WEYL’S CONSTRUCTION OF THE IRREDUCIBLE REGULAR REPRESENTATIONS OF THE COMPLEX CLASSICAL GROUPS
A Thesis Submitted to the Faculty of Graduate Studies and Research In Partial Fulfillment of the Requirements for the Degree of Master of Science In Mathematics University of Regina
By Mart´ınGabriel Chaktoura Regina, Saskatchewan June 2014
c Copyright 2014: Mart´ınGabriel Chaktoura
UNIVERSITY OF REGINA
FACULTY OF GRADUATE STUDIES AND RESEARCH
SUPERVISORY AND EXAMINING COMMITTEE
Martin Gabriel Chaktoura, candidate for the degree of Master of Science in Mathematics, has presented a thesis titled, Weyl’s Construction of the Irreducible Regular Representations of the Complex Classical Groups, in an oral examination held on June 10, 2014. The following committee members have found the thesis acceptable in form and content, and that the candidate demonstrated satisfactory knowledge of the subject material.
External Examiner: *Dr. Gerald Cliff, University of Alberta
Supervisor: Dr. Fernando Szechtman, Department of Mathematics & Statistics
Committee Member: **Dr. Bruce Gilligan, Department of Mathematics & Statistics
Committee Member: Dr. Allen Herman, Department of Mathematics & Statistics
Chair of Defense: Dr. Philip Charrier, Department of History
*Via Tele Conference **Not present at defense
Abstract
The main objective of this work is to construct all the regular representations of the complex classical groups. Since each of these groups is reductive, in the sense that every regular module is a direct sum of irreducible regular modules, it suffices to construct the irreducible ones. We achieve this by using Weyl’s method, which provides an explicit and concrete realization of each of the desired modules. Chapters 1 to 3 contain background information. The reader may choose to skip them and consult them only when necessary. Chapters 1 and 3 describe the relationship between the classical groups and their Lie algebras, with emphasis on the representation theory aspects of such objects. Chapter 2 summarizes many important results on the representation theory of complex semisimple Lie algebras. The actual work begins in Chapter 4, where we use Young diagrams and tableaux to construct all the irreducible complex representations of the symmetric groups. Surprisingly enough, these modules serve as building blocks to obtain all the desired representations of the complex classical groups, which is done in Chapters 5 through 7.
i Acknowledgments
I would like to express my deep gratitude to Dr. Natalio H. Guersenzvaig, who encouraged me to continue my studies in Mathematics. It was a great privilege to work under the supervision of Dr. Fernando Szechtman. I sincerely appreciate his guidance, support, and insightful ideas. Special thanks are extended to my profes- sors in Regina: Dr. Douglas Farenick, Dr. Bruce Gilligan, Dr. Allen Herman, and Dr. Karen Meagher. Their lectures were wonderful and inspiring. My thanks are due Dr. Juliana Erlijman and Dr. Leandro Cagliero for providing me with highly valuable references and comments. I am truly grateful towards the Faculty of Graduate Stud- ies and Research of the University of Regina and the Government of Saskatchewan for their financial support. Many kind people were always willing to help me along the way, both inside and outside university. I specially want to thank: the Argeramis, the Filomenos, the Londo˜nos,the Zemlaks, Felipe Ramos, Rebeca Casas, Guillermo Vidal, and Gustavo Krimker.
ii Contents
Abstract i
Acknowledgments ii
Table of Contents iii
1 Closed Linear Groups 1 1.1 Examples ...... 2 1.2 Topological Groups ...... 5 1.3 The Exponential Map ...... 7 1.4 The Lie Algebra of a Closed Linear Group ...... 9 1.5 Closed Linear Groups as Lie Groups ...... 15 1.6 The Differential of a Smooth Homomorphism ...... 16
2 Representations of Semisimple Lie Algebras 19 2.1 Root Space Decomposition ...... 19 2.2 Highest Weight Modules ...... 27
3 Linear Algebraic Groups 32 3.1 Definition and Examples ...... 32
iii 3.2 Regular Functions ...... 35 3.3 Regular Representations ...... 38 3.4 Differential of a Regular Representation ...... 43 3.5 Multiplicative Jordan Decomposition ...... 51 3.6 Proofs of Connectedness ...... 55 3.7 Reductive Groups ...... 60
4 Irreducible Representations of Sd 63 4.1 Young Diagrams and Tableaux ...... 64
4.2 Irreducible Modules for Sd ...... 66
5 Irreducible Regular Representations of SL(n) 73 5.1 The Schur Functor ...... 73 5.2 Finding Maximal Vectors ...... 80 5.3 Facts on Matrix Rings ...... 85 5.4 Facts on G-homomorphisms ...... 87 5.5 Proof of Irreducibility ...... 90
6 Irreducible Regular Representations of GL(n) 98 6.1 Irreducible Modules for gl(n)...... 99 6.2 Rational Characters ...... 102 6.3 Irreducible Regular Modules for GL(n)...... 105
7 The Symplectic and Orthogonal Cases 108 7.1 The Construction ...... 109 7.2 The Symplectic Case ...... 116
iv 7.3 Admissible and Associated Partitions ...... 121 7.4 The Odd Orthogonal Case ...... 123 7.5 On the Representations of a Subgroup of Index 2 ...... 131 7.6 The Even Orthogonal Case ...... 135
Bibliography 146
Appendix A Facts on Invariant Theory 148 A.1 Polynomial Invariants ...... 148 A.2 Polarization Maps ...... 151 A.3 General Strategy ...... 153 A.4 Polynomial Invariants of SL(n, F ) and GL(n, F )...... 161 A.5 Polynomial Invariants of Sp(n, F )...... 164 A.6 Polynomial Invariants of the Orthogonal Groups ...... 171 A.7 Useful Consequences ...... 176
v Chapter 1
Closed Linear Groups
Every complex classical group is a closed linear group, i.e., a closed subgroup of some GL(n, C). Each closed linear group can be assigned a Lie algebra, which is the tangent space at the identity element. This assignment is functorial in the sense that each smooth homomorphism between closed linear groups gives a homomorphism between the corresponding Lie algebras satisfying certain properties. The Lie algebra of a closed linear group encapsulates a great deal of information about the group. This relationship is particularly strong when the group is connected. Due to their linear structure, Lie algebras are much easier to study than closed linear groups and hence they become a valuable tool in the study of the structure and representation theory of closed linear groups. In this chapter we define the Lie algebra of a closed linear group and we record some important relations between them. The exponential map plays a central role, as it provides a powerful link between the algebra and the group. For the sake of conciseness, many proofs are omitted. In each case, a suitable reference is suggested.
1 1.1 Examples
The set of all n × n matrices with complex entries will be denoted by Mn(C). It is a vector space of dimension n2 over C with respect to usual matrix addition and mul- tiplication of a matrix by a scalar. If, in addition, we consider matrix multiplication,
n2 then Mn(C) becomes a C-algebra. The vector spaces Mn(C) and C are easily seen
2 to be isomorphic. Considering the standard (Euclidean) topology on Cn , we use this identification to induce a topology on Mn(C), called the standard topology on Mn(C).
Let GL(n) stand for the group of all invertible matrices in Mn(C). The continuity of the determinant function det : Mn(C) → C implies that
−1 GL(n) = det (C \{0})
is open in Mn(C). From the topological point of view, we always regard GL(n) as a subspace of Mn(C).
Definition 1.1.1. A closed linear group is a closed subgroup of GL(n).
Example 1.1.2. GL(n) is a closed linear group. In particular, so is C× = GL(1).
Example 1.1.3. The special linear group
SL(n) = {g ∈ GL(n) : det g = 1} is a closed linear group. Indeed, SL(n) = det−1({1}) is closed in GL(n) because {1} is closed in C.
2 Example 1.1.4. Let ω be a nondegenerate bilinear form on Cn. Let
n G(ω) = {g ∈ Mn(C): ω(gv, gw) = ω(v, w) for all v, w ∈ C } be the isometry group of ω. If Γ stands for the Gram matrix of ω relative to the
n standard basis of C , then G(ω) is the set of all g ∈ Mn(C) such that
gtΓg = Γ. (1.1)
Note that (1.1) implies det g 6= 0. In addition, (1.1) is equivalent to a set of n2 quadratic equations in the entries of g. Then G(ω) can be seen as the intersection of
2 the zero sets of n continuous functions Mn(C) → C. Thus G(ω) is closed in Mn(C) and hence in GL(n). Therefore G(ω) is a closed linear group.
Given p ∈ GL(n), define a nondegenerate bilinear form ω0 on Cn by
ω0(v, w) = ω(pv, pw).
So p−1G(ω)p = G(ω0). It follows that isometry groups of equivalent bilinear forms are always isomorphic. If ω is skew-symmetric then n is necessarily even, say n = 2`, and there is a basis of the space where ω has Gram matrix
0 I` J − = . (1.2) ` −I` 0
− When Γ is actually equal to J` , the group G(ω) is called symplectic and we denote
3 it by Sp(n). Suppose now that ω is symmetric. If n = 2` is even and Γ is the matrix
0 I` J + = , (1.3) ` I` 0 then G(ω) = O(n) is the even orthogonal group. If n = 2` + 1 is odd and Γ is
1 0 0 S` = 0 0 I , (1.4) ` 0 I` 0 then G(ω) = O(n) is the odd orthogonal group. Note that there is always a basis of
n + C relative to which ω has Gram matrix J` or S`.
Example 1.1.5. The special orthogonal group SO(n) is defined as O(n)∩SL(n). It is a closed linear group because O(n) and SL(n) are. In addition, SO(n) is a subgroup of index 2 in O(n).
Definition 1.1.6. The complex classical groups are GL(n), SL(n), Sp(n), O(n), and Sp(n). We just call them the classical groups.
∗ Example 1.1.7. For each matrix g ∈ Mn(C), let g denote the conjugate transpose of g. We say that g is unitary when g∗g = I. In this case g is invertible and g−1 = g∗.
The unitary group U(n) is the set of all n × n unitary matrices in Mn(C). The condition g∗g = I is equivalent to a set of 2n2 polynomial equations in the real and imaginary parts of the entries of g. Observe, however, that these equations are not polynomial in the entries of g because they involve complex conjugation. Reasoning
4 like in the previous example, we see that U(n) is a closed linear group. Finally, observe that U(1) is just the unit circle S1 in C×.
1.2 Topological Groups
Each subgroup G of GL(n) inherits a topology from GL(n). Moreover, the group operations in G turn out to be continuous maps. This compatibility between the algebraic and topological structures can be exploited to obtain a great amount of information about G. The natural generalization of such a situation leads to the notion of a topological group. Although we are only interested in the case of closed linear groups, in this section we study some useful properties of topological groups form an abstract point of view. Proofs are omitted. The reader may consult [HR79] or any textbook addressing the basic facts about topological groups.
Definition 1.2.1. Let G be a group that is also a Hausdorff topological space. Let
µ : G × G → G , η : G → G be the maps given by (x, y) 7→ xy and x 7→ x−1, respectively. We say that G is a topological group if both µ and η are continuous.
Example 1.2.2. (C, +) and (C×, ·) are topological groups.
Example 1.2.3. GL(n) is a topological group. Indeed, matrix multiplication is continuous because it is polynomial in each matrix entry. In addition, for each g in GL(n) we have A(g) g−1 = , det(g)
5 where A(g) stands for the adjoint matrix of g. This shows that matrix inversion is rational in each matrix entry, and hence continuous.
Lemma 1.2.4. Every subgroup of a topological group is a topological group with re- spect to the subspace topology.
Example 1.2.5. Using the previous lemma together with Examples 1.2.2 and 1.2.3 we obtain many examples of topological groups. In particular, any closed linear group is a topological group.
Proposition 1.2.6. Let G be a topological group and fix g ∈ G. Let Lg : G → G and
Rg : G → G be the left and right translations given by Lg(x) = gx and Rg(x) = xg, respectively. Then Lg and Rg are homeomorphisms.
Corollary 1.2.7. Let G be a topological group and fix g ∈ G. Then the inner auto- morphism τ : G → G defined by τ(x) = gxg−1 is a homeomorphism.
Lemma 1.2.8. Suppose G is a group with a Hausdorff topology. Let f : G × G → G be the map sending (x, y) 7→ xy−1. Then G is a topological group if and only if f is continuous.
Suppose G is a topological group with identity element e.
Definition 1.2.9. The connected component of G containing e is denoted by Ge and called the identity component of G.
Proposition 1.2.10. Ge is a normal subgroup of G. In addition, the cosets of Ge are the connected components of G.
Proposition 1.2.11. If U is an open neighbourhood of e in G such that U ⊂ Ge, then U generates Ge.
6 1.3 The Exponential Map
In this section we introduce the matrix exponential map, which is one of the most powerful tools in the study of closed linear groups. For any unproven statements, we refer the reader to [Kna02] or [GW09].
Consider the Euclidean norm || · || on the C-vector space Cn. Then Cn becomes
n a Banach space. Given a matrix A ∈ Mn(C), the function fA : C → R given by v 7→ ||Av|| is continuous. Since the closed unit ball in Cn is compact, we can define
n ||A|| = max{||Av|| : v ∈ C , ||v|| ≤ 1}.
This provides a norm on the C-vector space Mn(C). It can be shown that Mn(C) is a Banach algebra under this norm.
1 n Lemma 1.3.1. Let A ∈ Mn(C). Then the series of general term xn = n! A , n ≥ 0, is absolutely convergent.
Proof. It suffices to prove that the partial sums
k X Sk = ||xn|| n=1 are bounded above. For each k ∈ N,
k k X 1 X 1 S = ||An|| ≤ ||A||n ≤ e||A||. k n! n! n=0 n=0
Definition 1.3.2. Given a matrix A ∈ Mn(C), the previous lemma enables us to
7 define ∞ X 1 exp A = An. n! n=0
Thus we obtain a map exp : Mn(C) → Mn(C), called the matrix exponential map, or just the exponential map.
Proposition 1.3.3. Let A, B ∈ Mn(C). Then:
(i) If A and B commute, then (exp A)(exp B) = exp(A + B).
(ii) exp A is nonsingular.
(iii) The map ϕA : C → Mn(C) sending z 7→ exp(zA) is indefinitely differentiable and d exp(zA) = A exp(zA). dz
(iv) det(exp A) = etrA.
(v) The map Mn(C) → Mn(C) given by X 7→ exp X is analytic.
Note 1.3.4. Suppose that V is a finite dimensional vector space over C. If || · || is any norm on V , then End(V ) becomes a Banach algebra under the operator norm defined by ||T || = max{||T (v)|| : v ∈ V, ||v|| ≤ 1}.
Proceeding as above, we can define
∞ X 1 exp T = T n n! n=0 for T ∈ End(V ). Given a basis B of V , let MB : End(V ) → Md(C) be the map sending each endomorphism of V to its matrix relative to B. This map is a continuous algebra
8 isomorphism with continuous inverse. In particular,
exp ◦MB = MB ◦ exp .
Lemma 1.3.5. There exists an open ball U around 0 in Mn(C) and an open neigh- bourhood V of I in Mn(C) such that exp : U → V is bijective and analytic, with analytic inverse.
Given a matrix X ∈ Mn(C), consider the map ϕ : R → GL(n) given by
ϕ(t) = exp(tX).
By Proposition 1.3.3, ϕ is continuous and satisfies ϕ(s + t) = ϕ(s)ϕ(t) for all t, s ∈ R. Thus ϕ is a continuous homomorphism from the additive group of all real numbers into GL(n). It is a fundamental fact in Lie theory that every continuous homomorphism
R → GL(n) is obtained in this way.
Theorem 1.3.6. Let ϕ : R → GL(n) be a continuous group homomorphism. Then there exists a unique matrix X in Mn(C) such that ϕ(t) = exp(tX) for all t ∈ R.
1.4 The Lie Algebra of a Closed Linear Group
In this section we associate a Lie algebra to each closed linear group. For any unproven statements, see [Kna02] or [GW09].
Definition 1.4.1. Let G be subgroup of GL(n). We say that a matrix X ∈ Mn(C) is a tangent vector to G at I when there exists a smooth curve c : R → GL(n) with
9 c(R) ⊂ G, c(0) = I and c0(0) = X. The set of all such tangent vectors is denoted by Lie(G) and called the Lie algebra of G. The following result justifies the terminology.
Proposition 1.4.2. Let G be a subgroup of GL(n). Consider the matrix space Mn(C) as the Lie algebra gl(n) under the usual bracket [X,Y ] = XY − YX. Then Lie(G) is a real subalgebra of gl(n).
Note 1.4.3. Observe that Mn(C) = gl(n) as sets. However, we prefer the notation
Mn(C) when considering the algebra structure given by matrix multiplication, and we use gl(n) when we refer to the Lie algebra structure given by the commutator.
Note 1.4.4. In Proposition 1.4.2, Lie(G) may not be a complex subspace of gl(n) even if G contains matrices with complex entries. For example, let’s compute the Lie algebra u(n) of the unitary group U(n).
Let c : R → GL(n) be a smooth curve into U(n) with c(0) = I and c0(0) = X. We have c(t)∗c(t) = I for all t ∈ R. So, differentiating at t = 0, we obtain
c0(0)∗c(0) + c(0)∗c0(0) = 0, which means X∗+X = 0. Thus X is skew-Hermitian. Conversely, we claim that every skew-Hermitian matrix is in u(n). Indeed, let X ∈ gl(n) be such that X∗ + X = 0.
We use the exponential map to define a suitable curve. Let d : R → GL(n) be the map given by d(t) = exp(tX). (1.5)
It is clear that d is smooth, d(0) = I, and d0(0) = X. In addition, for all t ∈ R,
d(t)∗d(t) = [exp(tX)]∗ exp(tX) = exp(tX∗) exp(tX) = exp[t(X∗ + X)] = exp 0 = I.
10 Hence X ∈ u(n). Therefore,
u(n) = {X ∈ gl(n): X∗ + X = 0}.
It follows that u(n) is not closed under scalar multiplication by complex numbers.
For instance, note that u(1) = iR.
Example 1.4.5. The Lie algebra of GL(n) is gl(n). Indeed, for X ∈ gl(n), the smooth curve d : R → GL(n) defined by (1.5) has d(0) = I and d0(0) = X.
Before we compute the Lie algebra of SL(n), we need the following result.
Lemma 1.4.6. Let c : R → GL(n) be a smooth curve such that c(0) = I. Then
d det c(t)| = tr c0(0). dt t=0
Example 1.4.7. Now we are able to compute sl(n), the Lie algebra of SL(n). Let c : R → GL(n) be a smooth curve into SL(n) with c(0) = I and c0(0) = X. Since the function t 7→ det c(t) is constant, the previous lemma gives
d tr X = det c(t)| = 0. dt t=0
Conversely, we claim that any matrix with null trace is in sl(n). Let X ∈ gl(n) be such that tr X = 0. Let d : R → GL(n) be the curve defined by (1.5). Applying Proposition 1.3.3, we have
det(d(t)) = etr(tX) = e0 = 1.
11 Therefore d0(0) = X ∈ sl(n). A basis of sl(n) as a complex vector space consists of the matrices
• eij with 1 ≤ i, j ≤ n and i 6= j, and
• eii − ei+1,i+1 with 1 ≤ i < n.
We shall always consider these matrices as the standard basis of sl(n). In particular, the dimension of sl(n) over C is n2 − 1.
Example 1.4.8. Next we compute the Lie algebra g of G(ω), where ω is a non- degenerate bilinear form on Cn. Let Γ be the Gram matrix of ω with respect to the canonical basis of Cn. Suppose that c : R → GL(n) is a smooth curve into G(ω) with c(0) = I and c0(0) = X. We have c(t)tΓc(t) = Γ for all t ∈ R. Differentiating at t = 0 we obtain c0(0)tΓc(0) + c(0)tΓc0(0) = 0 that is, XtΓ + ΓX = 0 (1.6)
Conversely, let X be a matrix in gl(n) satisfying (1.6). Let d : R → GL(n) be the smooth curve given by (1.5). Using (1.6), we have
Γd(t)Γ−1 = exp(tΓXΓ−1) = exp(−tXt) = [d(t)t]−1. for all t ∈ R. It follows that the image of d is contained in G(ω). So d0(0) = X ∈ g. In conclusion, g consists of all matrices X ∈ gl(n) for which (1.6) holds. In other
12 words, g is the Lie algebra preserving ω, i.e.,
n g = L(ω) = {X ∈ gl(n): ω(Xv, w) + ω(v, Xw) = 0 for all v, w ∈ C }.
− In particular, suppose that n = 2` and Γ = J` , as defined in (1.2). So we have G(ω) = Sp(n) and g is the symplectic Lie algebra sp(n). A simple calculation based on (1.6) shows that
AB sp t t (n) = : A, B, C ∈ M`(C),B = B,C = C . C −At
It follows that a basis of sp(n) as a vector space over C is given by the matrices
• eii − e`+i,`+i (1 ≤ i ≤ `),
• eij − e`+j,`+i (1 ≤ i, j ≤ n and i 6= j),
• ei,`+i (1 ≤ i ≤ `),
• e`+i,i (1 ≤ i ≤ `),
• ei,`+j + ej,`+i (1 ≤ i < j ≤ `), and
• e`+i,j + e`+j,i (1 ≤ i < j ≤ `).
In particular, sp(n) has dimension 2`2 + ` over C. + Suppose now that n = 2` is even and Γ = J` as defined in (1.3). Then G(ω) = O(n) and g is the even orthogonal Lie algebra so(n). Using (1.6) we obtain
AB so t t (n) = : A, B, C ∈ M`(C),B = −B,C = −C . C −At
13 Thus a basis for so(n) over C consists of the matrices
• eii − e`+i,`+i (1 ≤ i ≤ `),
• eij − e`+j,`+i (1 ≤ i, j ≤ n and i 6= j),
• ei,`+j − ej,`+i (1 ≤ i < j ≤ `), and
• e`+i,j − e`+j,i (1 ≤ i < j ≤ `).
Hence the complex dimension of so(2`) is 2`2 − `.
Finally, suppose that n = 2` + 1 is odd and Γ = S` as in (1.4). So G(ω) = O(n) and g is the odd orthogonal Lie algebra so(n). We have
t t 0 −S1 −S2 ` t t so(n) = S A A : Ai ∈ M`(C),Sj ∈ C ,A2 = −A2,A3 = −A3 . 2 1 2 t S1 A3 −A1
For convenience, whenever we work with so(2` + 1), the canonical basis of C2`+1 will be written as {e0, e1, . . . , e2`}, that is, with the indices running from 0 through 2`.
Similarly, we will use the notation {eij : 0 ≤ i, j ≤ 2`} for the canonical basis of gl(2` + 1). A basis for so(2` + 1) is formed by the matrices
• eii − e`+i,`+i (1 ≤ i ≤ `),
• eij − e`+j,`+i (1 ≤ i, j ≤ n and i 6= j),
• ei,`+j − ej,`+i (1 ≤ i < j ≤ `),
• e`+i,j − e`+j,i (1 ≤ i < j ≤ `),
• e0i − e`+i,0 (1 ≤ i ≤ `), and
14 • e0,`+i − ei0 (1 ≤ i ≤ `).
Consequently, so(2` + 1) has dimension 2`2 + ` over C.
Example 1.4.9. Equation (1.6) can be written as X = −Γ−1XtΓ. Taking traces we deduce tr X = 0. Thus the curve d of the previous example satisfies d(R) ⊂ SL(n). It follows that Lie(SO(n)) = Lie(O(n)) = so(n).
In the previous examples we observe an interesting pattern: if G is a closed linear group with Lie algebra g, then X ∈ g implies exp(tX) ∈ G for all t ∈ R. This fact holds in general and is the main content of the following result.
Theorem 1.4.10. Let G be a closed subgroup of GL(n) and let g = Lie(G). Then
g = {X ∈ Mn(C) : exp(tX) ∈ G for all t ∈ R}.
In particular exp(g) ⊂ G.
1.5 Closed Linear Groups as Lie Groups
Theorem 1.5.1. Let G be a closed subgroup of GL(n) and let g = Lie(G). Then there exists an open neighbourhood B of 0 in g such that exp B is open in G and exp : B → exp B is a homeomorphism. Hence exp g contains an open neighbourhood of I in G. Moreover, G is a regular submanifold of GL(n). With respect to the differentiable structure that G inherits from GL(n), G is a Lie group of dimension
−1 k = dimR g. Finally, (exp B, exp ) is a compatible chart in G around I, where g is identified with the Euclidean space Rk.
15 Proof. See [Kna02] or [GW09].
Corollary 1.5.2. Let G be a closed subgroup of GL(n) with Lie algebra g. Then exp g generates the identity component GI .
Proof. Because exp is continuous and g is connected, exp g is connected. Then exp g is contained in GI . The result follows from Theorem 1.5.1 and Proposition 1.2.11.
Note 1.5.3. Suppose that G is a closed subgroup of GL(n). Since G is a regular submanifold of GL(n), a curve c : R → G is smooth as a map into G if and only if it is smooth as a map into GL(n).
1.6 The Differential of a Smooth Homomorphism
In this section, G ⊂ GL(n) and H ⊂ GL(m) are closed linear groups with Lie algebras g and h, respectively. Let π : G → H be a smooth homomorphism. For any unproven statements, we refer the reader to [Kna02].
