORY
Case (b): suppose j, k are in the bottom row of D.
Case (c): suppose j, k are in distinct rows of D. �
Theorem 6.42. The diagram (∆) is commutative.
Proof . We must show that for all diagrams D ∈ Br(n),
µ(D) = Aν(D).
We have seen that the Brauer algebra
Br(n) =⊕s
KSra(s)KSr,
where a(s) is the diagram defined on p96.
i.e. every diagram is of the form (σ, τ)a(s) for some σ, τ ∈ Sr. By Lemma6.40,
µ((σ, τ) ·D) = (σ, τ) · µ(D)
andAν((σ, τ)D) = (σ, τ)Aν(D).
Hence it suffices to take D = a(s). As a warm-up, we take D = id. Then
Aν(D) = A(tid)
=∑
ξ(bi1 , bi1)⊗ ξ(bi2 , bi2)⊗ · · ·⊗= id
= µ(a(0)) = µ(D).
Having done the case s = 0, we now proceed to general s. Note that weobserved above that
e =n∑
i,j=1
ξ(bi, bj)⊗ ξ(bi, bj).
Therefore,
µ(a(s)) =∑i1,...,isj1,...,js
k1,...,kr−2s
ξ(bi1 , bj1)⊗· · ·⊗ξ(bir , bjr)⊗ξ(bk1 , bk1)⊗· · ·⊗ξ(bkr−2s , bkr−2s).
This equals A(tD), where D = a(s). �
Corollary 6.43 (First fundamental theorem, first formulation). The map µ :Br(n)→ End(V ⊗r) has image EndO(V )V
⊗r.
Proof . The commutativity of the diagram (∆) shows that
A(imν) = imµ.
But by the first fundamental theorem in the form we already have,
imν = (T 2r(V ))O(V ),
INVARIANT THEORY 101
and since A is an O(V )-isomorphism, the result follows. �
6.8. The second fundamental theorem
It is a consequence of Theorem 6.42 that ker ν = kerµ. We start by findingsome elements of kerµ. Since for v,w ∈ T r(V ) the map v ⊗w 7→ det(vi, vj)is zero if r > n, i.e. ∑
σ∈Sr
(sgn σ)µ(σ) ∈ kerµ.
Take r = n+ 1 (it is easy to show that if r ≤ n we have an isomorphism). Let{1, . . . , 2r} be split into two disjoint subsets S, S ′ of equal cardinality, i.e.
S = {i1 < . . . < ir}, S ′ = {j1 < . . . < jr}.Then the element ∑
π∈Sr
(sgn π)r∏
k=1
(vik , vjπk) = 0
by its interpretation as the determinant of the inner products of a set of linearlyindependent vectors (there are r > n = dimV of them).
Proposition 6.44. Let S, S ′ be as above. Let D(S, π) be the diagram withedges (ik, jπk) for k = 1, . . . , r. Then the element
K(S) =∑π∈Sr
(sgn π)D(S, π) ∈ kerµ.
Conjecture 6.45. Let
S ={
1, 2, . . . ,[r
2
],[r
2
]+ 1 + r,
[r2
]+ 2 + r, . . . , 2r
}1.
Then K(S) generates kerµ in Br(n).
Remark 6.46. This is true for n = 2 and 3 (n = 2 is fairly easy). The resultfor n = 3 was proved by Lehrer and Zhang. The proof for general n is still anopen problem.
Exercise 6.47. Prove that γ0 is O(V )-invariant, using matrices.
Exercise 6.48. Prove Proposition 6.36.
Exercise 6.49. Prove cases (b) and (c) in the proof of Lemma 6.40.
1Notice that S has r elements, and S′ is defined as its complement in {1, 2, . . . , 2r}.
Appendix A
Semisimple modules and double centraliser theory
In this appendix we give a brief survey of relevant results we need in Chapters3 and 4 from the theory of semisimple modules and double centraliser theory.
A1. Semisimple modules
Let A be a ring, and recall the following definition of an A-module.
Definition A.1. An A-module is an abelian group M with an additionaloperation A×M →M : (a,m) 7→ am called the A-action which satisfies
a(m+ n) = am+ an, for all a ∈ A,m, n ∈M(a+ b)m = am+ bm, for all a, b ∈ A,m ∈M
(ab)m = a(bm), for all a, b ∈ A,m ∈M1m = m, for all m ∈M.
We shall assume in this appendix that all our modulesM satisfy the descendingchain condition (DCC), i.e. that any chain of submodules
M0 ⊇M1 ⊇M2 ⊇ · · ·
stabilises, so there exists some N such that MN = MN+k for all k ≥ 1.
Definition A.2. M is a simple module if its only submodules are 0 and M .
Proposition A.3. Let A be a ring and M an A-module. Then the followingare equivalent.
(i) M is the sum of a family of simple submodules {Mα}.(ii) M is the direct sum of a family of simple submodules.(iii) For any submodule M0 of M , there exists a submodule M1 such that
M = M0 ⊕M1.
Proof . (i)⇒(ii) Suppose that
M =∑α∈S
Mα,
102
INVARIANT THEORY 103
and let T be a maximal subset of S such that the sum∑
α∈T Mα is direct, i.e.that for all α0 ∈ T , (∑
α6=α0α∈T
Mα
)∩Mα0 = 0.
Then M ′ = ⊕α∈TMα is a submodule of M . Let β ∈ S. We want to show thatMβ ⊆ M ′. If not, then Mβ ∩M ′ = 0 because Mβ is simple. So T t {β} ) Tis a larger subset of S with corresponding sum of Mα direct, contradicting themaximality of T . Hence we must have Mβ ⊆M ′.
(ii)⇒(iii) Let
M =⊕α∈T
Mα,
and suppose M0 is a submodule of M . If M0 6= M there exists some α suchthatM0∩Mα = 0. Let T ′ be a subset of T , maximal subject to the requirementthat the sum
M0 +∑β∈T ′
Mβ
is direct. Then we can argue as above to show that M0 +∑
β∈T ′Mβ = M , andwriting M1 =
∑β∈T ′Mβ we get M = M0 ⊕M1.
(iii)⇒(i) First note that by the descending chain condition, any nonzero sub-module of M contains a simple submodule. Let M0 be the sum of all thesimple submodules of M . Then M = M0 ⊕M1 by assumption, and if M2 is asimple submodule of M1 then M2 ⊂M0. So M1 = 0. �
Definition A.4. If any and hence all of the conditions of Proposition A.3 aresatisfied we say M is a semisimple module.
Remark A.5. There exist many modules which are not semisimple, for ex-ample F2[Z/2Z], the group algebra for Z/2Z over the two-element field.
Lemma A.6. Any submodule or quotient module of a semisimple module issemisimple.
Remark A.7. For semisimple modules, a submodule is always isomorphic toa quotient module, for ifM0 ⊂M is a submodule thenM = M0⊕M1 for someM1 and hence M0
∼= M/M1.
A2. Double centraliser theory
Let M be a semisimple A-module and writeB = EndA(M) = {f : M →M | f(am) = af(m) for all a ∈ A, m ∈M}.
This is a ring under composition and pointwise addition of functions, withidentity the identity map. Also, M is a B-module.
104 INVARIANT THEORY
Consider the ring EndB(M). Then we have a ring homomorphism
µ : A→ EndB(M)
a 7→ (ξa : m 7→ am).
We say that the pair (A,M) has the double centraliser property if µ is surjec-tive, that is, if
B = EndA(M) and µ(A) = EndB(M)
where µ(A) is the image of A in End(M). The reason for the name is becausethe endomorphisms defined by A have precise centraliser the endomorphismsdefined by B, and vice versa.
Lemma A.8. Let M be a semisimple A-module and let f ∈ EndB(M), whereB = EndA(M) and let x ∈ M . Then there exists some a ∈ A such thatf(x) = ax.
Proof . Since Ax is a submodule of M and M is semisimple, we know thatM = Ax ⊕ M ′ for some submodule M ′ of A. Let p : M → Ax be thecanonical projection map with kernel M ′. Then p ∈ EndA(M) = B (sincep(a′(ax + m′)) = p(a′ax + a′m′) = a′ax = a′p(ax + m′)). Since fp = pf , wehave
fp(x) = f(x) = p(f(x)) ∈ Ax,i.e. f(x) = ax for some a. �
Theorem A.9 (Jacobson). LetM be a semisimple A-module and B = EndA(M).Let x1, . . . , xn be any finite set of elements of M . Let f ∈ EndB(M). Thenthere exists a ∈ A such that f(xi) = axi for all i.
Proof . WriteM (n) = M ⊕M ⊕ · · · ⊕M︸ ︷︷ ︸
n times,
where the A-action on M (n) is given by a · (m1, . . . ,mn) = (am1, . . . , amn).Write, for m ∈M ,
m(i) = (0, . . . , 0,m, 0, . . . , 0) (ith position).
Then M (n) 3 (m1, . . . ,mn) =∑n
i=1 m(i)i . Let φ : M (n) → M (n) be an element
of EndA(M (n)). Write
φ(m(i)) = (φi1(m), φi2(m), . . . , φin(m)),
where φij : M →M . Then
φ(m1, . . . ,mn) = φ( n∑i=1
m(i)i
)=
n∑i=1
(φi1(mi), . . . , φin(mi)
).