Given X ∈ g, there is a smooth curve c : R → G such that c(0) = I and c0(0) = X. Then π ◦ c : R → H is a smooth curve with (π ◦ c)(0) = I. So (π ◦ c)0(0) ∈ h. By defining dπ(X) := (π ◦ c)0(0), we obtain a map dπ : g → h, called the differential of π. Of course, we must show that this definition is correct, i.e., it does not depend on the choice of c. Suppose that d : R → G is another smooth curve such that d(0) = I
16 and d0(0) = X. Then
d (π ◦ d)0(0) = π(d(t)c(t)−1c(t))| dt t=0 d = π(d(t)c(t)−1)π(c(t))| dt t=0 d d = π(d(t)c(t)−1)| · π(c(0)) + π(d(0)c(0)−1) · π(c(t))| dt t=0 dt t=0 d = π(d(t)c(t)−1)| + (π ◦ c)0(0). dt t=0
So it is enough to prove that the smooth curve γ : R → G given by γ(t) = d(t)c(t)−1, which has γ(0) = I and γ0(0) = 0, is such that (π ◦ γ)0(0) = 0. Now we refer matters to local coordinates, using the chart (exp B, exp−1) of Theorem 1.5.1. (We identify g with an Euclidean space Rk.) By the continuity of γ, there exists ε > 0 such that γ((−ε, ε)) ⊂ exp B. Then
π ◦ γ = (π ◦ exp) ◦ (exp−1 ◦γ):(−ε, ε) → GL(m) is a smooth map. By the Chain Rule, we have (exp−1 ◦γ)0(0) = 0 because γ0(0) = 0. So, using the Chain Rule again, we obtain (π ◦ γ)0(0) = 0, as desired. Therefore dπ is well defined.
Proposition 1.6.1. dπ : g → h is a homomorphism of real Lie algebras.
Lemma 1.6.2. Let ι : G → G be the identity map. Then dι is the identity on g.
Lemma 1.6.3. Suppose that π1 : G1 → G2 and π2 : G2 → G3 are smooth homo- morphisms between closed linear groups. Then d(π2 ◦ π1) = dπ2 ◦ dπ1.
Corollary 1.6.4. Suppose that π : G → H is an isomorphism of Lie groups, i.e., a
17 smooth group isomorphism with smooth inverse. Then dπ : g → h is an isomorphism of Lie algebras. Moreover (dπ)−1 = d(π−1).
Note 1.6.5. Given X ∈ g, the map c : R → G sending t 7→ exp(tX) is a smooth curve with c(0) = I and c0(0) = X. Hence
d dπ(X) = π(exp(tX))| . dt t=0
Proposition 1.6.6. The diagram
G −−−→π H x x exp exp g −−−→dπ h is commutative. In other words,
π(exp X) = exp dπ(X) for every X ∈ g. Moreover, dπ is the only R-linear map g → h with this property.
Proof. Given X ∈ g, the map ϕ : R → GL(m) defined by ϕ(t) = π(exp(tX)) is a continuous group homomorphism. Applying Theorem 1.3.6, there exists a unique matrix µ(X) ∈ Mm(C) such that ϕ(t) = exp(tµ(X)) for all t ∈ R. Differentiating at t = 0 yields dπ(X) = µ(X).
18 Chapter 2
Representations of Semisimple Lie Algebras
In this chapter we summarize many fundamental facts on the representation theory of finite dimensional complex semisimple Lie algebras. Special emphasis is given to the case of the semisimple classical Lie algebras. Proofs are omitted and the reader is refered to [Hum78]. Throughout the chapter, L denotes a nonzero finite dimensional complex Lie al- gebra. All modules for L are assumed to be finite dimensional over C.
2.1 Root Space Decomposition
Definition 2.1.1. A subalgebra H of L is called toral if adx : L → L is diagonalizable for all x ∈ H. A toral subalgebra H of L is maximal when it is not properly contained in a toral subalgebra of L.
Example 2.1.2. The complex classical Lie algebras are sl(n), sp(n), and so(n) with n > 1, as defined in Chapter 1. We just call them the classical Lie algebras. They
19 are all simple, with the exceptions of so(2) —which is 1-dimensional— and so(4) —which is isomorphic to so(3) ⊕ so(3)—. In particular, every classical Lie algebra except for so(2) is semisimple. In any classical Lie algebra, the subalgebra consisting of all diagonal matrices is maximal toral.
Lemma 2.1.3. Every toral subalgebra is abelian.
Suppose that H is a toral subalgebra of L and consider the adjoint representa- tion ad : L → gl(L). By definition, ad(H) consists of diagonalizable operators. In addition, they commute with each other because
[adh1, adh2] = ad[h1, h2] = ad0 = 0
for all h1, h2 ∈ H. Therefore the action of H on L (via ad) is simultaneously diago- nalizable. It follows that L decomposes as
M Lα, α∈H∗
where Lα is the subspace of L defined by
Lα = {x ∈ L :[h, x] = α(h)x for all h ∈ H}.
∗ The set of all those nonzero linear functionals α ∈ H for which Lα 6= 0 will be denoted by Φ. Then M L = L0 ⊕ Lα. (2.7) α∈Φ It may happen that the only toral subalgebra of L is {0}. In this case (2.7) provides no information at all. However, when L is semisimple the situation is different and
20 this decomposition becomes a powerful tool in the study of the structure and the representations of L.
Lemma 2.1.4. Suppose that L is semisimple. Then L has a nonzero toral subalgebra. Thus L has at least one nonzero maximal toral subalgebra.
Assume henceforth that L is semisimple and fix a maximal toral subalgebra H of L. Then (2.7) is the root space decomposition of L induced by H. The elements of Φ are called roots. Let κ be the Killing form on L, i.e., the bilinear form L × L → C defined by κ(x, y) = tr(adx ◦ ady). So κ is nondegenerate. The following theorem summarizes many important facts on the root space decomposition of L induced by H.
Theorem 2.1.5. (a) Φ spans H∗.
(b) Given α ∈ Φ, the only scalar multiples of α that are roots are ±α.
∗ (c) If α, β ∈ H then [Lα,Lβ] ⊂ Lα+β.
(d) If α, β ∈ Φ and α 6= −β, then [Lα,Lβ] = Lα+β.
(e) H = L0.
(f) The restriction of κ to H is nondegenerate.
(g) dim Lα = 1 for all α ∈ Φ.
Since the restriction of κ to H is nondegenerate, we have a linear isomorphism
∗ H → H defined by h 7→ κ(h, −). We adopt the notation θ 7→ tθ for the inverse
∗ mapping. Thus, given θ ∈ H , tθ is the unique element in H such that θ(h) = κ(tθ, h) for all h ∈ H.
21 Lemma 2.1.6. Let α ∈ Φ. Then:
(a) [Lα,L−α] is one dimensional with basis {tα}.
(b) There exists a unique hα ∈ [Lα,L−α] such that α(hα) = 2.
(c) Given α ∈ Φ and 0 6= xα ∈ Lα there exists a unique yα ∈ L−α such that
[hα, xα] = 2xα, [hα, yα] = −2yα, [xα, yα] = hα.
In particular, span{xα, hα, yα} = Lα ⊕ [Lα,L−α] ⊕ L−α is a subalgebra of L isomorphic to sl(2).
Definition 2.1.7. Let ∆ be a basis of H∗ consisting of elements of Φ. We say that ∆ is a base or a fundamental system of Φ if for every α ∈ Φ the coordinates of α in ∆ are all non-negative integers or all non-positive integers. In such a case, the elements of ∆ are called fundamental roots or simple roots. A root α ∈ Φ is said to be positive when its coordinates in ∆ are non-negative, and negative when these coordinates are non-positive. The set of all positive (resp. negative) roots is denoted by Φ+ (resp. Φ−).
Theorem 2.1.8. Φ has a fundamental system.
Let ∆ = {α1, . . . , α`} be a fundamental system of Φ. By means of the isomorphism
∗ H → H induced by the Killing form, we see that {tα1 , . . . , tα` } is a basis of H. It follows that also {hα1 , . . . , hα` } is a basis of H. Let {λ1, . . . , λ`} be the corresponding
∗ dual basis of H . As we shall see in the next section, the linear functionals λi play a central role in the study of the irreducible representations of L.
22 Root space decomposition of sl(n) Let L = sl(n) with n ≥ 2. Let H be the maximal toral subalgebra of L consisting of all diagonal matrices. We have dim H = n − 1. The corresponding set of roots is
Φ = {εi − εj : i 6= j}
where, for 1 ≤ i ≤ n, εi : H → C is the linear functional given by εi(h) = hii. In addition Lεi−εj = Ceij for i 6= j. The set
∆ = {ε1 − ε2, ε2 − ε3, . . . , εn−1 − εn}
is a base of Φ. For each index i between 1 and n − 1 we write αi = εi − εi+1. Hence
hαi = eii − ei+1,i+1.
We choose xαi = ei,i+1 and this forces yαi = ei+1,i. Finally, observe that
λi = ε1 + ··· + εi, 1 ≤ i ≤ n − 1.
Root space decomposition of sp(2`) Let L = sp(2`), where ` ≥ 1. Let H be the maximal toral subalgebra of L consisting of all diagonal matrices. Then
Φ = {εi − εj : i 6= j} ∪ {εi + εj : i ≤ j} ∪ {−εi − εj : i ≤ j},
where, for 1 ≤ i ≤ `, εi : H → C is defined by εi(h) = hii. In addition, we have
23 • Lεi−εj = C(eij − e`+j,`+i) for i 6= j,
• Lεi+εj = C(ei,`+j + ej,`+i) for i ≤ j, and
• L−εi−εj = C(e`+i,j + e`+j,i) for i ≤ j.
A base of Φ is
∆ = {ε1 − ε2, ε2 − ε3, . . . , ε`−1 − ε`, 2ε`}.
We call α` = 2ε` and αi = εi − εi+1 for 1 ≤ i < `. Then hα` = e`` − e2`,2` and
hαi = eii − ei+1,i+1 − e`+i,`+i + e`+i+1,`+i+1
for 1 ≤ i < `. We choose xα` = e`,2` and so yα` = e2`,`. For 1 ≤ i < ` we take
xαi = ei,i+1 − e`+i+1,`+i and hence
yαi = ei+1,i − e`+i,`+i+1.
Finally, for 1 ≤ i ≤ `,
λi = ε1 + ··· + εi.
Root space decomposition of so(2`) Let L = so(2`), where ` ≥ 2. Let H be the maximal toral subalgebra of L consisting of all diagonal matrices. We have
Φ = {εi − εj : i 6= j} ∪ {εi + εj : i < j} ∪ {−εi − εj : i < j},
24 where the linear functionals εi are defined like in the preceding example. Moreover,
• Lεi−εj = C(eij − e`+j,`+i) for i 6= j,
• Lεi+εj = C(ei,`+j − ej,`+i) for i < j, and
• L−εi−εj = C(e`+i,j − e`+j,i) for i < j.
A fundamental system for Φ is
∆ = {ε1 − ε2, ε2 − ε3, . . . , ε`−1 − ε`, ε`−1 + ε`}.
We write α` = ε`−1 + ε` and αi = εi − εi+1 for 1 ≤ i < `. The elements hαi with 1 ≤ i < ` are the same as in the symplectic case. But now we have
hα` = e`−1,`−1 + e`` − e2`−1,2`−1 − e2`,2`
We choose xα` = e`−1,2` − e`,2`−1 and
xαi = ei,i+1 − e`+i+1,`+i
for 1 ≤ i < `. This forces yα` = e2`,`−1 − e2`−1,` and
yαi = ei+1,i − e`+i,`+i+1 for 1 ≤ i < `. Finally
1 1 λ`−1 = 2 (ε1 + ··· + ε`−1 − ε`), λ` = 2 (ε1 + ··· + ε`−1 + ε`),
25 and, for 1 ≤ i < ` − 1,
λi = ε1 + ··· + εi.
Root space decomposition of so(2` + 1) Let L = so(2`+1), where ` ≥ 1. In this case we index the rows and columns of the elements of L with the integers 0, 1,..., 2`. Let H be the maximal toral subalgebra of L consisting of all diagonal matrices. Then
Φ = {εi − εj : i 6= j} ∪ {εi + εj : i < j} ∪ {−εi − εj : i < j} ∪ {±εi}
where, for 1 ≤ i ≤ `, εi : H → C sends h 7→ hii. In addition,
• Lεi−εj = C(eij − e`+j,`+i) for i 6= j,
• Lεi+εj = C(ei,`+j + ej,`+i) for i < j,
• L−εi−εj = C(e`+i,j + e`+j,i) for i < j,
• Lεi = C(e0,`+i − ei,0), and
• L−εi = C(e0,i − e`+i,0).
The set
∆ = {ε1 − ε2, ε2 − ε3, . . . , ε`−1 − ε`, ε`}
is a base of Φ. Let α` = ε` and αi = εi − εi+1 for 1 ≤ i < `. It follows that
hα` = 2(e`` − e2`,2`)
26 and, for 1 ≤ i < `,
hαi = eii − ei+1,i+1 − e`+i,`+i + e`+i+1,`+i+1.
We choose xαi = ei,i+1 − e`+i+1,`+i (1 ≤ i < `) and xα` = e0,2` − e`,0. This implies that yαi = ei+1,i − e`+i,`+i+1 (1 ≤ i < `) and yα` = 2(e2`,0 − e0,`). To conclude, we have
1 λ` = 2 (ε1 + ··· + ε`) and, for 1 ≤ i < `,
λi = ε1 + ··· + εi.
Whenever we work with any semisimple classical Lie algebra, we will keep the
notation used in the previous examples as well as the choices of H, ∆, and xαi .
2.2 Highest Weight Modules
In this section we assume that L is semisimple. Let R : L → gl(V ) be a complex representation of L.
Theorem 2.2.1. Let x ∈ L. If ad x : L → L is diagonalizable (resp. nilpotent), then R(x): V → V is diagonalizable (resp. nilpotent).
Theorem 2.2.2. Suppose that L is a semisimple subalgebra of gl(n). If x ∈ L is diagonalizable (resp. nilpotent), then ad x is diagonalizable (resp. nilpotent).
Fix a maximal toral subalgebra H of L and let Φ be the corresponding set of
+ roots. For each α ∈ Φ we fix xα ∈ Lα − {0} and let yα be the only element of L−α
27 such that [xα, yα] = hα. Let ∆ = {α1, . . . , α`} be a base of Φ. As before, the dual
basis of {hαi } will be denoted by {λαi }. Since L is semisimple and H is toral, Theorem 2.2.1 ensures that R(h) is diago- nalizable for all h ∈ H. In addition, the elements of R(H) commute with each other because [R(h),R(h0)] = R([h, h0]) = R(0) = 0 for all h, h0 ∈ H. It follows that the operators R(h) with h ∈ H are simultaneously diagonalizable. Thus V decomposes as
M V = Vµ, µ∈H∗ where
Vµ = {v ∈ V : hv = µ(h)v for all h ∈ H}.
∗ Let Π stand for the set of those µ ∈ H such that Vµ 6= 0. Then
M V = Vµ. µ∈Π
Observe that Π is finite because dim V < ∞. The elements of Π are called weights.
If µ ∈ Π, Vµ is said to be a weight space.
Definition 2.2.3. Let v ∈ V . We say that v is a maximal vector of V (with respect to ∆) if
• v is a common eigenvector for the action of H on V , and
+ • xα · v = 0 for all α ∈ Φ .
28 Lemma 2.2.4. If V 6= 0, then V has (at least) one maximal vector.
Lemma 2.2.5. Suppose that v is a common eigenvector for the action of H on V .
Then v is a maximal vector if and only if xα · v = 0 for all α ∈ ∆.
Theorem 2.2.6. Let v be a maximal vector of V with weight λ. Let U be the L- submodule of V generated by v, i.e., U is the intersection of all the L-submodules of
V containing v. Then U is irreducible and dim Uλ = 1.
Definition 2.2.7. We denote by D(∆) the set of all linear combinations of λ1, . . . , λ` with non-negative integer coefficients, i.e.,
( ` ) X D(∆) = ciλi : ci ∈ N0, 1 ≤ i ≤ ` . i=1
The elements of D(∆) are called dominant integral weights. For ν, µ ∈ H∗ we write
ν ≤∆ µ when there are non-negative integers b1, . . . , b` such that
µ − ν = b1α1 + ··· + b`α`.
∗ It is easy to check that ≤∆ is a partial order on H .
Theorem 2.2.8. Suppose that V is an irreducible L-module. Then V has a unique maximal vector up to scaling. Let λ be the corresponding weight. Then µ ≤∆ λ for all µ ∈ Π. In addition λ ∈ D(∆).
Definition 2.2.9. With the notation of the previous theorem, we say that λ is the highest weight of V .
Theorem 2.2.10. Suppose that V1 and V2 are irreducible L-modules with highest weights µ1 and µ2, respectively. Then V1 ' V2 as L-modules if and only if µ1 = µ2.
29 Definition 2.2.11. Suppose that V is an irreducible L-module with highest weight
λ. Then, with some abuse of notation, we write V = V (λ). If λ = λi for some i, V is called the i-th fundamental module for L.
The previous theorems show that there is an injection from the set of isomorphism classes of finite dimensional irreducible L-modules into the set of all dominant integral weights. By considering the universal enveloping algebra of L, it can be shown that this map is also surjective. The interested reader can find this proof in [Hum78]. A different approach is presented in [GW09] for the case of the classical Lie algebras. In first place they construct all the fundamental modules, and then they use them to prove the existence of V (λ) for each λ ∈ D(∆). In this work we show that Weyl’s method provides explicit and concrete con- structions of most of the finite dimensional irreducible modules for the classical Lie algebras. In fact, we obtain all the desired representations of sl(n) and sp(n), but only half of them in the orthogonal case.
Example 2.2.12. Let L be a semisimple classical Lie algebra. In this example we characterize D(∆) in terms of the coordinate functions on the diagonal subalgebra H of L. Suppose first that L is sl(` + 1) or sp(2`) with ` ≥ 1. We have
c1λ1 + ··· + c`λ` = γ1ε1 + ··· + γ`ε` (2.8)
with γi = ci + ··· + c` for 1 ≤ i ≤ `. Equivalently,
c1 = γ1 − γ2 , c2 = γ2 − γ3 , . . . , c`−1 = γ`−1 − γ` , c` = γ`.
30 Hence D(∆) consists of the linear functionals
γ1ε1 + ··· + γ`ε` (2.9)
where the γi are integers such that γ1 ≥ · · · ≥ γ` ≥ 0. Secondly assume L = so(2` + 1) with ` ≥ 1. In this case, Equation (2.8) holds
c` c` with γ` = 2 and γi = ci + ··· + c`−1 + 2 for 1 ≤ i < `. Equivalently,
c1 = γ1 − γ2 , c2 = γ2 − γ3 , . . . , c`−1 = γ`−1 − γ` , c` = 2γ`.
Thus D(∆) consists of the elements (2.9) where γ1 ≥ · · · ≥ γ` ≥ 0 and one of the following conditions holds:
(a) γi is integer for all i, (b) 2γi is an odd integer for all i. Finally let L = so(2`) with ` > 1. Then we have (2.8) with
c`−1+c` c`−1+c` γ1 = c1 + ··· + c`−2 + 2 , γ2 = c2 + ··· + c`−2 + 2 ,...,
c`−1+c` c`−c`−1 γ`−1 = 2 , γ` = 2 or, equivalently,
c1 = γ1 − γ2 , . . . , c`−2 = γ`−2 − γ`−1 , c`−1 = γ`−1 − γ` , c` = γ`−1 + γ`.
It follows that D(∆) consists of the elements (2.9) where γ1 ≥ · · · ≥ γ`−1 ≥ |γ`| and one of the conditions (a), (b) is satisfied.
31 Chapter 3
Linear Algebraic Groups
The classical groups are not only closed linear groups but also linear algebraic groups, in the sense that they are groups of invertible matrices defined by polynomial equations in the matrix entries. In this chapter we take advantage of this algebraic description of the classical groups to obtain further information about their structure, topology and representation theory.
3.1 Definition and Examples
Let P(Mn(C)) stand for the algebra of polynomial functions on Mn(C), that is,
P(Mn(C)) = C[z11, z12, . . . , znn],
where the zij are the matrix entry functions on Mn(C). A subgroup G of GL(n) is a linear algebraic group if there exists a subset A of
P(Mn(C)) such that
G = {g ∈ GL(n): f(g) = 0 for all f ∈ A}.
32 Since each element of P(Mn(C)) is a continuous function, G can be seen as the intersection between GL(n) and the zero sets of a collection of continuous functions
Mn(C) → C. Therefore, every linear algebraic group is a closed linear group. In particular, all the results from Chapter 1 apply to linear algebraic groups.
Example 3.1.1. By taking A = {0} we see that GL(n) itself is a linear algebraic group. In particular, C× = GL(1) is a linear algebraic group.
Example 3.1.2. SL(n) is a linear algebraic group with A = {det −1}.
Example 3.1.3. Let ω be a nondegenerate bilinear form on Cn with Gram matrix Γ relative to the canonical basis of Cn. Then
G(ω) = {g ∈ GL(n): gtΓg = Γ}.
As observed in Example 1.1.4 of Chapter 1, the condition gtΓg = Γ is equivalent to a set of quadratic equations in the entries of g. This shows that G(ω) is a linear algebraic group.
Example 3.1.4. SO(n) is a linear algebraic group because O(n) and SL(n) are.
Note 3.1.5. Every classical group is a linear algebraic group.
Proposition 1.4.2 and Note 1.4.4 proved that the Lie algebra of a closed linear group is a real subalgebra of gl(n) but it may not be closed under scalar multiplication by complex numbers. However, Examples 1.4.5 through 1.4.9 showed that the Lie algebras of the groups GL(n), SL(n), Sp(n), O(n), and SO(n) —which are linear algebraic groups— are complex linear. In general, we have the following result.
33 Proposition 3.1.6. Let G ⊂ GL(n) be a linear algebraic group with Lie algebra g. Then
g = {X ∈ Mn(C) : exp(zX) ∈ G for all z ∈ C}.
In particular, g is a complex subalgebra of gl(n).
Proof. Given z ∈ C and X ∈ g, we claim that exp(zA) ∈ G. Indeed, let f ∈ A where A is a set of polynomial functions defining G. By Lemma 1.4.10 of Chapter 1, the map ϕ : R → Mn(C) sending t 7→ f(exp(tX)) is identically 0. In particular, all the derivatives of ϕ are 0.
Now consider the map φ : C → Mn(C) given by
φ(z) = f(exp(tX)).
Since f ∈ P(Mn(C)) and exp is complex analytic, we deduce that φ is complex analytic. (See, for instance, Proposition 9.3.2 of [Die69]). Hence φ is holomorphic.
We have φ(t) = ϕ(t) = 0 and φ(k)(t) = ϕ(k)(t) = 0 for all k ∈ N and t ∈ R. In particular, φ(0) = φ(k)(0) = 0 for all k ∈ N. Since φ is complex analytic, it follows that φ = 0. Thus exp(zX) ∈ G. Finally, the smooth curve c : R → GL(n) sending t 7→ exp(tzX) has c(R) ⊂ G and c(0) = I. Therefore c0(0) = zX ∈ g.
Note 3.1.7. In Example 1.4.4 of Chapter 1 we saw that the Lie algebra of the unitary group U(n) is not closed under scalar multiplication by complex numbers. Therefore, U(n) is not a linear algebraic group.
34 3.2 Regular Functions
Definition 3.2.1. We say that a function f : GL(n) → C is regular if it is a polyno- mial with complex coefficients in the matrix entry functions zij (1 ≤ i ≤ n, 1 ≤ j ≤ n)
1 −1 and in the function det = det . The set of all such functions will be denoted by O[GL(n)]. So, by definition,
−1 O[GL(n)] = C[z11, z12, . . . , znn, det ].
Notice that O[GL(n)] is an associative and commutative C-algebra with 1.
Definition 3.2.2. Let G ⊂ GL(n) be a linear algebraic group. A function f : G → C is regular if it is the restriction of a regular function GL(n) → C. The set of all regular functions on G is denoted by O[G]. The restriction map from O[GL(n)] to O[G] is an epimorphism of C-algebras with kernel
IG = {f ∈ O[GL(n)] : f|G = 0}.
Consequently,
O[G] 'O[GL(n)]/IG.
Lemma 3.2.3. Let G, H be linear algebraic groups such that H ⊂ G. If f : G → C is a regular function, then so is f|H .
Definition 3.2.4. We say that a map ϕ : G → H between linear algebraic groups is regular if f ◦ ϕ : G → C is regular for all f ∈ O[H].
Lemma 3.2.5. The identity map on a linear algebraic group is regular. If ϕ : G → H and ψ : H → K are regular maps between linear algebraic groups, then ψ◦ϕ is regular.
35 Lemma 3.2.6. Let ϕ : G → H be a regular map between linear algebraic groups. If
K is an algebraic group such that K ⊂ G, then ϕ|K is regular. If K is an algebraic group such that ϕ(G) ⊂ K ⊂ H, the co-restriction ϕ|K is regular.
Example 3.2.7. Let G be a linear algebraic group. For fixed g ∈ G, the maps
Lg : G → G and Rg : G → G respectively defined by x 7→ gx and x 7→ xg are regular. According to the previous lemma, it suffices to prove this claim when G = GL(n). ˜ ˜ Indeed, suppose G = GL(n) and let f ∈ O[G]. Define f = f◦Lg, so that f(x) = f(gx) for all g ∈ G. Since f is regular, det(gx) = (det g)(det x), and the entries of gx linear ˜ combinations of the entries of x, it follows that f is regular. Hence Lg is regular.
Likewise we see that Rg is regular.
Example 3.2.8. Let g ∈ G where G is a linear algebraic group. Since Lg−1 and Rg are regular, we can define maps L(g): O[G] → O[G] and R(g): O[G] → O[G] by
L(g)f = f ◦ Lg−1 and R(g)f = f ◦ Rg. It is straightforward to check that L(g) and R(g) are algebra isomorphisms.
Lemma 3.2.9. Let G ⊂ GL(n) and H ⊂ GL(m) be linear algebraic groups. Suppose that ϕ : G → H is a regular map. Then ϕ is smooth as a map between (real) manifolds.
Proof. Recall that, since linear algebraic groups are closed linear groups, they have a (real) differentiable structure according to Theorem 1.5.1 of Chapter 1.
Each coordinate function zij ◦ ϕ of ϕ is regular by definition. Thus zij ◦ ϕ = Fij|G
2 for some regular function Fij : GL(n) → C, which is smooth as a function of 2n real variables because it is rational in those variables. So the map F : GL(n) → Mm(C) given by
F (g) = (Fij(g))1≤i,j≤m
36 H is smooth. We have ϕ = F |G . Since G is a regular submanifold of GL(n) and H is a regular submanifold of Mm(C), it follows that ϕ is smooth.
Definition 3.2.10. Let G and H be linear algebraic groups. An algebraic group homomorphism ϕ : G → H is a group homomorphism that is regular. We say that G and H are isomorphic as algebraic groups if there exists a bijective algebraic group homomorphism ϕ : G → H with regular inverse. In such a case, ϕ is called an isomorphism of algebraic groups.
Example 3.2.11. Let G = GL(n). Given g ∈ G, consider the map ϕ : G → G
−1 sending x 7→ gxg . Clearly ϕ is a group automorphism. Since ϕ = Lg ◦ Rg−1 and
−1 −1 ϕ = Rg ◦ Lg−1 , we deduce that both ϕ and ϕ are regular.
Example 3.2.12. A calculation reveals that SO(2) consists of the matrices
z 0 (3.10) 0 z−1 with z ∈ C×. So the map C× → SO(2) sending each z to (3.10) is an isomorphism of linear algebraic groups.
Let G, H be linear algebraic groups. Suppose G ⊂ GL(n) and H ⊂ GL(m). We identify the group-theoretic direct product G × H with the subset
g 0 K = : g ∈ G, h ∈ H 0 h of GL(n+m) by means of the group isomorphism G×H → K given by (g, h) 7→ g⊕h.
37 Lemma 3.2.13. G × H is a linear algebraic group.
Proof. The polynomial equations defining G and H extend to polynomial equations on Mn+m(C). In addition, set to zero the matrix entry functions corresponding to the upper right block and the lower left block.
Theorem 3.2.14. Let G be a linear algebraic group. Then the maps µ : G × G → G and η : G → G given by multiplication and inversion are regular. If f ∈ O[G] then
0 00 there exist an integer p and fi , fi ∈ O[G] for 1 ≤ i ≤ p such that
p X 0 00 f(gh) = fi (g)fi (h) (3.11) i=1 for all g, h ∈ G.
Proof. See [GW09].
Note 3.2.15. The previous theorem shows again that, for fixed g ∈ G, the maps
Lg : G → G and Rg : G → G given by left and right multiplication by g are regular.
3.3 Regular Representations
Throughout this section, G denotes a linear algebraic group. Let ρ : G → GL(V ) be a representation of G on a complex vector space V of finite dimension d. On occasions we may write (ρ, V ) instead of ρ to emphasize the vector space upon which
G is acting. For each basis B of V , let MB : End(V ) → Md(C) be the algebra isomorphism sending each endomorphism of V to its matrix relative to B. We say that ρ is regular if there exists a basis B of V for which the map
MB ◦ ρ : G → GL(d)
38 is regular. This definition is independent of the choice of the basis because, for fixed p ∈ GL(d), conjugation by p is a regular map from GL(d) to itself. A G-module is called regular when the corresponding representation is regular. For v ∈ V and α ∈ V ∗, we sometimes denote α(v) by hα, vi.
Lemma 3.3.1. Keep the notation as above. The following statements are equivalent: (1) ρ is regular;
(2) the function G → C sending g 7→ hα, ρ(g)(v)i is regular for all v ∈ V and α ∈ V ∗; (3) the entry functions of ρ relative to any basis of V are regular; (4) there exists a basis of V relative to which each entry function of ρ is regular.
∗ Proof. Let B = {v1, . . . , vd} be a basis of V and let B = {α1, . . . , αd} be the dual basis. Given indices i, j between 1 and d, a calculation shows that the function G → C sending g 7→ hαi, ρ(g)(vj)i is equal to zij ◦MB ◦ρ. This implies the equivalence between
(2), (3) and (4). By composing MB ◦ ρ with each of the matrix entry functions zij we see that (1) implies (3). Finally suppose that (3) holds and let’s deduce (1). Let f ∈ O[GL(d)]. Then f ◦ MB ◦ ρ is a polynomial in the functions xij ◦ MB ◦ ρ and in
1 h : G → C, where h(g) = det(ρ(g)) . By (3), we only need to check that h is regular. As h(g) = det(ρ(g−1)), we have h = k ◦ η, where η(g) = g−1 and k(g) = det(ρ(g)). We already know that η is regular. In addition (3) implies that k is regular. Thus h is regular.
Note 3.3.2. Regular representations of G are often called rational representations, since each entry function is a rational function of the matrix entries zij : G → C. However, the only denominators that occur are powers of det, so these functions are defined everywhere on G.
39 Lemma 3.3.3. Let (ρ, V ) be a regular representation of G. Suppose that W is a
G-invariant subspace of V . Let ρ1 : G → GL(W ) be the representation defined by ρ1(g) = ρ(g)|W , and let ρ2 : G → GL(V/W ) be the representation defined by
ρ2(g)(v + W ) = ρ(g)(v) + W . Then ρ1 and ρ2 are regular.
Proof. Take a basis B1 = {w1, . . . , ws} of W and complete it to obtain a basis
B = {w1, . . . , ws, v1, . . . , vr}
of V . Then B2 = {v1 + W, . . . , vr + W } is a basis of V/W . So, for all g ∈ G,
(M ◦ ρ )(g) ∗ B1 1 (MB ◦ ρ)(g) = . 0 (MB2 ◦ ρ2)(g)
Since the matrix entry functions of MB ◦ ρ are regular by assumption, the same
happens to MB1 ◦ ρ1 and MB2 ◦ ρ2.
Example 3.3.4. Let (ρ, Cn) be the natural representation of G ⊂ GL(n). Thus ρ(g)(v) = gv for all g ∈ G and v ∈ Cn. If B denotes the standard basis of Cn, then
MB ◦ ρ is the identity map. So ρ is regular.
Example 3.3.5. Let (ρ1,V1) and (ρ2,V2) be regular representations of G. Let ρ be the representation of G on V1 ⊕ V2 given by ρ(g) = ρ1(g) ⊕ ρ2(g). Take a basis B1 of
V1 and a basis B2 of V2, and use them to construct a basis B of V1 ⊕ V2. Then
(M ◦ ρ )(g) 0 B1 1 (MB ◦ ρ)(g) = 0 (MB2 ◦ ρ2)(g) for all g ∈ G. It follows that ρ is regular.
40 Example 3.3.6. Let (ρ, V ) be a regular representation of G. Let (ρ∗,V ∗) be the representation of G given by ρ∗(g) = ρ(g−1)t. Thus ρ∗(g)(α) = α ◦ ρ(g−1) for all g ∈ G and α ∈ V ∗. We will use part (2) of Lemma 3.3.1 to show that ρ∗ is regular.
Let v ∈ V ∗∗ and α ∈ V ∗. Let f : G → C be defined by
f(g) = hv, ρ∗(g)(α)i.
There exists w ∈ V such that v(β) = β(w) for all β ∈ V ∗. Then, for all g ∈ G,
f(g) = (α ◦ ρ(g−1))(w) = hα, ρ(g−1)(w)i.
So f = h ◦ η where η is the inversion in G and h : G → C sends g 7→ hα, ρ(g)(w)i. Since both h and η are regular, we deduce that so is f.
Example 3.3.7. Let (ρ1,V1) and (ρ2,V2) be regular representations of G. Let ρ be the representation of G on V1 ⊗V2 given by ρ(g) = ρ1(g)⊗ρ2(g). Let B1 = {v1, . . . , vn} and B2 = {w1, . . . , wm} be basis of V1 and V2, respectively. Then
B = {v1 ⊗ w1, . . . v1 ⊗ wm, v2 ⊗ w1, . . . , v2 ⊗ wm, . . . , vn ⊗ wm}
is a basis of V1 ⊗ V2. For each g ∈ G, the matrix (MB ◦ ρ)(g) is the Kronecker product
of the matrices (MB1 ◦ ρ1)(g) and (MB2 ◦ ρ2)(g). Recall that the Kronecker product
41 of two matrices P = (pij) ∈ Mn(C) and Q ∈ Mm(C) is the nm × nm matrix
p11Q p12Q ··· p1nQ p21Q p22Q ··· p2nQ P ⊗ Q = . . . . . . . .. . pn1Q pn2Q ··· pnnQ
It follows that ρ is regular.
Example 3.3.8. Suppose G ⊂ GL(n). Let (π, Mn(C)) be the representation of G defined by π(g)(A) = gAg−1. Let ρ be the natural representation of G. From the theory of representations of abstract groups, we know that π is equivalent to the representation ρ0 : G → GL((Cn)∗ ⊗ Cn) given by ρ0(g) = ρ∗(g) ⊗ ρ(g). The previous examples show that ρ0 is regular. Since conjugation by a fixed invertible matrix is a regular map, it follows that π is regular.
Let g ∈ G and X ∈ g = Lie(G). So there exists a smooth curve c : R → G such that c(0) = I and c0(0) = X. Then the map d : R → GL(n) sending t 7→ gc(t)g−1 is also a smooth curve into G. Since d(0) = I, we have d0(0) = gXg−1 ∈ g. Thus g is invariant under π. It follows that the representation π0 : G → GL(g) defined by
0 0 π (g) = π(g)|g is regular. We call π the adjoint representation of G.
Definition 3.3.9. Let ρ be a representation of G on an infinite dimensional complex vector space V . We say that ρ is locally regular if each finite dimensional subspace E of V is contained in a finite dimensional G-invariant subspace W of V such that the restriction of ρ to W is regular.
Example 3.3.10. Let R : G → GL(O[G]) be the representation sending x 7→ R(x). We claim that R is locally regular. Indeed, let E be a subspace of O[G] with basis
42 1 rj 1 rj {f1, . . . , fk}. For each j, Theorem 3.2.14 yields regular functions pj , . . . , pj , qj , . . . , qj on G such that rj X i i fj(xy) = pj(x)qj(y) i=1 for all x, y ∈ G. Then rj X i i R(x)fj = qj(x)pj i=1 for all x ∈ G. So W = span{R(x)fj : x ∈ G, 1 ≤ j ≤ k} is a G-invariant subspace of
i O[G] contained in T = span{pj : 1 ≤ j ≤ k, 1 ≤ i ≤ rj}, which is finite dimensional. In addition E ⊂ W . Now we apply part (2) of Lemma 3.3.1 to show that R acts regularly on W . Let h ∈ W and α ∈ W ∗. Let f : G → C be the function given by f(g) = hα, R(g)hi. It suffices to consider the case where h = R(x)fj for some j and x ∈ G. Since W ⊂ T , we can extend α to a linear functional β ∈ T ∗. A calculation shows that rj X i i f = β(pj)(qj ◦ Rx) i=1 Consequently, f is regular. Likewise, the representation (L, O[G]) of G sending g 7→ L(g) is locally regular.
3.4 Differential of a Regular Representation
In this section, G denotes a linear algebraic group with Lie algebra g. Suppose that ρ is a regular representation of G on a complex vector space V of finite dimension d. We want to introduce the differential of ρ, which will be a representation of g, that is, Lie algebra homomorphism
dρ : g → gl(V ),
43 The relation between ρ and dρ provides an important and extremely useful link be- tween the representation theories of G and g. As we will show in this section, such a relation is particularly strong when G is connected.
Given a basis B of V , the group homomorphism MB ◦ ρ : G → GL(d) is regular by definition. In particular, it is smooth by Lemma 3.2.9. Therefore we can consider its differential
d(MB ◦ ρ): g → gl(d),
−1 as defined in Chapter 1. In order to go back to gl(V ), we apply MB and hence we obtain a Lie algebra homomorphism
−1 MB ◦ d(MB ◦ ρ): g → gl(V ).
By definition, we adopt
−1 dρ := MB ◦ d(MB ◦ ρ).
Of course, we must prove that this construction is independent from the choice of B. Let B0 be another basis of V . So there exists an invertible matrix p ∈ GL(d) such that
−1 −1 (MB0 ◦ MB )(X) = pXp
44 for all X ∈ gl(d). Thus, for all X ∈ g,
d d(M 0 ◦ ρ)(X) = (M 0 ◦ ρ)(exp(tX))| B dt B t=0 d −1 = (M 0 ◦ M ◦ M ◦ ρ)(exp(tX))| dt B B B t=0 d = p(M ◦ ρ)(exp(tX))p−1 | dt B t=0 d = p (M ◦ ρ)(exp(tX))| p−1 dt B t=0
−1 = pd(MB ◦ ρ)(X)p
−1 = (MB0 ◦ MB ◦ d(MB ◦ ρ))(X).
Consequently,
−1 −1 MB0 ◦ d(MB0 ◦ ρ) = MB ◦ d(MB ◦ ρ), as desired. In Chapter 1 we showed that the differential of a smooth group homomorphism is always R-linear. In the case of linear algebraic groups we can say something more.
Proposition 3.4.1. Let π : G → H be a regular homomorphism between linear algebraic groups. Then dπ is C-linear.
Proof. Let z0 ∈ C and X ∈ g. Suppose H ⊂ GL(m). Let φ : C → GL(m) be the map given by φ(z) = π(exp(zX)). Observe that φ is well defined by Proposition 3.1.6. In addition, φ is complex analytic because π is regular and exp is complex analytic. Let
ϕ = φ|R. Then
d d dπ(z X) = ϕ(z t)| = φ(z z)| = z φ0(0) = z ϕ0(0) = z dπ(X). 0 dt 0 t=0 dz 0 z=0 0 0 0
45 Corollary 3.4.2. The differential of any regular representation of G is C-linear.
Proposition 3.4.3. Let (ρ, V ) be a regular representation of G. Then,
ρ(exp X) = exp dρ(X) for all X ∈ g.
0 Proof. Let X ∈ g. Take a basis B of V and set ρ = MB ◦ ρ. Then we have
ρ0(exp X) = exp dρ0(X)
−1 by Proposition 1.6.6. So, applying MB in both sides and keeping in mind Note 1.3.4, we get
−1 0 −1 0 ρ(exp X) = MB (exp dρ (X)) = exp(MB (dρ (X))) = exp dρ(X).
Example 3.4.4. Let (ρ, Cn) be the natural representation of G. Then Lemma 1.6.2 implies that dρ is the natural representation of g, i.e., dρ(X)(v) = Xv for all X ∈ g and v ∈ Cn.
Example 3.4.5. For i ∈ {1, 2}, let (ρi,Vi) be a regular representation of G. Let ρ be the regular representation of G on V = V1 ⊕ V2 defined by ρ(g) = ρ1(g) ⊕ ρ2(g). We
fix bases of V1 and V2, and we use them to obtain a basis of V in the obvious way.
In order to clarify notation, we identify each element of GL(V1), GL(V2) and GL(V ) with its matrix in the the corresponding basis. Let X ∈ g and let c : R → G be a
46 smooth curve with c(0) = I and c0(0) = X. Then
d dρ(X) = (ρ ◦ c)0(0) = [(ρ ◦ c)(t) ⊕ (ρ ◦ c)(t)]| = (ρ ◦ c)0(0) ⊕ (ρ ◦ c)0(0) dt 1 2 t=0 1 2
Therefore,
dρ(X) = dρ1(X) ⊕ dρ2(X).
Example 3.4.6. Let (ρ, V ) be a regular representation of G. Let (ρ∗,V ∗) be the regular representation of G given by ρ∗(g) = ρ(g−1)t. Take a basis B of V and let B∗ be the dual basis of V ∗. As above, we identify each element of GL(V ) and GL(V ∗) with its matrix in the corresponding basis. Then, for all X ∈ g,
d d d t dρ∗(X) = ρ∗(exp(tX))| = ρ(exp(−tX))t| = − ρ(exp(tX))| . dt t=0 dt t=0 dt t=0
Thus dρ∗(X) = −dρ(X)t.
Example 3.4.7. For i ∈ {1, 2} let (ρi,Vi) be a regular representation of G and let Bi be a basis of Vi. Then {v1 ⊗v2 : vi ∈ Bi} is a basis of V = V1 ⊗V2. Let ρ : G → GL(V ) be the regular representation defined by ρ(g) = ρ1(g) ⊗ ρ2(g). As before, we identify each element of GL(V1), GL(V2) and GL(V ) with its matrix in the corresponding basis. Given X ∈ g, let c : R → G be a smooth curve with c(0) = I and c0(0) = X. Then
d dρ(X) = (ρ ◦ c)0(0) = [(ρ ◦ c)(t) ⊗ (ρ ◦ c)(t)]| dt 1 2 t=0 0 0 = (ρ1 ◦ c) (0) ⊗ (ρ2 ◦ c)(0) + (ρ1 ◦ c)(0) ⊗ (ρ2 ◦ c) (0).
47 Therefore
dρ(X) = dρ1(X) ⊗ I + I ⊗ dρ2(X).
Example 3.4.8. Let π : G → GL(g) be the adjoint representation of G, as defined in Example 3.3.8. Let B be a basis of g. For each Y ∈ g, let [Y ]B be the vector of coordinates of Y in B. Then, for all X,Y ∈ g,
[dπ(X)(Y )]B = (MB ◦ dπ)(X)[Y ]B d = (M ◦ π)(exp(tX))| [Y ] dt B t=0 B d = [(M ◦ π)(exp(tX))[Y ] ] | dt B B t=0 d = [π(exp(tX))(Y )] | dt B t=0 d = [exp(tX)Y exp(−tX)] | dt B t=0 d = [exp(tX)Y exp(−tX)]|t=0 dt B
= [XY − YX]B.
So dπ(X)(Y ) = [X,Y ]. Therefore, dπ is just the adjoint representation of g.
Lemma 3.4.9. Let (ρ1,V1) and (ρ2,V2) be regular representations of G that are equiv- alent to each other. Then their differentials are equivalent.
−1 Proof. There exists a linear isomorphism P : V1 → V2 such that ρ2(g) = P ρ1(g)P for all g ∈ G. As we did in the previous examples, we take basis of V1 and V2, and we identify each linear endomorphism with a matrix. Let X ∈ g and let c : R → G be a smooth curve with c(0) = I and c0(0) = X. Then
d dρ (X) = (ρ ◦c)0(0) = [P (ρ ◦c)(t)P −1]| = P (ρ ◦c)0(0)P −1 = P dρ (X)P −1. 2 2 dt 1 t=0 1 1
48 Theorem 3.4.10. Suppose that G is connected. For i ∈ {1, 2}, let (ρi,Vi) be regular representations of G. Then ρ1 and ρ2 are equivalent if and only if their differentials are.
Proof. In view of the previous lemma, we only need to prove one implication. Suppose
−1 that there exists a linear isomorphism P : V1 → V2 such that dρ2(X) = P dρ1(X)P for all X ∈ g. Then, using Lemma 3.4.3, we have
−1 −1 −1 ρ2(exp X) = exp(P dρ1(X)P ) = P (exp dρ1(X))P = P ρ1(exp X)P for all X ∈ g. Since G is connected, Lemma 1.5.2 ensures that exp g generates G. It
−1 follows that ρ2(g) = P ρ1(g)P for all g ∈ G.
Lemma 3.4.11. Let (ρ, V ) be a regular representation of G. Suppose that W is a G-invariant subspace of V . Then W is also g-invariant.
Proof. As usual, we fix a basis of V and we identify each element of GL(V ) with a matrix. Moreover, we identify each element of V with the corresponding vector of coordinates. Let X ∈ g and w ∈ W . Let c : R → G be a smooth curve such that c(0) = I and c0(0) = X. Then
d dρ(X)w = (ρ ◦ c)0(0)w = [ρ(c(t))w]| . dt t=0
Since ρ(c(t))w ∈ W for all t and W is a closed subspace of V , we deduce dρ(X)w ∈ W .
Theorem 3.4.12. Suppose that G is connected and let (ρ, V ) be a regular represen- tation of G. Let W be a subspace of V . Then W is G-invariant if and only if W is
49 g-invariant.
Proof. By the previous lemma, we only have to prove one implication. Suppose that W is g-invariant. Let X ∈ g and w ∈ W . Then
" ∞ # ∞ X 1 X 1 ρ(exp X)(w) = (exp dρ(X))(w) = dρ(X)k (w) = dρ(X)k(w) ∈ W. k! k! k=0 k=0
Since G is connected, it is generated by exp g. Therefore W is G-invariant.
The following result shows that the differential is well-behaved with respect to invariant subspaces.
Proposition 3.4.13. Let (ρ, V ) be a regular representation of G. Suppose that W is a G-invariant subspace of V . Let ρW : G → GL(W ) be the regular representation defined by ρW (g) = ρ(g)|W . Then dρW (X) = dρ(X)|W for all X ∈ g.
Proof. Take a basis of W and complete it to obtain a basis of V . As usual, we identify each endomorphism with its matrix in the corresponding basis. Let X ∈ g and let c : R → G be a smooth curve such that c(0) = I and c0(0) = X. Then
(ρ ◦ c)(t) ∗ W (ρ ◦ c)(t) = . 0 ∗
Differentiating at t = 0 we obtain
dρ (X) ∗ W dρ(X) = . 0 ∗
Hence dρ(X)|W = dρW (X).
50 To conclude this section, we observe that the differential is also well-behaved with respect to restrictions of the domain.
Proposition 3.4.14. Let (ρ, V ) be a regular representation of G. Suppose that H is a linear algebraic group contained in G with Lie algebra h. Then d(ρ|H) = (dρ)|h.
3.5 Multiplicative Jordan Decomposition
If G is a linear algebraic group, we have just seen that the relationship between a regular representation of G and its differential is particularly deep when G is con- nected. Therefore, our next goal is to show that the groups GL(n), SL(n), Sp(n) and SO(n) are connected. In this section we introduce a powerful tool from linear algebra, known as the multiplicative Jordan decomposition. In the next section we will use such decomposition to establish the connectedness of the desired groups. A matrix or linear endomorphism u is called unipotent when u − I is nilpotent.
Suppose that A is a nilpotent element of Mn(C) or End(V ), where V is a vector space of finite dimension n. Then An = 0 and hence, by definition,
A2 An−1 exp A = I + Y where Y = A + + ··· + . 2! (n − 1)!
Observe that exp A is unipotent because Y is nilpotent. Now suppose that u is a unipotent element of Mn(C) or End(V ). So det u = 1 because all the eigenvalues of u are equal to 1. In particular u is invertible. Moreover, since A = u − I is nilpotent, we have A2 An−1 log u = A − + ··· + (−1)n 2 n − 1 by definition. Then log u is nilpotent because A is.
51 Lemma 3.5.1. Let N (resp. U) be the set of nilpotent (resp. unipotent) elements in Mn(C) or End(V ). Then the polynomial map N → U sending A 7→ exp A is a bijection with polynomial inverse u 7→ log u.
Proof. It follows from the discussion above together with the substitution principle.
Lemma 3.5.2. Let G be a linear algebraic group with Lie algebra g. Let A ∈ Mn(C) be nilpotent. Then A ∈ g if and only if exp A ∈ G.
Proof. One of the implications holds by Theorem 1.4.10. Suppose now that exp A ∈
G. Given a polynomial function f ∈ P(Mn(C)) that vanishes on G, we consider the function ϕ : C → C defined by ϕ(z) = f(exp(zA)). Observe that ϕ is polynomial in z because A is nilpotent. Since G is a group, we have exp(nA) = (exp A)n ∈ G for all n ∈ N, and hence f(N) = 0. Keeping in mind that f is polynomial, we deduce f = 0. Therefore exp(zA) ∈ G for all z ∈ C, which implies A ∈ g.
Theorem 3.5.3. Let G denote GL(n) or GL(V ), where V is a vector space of finite dimension n. Let g ∈ G. Then there exist s, u ∈ G such that (1) g = su = us, (2) s is diagonalizable and u is unipotent. Properties (1) and (2) uniquely determine u and s. Furthermore, there is a poly- nomial φ ∈ C[X] such that s = φ(g).