INVARIANT THEORY 105
Hence we have essentially decomposed a map M (n) → M (n) into an n × nmatrix of maps M → M . We claim that φ ∈ EndA(M (n)) if and only ifφij ∈ EndA(M) for all i, j.
Since φ(am(i)) = aφ(m(i)), it follows by the definition of the action that foreach i = 1, . . . , n, we have
(φi1(am), . . . , φin(am) = a(φi1(m), . . . , φin(m))
= (aφi1(m), . . . , aφin(m)),
whence φij(am) = aφij(m) for all i, j, i.e. φij ∈ EndA(M). Conversely, givenfunctions ψij ∈ EndA(M) for all i, j, we must show that the map ψ defined by
ψ(m1, . . . ,mn) =n∑i=1
(ψi1(mi), . . . , ψin(mi)
)lies in EndA(M (n)).
So we have shown that EndA(M (n)) ∼= Matn(EndA(M)). Now let f ∈ EndB(M).Then the map f : M (n) →M (n) defined by
f(m1, . . . ,mn) =(f(m1), . . . , f(mn)
)lies in EndA(M (n)). Let m = (m1, . . . ,mn) ∈M (n) – then
ψ(am) = ψ(a · (m1, . . . ,mn))
= ψ(am1, . . . , amn)
=n∑i=1
(ψi1(ami), . . . , ψin(ami)
)=
n∑i=1
(aψi1(mi), . . . , aψin(mn)
)=
n∑i=1
a(ψi1(mi), . . . , ψin(mn)
)= aψ(m)
by definition of the A-action on M (n), and since ψij ∈ EndA(M) for alli, j. Hence ψ ∈ EndAM
(n). We now apply Lemma A.8 to the elementx = (x1, . . . , xn) ∈ M (n). Then if we define a function f ∈ EndAM
(n) byf(m1, . . . ,mn) = (f(m1), . . . , f(mn)), we have
f(x1, . . . , xn) = a · (x1, . . . , xn),
for some a ∈ A, since f ∈ EndAM(n), by Lemma A.8. But f(x1, . . . , xn) =
(f(x1), . . . , f(xn)) and so
(f(x1), . . . , f(xn)) = (ax1, . . . , axn)
by definition of our A-action on M (n), and we are done. �
106 INVARIANT THEORY
Corollary A.10 (Double centraliser theorem). Let Z(A) be the centre of thering A, i.e.
Z(A) = {z ∈ A | az = za for all a ∈ A}.Let M be a semisimple A-module which is finitely-generated as a Z(A)-module.Let B = EndA(M); then (A,M) has the double centraliser property, i.e. themap µ : A→ EndB(M) defined above is surjective.
Proof . We are given that M is a finitely-generated Z(A)-module, hence thereexist x1, . . . , xn ∈ M such that M = Z(A)x1 + Z(A)x2 + . . . + Z(A)xn. Letf ∈ EndB(M). We must show that f = ξa for some a ∈ A (using the notationof p104. By Theorem A.9, since M is a semisimple A-module, there existsa ∈ A such that f(xi) = axi for all i. But any element x ∈ M is of the formx = z1x1 + z2x2 + . . .+ znxn with zi ∈ Z(A). Therefore
f(x) = f(z1x1) + . . .+ f(znxn)
= z1f(x1) + . . .+ znf(xn)
= z1ax1 + . . .+ znaxn
= a(z1x1 + . . .+ znxn)
= ax
= ξa(x)
since zi ∈ Z(A). Hence f = ξa and so µ is surjective. �
Example A.11. We give some simple examples of pairs (A,M) which havethe double centraliser property.
(i) If A is a K-algebra with identity, then K = K1 ⊆ Z(A) and if M is asemisimple A-module which is finite-dimensional as a K-vector space,then in particularM is a finitely-generated Z(A)-module and so (A,M)has the double centraliser property by Corollary A.10.
(ii) If G is a finite group and K has characteristic zero, then any finite-dimensionalKG-module (A = KG is the group ring overK) is semisim-ple by Maschke’s theorem – if M is a finite-dimensional KG-moduleandM1 ⊆M is a submodule, then there is a KG-invariant complementM2 such that M = M1 ⊕M2.
(iii) If V is a K-vector space of dimension n, then M = V ⊗r is a KSr-module, and is semisimple (by (ii)). Therefore if B = EndKSrV
⊗r,then µ : KSr → EndBV
⊗r is surjective. This is the result which weuse in Chapter 3.
A3. Semisimple rings
We include, mainly for the sake of completeness, a brief discussion of semisim-ple rings. In particular we give the structure theorem for semisimple algebras.
INVARIANT THEORY 107
Definition A.12. The ring A is said to be semisimple if A regarded as aleft-module over itself, denoted AA, is a semisimple A-module.
Lemma A.13. If A is a semisimple ring, then every A-module is semisimple.