Proof. If g = s + N is the additive Jordan decomposition of g with s diagonalizable and N nilpotent, take u = I + s−1N. Uniqueness follows from the uniqueness of the additive Jordan decomposition.
52 For each C ∈ Mn(C), let fC : Mn(C) → C be the function sending D 7→ tr(DC).
The restriction of fC to GL(n) will also be denoted by fC . A short calculation reveals
that feij = zji.
Theorem 3.5.4. Let G ⊂ GL(n) be a linear algebraic group whose Lie algebra g is semisimple. If g ∈ G and g = su is its multiplicative Jordan decomposition, then s ∈ G and u ∈ G.
k Proof. Let P (Mn(C)) be the subspace of P(Mn(C)) consisting of those polynomial functions that are homogeneous of degree k. For each m ∈ N, let Wm denote the finite dimensional subspace of O[GL(n)] defined by
m M k Wm = P (Mn(C)). k=0
Clearly Wm is invariant under the operators R(g) with g ∈ GL(n). Let g ∈ G and let g = su be its multiplicative Jordan decomposition, with s diagonalizable and u unipotent. Then R(g) = R(s)R(u) and hence
R(g)|Wm = R(s)|Wm R(u)|Wm . (3.12)
We claim that this is the multiplicative Jordan decomposition of R(g)|Wm . The operators R(s)|Wm and R(u)|Wm commute because s and u commute. In Example 3.3.10 we saw that the representation R : GL(n) → GL(O[GL(n)]) is locally regular. So Wm is contained in a G-invariant finite dimensional subspace Tm of O[GL(n)] such that the restriction of R to Tm is regular. Applying Lemmas 3.5.1
53 and 3.5.2, there exists a nilpotent matrix X ∈ g such that u = exp X. Then
R(u) = exp dR(X) (3.13)
by Proposition 3.4.3, where R is understood to act on Tm. Since X is nilpotent and g is a semisimple subalgebra of gl(n), Theorem 2.2.2 ensures that dR(X) is nilpotent.
Thus, Equation (3.13) and Lemma 3.5.1 imply that R(u) is unipotent on Tm, and hence on Wm.
To prove that R(s)|Wm is diagonalizable, we can assume that s is diagonal (because for any basis B of Mn(C), the set {fC : C ∈ B} generates P(Mn(C)) as an algebra).
So suppose s = diag[d1, . . . , dn]. A simple calculation reveals that
R(s)feij = difeij .
Since R(s) is linear and preserves products, and Wm is spanned by functions that
are products of elements from the set {feij : 1 ≤ i ≤ n, 1 ≤ j ≤ n}, it follows that
R(s)|Wm is diagonalizable.
Therefore (3.12) is the multiplicative Jordan decomposition of R(g)|Wm . Let f ∈A, where A is a set of polynomial functions defining G. Choose m large enough so that
f ∈ Wm. Since R(s)|Wm is a polynomial in R(g)|Wm and R(g) preserves IG, we deduce R(s)f ∈ IG. In particular 0 = R(s)f(I) = f(s), whence s ∈ G. Finally, u = s−1g ∈ G.
54 3.6 Proofs of Connectedness
In this section we prove that the complex groups GL(n), SL(n), Sp(n) and SO(n) are connected. Let G be any of these groups. The diagonal subgroup of G will be denoted by H. If G = GL(n), then
× H = {diag[d1, . . . , dn]: di ∈ C }.
If G = SL(n), we have
−1 × H = {diag[d1, . . . , dn−1, (d1 ··· dn−1) ]: di ∈ C }.
Now suppose that n = 2` is even and G is Sp(n) or SO(n). Then
−1 −1 × H = {diag[d1, . . . , d`, d1 , . . . , d` ]: di ∈ C }.
If n = 2` + 1 is odd and G = SO(n),
−1 −1 × H = {diag[1, d1, . . . , d`, d1 , . . . , d` ]: di ∈ C }.
We begin by considering the general linear group GL(n). Although this case is quite simple, it will give us a clue of how to proceed in the other situations.
Theorem 3.6.1. GL(n) is connected.
Proof. Take an element g ∈ GL(n) and consider its multiplicative Jordan decompo- sition g = su with s diagonalizable and u nilpotent. By Lemma 3.5.1, there exists a nilpotent matrix A in gl(n) such that u = exp A. In addition, s can be expressed as
55 php−1 with p ∈ GL(n) and h ∈ H. If
h = diag[d1, . . . , dn], then h = exp h0 with
0 h = diag[ln d1,..., ln dn].
Here ln di represents any natural logarithm of di. Therefore,
g = su = p(exp h0)p−1(exp A) = (exp(ph0p−1))(exp A).
Thus exp(gl(n)) generates GL(n). Hence GL(n) is connected by Corollary 1.5.2.
Suppose now that G is any of the groups SL(n), SO(n), or Sp(n). We want to apply the same argument as above to show that G is connected. Assume n>1 because the case n = 1 is trivial. Moreover, SO(2) is connected because it is homeomorphic to C× = GL(1). So we take n > 2 in the orthogonal case. Let g be the Lie algebra of G, and let h stand for the diagonal subalgebra of g. Given g ∈ G, we write its multiplicative Jordan decomposition g = su with s diago- nalizable and u unipotent. The assumptions on n ensure that g is a semisimple Lie algebra. Thus, by Theorem 3.5.4, both s and u are in G. Applying Lemmas 3.5.1 and 3.5.2, there is a nilpotent matrix A ∈ g such that u = exp A. Now we come to the crucial step. Assume for now that it is possible to express s as php−1 with p ∈ G and h ∈ H. Then we can find h0 ∈ h such that h = exp h0. So, computing as before,
g = (exp(ph0p−1))(exp A).
56 We have php−1 ∈ g by Example 3.3.8. Thus we see that exp g generates G, and hence G is connected. Therefore, the proof reduces to the following result.
Lemma 3.6.2. Let g ∈ G be a diagonalizable matrix. Then pgp−1 ∈ H for some p ∈ G.
Proof. Suppose first that G = SL(n). There exists an invertible matrix p such that
−1 pn −1 −1 −1 pgp ∈ H. Setting p0 = det p p, we have p0gp0 = pgp ∈ H with det p0 = 1. Before we consider the symplectic and orthogonal groups, suppose that g preserves a nondegenerate bilinear form ω on Cn. There is an eigenspace decomposition
n M C = Vλ, (3.14)
n with Vλ ={v ∈C : gv =λv} for each λ∈C. Moreover ω(u, v)=ω(gu, gv)=λµω(u, v) for all u ∈ Vλ and v ∈ Vµ. Hence
ω(Vλ,Vµ) = 0 whenever λµ 6= 1. (3.15)
Let µ ∈ C with Vµ 6= 0. Notice that µ 6= 0. We claim that
dim Vµ = dim V1/µ. (3.16)
∗ Let ϕ : Vµ → V1/µ be the linear map sending v 7→ ω(v, −). Using (3.14), (3.15), and the fact that ω is non-degenerate, we deduce that ϕ is injective. Thus
∗ ∗ Vµ ,→ V1/µ ' V1/µ ,→ Vµ ' Vµ,
57 which implies (3.16). Let µ1, . . . , µm be the distinct eigenvalues of g that are not ±1.
−1 From (3.16) we see that m = 2r is even and we can take µi = µr+i for 1 ≤ i ≤ r. Then
det g = (−1)dim V−1 . (3.17)
For 1 ≤ i ≤ r define Wi = Vµi ⊕ V1/µi . Then (3.14) and (3.15) imply
n C = V1 ⊕ V−1 ⊕ W1 ⊕ · · · ⊕ Wr, (3.18) where the summands are ω-orthogonal to each other and the restriction of ω to each summand is non-degenerate. For each i such that 1 ≤ i ≤ r, take a basis {wi , . . . , wi } 1 si of V and let {ui , . . . , ui } be the dual basis of V with respect to ω. Thus µi 1 si 1/µi
i i ω(wj, uk) = δjk. (3.19)
The union of these basis is a basis BWi of Wi. Now suppose that n = 2` is even and that ω is the skew-symmetric bilinear form
n − whose Gram matrix in the canonical basis of C is J` , as defined in Chapter 1.
Assume g ∈ Sp(n). If V1 6= 0 then ω|V1 is non-degenerate and hence, by elementary linear algebra, dim V1 is even and there exists a symplectic basis BV1 for V1. Likewise, if V−1 6= 0 then dim V−1 is even we have a symplectic basis BV−1 for V−1. Using (3.19)
together with the facts that ω is skew-symmetric and each of the subspaces Vµi , V1/µi are ω-isotropic, we see that the Gram matrix of ω|Wi with respect to BWi looks like
0 I . −I 0
58 n By (3.18), BV1 ∪ BV−1 ∪ BW1 ∪ · · · ∪ BWr is a basis of C . By reordering vectors we
n − obtain a basis B = {v1, . . . , vn} of C relative to which ω has Gram matrix J` and
λjvj for 1 ≤ j ≤ ` gvj = −1 λj v`+j for ` + 1 ≤ j ≤ n,
where λ1, . . . , λ` are nonzero complex numbers. Let p ∈ GL(n) be the only invertible matrix such that pvi = ei for 1 ≤ i ≤ n. Clearly p ∈ Sp(n) and
−1 −1 −1 pgp = diag[λ1, . . . , λ`, λ1 , . . . , λ` ] ∈ H.
Now suppose that n = 2` is even but the Gram matrix of ω relative to the canonical
+ basis is J` . Assume g ∈ SO(n). Then (3.17) shows that dim V−1 is even. Thus (3.18) implies that so is dim V1. At this point, the argument continues like in the symplectic case, mutatis mutandis. When defining p we must ensure that det p = 1. This can be achieved, if necessary, by interchanging two eigenvectors with reciprocal eigenvalues. Finally suppose that n = 2` + 1 is odd and the Gram matrix of ω relative to the canonical basis is S`. Assume g ∈ SO(n). It follows from (3.17) that dim V−1 is
even, and hence dim V1 odd. Since ω|V1 is nondegenerate and symmetric, it has Gram matrix 1 0 0 0 0 I 0 I 0
relative to a basis BV1 of V1. Now proceed like in the previous cases.
Corollary 3.6.3. The groups SL(n), Sp(n), and SO(n) are connected for all n ∈ N.
59 3.7 Reductive Groups
Definition 3.7.1. A linear algebraic group G is reductive if every regular represen- tation of G is completely reducible.
Lemma 3.7.2. Let G and H be linear algebraic groups with H ⊂ G ⊂ GL(n). Assume that H is reductive and has finite index in G. Then G is reductive.
Proof. This is a direct generalization of the proof of Maschke’s Theorem, which states that every finite dimensional representation of a finite group K (over a field F such that char(F ) - |K|) is completely reductive. We refer the reader to [GW09].
Lemma 3.7.3. Let ρ : C× → GL(n) be a regular homomorphism. For each p ∈ Z
p × define Ep = {v ∈ C : ρ(z)v = z v for all z ∈ C }. Then
n M C = Ep, p∈Z and hence the matrices ρ(z) with z ∈ C× are simultaneously diagonalizable. In partic- ular, C× and SO(2) (which are isomorphic are linear algebraic groups) are reductive.
× −1 Proof. Since O[C ] = C[z, z ], the matrix entry functions zij ◦ ρ are Laurent poly- nomial functions. Thus there exists a positive integer r and matrices T−r,...,Tr in
Mn(C) such that r X p ρ(z) = z Tp (3.20) p=−r for all z ∈ C×. Given z, w ∈ C×, the relation ρ(z)ρ(w) = ρ(zw) translates into
r r r X X p q X k k z w TpTq = z w Tk. p=−r q=−r k=−r
60 Since this holds for all z, w ∈ C×, we deduce
2 TpTq = 0 if p 6= q and Tp = Tp for − r ≤ p ≤ r. (3.21)
In addition, ρ(1) = I means r X I = Tp (3.22) p=−r
n Set Wp = TpC for −r ≤ p ≤ r. Applying (3.21), (3.22), and a basic result from linear algebra, we obtain r n M C = Wp. p=−r
It only remains to prove that Wp = Ep for all p. Fix an integer p between −r and r. Let v ∈ Wp. Then Tpv = v. So, using (3.20) and (3.21), we have
r X q p p ρ(z)v = z TqTpv = z Tpv = z v q=−r
× for all z ∈ C . Thus v ∈ Ep. Conversely, suppose v ∈ Ep. Then (3.20) gives
r p X q z v = z Tqv q=−r
× for all z ∈ C . Therefore v = Tpv ∈ Wp.
Theorem 3.7.4. The groups SL(n), Sp(n), O(n), and SO(n) are reductive.
Proof. We have just shown that SO(2) is reductive. Suppose that G is any of the groups SL(n), Sp(n), or SO(n+1), with n > 1. Let (ρ, V ) be a regular representation of G and let W be a G-invariant subspace of V . Consider the representation dρ of g = Lie(G). By Lemma 3.4.11, W is g-invariant. Since g is a finite dimensional
61 complex semisimple Lie algebra, Weyl’s theorem yields a g-invariant subspace Z of V such that V = W ⊕ Z. Theorem 3.4.12 and Corollary 3.6.3 imply that Z is also G-invariant. Therefore ρ is completely reducible. Finally, applying Lemma 3.7.2 to the subgroup SO(n) of O(n), it follows that O(n) is reductive.
Note 3.7.5. The group GL(n) is also reductive. However, we can not prove this fact using the Lie group-Lie algebra relationship because it is not true that every finite dimensional complex representation of gl(n) is completely reducible. As an example, consider the representation R : gl(n) → gl(C2) sending each X to the linear endomorphism with matrix 0 tr(X) 0 0 relative to the canonical basis of C2. Then C × {0} is a gl(n)-invariant subspace that lacks of a gl(n)-invariant complement. To see that GL(n) is reductive we use the fact that the direct product of two reductive linear algebraic groups is reductive.
Since GL(n) is a homomorphic image of C× × SL(n) via (z, g) 7→ zg, it follows that GL(n) is reductive because C× and SL(n) are. The reader can find the details of this argument in [GW09], Proposition 4.2.6.
62 Chapter 4
Irreducible Representations of Sd
Young diagrams were introduced in 1901 by the British mathematician Alfred Young. A few years later, they were used by the famous German mathematician Ferdinand Georg Frobenius to build all the irreducible complex representations of the symmetric groups. In this chapter we describe the construction, following the expositions given in Section 5.4 of [Jac89] and Chapter 4 of [FH91]. Although the symmetric groups seem to have little relation with the classical groups, the subsequent chapters will show that the irreducible modules for the symmetric groups are a key ingredient in Weyl’s construction of the irreducible regular modules for the classical groups. Throughout this chapter we fix a positive integer d. The group of permutations of the set {1, . . . , d} will be called the symmetric group on d symbols and denoted by
Sd. Representations of Sd are always assumed to be complex.
63 4.1 Young Diagrams and Tableaux
A partition of d is a n-tuple γ = (γ1, γ2, . . . , γn) of integers such that
n X d = γi and γ1 ≥ γ2 ≥ · · · ≥ γn ≥ 0. i=1
For convenience, we allow partitions to end in one or more zeros. However, we consider two partitions as equal if they only differ by zeros at the end.
Each partition γ = (γ1, γ2, . . . , γn) of d is associated to a Young diagram, which consists of d empty boxes arranged in n rows. The rows are lined up on the left and the number of boxes in the row i is γi for 1 ≤ i ≤ n. For instance, the partition (3, 2, 1, 1) of 7 corresponds to the Young diagram
The partition µ = (µ1, . . . , µk) of d given by the columns of the Young diagram of γ is said to be the conjugate partition of γ. Without reference to any picture, we have k = γ1 and µi is the number of elements γj which are greater than or equal to i. Notice that the Young diagram of µ is the transpose of the diagram of γ. For example, the conjugate partition of (3, 2, 1, 1) is (4, 2, 1) and the corresponding diagram is
By distributing the numbers 1, . . . , d in the Young diagram of γ in such a way that each number appears in exactly one box, we obtain a Young tableau of shape γ.
64 For instance, 1 4 6 3 2 (4.23) 5 7 is a Young tableau of shape (3, 2, 1, 1). Fix a Young tableau T of shape γ. We define
R(T ) (resp. C(T )) to be the subgroup of Sd consisting of those permutations that stabilize the sets given by the rows (resp. columns) of T . For example, if T is (4.23) then
R(T ) = h(1 4), (1 6), (2 3)i ' S3 × S2. and
C(T ) = h(1 3), (1 5), (1 7), (2 4)i ' S4 × S2.
Notice that, in general, we have
R(T ) ' Sγ1 × · · · × Sγn and C(T ) ' Sµ1 × · · · × Sµk , where µ is the conjugate partition of γ.
We define the following elements of the group algebra C[Sd] of Sd:
X X X X aT = σ, bT = (sgτ)τ, cT = aT bT = (sgτ)στ. (4.24) σ∈R(T ) τ∈C(T ) σ∈R(T ) τ∈C(T )
It is clear that aT and bT are nonzero. We claim that cT 6= 0 as well. Observe first that R(T ) ∩ C(T ) = 1. Indeed, if σ ∈ R(T ) ∩ C(T ) and i ∈ {1, . . . , d}, then σ(i) = i because σ stabilizes both the row and the column containing i. Thus, if σ1τ1 = σ2τ2
−1 −1 with σj ∈ R(T ) and τj ∈ C(T ), we have σ2 σ1 = τ2τ1 = 1, whence σ1 = σ2 and
τ1 = τ2. So the elements στ with σ ∈ R(T ) and τ ∈ C(T ) are all different to each
65 other. This implies cT 6= 0. The element cT is called a Young symmetrizer and will have central importance in all that follows.
4.2 Irreducible Modules for Sd
The group algebra C[Sd] is an Sd-module, with the action given by left multipli- cation. If T is any Young tableau, then C[Sd]cT is easily seen to be a Sd-submodule of C[Sd]. It turns out that C[Sd]cT is irreducible and, moreover, any irreducible representation of Sd can be obtained in this way.
0 Theorem 4.2.1. C[Sd]cT is an irreducible module for Sd. If T and T are Young
0 tableaux, then C[Sd]cT ' C[Sd]cT 0 as Sd-modules if and only if T and T correspond to the same partition of d. Furthermore, any irreducible module for Sd is isomorphic to C[Sd]cT for some tableau T .
We prove this result through a series of simple lemmas.
Given ρ ∈ Sd and a tableau T , ρT will stand for the tableau we obtain by applying ρ to the entries of T .
Lemma 4.2.2. Let T be a Young tableau. Call a = aT , b = bT , and c = cT . Then:
(i) σa = a = aσ for all σ ∈ R(T ) and τb = (sgτ)b = bτ for all τ ∈ C(T );
−1 −1 (ii) R(ρT ) = ρR(T )ρ and C(ρT ) = ρC(T )ρ for all ρ ∈ Sd;
−1 −1 (iii) aρT = ρaρ and bρT = ρbρ for all ρ ∈ Sd.
−1 (iv) cρT = ρcρ for all ρ ∈ Sd.
66 −1 Proof. (i) is a direct consequence of (4.24). Let ρ ∈ Sd. Clearly ρR(T )ρ ⊂ R(ρT ). Thus
ρR(T )ρ−1 ⊂ R(ρT ) = ρ(ρ−1R(ρT )ρ)ρ−1 ⊂ ρR(ρ−1ρT )ρ−1 = ρR(T )ρ−1, which implies R(ρT ) = ρR(T )ρ−1. Likewise we get C(ρT ) = ρC(T )ρ−1. Part (iii) follows from (4.24) and (ii). Finally, (iv) is a consequence of (iii).
Note 4.2.3. Although this will follow later from a more abstract argument, we can use part (iv) of the previous lemma to prove that tableaux with the same shape give
0 isomorphic representations of Sd. Indeed, suppose that T and T are tableaux with
0 the same shape. Then T = ρT for some ρ ∈ Sd. Letting A = C[Sd], we have
−1 −1 AcT 0 = AρcT ρ = AcT ρ .
−1 Thus the map AcT → AcT 0 sending x 7→ xρ is an isomorphism of Sd-modules.
We order the partitions of d lexicographically: if γ and µ are partitions of d, then
γ > µ means that γi > µi at the first position i where γ and µ differ.
Lemma 4.2.4. Let γ and µ be partitions of d such that γ ≥ µ. Let T and T 0 be tableaux of shapes γ and µ, respectively. Suppose that no two numbers appearing in the same row of T are in the same column of T 0. Then γ = µ and T 0 = στT for some σ ∈ R(T ) and τ ∈ C(T ).
Proof. Suppose, if possible, that γ1 > µ1. This means that the number of columns of T 0 is less than the length of the first row of T . So there are two numbers in the first row of T that appear in the same column of T 0. This contradiction shows that
67 0 γ1 = µ1. Since each column of T contains exactly one element from the first row of
0 0 T , there exists τ1 ∈ C(T ) such that τ1T and T have the same numbers in the first row. Since the numbers in the second row of T do not appear in the first row of T 0, the
0 0 same argument shows that γ2 = µ2 and there exists τ2 ∈ C(τ1T ) = C(T ) such that
0 τ2τ1T and T have the same elements in each of the rows 1 and 2. Continuing in this way we see that γ = µ and there exists τ 0 ∈ C(T 0) such that τ 0T 0 and T have the same elements in each row. Then τ 0T 0 = σT for some σ ∈ R(T ). Since τ 0 ∈ C(T 0) = C(τ 0T 0) = C(σT ) = σC(T )σ−1, we can write τ 0 = στσ−1 with τ ∈ C(T ). So στσ−1T 0 = σT , whence T 0 = στ −1T .
Lemma 4.2.5. Let γ and µ be partitions of d such that γ > µ. Let T and T 0 be tableaux of shapes γ and µ, respectively. Then cT 0 C[Sd]cT = 0.
Proof. Applying Lemma 4.2.4, there exists a transposition π = (i j) that belongs to both R(T ) and C(T 0). Thus, using part (i) of Lemma 4.2.2,
bT 0 aT = bT 0 (πaT ) = (bT 0 π)aT = −bT 0 aT .
Hence bT 0 aT = 0. If ρ is an arbitrary element of Sd, then ρT is a tableau of shape γ,
−1 and so bT 0 aρT = 0. Since aρT = ρaT ρ , we deduce bT 0 ρaT = 0 for all ρ ∈ Sd.
Lemma 4.2.6. Let γ be a partition of d and let T be a tableau of shape γ. Then, an element x ∈ C[Sd] satisfies σxτ = (sgτ)x for all σ ∈ R(T ) and τ ∈ C(T ) if and only if x is a scalar multiple of cT .
68 Proof. For all σ ∈ R(T ) and τ ∈ C(T ),
σcT τ = (σaT )(bT τ) = aT (sgτ)bT = (sgτ)cT .
So any scalar multiple of cT satisfies the conditions. Conversely, let
X x = kρρ
ρ∈Sd be an element of C[Sd] satisfying the conditions. Since Sd is a basis of C[Sd], we have
(sgτ)kρ = kσρτ (4.25)
for all σ ∈ R(T ) and τ ∈ C(T ). In particular kστ = (sgτ)k1 for σ ∈ R(T ) and
τ ∈ C(T ). Therefore, if we can show that kρ = 0 whenever ρ∈ / R(T )C(T ), it will follow that x = k1cT . So let ρ ∈ Sd \ R(T )C(T ). Then ρT 6= στT for all σ ∈ R(T ) and τ ∈ C(T ). By Lemma 4.2.4, there exists a transposition π ∈ R(T ) ∩ C(ρT ). Since C(ρT ) = ρC(T )ρ−1, we can write π = ρπ0ρ−1 for some transposition π0 ∈ C(T ).
0 Thus ρ = πρπ . Using (4.25) we deduce kρ = −kρ, which implies kρ = 0.
Lemma 4.2.7. Let γ be a partition of d and let T be a tableau of shape γ. Then there exists α ∈ C \{0} such that αcT is idempotent. In addition cT C[Sd]cT = CcT .
Proof. Given y ∈ C[Sd], the element x = cT ycT satisfies σxτ = (sgτ)x for every
2 σ ∈ R(T ) and τ ∈ C(T ). So Lemma 4.2.6 yields cT ycT ∈ CcT . In particular cT = kcT for some k ∈ C. It only remains to show that k 6= 0.
2 Suppose, if possible, that cT = 0. Then the linear map θ : C[Sd] → C[Sd] sending z 7→ cT z is nilpotent, and hence it has null trace. Given ρ ∈ Sd, consider the matrix
69 of the map z 7→ ρz relative to the basis Sd of C[Sd]. We see that the trace of this map is 0 if ρ 6= 1, and d! if ρ = 1. Since the coefficient of the identity permutation in cT is 1, it follows that 0 = tr(θ) = d!, a contradiction.
Finally, we put the pieces together to prove Theorem 4.2.1. Let γ be a partition of d and let T be a tableau of shape γ. Set A = C[Sd]. By Lemma 4.2.7, there is a
2 nonzero scalar multiple eT of cT such that eT = eT and eT AeT = CeT . In particular eT AeT is a field with identity element eT . By contradiction, suppose that AeT is not irreducible. Applying Maschke’s Theorem, there are non-zero A-submodules J, K of
AeT such that AeT = J ⊕ K. We have eT = j + k with j ∈ J and k ∈ K. Observe that both j and k are nonzero, for otherwise eT would be in J or in K. Write j = reT and k = seT with r, s ∈ A. Thus jeT = j and keT = k because eT is idempotent. In addition
2 j + k = eT = eT = eT (j + k) = eT j + eT k.
Since the sum J ⊕ K is direct, we deduce j = eT j and k = eT k. Thus
2 j = jeT = j(j + k) = j + jk.
2 Since {j, j } ⊂ J and jk ∈ K, we deduce jk ∈ J ∩ K = 0 . So jk = 0. But j = eT jeT and k = eT keT are in eT AeT . Thus eT AeT has divisors of zero, a contradiction. Now let T 0 be a tableau of shape µ, where µ is a partition of d different from γ. Without loss of generality, we can assume γ < µ. As before, there is a scalar multiple eT 0 of cT 0 that is idempotent. Set J = AeT and K = AeT 0 . Lemma 4.2.5 gives JK = 0. Suppose, if possible, that there exists an isomorphism f : J → K of
70 A-modules. Since eT 6= 0,
2 0 6= f(eT ) = f(eT ) = eT f(eT ) ∈ JK.
Thus JK 6= 0, a contradiction. Consequently AcT 6' AcT 0 as Sd-modules.
So far we have constructed p(d) irreducible representations of Sd, where p(d) is the number of partitions of d. But p(d) is also the number of conjugacy classes in
Sd. In addition, the number of complex irreducible representations of a finite group G cannot exceed the number of conjugacy classes in G (actually, these numbers coincide). Therefore, representations corresponding to tableaux of the same shape must be isomorphic and we have built a full set of representatives of the equivalence classes of irreducible representations of Sd.
Note 4.2.8. As we have just shown, the isomorphism type of the module AcT depends only on the partition γ and not on the particular choice of the tableau T . Thus the irreducible representations of Sd can be indexed by the partitions of d or, equivalently, by the Young diagrams with d boxes. For this reason, the elements aT , bT , cT are frequently denoted by aγ, bγ, cγ. Likewise, we usually write R(γ) and C(γ) instead of R(T ) and C(T ).
Note 4.2.9. Let γ = (γ1, . . . , γk) be a partition of d with γk 6= 0. Although we shall not need this in the sequel, we mention that there is a formula due to Frobenius that gives an explicit expression for the character χγ of the irreducible Sd-module
Vγ = C[Sd]cγ. Since Vγ is finite dimensional and the ground field C has characteristic 0, we have
χγ(1) = tr(id|Vγ ) = dim Vγ.
71 Using Frobenius’ formula in this case we obtain
d! Y dim V = (` − ` ), γ Qk i j i=1(`i!) i `1 = γ1 + k − 1 , `2 = γ2 + k − 2 , . . . , `k = γk. The interested reader can find more details in Sections 4.1 and 4.3 of [FH91]. 72 Chapter 5 Irreducible Regular Representations of SL(n) Here we follow Weyl’s method —which uses Young diagrams and tableaux— to construct all the irreducible regular modules for SL(n). At the Lie algebra level, we obtain all the finite dimensional irreducible representations of sl(n). For the sake of clarity, the first section is devoted to state the main results and to provide a variety of examples. In the remaining sections we prove the assertions made in the first one. 5.1 The Schur Functor Throughout this section, the ground field of all vector spaces is assumed to be C. Suppose that V is a module for a group G. Let d be a positive integer. For each ⊗d ⊗d σ ∈ Sd, the map πσ : V → V sending v1 ⊗ · · · ⊗ vd 7→ vσ(1) ⊗ · · · ⊗ vσ(d) is an isomorphism of G-modules, i.e., a linear isomorphism that commutes with the action ⊗d of G. Thus the formula v · σ = πσ(v) defines a right action of Sd on V , which 73 commutes with the left action of G. So there is no ambiguity in the notation g · v · σ ⊗n ⊗d for g ∈ G, v ∈ V and σ ∈ Sd. Clearly, the action of Sd on V extends to a right action of the group algebra C[Sd] that still commutes with the left action of G. ⊗d Now let γ be a partition of d. The discussion above shows that V · cγ is a G-submodule of V ⊗d. We adopt the notation ⊗d SγV = V · cγ. Each homomorphism ϕ : V → W of G-modules induces a homomorphism Sγϕ : SγV → SγW of G-modules sending v1 ⊗ · · · ⊗ vd · cγ 7→ ϕ(v1) ⊗ · · · ⊗ ϕ(vd) · cγ. It follows that Sγ is a functor from G-modules to G-modules, called the Schur functor. Note 5.1.1. To be strict, we should fix a tableau T of shape γ and write ST instead ⊗d ⊗d of Sγ. However, we will show latter in this chapter that V · cT ' V · cT 0 as G-modules if T and T 0 are tableaux of the same shape. This justifies the abuse of notation. Now suppose G = GL(n) and let V = Cn be the natural module for G. Let ρ be the corresponding representation of G on V ⊗d and let dρ : g → gl(V ⊗d) be its differential, where g = gl(n). Recall that the action of a matrix X ∈ g on any 74 elementary tensor v = v1 ⊗ v2 ⊗ · · · ⊗ vd is given by dρ(X)(v) = Xv1 ⊗ v2 ⊗ · · · ⊗ vd + v1 ⊗ Xv2 ⊗ · · · ⊗ vd + ··· + v1 ⊗ v2 ⊗ · · · ⊗ Xvd. ⊗d Notice that the left action of g on V commutes with the right action of C[Sd]. It follows that SγV is g-invariant. Observe, however, that also Lemma 3.4.11 implies ⊗d that SγV is a g-submodule of V . Now we are able to state the main result of this chapter. For the remainder of the section, V stands for the natural module Cn for SL(n). We assume n > 1. Theorem 5.1.2. (i) If γ is a partition of a positive integer, then SγV 6= 0 if and only if γn+1 = 0. (ii) Suppose that γ is a partition such that γn+1 = 0. Then SγV is an irreducible module for SL(n) and sl(n). Moreover, SγV is the irreducible sl(n)-module with highest weight (γ1 − γn)ε1 + (γ2 − γn)ε2 + ··· + (γn−1 − γn)εn−1 = γ1ε1 + ··· + γnεn. (5.26) (iii) Every finite dimensional irreducible module for sl(n) is isomorphic to SγV for a suitable partition γ. (iv) Every irreducible regular module for SL(n) is isomorphic to SγV for some γ. (v) If λ and µ are partitions such that γn+1 = µn+1 = 0, then SγV ' SµV over SL(n) (resp. sl(n)) if and only if the difference λi − µi is constant. 75 (vi) The map γ 7→ SγV gives a bijection between the set of all partitions γ such that γn = 0 and the set of isomorphism classes of non-trivial irreducible regular reg- ular modules for SL(n) (resp. non-trivial finite dimensional irreducible modules for sl(n)). (vii) Let d be a positive integer. For each partition γ of d, let dγ be the dimension of the complex vector space C[Sd]cγ considered in Chapter 4. Then the tensor power V ⊗d decomposes as a module over SL(n) (resp. sl(n)) as ⊗d M dγ V ' [SγV ] , γ where γ runs over all partitions of d. Note 5.1.3. Observe that parts (i) and (ii) imply assertions (iii) through (vi). Indeed, suppose that (i) and (ii) are true. Since every dominant integral weight for sl(n) has the form (5.26) with γ1 ≥ · · · ≥ γn ≥ 0 and γ1 > 0, we see that (iii) follows from (ii) and Theorem 2.2.10. (In the case of the trivial sl(n) module we can take γi = 1 for all i.) Now let U be a regular irreducible SL(n)-module. By Theorem 3.4.12 and Corol- n lary 3.6.3, U is irreducible over sl(n). So U ' Sγ(C ) as sl(n)-modules for some n partition γ. Applying Theorem 3.4.10, we deduce U ' Sγ(C ) as SL(n)-modules. Thus (iv) holds. For sl(n), part (v) is a consequence of (ii) and Theorem 2.2.10. The assertion for SL(n) follows applying Theorem 3.4.10 and Corollary 3.6.3. 76 We prove part (vi) for sl(n). The map γ 7→ SγV is well defined by (ii) and injective by (v). Now let U be a non-trivial finite dimensional sl(n)-module. The corresponding highest weight has the form γ1ε1 + ··· + γn−1εn−1 with γ1 ≥ · · · ≥ γn−1 ≥ 0. Since U is not trivial, we have γ1 > 0. Then U ' SγV with γ = (γ1, . . . , γn−1). Again, the case of SL(n) follows by Theorem 3.4.10 and Corollary 3.6.3. In conclusion, it suffices to prove (i), (ii), (vii) in order to obtain Theorem 5.1.2. We will complete the proof in Sections 5.2 and 5.5. The remainder of this section is devoted to provide examples. Example 5.1.4. Let d be an integer such that 1 ≤ d ≤ n and consider the partition γ = (1, 1,..., 1) of d. The corresponding Young diagram consists of a single column with d boxes. Thus, the associated Young symmetrizer is given by X cγ = sg(τ)τ, τ∈Sk d which implies SγV = Λ V . Consequently, for 1 ≤ d < n, the d-th exterior power of V is the fundamental sl(n)-module with highest weight λk = ε1 + ··· + εd. Now we look at the case d = n. By Theorem 5.1.2, the highest weight of sl(n) acting n n on Λ V is ε1 + ··· + εn = 0. Then Λ V must be the trivial sl(n)-module. We can n also see this directly. In fact, Λ V is linearly spanned by v = e1 ∧ ... ∧ en and sl(n) acts like 0 on v. Example 5.1.5. Let d ∈ N. The partition γ = (d) gives a Young diagram with d 77 boxes and only one row. The corresponding symmetrizer is X cγ = σ. σ∈Sd d d Therefore SγV = Sym (V ). The highest weight of sl(n) acting on Sym (V ) is dε1. There is another familiar realization of this module. Let C[X1,...,Xn] be the poly- nomial algebra in n commuting variables and let Hd be the subspace consisting of the homogeneous polynomials of degree d. We have a representation R : gl(n) → gl(Hd) defined on the canonical basis by ∂ R(eij) = mXi ◦ . ∂Xj (Differentiation with respect to Xj followed by multiplication by Xi.) It is well known d that Hd is an irreducible sl(n)-module with highest weight dε1. Thus Sym (V ) ' Hd as sl(n)-modules. This identification can be seen explicitly by mapping each monomial a1 a2 an X1 X2 ··· Xn with a1 + a2 + ··· + an = d to the element d e1 ··· e1 e2 ··· e2 ··· en ··· en ∈ Sym V. | {z } | {z } | {z } a1 times a2 times an times Example 5.1.6. Consider the partition γ = (2, 1, 1,..., 1) of n. For instance, the Young diagram corresponding to the case n = 4 is 78 By Theorem 5.1.2, SγV is the irreducible sl(n)-module with highest weight 2ε1 + ε2 + ··· + εn−1 = ε1 − εn. It follows that SγV is isomorphic to the adjoint module, i.e., SγV ' sl(n) over sl(n). d Example 5.1.7. Let d ∈ N. Since Sym V is irreducible over sl(n), the same occurs to (SymdV )∗. Since (SymdV )∗ and Symd(V ∗) are naturally isomorphic as sl(n)-modules, the latter is also irreducible over sl(n). Let {e1, . . . , en} be the canonical basis of V ∗ ∗ ∗ ∗ and let {e1, . . . , en} be the corresponding dual basis of V . Then en is a maximal ∗ vector of V with weight −εn = ε1 + ··· + εn−1. Hence ∗ ∗ ∗ enen ··· en | {z } d times d ∗ is a maximal vector of Sym (V ) with weight −dεn = d(ε1 + ··· + εn−1). Therefore d ∗ Sym (V ) ' SγV , where γ is the partition (d, . . . , d) of d(n − 1). This corresponds to the rectangular Young diagram with n − 1 rows and d columns. Example 5.1.8. Suppose 1 ≤ d ≤ n. Since (ΛdV )∗ ' Λd(V ∗) over sl(n), reasoning as above we see that Λd(V ∗) is the irreducible sl(n)-module with highest weight d ∗ n−d 0 ε1 + ··· + εn−d. Hence Λ (V ) ' Λ V over sl(n), where we take Λ V = C. Note 5.1.9. Theorem 5.1.2 shows that there is a bijective correspondence between the irreducible regular representations of SL(n) and the finite dimensional irreducible representations of sl(n). Thus, if U is an irreducible regular module for SL(n) and the highest weight of sl(n) acting on U is λ, it is often said that U is the irreducible SL(n)-module with highest weight λ. 79 5.2 Finding Maximal Vectors Lemma 5.2.1. Suppose k > n. Then Λk(Cn) = 0. Proof. The exterior power Λk(Cn) is linearly spanned by the vectors ei1 ∧ ei2 ∧ · · · ∧ eik (5.27) with 1 ≤ i1, i2, . . . , ik ≤ n. Since k > n, each vector (5.27) has eij = eij0 for some k n indices ij 6= ij0 . Thus Λ (C ) = 0. n Corollary 5.2.2. Let γ be a partition such that γn+1 6= 0. Then Sγ(C ) = 0. n P Proof. Let V = C and d = γi. Given an elementary tensor v = v1 ⊗ · · · ⊗ vd ⊗d in V , it suffices to show that v · bγ = 0. Without loss of generality, we fill in the corresponding Young diagram with the numbers 1, . . . , d in consecutive order down the columns, from left to right. For instance, if γ = (3, 2, 1, 1) we get 1 5 7 2 6 (5.28) 3 4 Recall that C(γ) ' Sµ1 × · · · × Sµk where µ is the conjugate partition of γ. Thus 0 v · bγ = (v1 ∧ · · · ∧ vµ1 ) ⊗ v , 0 ⊗(d−µ1) where v is some element of V . Since µ1 > n, the expression in parentheses is 0 by Lemma 5.2.1. n Let V = C . Let γ = (γ1, . . . , γk) be a partition of a positive integer d such that 80 k ≤ n. Let T be a tableau of shape γ. We define eT = ei1 ⊗ ei2 ⊗ · · · ⊗ eid where, for 1 ≤ j ≤ d, we have ij = t if j occurs in the t-th row of T . Thus the numbers in the first row of T indicate the tensor positions in eT containing e1, the numbers in the second row of T indicate the tensor positions in eT containing e2, and so on. For example, if T is (5.28) then eT = e1 ⊗ e2 ⊗ e3 ⊗ e4 ⊗ e1 ⊗ e2 ⊗ e1. If 1 2 3 1 5 6 T 0 = 4 5 and T 00 = 3 4 , 6 2 7 7 then eT 0 = e1 ⊗ e1 ⊗ e1 ⊗ e2 ⊗ e2 ⊗ e3 ⊗ e4 and eT 00 = e1 ⊗ e3 ⊗ e2 ⊗ e2 ⊗ e1 ⊗ e1 ⊗ e4. ⊗d Lemma 5.2.3. Keep the above setup. Then eT · cT 6= 0. In particular V · cT 6= 0. ⊗d Regarding V · cT as a module for gl(n), we have ers · eT · cT = 0 for 1 ≤ r < s ≤ n. Moreover, if h is a diagonal element of gl(n), then h · eT · cT = (γ1h11 + ··· + γkhkk)(eT · cT ). (5.29) Proof. Equation (5.29) is clear from the definition of eT . By definition of R(T ) and eT , we have eT · σ = eT for all σ ∈ R(T ). So eT · aT 81 is a nonzero scalar multiple of eT , namely eT · aT = |R(T )|eT . Thus it will suffice to show that eT · bT 6= 0 and ers · eT · bT = 0 for 1 ≤ r < s ≤ n. Suppose first that T = T (γ), where the boxes in T (γ) are numbered consecutively down the columns, from left to right. For instance, if γ = (3, 2, 1, 1) then T (γ) is (5.28). Let µ = (µ1, . . . , µ`) be the conjugate partition of γ. Thus eT = (e1 ⊗ · · · ⊗ eµ1 ) ⊗ (e1 ⊗ · · · ⊗ eµ2 ) ⊗ · · · ⊗ (e1 ⊗ · · · ⊗ eµ` ). Therefore, recalling that C(T ) ' Sµ1 × Sµ2 × · · · × Sµ` , we have eT · bT = ω1 ⊗ ω2 ⊗ · · · ⊗ ω`, with ωi = e1 ∧· · ·∧eµi for 1 ≤ i ≤ `. Since ωi 6= 0 for all i, we deduce that eT ·bT 6= 0. (This argument works for tensor products between finite dimensional vector spaces, but fails in general for tensor products between modules over an arbitrary ring.) If 1 ≤ r < s ≤ n, then ers · eT · bT = 0 because ers kills each of the ωi. Now suppose that T is any tableau of shape γ. So there exists σ ∈ Sd such that T = σT (γ). We claim that eT · σ = eT (γ). Indeed, write eT = ek1 ⊗ · · · ⊗ ekd and eT (γ) = ei1 ⊗ · · · ⊗ eid . Then eT · σ = ekσ(1) ⊗ · · · ⊗ ekσ(d) 82 By definition, for 1 ≤ j ≤ d and 1 ≤ t ≤ n, we have ij = t ⇔ j occurs in the t-th row of T (γ) ⇔ σ(j) occurs in the t-th row of σT (γ) = T ⇔ kσ(j) = t. Thus ij = kσ(j) for 1 ≤ j ≤ d, which implies eT · σ = eT (γ). In addition, Lemma 4.2.2 −1 gives bT = σbT (γ)σ . Consequently, −1 eT · bT = (eT (γ) · bT (γ)) · σ 6= 0 and, for 1 ≤ r < s ≤ n, −1 ers · eT · bT = ers · (eT (γ) · bT (γ)) · σ = 0. n Note 5.2.4. Corollary 5.2.2 and Lemma 5.2.3 imply that Sγ(C ) = 0 if and only if n γn+1 6= 0. Now suppose that γn+1 = 0. Lemma 5.2.3 shows that Sγ(C ), viewed as a module for sl(n), contains a maximal vector of weight γ1ε1 + ··· + γnεn, which is equal to (γ1 − γn)ε1 + (γ2 − γn)ε2 + ··· + (γn−1 − γn)εn−1 83 because ε1 + ··· + εn = 0 on the diagonal subalgebra of sl(n). Therefore, the proof of Theorem 5.1.2 will be n complete if we prove part (vii) and we show that Sγ(C ) is irreducible over sl(n) and SL(n) when γn+1 = 0. This will be achieved in Section 5.5. To conclude this section, we mention that the argument used in the proof of Lemma 5.2.3 gives a stronger result that will be useful in a latter chapter. However, we need to complicate the notation a little bit. n As before, let V = C and let T be a tableau of shape γ = (γ1, . . . , γk), where γ is a partition of a positive integer d such that k ≤ n. But now, instead of working with the elements e1, . . . , ek of V , we consider the vectors ej1 , . . . , ejk , where the j’s are indices such that 1 ≤ j1 < ··· < jk ≤ n. We define e = ei1 ⊗ ei2 ⊗ · · · ⊗ eid , where, for 1 ≤ α ≤ d, we have iα = jβ if α occurs in the β-th row of T . For example, if T is (5.28), n > 9, j1 = 2, j2 = 4, j3 = 8, and j4 = 9, then e = e2 ⊗ e4 ⊗ e8 ⊗ e9 ⊗ e2 ⊗ e4 ⊗ e2. ⊗d Lemma 5.2.5. Keep the above setup. Then e · cT 6= 0. Regarding V · cT as a module for gl(n), we have ejr,js · e · cT = 0 for 1 ≤ r < s ≤ k. Moreover, if h ∈ gl(n) is diagonal then h · e · cT = (γ1hj1j1 + ··· + γkhjkjk )(e · cT ). 84 In the following two sections we develop some auxiliary results that will play a role in the proof of the irreducibility statements of Theorem 5.1.2. 5.3 Facts on Matrix Rings n Lemma 5.3.1. Let F be a field. Consider the matrix ring Mn(F ) acting on F by n left multiplication. Then F is an irreducible Mn(F )-module. n Proof. Let S be a nonzero Mn(F )-submodule of F . Let x ∈ S such that xi 6= 0 for some 1 ≤ i ≤ n. For 1 ≤ j ≤ n we have ejix = xiej ∈ S, which implies ej ∈ S. Lemma 5.3.2. Let F be a field. Let c1, . . . , cn be elements of F , not all of them equal to 0. Let I be the subset of Mn(F ) defined by n I = {(c1v|c2v| · · · |cnv): v ∈ F }. (5.30) Then I is a minimal left ideal of Mn(F ). In addition, any minimal left ideal of Mn(F ) has this form. n Proof. Clearly I is closed under addition. For all A ∈ Mn(F ) and v ∈ F we have A(c1v|c2v| · · · |cnv) = (c1(Av)|c2(Av)| · · · |cn(Av)), (5.31) which proves that I is a left ideal of Mn(F ). Let J be a nonzero left ideal of Mn(F ) contained in I. Then J contains a matrix of the form (c1v|c2v| · · · |cnv) for some n nonzero v ∈ F . Since the action of Mn(F ) on J is given by (5.31), Lemma 5.3.1 yields J = I. 85 Now let K be a minimal left ideal of Mn(F ). Let x ∈ K such that xij 6= 0 for some i, j. Acting on the left through e1i, we see that K contains a matrix of the form k1 k2 ··· kn 0 0 ··· 0 . . . . . . .. . 0 0 ··· 0 with kj 6= 0. So K shares a nonzero matrix with the minimal left ideal ˜ n I = {(k1v|k2v| · · · |knv): v ∈ F }. Therefore K = I˜. Analogously we obtain the following result. Lemma 5.3.3. Let F be a field. Let c1, . . . , cn be elements of F , not all of them equal to 0. Let I be the subset of Mn(F ) defined by t n I = {(c1v|c2v| · · · |cnv) : v ∈ F }. Then I is a minimal right ideal of Mn(F ). In addition, any minimal right ideal of Mn(F ) has this form. Lemma 5.3.4. Let F be a field. Let I and J be minimal left (or right) ideals of Mn(F ). Then I and J are isomorphic as left (or right) modules over Mn(F ). Proof. We only consider the case where I and J are left ideals. The other case is 86 similar. By Lemma 5.3.2, I has the form (5.30). Likewise, n J = {(d1v|d2v| · · · |dnv): v ∈ F } for some d1, . . . , dn ∈ F , not all of them equal to 0. The map I → J sending (c1v|c2v| · · · |cnv) 7→ (d1v|d2v| · · · |dnv) is a (well defined) isomorphism of left Mn(F )-modules. L The following lemma will be applied latter to the case R = i Mni (C). L Lemma 5.3.5. Suppose a ring R has two-sided ideals I1,...,In such that R = i Ii. Assume that each of the ideals Ii, seen as a ring, has an identity element 1i. Let I be a minimal left (or right) ideal of R. Then I ⊂ Ii for some index i. Moreover, I is a minimal left (or right) ideal of Ii. Proof. Let x ∈ I − {0}. Write x = x1 + ··· + xn with xi ∈ Ii. We have xi 6= 0 for some index i. Since 1i · x = xi ∈ I ∩ Ii, we deduce that I ∩ Ii is a nonzero left ideal of R contained in I. The minimality of I implies I ⊂ Ii. The last assertion follows from the fact that any left ideal of Ii is a left ideal of R. 5.4 Facts on G-homomorphisms Lemma 5.4.1. Let U be a finite dimensional complex vector space. Suppose that U is an irreducible G-module for a group G. Then ⊕n ⊕n HomG(U ,U ) ' Mn(C) 87 as C-algebras. ⊕n ⊕n Proof. Given A ∈ Mn(C), we define a map ϕA : U → U by n n ! X X ϕA(v1, . . . , vn) = A1jvj,..., Anjvj . j=1 j=1 ⊕n ⊕n Then ϕA ∈ HomG(U ,U ). So we obtain a map ⊕n ⊕n ϕ : Mn(C) → HomG(U ,U ) sending A 7→ ϕA. It is straightforward to check that ϕ is a monomorphism of C- algebras. ⊕n ⊕n Now let f ∈ HomG(U ,U ). Fix an index j such that 1 ≤ j ≤ n. Let ⊕n hj : U → U be the injection sending u 7→ (0,..., 0, u, 0,..., 0), where u occupies the j-th position. Let fj = f ◦ hj. Let f1j, . . . , fnj be the component functions of fj, that is, fj(u) = (f1j(u), . . . , fnj(u)) with fij : U → U. Since fij ∈ HomG(U, U), Schur’s lemma implies that fij is multiplication by a scalar Aij ∈ C. By the linearity of f, we have f(v1, . . . , vn) = f1(v1) + ··· + fn(vn) ⊕n for all (v1, . . . , vn) ∈ U . It follows that f = ϕA with A = (Aij) ∈ Mn(C). Lemma 5.4.2. Let U be a nonzero finite dimensional complex vector space. Suppose 88 that U is a G-module for a finite group G. Write r M ⊕ni U = Ui i=1 where U1,...,Ur are irreducible G-modules such that Ui 6' Uj whenever i 6= j. Then r M ⊕ni ⊕ni HomG(U, U) = HomG(Ui ,Ui ). i=1 Proof. The inclusion ⊃ is trivial. Let f ∈ HomG(U, U). Fix indices i, j such that 1 ≤ i ≤ r and 1 ≤ j ≤ ni. Let Vij = {0} ⊕ · · · ⊕ {0} ⊕ Ui ⊕ {0} ⊕ · · · ⊕ {0}, where the only nonzero summand is in the (n1 + ··· + ni−1 + j)-th position. Let fij = f ◦ hij : Ui → U, where hij : Ui → Vij is the obvious isomorphism. Let γk` : Ui → Uk be the component functions of fij, that is, fij(u) = (γ11(u), . . . , γ1n1 (u), . . . , γr1(u), . . . , γrnr (u)) with γk` : Ui → Uk. Since γk` ∈ HomG(Ui,Uk), Schur’s lemma gives γk` = 0 for k 6= i. Consequently, f(Vi1 ⊕ · · · ⊕ Vini ) ⊂ Vi1 ⊕ · · · ⊕ Vini . 