Proof . Every A-module is a quotient of a free A-module, which is of the form
F =n⊕i=1
AA
for some n. The result then follows from Lemma A.6. �
Definition A.14. Let A be a semisimple ring. Then we call a (left) ideal Lof A minimal if it itself contains no nontrivial ideals.
Remark A.15. Notice that minimal ideals of a semisimple ring A are preciselythe simple submodules of A when considered as a left module over itself.
Lemma A.16. Suppose A is a semisimple ring and L a minimal left ideal ofA. If M is any simple A-module and M 6∼= L as an A-module, then LM = 0.
Proof . If LM 6= 0 then there exists m ∈ M such that Lm 6= 0, and since Mis simple, M = Lm. But then the map l 7→ lm is an A module isomorphismfrom L to M , which gives a contradiction. �
Corollary A.17. Any simple A-module is isomorphic to L for some minimalideal L.
Proof . A is isomorphic to a sum of minimal ideals since it is semisimple. Hencewe have
A =n⊕i=1
Li.
Let M be a simple A-module. Then if M 6∼= Li for all i, we have LiM = 0 forall i by Lemma A.16. Hence
AM =n⊕i=1
LiM = 0,
so M = 0 since 1 ∈ A. �
Let A be a semisimple ring. We know that A =∑
L L, where L are minimalleft ideals of A. Suppose the isomorphism classes of minimal left ideals L (asA-modules) are indexed by I, i.e. suppose {Mi}i∈I is a list of isomorphismclasses of minimal left ideals for A. Then for i ∈ I, let
Ai =∑L∼=Mi
L.
Then A =∑Ai and AiAj = 0 for i 6= j by simplicity, because if L 6∼= L′,
LL′ = 0 by Lemma A.16.
108 INVARIANT THEORY
Theorem A.18 (Structure theorem of semisimple algebras). Let A be a semisim-ple ring.
(i) The left ideals Ai are two-sided (i.e. they are also right ideals).(ii) There is a finite subset {1, . . . , r} of I such that
A =r⊕i=1
Ai, Ai = Aei,
where ei is a central idempotent of A (so e2i = ei and ei ∈ Z(A)).
(iii) The Ai have just one isomorphism class of simple modules (Ai is asimple ring)
(iv) If ei is the idempotent above, then ei is the unit of Ai and 1 = e1 + e2 +. . .+ er, and eiej = δijei.
Proof . We have Ai ⊆ AiA = Ai(∑Aj) ⊆ AiAi ⊆ Ai since Ai is a left ideal.
Hence equality holds throughout, and AiA ⊆ Ai, i.e. Ai is also a right ideal,and so is two-sided. Next, notice that since A =
∑Ai,
1 = e1 + . . .+ er,
for some r, where ei ∈ Ai. For any element a ∈ A,
a = a1 =∑
aei ∈r∑i=1
Ai,
and so A =∑r
i=1Ai. By Lemma A.16, the sum is direct.Each Ai is simple, since if M is a simple Ai-module, then LM 6= 0 for someminimal ideal L of A contained in Ai, and hence L ∼= M as A-modules andhence as Ai-modules.Finally, notice that 1 =
∑ei implies ej =
∑ejei = e2
j . The ei are obviouslycentral since a1 = 1a. �
Remark A.19. Suppose A ∼= Matn(D) where D is a skew-field, or divisionring. Then among the minimal left ideals are column matrices with but onenonzero column – clearly A is the direct sum of these.
Exercises
Exercise A.20. Prove Lemma A.6 using Proposition A.3.
Exercise A.21. Prove the converse of the claim in the proof of Theorem A.9.
Exercise A.22. Show that the algebra A has a single isomorphism class ofsimple modules if and only if A has no nontrivial two-sided ideals (i.e. the Aiabove are minimal two-sided ideals).
INVARIANT THEORY 109
Exercise A.23. Let A be a ring and let M be a finitely generated free A-module. Show that (A,M) has the double centraliser property. Is M alwayssemisimple? Give an example to justify your answer.
References
[1] M. Atiyah and I. Macdonald, Introduction to Commutative Algebra,Addison-Wesley, 1969.
[2] C. Curtis and I. Reiner, Representation Theory of Finite Groups and As-sociative Algebras, Interscience Publications, 1962.
[3] H. Kraft and C. Procesi, Classical Invariant Theory: A Primer, Prelimi-nary version, 1996.
[4] S. Lang, Algebra, Springer-Verlag Graduate Texts in Mathematics, 2002(3rd edition).
[5] B. Sagan, The Symmetric Group: representations, combinatorial algo-rithms and symmetric functions, Springer-Verlag, 2001.
[6] B.L. van der Waerden, Algebra, Springer-Verlag, 1991 (English translationof 1966 German edition).
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