89 Therefore r M ⊕ni ⊕ni f ∈ HomG(Ui ,Ui ). i=1 Corollary 5.4.3. With the notation the previous lemma, we have r M HomG(U, U) ' Mni (C). i=1 5.5 Proof of Irreducibility Let S be any finite group. (Later on we will set S = Sd.) Let U be a nonzero finite dimensional complex vector space. Suppose that U is a right S-module. Then U is also a right module over the group algebra A = C[S]. Let B = HomS(U, U). So U is a left B-module. The actions of B and A on U commute by definition. If W is any left A-module, then U ⊗A W becomes a B-module via b · (u ⊗ w) = b(u) ⊗ w for b ∈ B, u ∈ U, and w ∈ W . Lemma 5.5.1. (i) For any c ∈ A, the map U ⊗A Ac → Uc sending u ⊗ a 7→ u · a is an isomorphism of B-modules. (ii) If W = Ac is an irreducible left A-module, then U ⊗A W ' Uc is 0 or an irreducible left B-module. (iii) If {Ac1, . . . , Ack} is a complete set of representatives of isomorphism types of 90 complex irreducible S-modules and we let di = dimC Aci, then M di U ' (Uci) as modules over B. Proof. Given c ∈ A, it is readily checked that the map U ⊗A Ac → Uc sending u ⊗ a 7→ u · a is a homomorphism of B-modules. The inverse Uc → U ⊗A Ac is given by uc 7→ u ⊗ c. The latter map is well defined because, if u, u0 ∈ U are such that uc = u0c, then u ⊗ c = uc ⊗ 1 = u0c ⊗ 1 = u0 ⊗ c. For (ii), suppose that W = Ac is an irreducible left A-module. We identify A with 0 L a direct sum of matrix algebras A = i Mni (C) (see [FH91], Proposition 3.29). We consider first the case where U is an irreducible A-module. Since the regular representation of A contains a copy of each irreducible A-module, A0 has a minimal right ideal U 0 isomorphic to U and a minimal left ideal W 0 isomorphic to W . By 0 Lemma 5.3.5, there exist indices i, j such that U is a minimal right ideal of Mni (C) 0 and W is a minimal left ideal of Mnj (C). Suppose first that i 6= j. Let Ii be the 0 0 ni × ni identity matrix in Mni (C). So, for all u ∈ U and w ∈ W , u ⊗A0 w = (u · Ii) ⊗A0 w = u ⊗A0 (Ii · w) = u ⊗A0 0 = 0. 0 0 Thus U ⊗A0 W = 0 if i 6= j. Now assume i = j and write n = ni = nj. By Lemma 5.3.2, there exist complex numbers c1, . . . , cn such that 0 n W = {(c1v| · · · |cnv): v ∈ C }. 91 Similarly, Lemma 5.3.3 gives 0 t n U = {(d1v| · · · |dnv) : v ∈ C }, n for some d1, . . . , dn ∈ C. For all v, w ∈ C , we have t t (d1v| · · · |dnv) ⊗A0 (c1w| · · · |cnw) = (d1v| · · · |dnv) ⊗A0 [(0| · · · |0|w) · (c11| · · · |cn1)] t = [(d1v| · · · |dnv) · (0| · · · |0|w)] ⊗A0 (c11| · · · |cn1) = hv, wi(0| · · · |0|d) ⊗A0 (c11| · · · |cn1), t n t n where d = (d1 ··· dn) ∈ C and 1 = (1 ··· 1) ∈ C . Therefore 0 0 dimC(U ⊗A W ) = dimC(U ⊗A0 W ) ≤ 1. Thus U ⊗A W is 0 or an irreducible B-module. Now we move on to the general case in which U may not be an irreducible A-module. By Maschke’s theorem, U decomposes as M ⊕mi U = Ui i where the Ui are irreducible right A-modules, not isomorphic to each other. Then M ⊕mi U ⊗A W ' (Ui ⊗A W ) . i We claim that at most one of the factors Ui ⊗A W is different from 0. Indeed, let U¯ and Uˆ be irreducible right A-modules such that U¯ 6' Uˆ. By above, U¯ and Uˆ correspond to minimal right ideals of A0, say U¯ 0 and Uˆ 0. So there exist indices i, j 92 ¯ 0 ˆ 0 such that U (resp. U ) is a minimal right ideal of Mni (C) (resp. Mnj (C)). If i = j, Lemma 5.3.4 implies U¯ ' Uˆ as A0-modules, a contradiction. So i 6= j. It follows by ¯ ˆ above that U ⊗A W = 0 or U ⊗A W = 0. Therefore U ⊗A W = 0 or ⊕mk ⊕mk U ⊗A W ' (Uk ⊗A W ) ' C for some index k. We only need to consider the latter case. By Corollary 5.4.3, we have M B ' Mmi (C). i ⊕mk Notice that only the factor Mmk (C) of B acts on U ⊗A W (' C ), and the action is by left multiplication. So Lemma 5.3.1 implies that U ⊗A W is an irreducible B-module. Finally suppose {Ac1, . . . , Ack} is a complete set of representatives of isomorphism types of complex irreducible S-modules. It is a fact from the theory of representations of finite groups that the group algebra A decomposes as a module over itself as M di A ' (Aci) where di = dimC Aci. (See, for example, Corollary 2.18 of [FH91].) Thus, using part (i), we deduce M di M di M di U ' U ⊗A A ' U ⊗A (Aci) ' (U ⊗A Aci) ' (Uci) as modules over B. ⊗d Note 5.5.2. Consider the particular case S = Sd and U = V , where V is a nonzero 93 finite dimensional complex module for a group G. Then U is a right A = C[Sd]- module as described in Section 5.1. Let γ be a partition of d and let T,T 0 be tableaux of shape γ. By Theorem 4.2.1 we have AcT ' AcT 0 over A. Thus ⊗d ⊗d V ⊗A AcT ' V ⊗A AcT 0 as modules over B. Therefore, part (i) of Lemma 5.5.1 implies ⊗d ⊗d V · cT ' V · cT 0 (5.32) over B. Since the left action of G on V ⊗d commutes with the right action of A, (5.32) is also an isomorphism of G-modules. This justifies the abuse of notation indicated in Note 5.1.1. ⊗d n Henceforth we set S = Sd and U = V , where V = C is the natural GL(n)- module. Given linear endomorphisms f1, . . . , fd in EndV , we define the element ⊗d (f1 ⊗ · · · ⊗ fd)! ∈ End(V ) (5.33) as the sum of all permutations of f1 ⊗ · · · ⊗ fd. For example, if d = 3, (f1 ⊗ f2 ⊗ f3)! = f1 ⊗ f2 ⊗ f3 + f1 ⊗ f3 ⊗ f2 + f2 ⊗ f1 ⊗ f3+ + f2 ⊗ f3 ⊗ f1 + f3 ⊗ f1 ⊗ f2 + f3 ⊗ f2 ⊗ f1. Observe that gl(n) acts on V ⊗d by means of the operators (f ⊗ I ⊗ · · · ⊗ I)! with f ∈ EndV , all of which are in B. 94 Lemma 5.5.3. B is generated as an algebra by the operators (f ⊗ I ⊗ · · · ⊗ I)! with f ∈ EndV . In particular, a subspace Y of V ⊗d is B-invariant if and only if it is gl(n)-invariant. ⊗d ⊗d Proof. Since V is a right Sd-module, End(V ) becomes a right Sd-module by −1 ⊗d ⊗d (ϕ · σ)(u) = ϕ(u · σ) · σ , ϕ ∈ End(V ), σ ∈ Sd, u ∈ V . Note that B is just the subset of End(V ⊗d) consisting of those maps that are fixed ⊗d by Sd. In addition (EndV ) is a right Sd-module under the action (f1 ⊗ · · · ⊗ fd) · σ = fσ(1) ⊗ · · · ⊗ fσ(d), fi ∈ EndV, σ ∈ Sd. In this case, the set of fixed points is the symmetric power Symd(EndV ). The map ⊗d ⊗d (EndV ) → End(V ) sending f1 ⊗ · · · ⊗ fd 7→ f1 ⊗ · · · ⊗ fd is an isomorphism of Sd-modules. (Observe that the symbol f1 ⊗· · ·⊗fd has two different meanings here). Therefore, the subspace Symd(EndV ) of (EndV )⊗d corresponds to the subspace B of End(V ⊗d). It follows that B is linearly spanned by the operators (5.33). Thus it will suffice to prove that all such operators are in the subalgebra S of B generated by the set {(f ⊗ I ⊗ · · · ⊗ I)! : f ∈ EndV }. Proceeding by induction on j, we show next that (f1 ⊗ · · · ⊗ fj ⊗ I ⊗ · · · ⊗ I)! ∈ S for all f1, . . . , fj ∈ EndV . The case j = 1 is trivial. Now suppose that the claim holds for an integer j such that 1 ≤ j < d. Let f1, . . . , fj+1 ∈ EndV . Since both (fj+1 ⊗ I ⊗ · · · ⊗ I)! and (f1 ⊗ · · · ⊗ fj ⊗ I ⊗ · · · ⊗ I)! 95 are in S, so does their composition, which is equal to (d − 1)! times ((fj+1 ◦ f1) ⊗ · · · ⊗ fj ⊗ I ⊗ · · · ⊗ I)! + ··· + (f1 ⊗ · · · ⊗ (fj+1 ◦ fj) ⊗ I ⊗ · · · ⊗ I)!+ + (d − j)(f1 ⊗ · · · ⊗ fj ⊗ fj+1 ⊗ I ⊗ · · · ⊗ I)!. Since the first j terms of the previous sum are in S, so does the last one. This completes the induction. The case j = d implies the conclusion. Proposition 5.5.4. Let γ be a partition of d such that γn+1 = 0. Then SγV is irreducible over GL(n), gl(n), SL(n), and sl(n). Proof. We already know that γn+1 = 0 implies SγV 6= 0. As a consequence of Lemmas 5.5.1 and 5.5.3, SγV is an irreducible module over gl(n). Now we use the decomposition gl(n) = sl(n)⊕s, where s is the space of all n×n scalar matrices. Since ⊗d s acts on V by means of scalar operators, we deduce that SγV is also irreducible over sl(n). Finally, Lemma 3.4.11 shows that SγV is irreducible over the groups GL(n) and SL(n). Note 5.5.5. In the previous proposition we used the Lie Group-Lie Algebra relation- ship to deduce irreducibility over GL(n) and SL(n). However, this can be achieved in a purely algebraic fashion by proving the following result. ⊗d ⊗d Theorem 5.5.6. B = HomSd (V ,V ) is generated as an algebra by all g ⊗ · · · ⊗ g with g ∈ GL(V ). Proof. By Lemma 5.5.3, it suffices to see that we can generate all (f ⊗I ⊗· · ·⊗I)! with f ∈ EndV . Let T be the subspace of B spanned by all g ⊗ · · · ⊗ g, g ∈ GL(V ), and let W be the vector space B/T . Given f ∈ EndV , choose distinct complex numbers 96 α1, . . . , αd so that none of their opposites is an eigenvalue of f. Then f +αiI ∈ GL(V ) for all 1 ≤ i ≤ d and we have (f + αiI) ⊗ · · · ⊗ (f + αiI) ≡ 0 mod T, 1 ≤ i ≤ d. This gives d−1 v0 + αiv1 + ··· + αi vd−1 = 0, 1 ≤ i ≤ d, where 1 d−j j vj = (d−j)!j! (f ⊗ · · · ⊗ f ⊗ I ⊗ · · · ⊗ I)! + T, 0 ≤ j ≤ d − 1. Since the numbers αi are all different, the Vandermonde matrix V (α1, . . . , αd) is invertible. Hence v0 = ··· = vd−1 = 0, as required. It follows that a subspace of V ⊗d is B-invariant if and only if it is GL(n)-invariant. Thus SγV is irreducible over GL(n) if γn+1 = 0. Since the scalar matrices in GL(n) act on V ⊗d as scalar operators, the decomposition GL(n) = C× · SL(n) implies irreducibility over SL(n). Now we complete the proof of Theorem 5.1.2. It only remains to prove part (vii). We know from Theorem 4.2.1 that the set {Acγ}, where γ runs over all the partitions of d, is a complete set of representatives of isomorphism classes of complex irreducible Sd-modules. Let dγ = dimC Acγ Thus, applying part (iii) of Lemma 5.5.1, we deduce ⊗d M ⊗d dγ M dγ V ' [V · cγ] = [SγV ] , γ γ as modules over GL(n), gl(n), SL(n), and sl(n). 97 Chapter 6 Irreducible Regular Representations of GL(n) n As shown in the previous chapter, the modules Sγ(C ) with γn+1 = 0 provide all the irreducible regular representations of SL(n). Although these modules are also irreducible over GL(n), they are not all the irreducible regular modules for GL(n). In this chapter we complete the picture by tensoring with suitable 1-dimensional representations of GL(n). Naturally enough, the determinant function will play a central role. We begin by finding all the finite dimensional irreducible modules for gl(n). The second section introduces some technical tools, which will be useful in this chapter and in the next one. In the third section we construct and classify the irreducible regular representations of GL(n). The ground field of all vector spaces is assumed to be C. 98 6.1 Irreducible Modules for gl(n) Given a complex number k, we define a representation of gl(n) on the 1-dimensional vector space C by X · z = k tr(X)z, X ∈ gl(n), z ∈ C. Whenever we consider this action on C, we adopt the notation kT = C. Clearly, kT is irreducible over gl(n). In addition, different values of k produce non-isomorphic modules. In fact, suppose that there is a linear isomorphism P : kT → k0T that commutes with the action of gl(n). Then P (k tr(I)z) = k0 tr(I)P (z) for all z ∈ C, which implies k = k0. Proposition 6.1.1. Every finite dimensional irreducible gl(1)-module is isomorphic to kT for a suitable k ∈ C. Moreover, kT ' k0T if and only if k = k0. Proof. Let (R,W ) be a finite dimensional irreducible representation of gl(1). Schur’s Lemma says that gl(1) acts on W by means of scalar operators. So W is 1-dimensional and there exists k ∈ C such that R(z) = kzIW for all z ∈ C. Thus W ' kT . For the remainder of this section we assume n > 1 and we let V = Cn be the natural module for gl(n). Theorem 6.1.2. Let k ∈ C and let γ be a partition of d such that γn+1 = 0. Then SγV ⊗ kT ' SγV over sl(n). In particular, SγV ⊗ kT is irreducible over gl(n). The identity matrix I ∈ gl(n) acts on SγV ⊗ kT by multiplication by d + nk. Proof. The map SγV → SγV ⊗ kT sending w 7→ w ⊗ 1 is an isomorphism of sl(n)- modules. Applying Theorem 5.1.2, we deduce that SγV ⊗ kT is irreducible over sl(n) 99 ⊗d and, in particular, over gl(n). Since SγV ⊂ V , the scalar matrix I ∈ gl(n) acts on SγV ⊗ kT by multiplication by d + k tr(I) = d + nk. Lemma 6.1.3. Let (R,W ) be a finite dimensional irreducible representation of gl(n). Then the isomorphism type of R is determined by R|sl(n) and by the complex number p such that R(I) = pIW . Proof. The center of gl(n) is s, the space of all scalar matrices. Since gl(n) = s⊕sl(n), the result follows applying Schur’s Lemma. 0 Corollary 6.1.4. Let γ and µ be partitions such that γn+1 = µn+1 = 0. Let k, k ∈ C. Then 0 SγV ⊗ kT ' SµV ⊗ k T (6.34) over gl(n) if and only if 0 γ1 − µ1 = ··· = γn − µn = k − k. (6.35) In addition, kT ' k0T over gl(n) if and only if k = k0. Proof. We have already observed that the last assertion is true. P 0 P Let d = γi and d = µi. By Theorem 6.1.2, the identity matrix in gl(n) acts 0 0 0 on SγV ⊗ kT (resp. SµV ⊗ k T ) by multiplication by d + nk (resp. d + nk ). Suppose now that (6.35) holds. Then SγV ' SµV over sl(n) by part (v) of Theorem 5.1.2. So Theorem 6.1.2 gives (6.34) over sl(n). In addition 0 0 X 0 0 0 (d + nk) − (d + nk ) = (γi − µi) + n(k − k ) = n(k − k) − n(k − k ) = 0. Thus Lemma 6.1.3 implies (6.34) over gl(n). 100 Conversely, suppose (6.34) over gl(n). In particular, this is an isomorphism of sl(n)-modules. Hence Theorem 5.1.2 says that the difference γi − µi is constant. So the equality d + nk = d0 + nk0 can be written as 0 n(λ1 − µ1) = n(k − k), 0 whence λ1 − µ1 = k − k. Theorem 6.1.5. Consider the set n−1 Ω = {(γ1, . . . , γn−1, k) ∈ Z × C : γ1 ≥ · · · ≥ γn−1 ≥ 0}. Then the map sending SγV ⊗ kT if γ1 > 0 and γ = (γ1, . . . , γn−1), (γ1, . . . , γn−1, k) 7→ kT if γ1 = 0, gives a bijection between Ω and the set of isomorphism classes of finite dimensional irreducible gl(n)-modules. Proof. The map is well defined by Theorem 6.1.2 and injective by Corollary 6.1.4. Now let (R,W ) be a finite dimensional irreducible regular representation of gl(n). By Schur’s Lemma, R(I) = pIW for some p ∈ C. Since gl(n) = sl(n) ⊕ s, we deduce that W is irreducible over sl(n). The corresponding highest weight has the form γ1ε1 + ··· + γn−1εn−1 P p−d where the γi are integers such that γ1 ≥ · · · ≥ γn−1 ≥ 0. Let d = γi and k = n , 101 so that R(I) = (d + nk)IW . Therefore, Theorem 6.1.2 and Lemma 6.1.3 imply SγV ⊗ kT if γ1 > 0 and we take γ = (γ1, . . . , γn−1), W ' kT if γ1 = 0. as modules over gl(n). 6.2 Rational Characters Definition 6.2.1. Let G be a linear algebraic group. A rational character of G is a regular homomorphism χ : G → C×. Denote by X (G) the set of all rational characters of G. It has the natural structure of an abelian group under pointwise multiplication, i.e., (χ1χ2)(g) = χ1(g)χ2(g) for χ1, χ2 ∈ X (G) and g ∈ G. Definition 6.2.2. A linear algebraic group T is called an algebraic torus of rank ` when it is isomorphic to (C×)`, the direct product of ` copies of C×. Note that if G is any of the groups GL(`), SL(`+1), Sp(2`), SO(2`), or SO(2`+1), then the diagonal subgroup H is an algebraic torus of rank `. Lemma 6.2.3. Let T be an algebraic torus of rank `. Then X (T ) ' Z` as groups. Proof. To simplify the notation, we can assume T = (C×)`. For each element m = (m1, m2, . . . , m`) (6.36) 102 of Z`, we define a rational character φ(m): T → C× by m1 m2 m` φ(m)(z1, z2, . . . , z`) = z1 z2 ··· z` . Thus we obtain a group homomorphism φ : Z` → X (T ). To see that φ is injective, suppose that (6.36) is an element of Z` such that φ(m) = 1. For 1 ≤ i ≤ ` and z ∈ C×, we have 1 = φ(m)(1,...,z,..., 1) = zmi . So mi = 0 and hence m = 0. It follows that φ is injective. Now let χ ∈ X (T ) and fix × × an index i between 1 and `. Consider the function χi : C → C defined by χi(z) = χ(1,..., 1, z, 1,..., 1), where z occupies the i-th coordinate position. Since χi is a regular homomorphism, mi × Lemma 3.7.3 implies that there exists mi ∈ Z such that χi(z) = z for all z ∈ C . Therefore, for all w = (z1, . . . , z`) ∈ T , m1 m2 m` χ(w) = χ1(z1)χ2(z2) ··· χ`(z`) = z1 z2 ··· z` = φ(m1, . . . , m`)(w). It follows that φ is surjective. Proposition 6.2.4. Let G be any of the groups GL(`), SL(`+1), Sp(2`), SO(2`), or SO(2`+1). Let g be the Lie algebra of G. Denote by H the diagonal subgroup of G and let h stand for the diagonal subalgebra of g. Let (ρ, V ) be a regular representation of G. Suppose that v ∈ V is a common eigenvector for the action of h on V (via dρ). Let α ∈ h∗ be the corresponding weight. Then v is a common eigenvector for the action of 103 H on V . In addition, there exist integers m1, . . . , m` such that α = m1ε1 + ··· + m`ε` and m1 m2 m` ρ(h)(v) = (h11 h22 ··· h`` )v for all h ∈ H. Proof. Given h ∈ H, there exists X ∈ h such that h = exp X. Then " ∞ # ∞ X 1 X 1 ρ(h)(v)=ρ(exp X)(v)=(exp dρ(X))(v)= dρ(X)j (v) = dρ(X)j(v) . j! j! j=0 j=0 Since dρ(X)j(v) ∈ Cv for all j, we deduce that ρ(h)(v) is a scalar multiple of v. Consequently, there exists a function χ : H → C× such that ρ(h)(v) = χ(h)v for all h ∈ H. Keeping in mind that v 6= 0 and ρ preserves products, we see that χ is a group homomorphism. Moreover, χ is regular because ρ is. Thus χ is a rational character of H. Hence, the proof of Lemma 6.2.3 yields integers m1, . . . , m` such that m1 m2 m` χ(h) = h11 h22 ··· h`` . Let X ∈ h. Then exp(tX) ∈ H for all t ∈ R. Observe that, for all t, exp(tX) is diagonal with entries etXii . In the following calculation we fix a basis of V , and we n use it identify V ' C and End(V ) ' Mn(C). Therefore, d d d dρ(X)v = ρ(exp(tX))| v = [ρ(exp(tX))v]| = χ(exp(tX))v| = dt t=0 dt t=0 dt t=0 d = et(m1X11+···m`X``)v| = (m X + ··· m X )v dt t=0 1 11 ` `` = (m1ε1 + ··· + m`ε`)(X)v. Thus α = m1ε1 + ··· + m`ε`. 104 6.3 Irreducible Regular Modules for GL(n) Throughout this section, n is a fixed positive integer and V = Cn is the natural module for GL(n). We assume n > 1 because we have already classified the regular representations of GL(1) = C× in Section 3.7.3. For each integer k, we define a representation of GL(n) on the 1-dimensional complex vector space C via k g · z = (det g) z, g ∈ GL(n), z ∈ C. When we consider this action on C, we write Dk instead of C. It is clear that Dk is an irreducible regular module for GL(n). Lemma 6.3.1. Given k ∈ Z, the differential of Dk is kT , as defined in Section 6.1. Proof. Let ρ be the representation of GL(n) on Dk. We identify each element of gl(C) with its 1 × 1 matrix relative to the basis {1} of C. Then, for all X ∈ gl(n), d d d dρ(X) = ρ(exp(tX))| = [det(exp(tX))]k| = ektr(tX)| = ktr(X). dt t=0 dt t=0 dt t=0 k Let γ be a partition such that γn+1 = 0. Then SγV ⊗ D is a regular GL(n)- module because each of the factors is. In view of the previous lemma, the differential k k of SγV ⊗ D is SγV ⊗ kT . So Lemma 3.4.11 and Theorem 6.1.2 imply that SγV ⊗ D 0 is irreducible over GL(n). If k ∈ Z and µ is a partition with µn+1 = 0, Theorem k k0 3.4.10 and Corollary 6.1.4 show that SγV ⊗ D ' SµV ⊗ D over GL(n) if and only 0 k k0 0 if γi − µi = k − k for 1 ≤ i ≤ n. Similarly, D ' D over GL(n) if and only if k = k . 105 Theorem 6.3.2. Consider the set n Ω = {(γ1, . . . , γn−1, k) ∈ Z : γ1 ≥ · · · ≥ γn−1 ≥ 0}. Then the map sending k SγV ⊗ D if γ1 > 0 and γ = (γ1, . . . , γn−1), (γ1, . . . , γn−1, k) 7→ k D if γ1 = 0, gives a bijection between Ω and the set of isomorphism classes of irreducible regular GL(n)-modules. Proof. The map is well defined and injective by the discusion above. Now let (ρ, W ) be an irreducible regular representation of GL(n). By Theorems 3.4.12 and 3.6.1, W is irreducible over gl(n) and hence over sl(n). Let w be the maximal vector (unique up to scaling) for the action of sl(n) on W . The corresponding highest weight has the form γ1ε1 + ··· + γn−1εn−1 (6.37) where the γi are integers such that γ1 ≥ · · · ≥ γn−1 ≥ 0. Let X be a diagonal matrix in gl(n). So we can write X = Y + Z with tr(X) tr(X) Y = X − I ∈ sl(n) and Z = I ∈ s. n n 106 Then a calculation shows that dρ(X)(w) = [(γ1 + k)ε1 + ··· + (γn−1 + k)εn−1 + kεn](X)w for some scalar k ∈ C independent from X. Proposition 6.2.4 implies k ∈ Z. Since dρ(I) is a scalar operator, we must have dρ(I) = (γ1 + ··· + γn−1 + nk)IW = (d + nk)IW . P where d = γi. Summarizing, we have shown that W is the irreducible sl(n) module with highest weight (6.37) and that I ∈ gl(n) acts on W by multiplication by d + nk. Therefore, Theorem 6.1.2 and Lemma 6.1.3 imply SγV ⊗ kT if γ1 6= 0 and we take γ = (γ1, . . . , γn−1), W ' kT if γ1 = 0. as modules over gl(n). Since k ∈ Z, the result follows in view of Lemma 6.3.1 and Theorem 3.4.10. 107 Chapter 7 The Symplectic and Orthogonal Cases In this chapter we describe Weyl’s construction in the symplectic and orthogonal cases. We obtain all the irreducible regular representations of the groups Sp(n), SO(n), and O(n). At the Lie algebra level, this method produces all the irreducible finite dimensional modules for sp(n), but only half of the finite dimensional irreducible so(n)-modules. The odd orthogonal case is quite simpler than the even orthogonal case. In fact, if n is odd then det(−I) = −1 and hence O(n) = SO(n) ∪ (−I)SO(n). Since −I is a central element in O(n), this decomposition will make it very easy to relate the irreducible representations of O(n) and SO(n). However, when n is even, there is no central element in O(n) \ SO(n). In this situation, the relationship between the irreducible representations of O(n) and SO(n) is more involved. Before we deal with the even orthogonal case, we devote a whole section to introduce some special tools. We assume throughout that the ground field of all vector spaces is C. 108 7.1 The Construction Let V = Cn be the natural module for GL(n). Let ω : V × V → C stand for a bilinear form whose Gram matrix relative to the canonical basis B = {e1, . . . , en} of V is one of the following matrices 1 0 0 0 I` 0 I` , , 0 0 I . (7.38) ` −I` 0 I` 0 0 I` 0 Let G = G(ω) and g = Lie(G). So G = Sp(n) and g = sp(n), or G = O(n) and g = so(n) depending on whether ω is skew-symmetric or symmetric. Let d ∈ N. For each pair J = (p, q) of integers such that 1 ≤ p < q ≤ d, the form ω determines a linear map ⊗d ⊗(d−2) ΦJ : V → V sending v1 ⊗ · · · ⊗ vd 7→ ω(vp, vq)v1 ⊗ · · · ⊗ vˆp ⊗ · · · ⊗ vˆq ⊗ · · · ⊗ vd. The maps ΦJ are called contractions. The intersection of the kernels of all these contractions is denoted by V hdi if ω is skew-symmetric, and by V [d] if ω is symmetric. It is clear that V hdi and V [d] are invariant under the action of the symmetric group ⊗d Sd on V . Lemma 7.1.1. Let g ∈ G and J = (p, q) with 1 ≤ p < q ≤ d. Then ΦJ (g · v) = g · ΦJ (v) (7.39) 109 for all v ∈ V ⊗d. In particular, V hdi is invariant under Sp(n) and sp(n), and V [d] is invariant under O(n) and so(n). Proof. The deduction of (7.39) is just a calculation based on the fact that G preserves ω. Hence V hdi is Sp(n)-invariant and V [d] is O(n)-invariant. The remaining assertions follow from Lemma 3.4.11. Definition 7.1.2. Let γ be a partition of d. We define hdi [d] ShγiV = V ∩ SγV and S[γ]V = V ∩ SγV. ⊗d It follows from the previous lemma that ShγiV is an Sp(n)-submodule of V , and ⊗d S[γ]V is a O(n)-submodule of V . The remainder of this section is devoted to prove the following facts: • The isomorphism classes of ShγiV and S[γ]V do not depend on the particular tableau used to construct SγV . So it makes sense to index these modules with partitions rather than tableaux. • ShγiV (resp. S[γ]V ) is irreducible over Sp(n) (resp. O(n)) when it is not 0. • V hdi (resp. V [d]) decomposes over Sp(n) (resp. O(n)) as a direct sum of various ShγiV (resp. S[γ]V ). The dual basis of B relative to ω will be denoted by {u1, . . . , un}. So ω(ei, uj) = δij for all i, j. Let n X ψ = ei ⊗ ui. i=1 110 For each pair J = (p, q) of integers such that 1 ≤ p < q ≤ d, we define a linear map ⊗(d−2) ⊗d ΨJ : V → V by n X v1 ⊗ · · · ⊗ vd−2 7→ v1 ⊗ · · · ⊗ vp−1 ⊗ ei ⊗ vp ⊗ · · · ⊗ vq−2 ⊗ ui ⊗ vq−1 ⊗ · · · ⊗ vd−2, i=1 that is, we insert ei and ui respectively in the p-th and q-th factors of each elementary tensor and we sum over all possible values of i. Observe that ΦJ ◦ΨJ is multiplication by n. In particular, ΨJ is injective and ΦJ is surjective. We also define ⊗d ⊗d ϑJ = ΨJ ◦ ΦJ : V → V . Then ker ϑJ = ker ΦJ because ΨJ is injective. Let ( , ): V × V → C be the usual inner product in V . Then we have an inner product in V ⊗ V defined by (v1 ⊗ v2, w1 ⊗ w2) = (v1, w1)(v2, w2) ⊗d for all v1, v2, w1, w2 ∈ V . Likewise, we can define an inner product in V such that (v1 ⊗ · · · ⊗ vd, w1 ⊗ · · · ⊗ wd) = (v1, w1) ··· (vd, wd) (7.40) for vi, wi ∈ V . This inner product is Sd-invariant, in the sense that (v·σ, w·σ) = (v, w) ⊗d for all v, w ∈ V and σ ∈ Sd. 111 ⊥ Lemma 7.1.3. Let J = (p, q) as above. Then ker ΦJ = (imΨJ ) with respect to the inner product (7.40). Proof. We show first that (v ⊗ w, ψ) = ω(v, w) for all v, w ∈ V . Indeed, let n n X X v = ajej and w = bjuj j=1 j=1 be arbitrary elements of V . If we write out the vectors u1, . . . , un, we see that they form an orthonormal basis of V . Thus n n X X (v ⊗ w, ψ) = (v, ei)(w, ui) = aibi = ω(v, w). i=1 i=1 For the ease of notation we consider only the case J = (1, 2). But the argument applies in general. Let Φ = ΦJ and Ψ = ΨJ . Since Φ is surjective and Ψ is injective, we have dim(imΨ)⊥ = dim V ⊗d − dim(imΨ) = dim V ⊗d − dim V ⊗(d−2) = dim(ker Φ). So the equality ker Φ = (imΨ)⊥ will follow if we prove the inclusion ⊂. Let r X 1 d x = vj ⊗ · · · ⊗ vj ∈ ker Φ , y = w3 ⊗ · · · ⊗ wd j=1 ⊗(d−2) with w3, . . . , wd ∈ V . Since such vectors y span V , it suffices to verify that 112 (x, Ψ(y)) = 0. We achieve this by direct calculation: r n X X 1 2 3 d (x, Ψ(y)) = (vj , ei)(vj , ui)(vj , w3) ··· (vj , wd) j=1 i=1 r X 1 2 3 d = (vj ⊗ vj , ψ)(vj ⊗ · · · ⊗ vj , y) j=1 r X 1 2 3 d = ω(vj , vj )(vj ⊗ · · · ⊗ vj , y) j=1 = (Φ(x), y). The result follows because Φ(x) = 0. P Corollary 7.1.4. The orthogonal complement of J imΨJ with respect to the inner product (7.40) is V hdi if ω is skew-symmetric, and V [d] if ω is symmetric. Proof. If A1,A2,...,Ar are subspaces of a finite dimensional Hilbert space, then r !⊥ r \ X ⊥ Ai = Ai . i=1 i=1 Consequently, !⊥ \ X ⊥ X ker ΦJ = (ker ΦJ ) = imΨJ . J J J Lemma 7.1.5. Let γ be a partition of d. Then hdi [d] ShγiV = V · cγ , S[γ]V = V · cγ. Proof. We consider only the symplectic case; the orthogonal case is analogous. We hdi P know that V is invariant under the action of Sd. Since C = J imΨJ is the 113 hdi orthogonal complement of V with respect to the Sd-invariant inner product (7.40), we deduce that also C is Sd invariant. Thus ⊗d hdi hdi SγV = V · cγ = (V ⊕ C) · cγ = (V · cγ) ⊕ (C · cγ) hdi hdi with V · cγ ⊂ V and C · cγ ⊂ C. Now the result follows taking the intersection with V hdi. Corollary 7.1.6. Let γ be a partition of d. Then the isomorphism class of ShγiV (resp. S[γ]V ) as a module over Sp(n) (resp. O(n)) does not depend on the particular tableau of shape γ used to construct SγV . 0 Proof. Let T and T be tableaux of shape γ. By Theorem 4.2.1, C[Sd]cT and C[Sd]cT 0 hdi hdi are isomorphic as Sd-modules. Applying Lemma 5.5.1, we deduce V ·cT ' V ·cT 0 hdi hdi as B-modules, where B = HomSd (V ,V ) is the set of those linear endomorphisms hdi hdi hdi of V that commute with the action of Sd. In particular, V · cT and V · cT 0 are isomorphic over Sp(n). Now the conclusion follows from the previous lemma. The same considerations apply to the orthogonal case. Theorem 7.1.7. If γ is a partition of d, then ShγiV is 0 or an irreducible Sp(n)- module, and S[γ]V is 0 or an irreducible O(n)-module. For each partition γ of d, let dγ stand for the dimension of the complex vector space C[Sd]cγ defined in Chapter 4. Then we have hdi M dγ [d] M dγ V ' [ShγiV ] over Sp(n) and V ' [S[γ]V ] over O(n), γ γ where γ runs over all the partitions of d. 114 Proof. We consider only the symplectic case; the orthogonal case is completely analo- hdi hdi gous. Suppose that ShγiV 6= 0. Let B = HomSd (V ,V ). Since C[Sd]·cγ is an irre- hdi ducible left C[Sd]-module (see Theorem 4.2.1), Lemma 5.5.1 says that ShλiV = V ·cγ is an irreducible left B-module. Let ρ be the natural representation of Sp(n) on V . We claim that B is linearly spanned by the operators ρ(g) ⊗ · · · ⊗ ρ(g)|V hdi (7.41) with g ∈ Sp(n). These operators are clearly in B. Now let f ∈ B. As V ⊗d = V hdi ⊕C P ˜ ⊗d ⊗d with C = J imΨJ , we can extend f to a linear map f : V → V that is zero hdi ˜ on C. Since both V and C are Sd-invariant, a calculation shows that f commutes ˜ with the action of Sd. We show next that f also commutes with all the operators ϑJ . hdi Let J = (p, q) with 1 ≤ p < q ≤ d. Let v = v1 + v2 with v1 ∈ V and v2 ∈ C. hdi ˜ Keeping in mind that V ⊂ ker ΦJ = ker ϑJ and im(ϑJ ) ⊂ C ⊂ ker f, we have ˜ ˜ ˜ ˜ (f ◦ ϑJ )(v) = (f ◦ ϑJ )(v1 + v2) = f(0 + ϑJ (v2)) = 0 = ϑJ (f(v1)) = (ϑJ ◦ f)(v). By a result on invariant theory (Theorem A.7.1 of Appendix A), we deduce that f˜ is a linear combination of operators of the form ρ(g) ⊗ · · · ⊗ ρ(g) with g ∈ Sp(n). It follows that f is a linear combination of operators of the form (7.41). Consequently, a subspace of V hdi is Sp(n)-invariant if and only if it is B-invariant. hdi Since V · cγ is irreducible over B, it is also irreducible over Sp(n). Finally, applying Theorem 4.2.1 and part (iii) of Lemma 5.5.1, we obtain hdi M hdi dγ M dγ V ' [V · cγ] = [ShγiV ] γ γ 115 as modules over B and, in particular, over Sp(n). 7.2 The Symplectic Case In this section, V = Cn is the natural module for Sp(n) with n = 2` ≥ 2. The nondegenerate skew-symmetric bilinear form defining Sp(n) is denoted by ω. Our main result is the following. Theorem 7.2.1. Let γ be a partition of a positive integer. Assume γ`+1 = 0. Then ShγiV is an irreducible module for Sp(n). In addition, ShγiV is the irreducible sp(n)- module with highest weight γ1ε1 + ··· + γ`ε`. In particular, any non-trivial finite dimensional irreducible sp(n)-module is isomorphic to ShγiV for a unique partition γ such that γ`+1 = 0. Any non-trivial irreducible regular Sp(n)-module is isomorphic to ShγiV for a unique γ with γ`+1 = 0. Proof. Let ` ∈ N and set V = Cn with n = 2`. Let γ be a partition of a positive integer d such that γ`+1 = 0. In the Young diagram corresponding to γ, we write down the numbers 1, . . . , d in consecutive order down the columns, from left to right. In this way we obtain a tableau T . For example, if γ = (3, 2, 1, 1) then 1 5 7 T = 2 6 (7.42) 3 4 Using the notation of Section 5.2, we have eT = (e1 ⊗ · · · ⊗ eµ1 ) ⊗ · · · ⊗ (e1 ⊗ · · · ⊗ eµk ), 116 where µ is the conjugate partition of γ. For instance, if γ = (3, 2, 1, 1) then µ = (4, 2, 1), and hence eT = e1 ⊗ e2 ⊗ e3 ⊗ e4 ⊗ e1 ⊗ e2 ⊗ e1. hdi Observe that eT ∈ V because µ1 ≤ ` and {e1, . . . , e`} spans a totally isotropic subspace of V relative to ω. Applying Lemma 5.2.3, we deduce that eT · cT is a hdi maximal vector of the sp(n)-module V · cT . The corresponding weight is γ1ε1 + ··· + γ`ε`. (7.43) hdi hdi In particular V ·cT 6= 0. So Theorem 7.1.7 implies that V ·cT is an irreducible module for Sp(n). Since Sp(n) is connected (by Corollary 3.6.3), Theorem 3.4.12 hdi implies that V · cT is also irreducible over sp(n). Therefore ShγiV is the irreducible sp(n)-module with highest weight (7.43). Keeping in mind the highest weight the- ory for sp(n), it follows that each finite dimensional irreducible module for sp(n) is isomorphic to SγV for a unique partition γ with γ`+1 = 0. Now let U be an irreducible regular Sp(n)-module. By Theorem 3.4.12, U is irreducible over sp(n). So U ' ShγiV over sp(n) for a suitable partition γ with γ`+1 = 0. Applying Theorem 3.4.10, we obtain U ' ShγiV as Sp(n)-modules. Any other partition µ such that µ`+1 = 0 will give a module with a different highest weight and hence not isomorphic to U. Example 7.2.2. Consider the partition γ = (1, 1,..., 1) of d, where 1 ≤ d ≤ `. Then hdi d ShγiV = V ∩ Λ V 117 is the irreducible sp(n)-module with highest weight λd = ε1 + ··· + εd. So ShγiV is the d-th fundamental module V (λd) for sp(n). If d = 1 then ShγiV is just V and hence it has dimension n. Suppose now that d ≥ 2. A simple symmetry d hdi argument shows that a vector w ∈ Λ V is in V if and only if Φ(1,2)(w) = 0. Thus d ShγiV = ker Φ(1,2) ∩ Λ V . In addition, we can check by direct calculation that, for all w ∈ ΛdV , we have Φ(1,2)(w) = 2θ(w), where θ :ΛdV → Λd−2V is the linear map sending X r+s−1 v1 ∧ · · · ∧ vd 7→ (−1) ω(vr, vs)v1 ∧ · · · ∧ vˆr ∧ · · · ∧ vˆs ∧ · · · ∧ vd. r Therefore V (λd) = ShγiV = ker θ. It can be shown that θ is surjective. For a proof of this fact, the reader may consult [Car05]. It follows that, for d ≥ 2, the dimension of the d-th fundamental module for sp(n) is n n − . d d − 2 Moreover, using the first isomorphism theorem, we obtain d d−2 Λ V ' V (λd) ⊕ Λ V. 118 as modules for sp(n). Now, by induction, we easily decompose ΛdV as d Λ V ' V (λd) ⊕ V (λd−2) ⊕ V (λd−4) ⊕ · · · , where the last summand is the trivial sp(n)-module is d is even, and the natural sp(n)-module V (λ1) = V if V is odd. Example 7.2.3. Consider the partition γ = (d), where d ≥ 2. Then hdi d ShγiV = V ∩ Sym V d hdi is the irreducible sp(n)-module with highest weight dε1. We claim that Sym V ⊂ V . d Indeed, let v = v1v2 ··· vd be a typical spanning vector of Sym V , that is, X v = vσ(1) ⊗ vσ(2) ⊗ · · · ⊗ vσ(d) σ∈Sd with vi ∈ V . Given integers p, q such that 1 ≤ p < q ≤ n, let J = (p, q). Let τ denote the transposition (p q). Then Sd is the disjoint union of the alternating subgroup Ad and its coset Adτ. To simplify the notation, we assume p = 1 and q = 2. However, 119 the argument works in general. Since ω is skew-symmetric, X ΦJ (v) = ω(vσ(1), vσ(2))vσ(3) ⊗ · · · ⊗ vσ(d) σ∈Sd X = [ω(vσ(1), vσ(2))vσ(3) ⊗ · · · ⊗ vσ(d) + ω(vστ(1), vστ(2))vστ(3) ⊗ · · · ⊗ vστ(d)] σ∈Ad X = [ω(vσ(1), vσ(2)) + ω(vσ(2), vσ(1))]vσ(3) ⊗ · · · ⊗ vσ(d) σ∈Ad = 0. Thus v ∈ V hdi and hence SymdV ⊂ V hdi. It follows that d ShγiV = Sym V. In the special case d = 2, we see that Sym2V is the irreducible sp(n)-module with 2 highest weight 2ε1. Therefore Sym V is isomorphic to the adjoint module of sp(n), i.e., Sym2V ' sp(n) (7.44) as modules for sp(n). Since Sp(n) is connected, Theorem 3.4.10 and Example 3.4.8 imply that (7.44) is also an isomorphism of Sp(n)-modules, where the action of the symplectic group on its Lie algebra is given by conjugation: g · X = gXg−1. Note 7.2.4. Theorem 7.2.1 implies that there is a bijective correspondence between the irreducible regular representations of Sp(n) and the finite dimensional irreducible representations of sp(n). Thus, if U is an irreducible regular module for Sp(n) and the highest weight of sp(n) acting on U is λ, it is often said that U is the irreducible Sp(n)-module with highest weight λ. 120 Note 7.2.5. As a consequence of Theorem 7.2.1, we have ShγiV 6= 0 when γ`+1 = 0. The converse implication is also true: if γ`+1 6= 0, then ShγiV = 0. For a proof of this fact, we refer the reader to Section VI.3 of [Wey39], or to Exercise 17.18 of [FH91]. 7.3 Admissible and Associated Partitions In this section, n is a fixed natural number. We say that a partition γ of a positive integer is n-admissible if µ1 +µ2 ≤ n, where µ is the conjugate partition of γ. Geometrically, γ is n-admissible when the sum of the lengths of the first two columns of its Young diagram is less than or equal to n. Suppose that γ and γ0 are n-admissible. Let µ (resp. µ0) denote the conjugate 0 0 0 0 partition of γ (resp. γ ). Then γ and γ are n-associated if µ1 + µ1 = n and µi = µi for i > 1. In other words, γ and γ0 are n-associated when the sum of the lengths of the first columns of their Young diagrams is n and all the other columns of their diagrams have the same length. Clearly, this is a symmetric relation. For example, γ = (4, 2) and γ0 = (4, 2, 1, 1) are 6-associated. The corresponding Young diagrams are γ : , γ0 : . Notice that the partitionγ ˜ = (1,..., 1) of n is n-admissible but it lacks of an n-associate partition. Assume henceforth that γ is an n-admissible partition differ- ent fromγ ˜ with conjugate partition µ. We claim that γ has a unique n-associated partition. Uniqueness is obvious from the definition. To prove existence, we replace the first column of the Young diagram of γ (which has length µ1) by a column con- taining exactly n − µ1 boxes. The diagram obtained in this way corresponds to an 121 n-admissible partition that is n-associated to γ if and only if 0 < n − µ1, µ2 ≤ n − µ1, and n − µ1 + µ2 ≤ n. But these conditions are easily seen to be met. Therefore, γ has a unique n-associated partition, which we will denote by γ\. n Observe that γ is self-associated if and only if µ1 = 2 . In particular, γ cannot be self-associated if n is odd. The partitionγ ˜ is special in the sense that it is the only n-admissible partition not having an n-associate partition. We consider this case separately. Recall from Chapter 6 that for each k ∈ Z, the 1-dimensional vector space C becomes a regular GL(n)-module via g · z = (det g)kz. In this case we use the notation C = Dk. n n Proposition 7.3.1. Let V = C . Then S[˜γ]V = Λ V. In particular, S[˜γ]V is the trivial sl(n)-module, i.e., the irreducible sl(n)-module with highest weight 0. Moreover, S[˜γ]V 1 is the trivial SL(n)-module and S[˜γ]V ' D over O(n). Proof. It is clear that n S[˜γ]V = Λ V = C(e1 ∧ · · · ∧ en). Thus sl(n) acts as 0 on the 1-dimensional vector space ΛnV . Since SL(n) is generated by exp sl(n) (by Corollaries 3.6.3 and 1.5.2), Proposition 3.4.3 shows that SL(n) acts as the identity on ΛnV . In particular, ΛnV is the trivial SO(n)-module. Finally, we regard ΛnV as a module for the full orthogonal group. If n is odd, then −I acts on ΛnV by multiplication by −1, and hence the conclusion follows from the decomposition O(n) = SO(n) ∪ (−I)SO(n). (7.45) Suppose now that n is even. Let g0 be the element of O(n) obtained from the n × n n identity matrix by permuting the 2 -th and n-th columns. Thus g0en/2 = en, g0en = 122 n n en/2, and g0ei = ei for i∈ / { 2 , n}. So g0 acts on Λ V by multiplication by −1. Since det g0 = −1, we have O(n) = SO(n) ∪ (g0)SO(n), which implies the result. 7.4 The Odd Orthogonal Case Throughout this section, ` is a positive integer and we set n = 2` + 1. The nondegenerate symmetric bilinear form defining O(n) is denoted by ω, and V stands for the natural module Cn for O(n). Lemma 7.4.1. Let ρ be a representation of O(n) on a finite dimensional complex vector space W . Then W is irreducible over O(n) if and only if W is irreducible over SO(n). In such a case, ρ(I) is either IW or −IW . Proof. If W is irreducible over SO(n), then it is clearly irreducible over O(n). Con- versely, suppose that W is O(n)-irreducible. By Schur’s Lemma, ρ(−I) = IW for some ∈ C. Moreover, 2 = 1 because −I has order 2. Thus = ±1. Using (7.45) we deduce that any SO(n)-invariant subspace of W is also O(n)-invariant. Lemma 7.4.2. The following statements are equivalent for a regular O(n)-module W : (i) W is O(n)-irreducible; (ii) W is SO(n)-irreducible; (iii) W is so(n)-irreducible. Proof. This follows from Lemma 7.4.1, Theorem 3.4.12, and Corollary 3.6.3. Lemma 7.4.3. For i ∈ {1, 2}, let ρi be an irreducible representation of O(n) on a finite dimensional complex vector space Wi. Suppose that W1 ' W2 over SO(n). Then only one of the following statements holds: 123 (i) W1 ' W2 over O(n); 1 (ii) W1 ' W2 ⊗ D over O(n). Proof. Let ϕ : W1 → W2 be an isomorphism of SO(n)-modules. Then we have an 1 isomorphism of SO(n)-modules φ : W1 → W2 ⊗ D sending w 7→ ϕ(w) ⊗ 1. As a consequence of decomposition (7.45), the map ϕ (resp. φ) is an isomorphism of O(n)- modules if and only if it preserves the action of −I. According to Lemma 7.4.1, we have ρi(−I) = iIWi with {1, 2} ⊂ {±1}. Therefore, if 1 = 2, then ϕ is an O(n)- 1 isomorphism, but (ii) does not hold because −I acts on W2 ⊗ D by multiplication by −2. Similarly, if 1 = −2, then φ is an O(n)-isomorphism but (i) is false. Theorem 7.4.4. Let γ be an n-admissible partition different from γ˜, as defined in Section 7.3. Then: (i) S[γ]V and S[γ\]V are isomorphic over SO(n) and over so(n), but not over O(n). 1 (ii) S[γ]V ' S[γ\]V ⊗ D as modules for O(n). (iii) If γ`+1 = 0, S[γ]V is the irreducible so(n)-module with highest weight γ1ε1 + ··· + γ`ε`. (7.46) (iv) Any nontrivial irreducible regular SO(n)-module is isomorphic to S[η]V for a suitable n-admissible partition η, which is unique if we impose η`+1 = 0 (or η`+1 6= 0). (v) Every nontrivial irreducible regular representation (ρ, W ) of O(n) is isomorphic to S[η]V for a unique n-admissible partition η, which we determine in the follow- ing way: if the highest weight of so(n) acting on W (via dρ) is a1ε1 + ··· + a`ε` 124 a1+···+a` (where the ai are non-negative integers) and we set = (−1) , then we take γ˜ if a1 = 0, η = (a1, . . . , a`) if a1 6= 0 and ρ(−I) = IW , \ (a1, . . . , a`) if a1 6= 0 and ρ(−I) = −IW . P Proof. Suppose γ`+1 = 0 and let d = γi. Proceeding like in the proof of Theorem 7.2.1, we define eT and we deduce that eT · cT is a maximal vector for the action of [d] so(n) on V · cT . The corresponding weight is (7.46). In particular S[γ]V is nonzero and hence it is O(n)-irreducible by Theorem 5.3.1. Since Lemma 7.4.2 ensures that S[γ]V is so(n)-irreducible, we obtain (iii). Now we write γ = (γ1, . . . , γp) with γp 6= 0 (and p ≤ `). Then the n-associated partition of γ is \ γ = (γ1, . . . , γp, 1,..., 1). | {z } n−2p Let d0 be sum of the components of γ\, i.e., d0 = d+n−2p. In the Young diagram cor- responding to γ\, we write down the numbers 1, . . . , d0 in consecutive order down the 0 columns, from left to right. In this way we obtain a tableau T . Let µ = (µ1, . . . , µt) be the conjugate partition of γ\. Then, if we set e = (e1 ⊗ · · · ⊗ ep ⊗ e0 ⊗ ep+1 ⊗ e`+p+1 ⊗ ep+2 ⊗ e`+p+2 ⊗ · · · ⊗ e` ⊗ e2`)⊗ ⊗ (e1 ⊗ · · · ⊗ eµ2 ) ⊗ · · · ⊗ (e1 ⊗ · · · ⊗ eµt ), 125 we have e · cT 0 = k(e1 ∧ · · · ∧ ep ∧ e0 ∧ ep+1 ∧ e`+p+1 ∧ ep+2 ∧ e`+p+2 ∧ · · · ∧ e` ∧ e2`)⊗ ⊗ (e1 ∧ · · · ∧ eµ2 ) ⊗ · · · ⊗ (e1 ∧ · · · ∧ eµt ) 6= 0 for some constant k. Since ω has Gram matrix S` (as defined in Chapter 1), a 0 calculation shows that Φ(p,q)(e · cT 0 ) = 0 for 1 ≤ p < q < d . Hence [d0] ⊗d0 e · cT 0 ∈ V ∩ V · cT 0 . In addition, we verify by direct calculation that e · cT 0 is a maximal vector for the action of so(n) with highest weight γ1ε1 + ··· + γpεp + (εp+1 − εp+1) + (εp+2 − εp+2) + ··· + (ε` − ε`), which is equal to (7.46). Arguing as before we deduce that S[γ\]V is the irreducible so(n)-module with highest weight (7.46). Thus Theorem 2.2.10 implies S[γ]V ' S[γ\]V (7.47) as so(n)-modules. Since SO(n) is connected, this is also an isomorphism of SO(n)- modules by Theorem 3.4.10. Observe, however, that the matrix −I ∈ O(n) acts on ⊗d ⊗d0 d d0 S[γ]V ⊂ V (resp. S[γ\]V ⊂ V ) by multiplication by (−1) (resp. (−1) ). As 0 0 d − d = 2(` − p) + 1, we see that d and d have different parities. It follows that S[γ]V 126 and S[γ\]V are not isomorphic over O(n). Then, applying Lemma 7.4.3, we deduce 1 S[γ]V ' S[γ\]V ⊗ D \ over O(n). If γ`+1 6= 0, the previous arguments work interchanging γ and γ . So far we have proven (i), (ii), and (iii). Suppose now that W is a nontrivial irreducible regular SO(n)-module. Then W is also nontrivial and irreducible over so(n). Thus, applying Proposition 6.2.4 and keeping in mind the highest weight theory for so(n), the highest weight of so(n) acting on W has the form a1ε1 +···+a`ε` where the coefficients ai are integers such that a1 ≥ · · · ≥ a` ≥ 0 and a1 6= 0. Taking η = (a1, . . . , a`), we obtain W ' S[η]V ' S[η\]V over so(n) and SO(n), as a consequence of (i) and (iii). Any n-admissible partition different from η and η\ will give a module with a different highest weight, and hence not isomorphic to W . Finally, suppose that (ρ, W ) is a nontrivial irreducible regular representation of 1 O(n). If the restriction of ρ to SO(n) is trivial, then W ' D ' S[˜γ]V over O(n). Any other partition will give a module whose restriction to SO(n) is nontrivial. Assume now that ρ|SO(n) is not trivial. Applying part (iv), we obtain W ' S[η]V ' S[η\]V over SO(n) for a unique partition η = (a1, . . . , a`). By Lemma 7.4.1, ρ(I) = ±IW . In a1+···+a` addition, the action of −I on S[η]V is by multiplication by = (−1) . Since n is odd, I acts on S[η\]V by multiplication by −. Thus W is isomorphic over O(n) to 127 S[η]V or S[η\]V depending on whether ρ(I) = IW or ρ(I) = −IW , respectively. Example 7.4.5. Consider the partition γ = (1,..., 1) of d, where 1 ≤ d ≤ `. Then [d] d S[γ]V = V ∩ Λ V is the irreducible so(n)-module with highest weight ε1 + ··· + εd. Proceeding like in Example 7.2.3 (but in this case using the fact that the bilinear d [d] d form is symmetric), we see that Λ V ⊂ V , whence S[γ]V = Λ V. Therefore, for 1 ≤ d < `, the exterior power ΛdV is the d-th fundamental module for so(n). As explained in the following note, using the method of Theorem 7.4.4 we are not able to construct the last fundamental module for so(n), that is, the irreducible module 1 with highest weight λ` = 2 (ε1 + ··· + ε`). Note 7.4.6. Theorem 7.4.4 does not provide the irreducible representations of the Lie algebra so(n) whose highest weights have the form γ γ 1 ε + ··· + ` ε , 2 1 2 ` where the γi are odd integers such that γ1 ≥ · · · ≥ γ` ≥ 0. In fact, Proposition 6.2.4 reveals that this cannot be achieved by Weyl’s construction, since we can only obtain highest weights with integer coefficients. These modules can be obtained by considering the universal enveloping algebra of so(n), as it is done in Section 20.3 of [Hum78]. An alternative method uses Clifford algebras to build V (λ`) and then shows that the existence of the fundamental modules implies the existence of V (λ) for each dominant integral weight λ. This is explained with great detail in Section 6.2 and Proposition 5.5.19 of [GW09]. 128 Example 7.4.7. The Lie algebra so(n) is simple, which means that its adjoint module is irreducible. The corresponding highest weight is ε1 if n = 3, and ε1 + ε2 if n ≥ 5. 3 Since the natural module C for so(3) is irreducible with highest weight ε1, we have so(3) ' C3 over so(3). If n ≥ 5, Example 7.4.5 implies so(n) ' Λ2Cn over so(n). Example 7.4.8. Here we decompose the symmetric powers SymdV as direct sums of irreducible so(n)-modules. Precisely, we show that d Sym V ' V (dε1) ⊕ V ((d − 2)ε1) ⊕ V ((d − 4)ε1) ⊕ · · · , (7.48) where the last summand is the natural module V = V (ε1) is d is odd, and the trivial module C = V (0) if d is even. Let γ = (d) with d ≥ 2. Then [d] d S[γ]V = V ∩ Sym V is V (dε1), the irreducible so(n)-module with highest weight dε1. Proceeding like in Example 7.2.2, we see that S[γ]V is precisely the kernel of the contraction map d d−2 θd : Sym V → Sym V defined by X v1 ··· vd 7→ ω(vr, vs)v1 ··· vˆr ··· vˆs ··· vd. r This map is a homomorphism of so(n)-modules, because it is the restriction of Φ(1,2) to SymdV . Consider first the case d = 2. Let 1 v = 2 e0e0 + e1e`+1 + e2e`+2 + ··· e`e2`. 129 Since θ2(v) 6= 0, θ2 is surjective. Hence, by the first isomorphism theorem, 2 Sym V ' ker θ2 ⊕ C ' V (2ε1) ⊕ V (0). 3 Suppose now d = 3 and let E1 = e1v ∈ Sym V . Then E1 is a maximal vector of 3 Sym V with weight ε1. Since ker θ3 ' V (3ε1), we have E1 ∈/ ker θ3, and hence θ3(E1) is a maximal vector of V with weight ε1. So θ3(E1) generates V as a module for so(n), which proves that θ3 is surjective. Thus the first isomorphism theorem gives 3 Sym V ' ker θ3 ⊕ V ' V (3ε3) ⊕ V (ε1). Now fix d with d ≥ 4 and assume that (7.48) is true for smaller values of d. Set d−2i i d Ed−2i = e1 v ∈ Sym V d d for 1 ≤ i ≤ b 2 c. Then Ed−2i is a maximal vector of Sym V with weight (d − 2i)ε1. As before, Ed−2i cannot be in ker θd ' V (dε1). It follows that θd(Ed−2i) is a maximal d−2 vector of Sym V with weight (d − 2i)ε1. In view of the decomposition (7.48) of d−2 Sym V , we deduce that θd is surjective. Therefore d d−2 Sym V ' ker θd ⊕ Sym V ' V (dε1) ⊕ V ((d − 2)ε1) ⊕ V ((d − 4)ε1) ⊕ · · · . Observe that (7.48) can be used to compute the dimensions of the modules V (dε1). Note 7.4.9. Proposition 7.3.1 and Theorem 7.4.4 show that if γ is an n-admissible partition then S[γ]V 6= 0. The converse implication is also true. For a proof of this fact, the reader may consult Theorem 5.7.A of [Wey39], or Exercise 19.20 of [FH91]. 130 7.5 On the Representations of a Subgroup of Index 2 Let G be a group. If H is a subgroup of index 2, there is a close relationship between the irreducible representations of G and their restrictions to H. In this section we prove a few facts about this relationship and in the next section we apply them to the case G = O(n) and H = SO(n) with n even. Theorem 7.5.1. (Frobenius Reciprocity) Let H be any subgroup of G. Let V be a G G-module and let W be an H-module. Set U = IndH W . Then, as vector spaces, HomG(U, V ) ' HomH (W, V ). Proof. By definition, we have U = C[G] ⊗C[H] W and the action of G on U is given by x · (g ⊗ w) = (xg) ⊗ w for all x, g ∈ G and w ∈ W . Let ϕ : HomG(U, V ) → HomH (W, V ) be the linear map sending each f to the map w 7→ f(1 ⊗ w). Then ϕ is a linear isomorphism whose inverse sends each T to the map g ⊗ w 7→ g · T (w). Lemma 7.5.2. Let V be a G-module. Suppose that V1 and V2 are non-isomorphic irreducible G-submodules of V such that V = V1 ⊕V2. Then the only proper non-trivial submodules of V are V1 and V2. Proof. Let S be a proper non-trivial submodule of V . Then S is irreducible by the Jordan-H¨olderTheorem. Since V1 and V2 are not isomorphic, exactly one of the projections of S into V1 and V2 is non-zero, which implies that S = V1 or S = V2. Suppose that H is a normal subgroup of G. Let π : H → GL(V ) be a represen- tation of H. Given x ∈ G, we have x−1Hx = H and hence we can define a map 131 Rx : H → GL(V ) by −1 Rx(h) = R(x hx). Observe that Rx is a representation because it is a composition of two group homo- morphisms. It is straighforward to check Rx and Ry are equivalent if x, y are elements of G such that xH = yH. If H has index 2 in G and s ∈ G \ H, we say that Rs is the conjugate representation of R. Proposition 7.5.3. Suppose that H is a subgroup of index 2 in G. Let (ρ, V ) be a representation of G and set π = ρ|H . Then π is self-conjugate, i.e., isomorphic to its conjugate. Proof. Fix s ∈ G \ H. Then P = ρ(s−1) is a linear automorphism of V satisfying π(h) ◦ P = P ◦ πs(h) for all h ∈ H. Suppose that H is a subgroup of index 2 in G. We make the 1-dimensional complex vector space C into a G-module by z if g ∈ H, g · z = −z if g ∈ G \ H. When we considering this action, we write C0 instead of C. If U is any complex G-module, we set U 0 = U ⊗ C0. This definition makes sense even when U = C is the trivial G-module because C ⊗ C0 ' C0. Proposition 7.5.4. Let V be a finite dimensional irreducible G-module. Suppose G that G has a subgroup H of index 2 and set W = ResH V , i.e., W is the restriction of V from G to H. (Note that V = W as sets.) Then exactly one of the following holds: 132 (i) W is irreducible and V 6' V 0. 0 (ii) W is not irreducible and V ' V . In this case W = U1 ⊕ U2, where U1 and U2 are irreducible submodules of W of the same dimension. Moreover, U1 and U2 G are conjugate but not isomorphic. In addition V = IndH Ui for i ∈ {1, 2}. Proof. Let ρ (resp. ρ0) be the representation of G on V (resp. V 0). Suppose first that W is irreducible. Let d : G → C× be the group homomorphism defined by d(H) = 1 and d(G \ H) = −1. Then the mapρ ¯ : G → GL(V ) given byρ ¯(g) = d(g)ρ(g) is a representation. Since ρ0 andρ ¯ are equivalent, our problem reduces to show that ρ andρ ¯ are not equivalent. Suppose, if possible, that there exists P ∈ GL(V ) such that P ◦ ρ(g) =ρ ¯(g) ◦ P for all g ∈ G. Since ρ|H =ρ ¯|H and W is irreducible, Schur’s Lemma implies that P is a scalar operator. Thus ρ(g) = d(g)ρ(g) for all g ∈ G, whence d(G) = 1, a contradiction. Therefore V 6' V 0. Assume henceforth that W is reducible and let U1 be an irreducible submodule of W . Fix an element s ∈ G \ H and set U2 = sU1. So dim U1 = dim U2. Observe that U1 6= U2, for otherwise U1 would be hH, si = G-invariant and hence equal to W , a contradiction. −1 Given h ∈ H and u ∈ U1, we have h(su) = s(s hs)u. Since [G : H] = 2, this proves that U2 is H-invariant and that U1,U2 are conjugate. We show next that U2 is irreducible over H. Let T2 be a nonzero H-submodule of −1 T2. Let T1 = s · T2. Reasoning as above, we see that T1 is a nonzero H-submodule of U1. The irreducibility of U1 implies T1 = U1 and hence T2 = U2. It is clear that H preserves the sum U1 + U2. In addition s maps U1 onto U2 −1 and s maps U2 onto U1. As G = hH, si, it follows that U1 + U2 is G-invariant. Hence W = U1 + U2 because V is irreducible over G. In order to prove that the sum 133 is direct, we look at the intersection U1 ∩ U2, which is an H-submodule of both U1 and U2. Keeping in mind that U1 and U2 are distinct and irreducible, we must have U1 ∩ U2 = {0}. Consequently W = U1 ⊕ U2. −1 G Since V = U1 ⊕ g · U1 = s · U2 ⊕ U2 we deduce V ' IndH Ui for i ∈ {1, 2}. We 0 G 0 also have V ' IndH Ui for i ∈ {1, 2} because the restriction of V to H is isomorphic to W . Thus V ' V 0. Finally, we show that U1 6' U2. Applying Schur’s Lemma and Theorem 7.5.1, 1 = dim HomG(V,V ) = dim HomH (U1,V ). If it were U1 ' U2, we would have dim HomH (U1,V ) ≥ 2, a contradiction. Let H be any subgroup of G. Then G acts on the set G/H = {gH : g ∈ G} by left multiplication. The vector space C[G/H] becomes a G-module by extending this action in the obvious way. Proposition 7.5.5. Let V be a finite dimensional G-module. Suppose that is H a subgroup of finite index n in G. Then, as modules over G, G G IndH (ResH V ) ' C[G/H] ⊗C V If [G : H] = 2 then G G 0 IndH (ResH V ) ' V ⊕ V . Proof. The C-linear map G G C[G/H] ⊗C V → IndH (ResH V ) = C[G] ⊗C[H] V 134 given by gH ⊗ v 7→ g ⊗ g−1v is an isomorphism of G-modules. Suppose now that H has index 2 in G and fix s ∈ G \ H. Let X = H + sH and Y = H − sH. Then CX (resp. CY ) is a G-submodule of C[G/H] isomorphic to C (resp. C0). In addition C[G/H] = CX ⊕ CY . Therefore G G 0 0 IndH (ResH V ) ' C[G/H] ⊗C V ' (C ⊕ C ) ⊗C V ' V ⊕ V . 7.6 The Even Orthogonal Case Throughout this section, ` is a positive integer and we set n = 2`. Let V = Cn be the natural module for O(n). Recall that the partition (1,..., 1) of n is denoted byγ ˜ and that we have already dealt with it in Proposition 7.3.1. Recall also that g0 ∈ O(n) is the matrix such that g0e` = e2`, g0e2` = e`, and g0ei = ei for i 6= {`, 2`}. Since det g0 = −1, we have O(n) = SO(n) ∪ (g0)SO(n). (7.49) Theorem 7.6.1. Suppose n ≥ 4. Let γ be an n-admissible partition different from γ˜. Then S[γ]V is irreducible as a module over O(n). In addition: (a) Suppose that γ` = 0. Then S[γ]V is the irreducible so(n)-module with highest weight γ1ε1 + ··· + γ`−1ε`−1. (7.50) We have S[γ]V ' S[γ\]V as modules over SO(n) and over so(n), but not over O(n). In fact, the element g0 ∈ O(n) fixes the maximal vector of S[γ]V but maps the maximal vector of S[γ\]V to its opposite. 135 (b) Suppose that γ` 6= 0 and γ`+1 = 0, i.e, γ is n-associated to itself. Then S[γ]V is not irreducible over SO(n) (resp. so(n)). In fact, we have S[γ]V = U1 ⊕ U2, where U1 is the irreducible so(n)-module with highest weight γ1ε1 +···+γ`ε` and U2 is the irreducible so(n)-module with highest weight γ1ε1 +···+γ`−1ε`−1 −γ`ε`. O(n) Moreover S[γ]V = IndSO(n)Ui for i ∈ {1, 2}. P Proof. Let d = γi. Suppose first that γ` = 0. We define T and eT like in the proof of Theorem 7.2.1. [d] Then u = eT · cT is a maximal vector for the action of so(n) on V · cT with highest weight (7.50). Observe that g0 fixes u because e` does not appear in eT . Let U be the [d] so(n)-submodule of V · cT generated by u. Thus U is the complex span of the set S = {u} ∪ {X1 ··· Xku : k ∈ N,Xi ∈ so(n)}. −1 If X is in so(n) then so is g0Xg0 by Example 3.3.8. It follows that g0 preserves S and hence U. Theorem 3.4.12 and Corollary 3.6.3 imply that U is invariant under the action of SO(n). Consequently, the decomposition (7.49) shows that O(n) preserves [d] [d] U. Since V · cT is O(n)-irreducible (by Theorem 7.1.7), we deduce U = V · cT . [d] Thus V ·cT is irreducible over so(n). Therefore S[γ]V is the irreducible so(n)-module with highest weight (7.50). \ Next we consider the partition γ . If γ = (γ1, . . . , γp) with γp 6= 0 (and p < `), then \ γ = (γ1, . . . , γp, 1,..., 1). | {z } 2`−2p Let d0 be sum of the components of γ\. In the Young diagram corresponding to γ\, we write down the numbers 1, . . . , d0 in consecutive order down the columns, from left 136 0 to right. In this way we obtain a tableau T . Let µ = (µ1, . . . , µt) be the conjugate partition of γ\. Then, if we set e = (e1 ⊗ · · · ⊗ ep ⊗ ep+1 ⊗ e`+p+1 ⊗ ep+2 ⊗ e`+p+2 ⊗ · · · ⊗ e` ⊗ e2`)⊗ ⊗ (e1 ⊗ · · · ⊗ eµ2 ) ⊗ · · · ⊗ (e1 ⊗ · · · ⊗ eµt ), we have e · cT 0 = k(e1 ∧ · · · ∧ ep ∧ ep+1 ∧ e`+p+1 ∧ ep+2 ∧ e`+p+2 ∧ · · · ∧ e` ∧ e2`)⊗ ⊗ (e1 ∧ · · · ∧ eµ2 ) ⊗ · · · ⊗ (e1 ∧ · · · ∧ eµt ) 6= 0 for some constant k. Reasoning like in the proof of Theorem 7.4.4, e·cT 0 is a maximal [d0] vector for the action of so(n) on V · cT 0 with highest weight (7.50). Notice that g0 · (e · cT 0 ) = −e · cT 0 [d0] because g0 maps e` ↔ e2`. Hence, arguing as above, we see that V ·cT 0 is irreducible over so(n). Consequently, S[γ\]V is the irreducible so(n)-module with highest weight (7.50). Therefore S[γ]V ' S[γ\]V over so(n) and over SO(n), by Theorems 2.2.10 and 3.4.10. However, this is not an isomorphism of O(n)-modules because g0 acts differently on the maximal vectors. Now we assume γ` 6= 0 and γ`+1 = 0. We claim that S[γ]V , viewed as a module for so(n), contains two linearly independent maximal vectors. In the Young diagram corresponding to γ, we write down the numbers 1, . . . , d in consecutive order down 137 the columns, from left to right. In this way we obtain a tableau T . Define eT like in Lemma 5.2.3. Set j` = 2` and jk = k for 1 ≤ k < `, and define e like in Lemma 5.2.5. For example, if ` = 4 and T is 1 5 7 2 6 , 3 4 then eT = e1 ⊗ e2 ⊗ e3 ⊗ e4 ⊗ e1 ⊗ e2 ⊗ e1 and e = e1 ⊗ e2 ⊗ e3 ⊗ e8 ⊗ e1 ⊗ e2 ⊗ e1. [d] Regarding V · cT as a module over so(n), Lemmas 5.2.3 and 5.2.5 imply that eT · cT is a maximal vector with highest weight γ1ε1 + ··· + γ`ε`. (7.51) and e · cT is a maximal vector with highest weight γ1ε1 + ··· + γ`−1ε`−1 − γ`ε`. (7.52) It follows that S[γ]V has non-isomorphic irreducible so(n)-submodules U1 and U2 with highest weights (7.51) and (7.52). In particular S[γ]V is not irreducible over SO(n). Since S[γ]V is O(n)-irreducible (by Theorem 7.1.7), Lemma 7.5.2 and Proposition 7.5.4 give S[γ]V = U1 ⊕ U2. Finally, Proposition 7.5.4 implies O(n) O(n) S[γ]V = IndSO(n)U1 = IndSO(n)U2. 138 Note 7.6.2. Theorem 7.6.1 and Proposition 7.3.1 provide all the irreducible regular representations of SO(n) with n ≥ 4. Indeed, let W be an irreducible regular module for SO(n). Denote by λ the highest weight of so(n) and write λ = a1ε1 + ··· + a`ε`. The highest weight theory for so(n) together with Proposition 6.2.4 imply that the ai are integers and a1 ≥ · · · ≥ a`−1 ≥ |a`|. If λ = 0 then n W ' S[˜γ]V = Λ V. Assume now that λ 6= 0 and let γ be the partition (a1, . . . , a`−1, |a`|). If a` = 0 then W ' S[γ]V ' S[γ\]V. If a` 6= 0 then W is isomorphic to one of the two direct summands of S[γ]V . Observe, however, that Weyl’s method is unable to construct all the finite dimen- sional irreducible representations of so(n), but only half of them. The comments we made in Note 7.4.6 about the odd orthogonal case apply to the even orthogonal case as well, with the only difference that in the latter situation we use Clifford algebras to construct both V (λ`−1) and V (λ`). Note 7.6.3. Theorem 7.6.1 does not describe the regular irreducible representa- tions of SO(2). However, it is easy to obtain such description by hand. Recall that SO(2) ' C× as linear algebraic groups. Let ρ : C× → GL(W ) be a regular irre- ducible representation. By Lemma 3.7.3, W must be 1-dimensional and there exists 139 an integer p such that, for all z ∈ C×, ρ(z) is multiplication by zp. Conversely, given × p p ∈ Z, it is readily verified that the map ρp : C → GL(C) defined by ρp(z)(a) = z a is a regular irreducible representation. Let p, q ∈ Z such that ρp and ρq are equivalent, i.e., P ◦ ρp(z) = ρq(z) ◦ P for some P ∈ GL(C). Evaluation at 1 yields zp = zq for all z 6= 0, whence p = q. Thus, the regular irreducible representations of SO(2) can be indexed by the set Z. Note 7.6.4. Since so(2) ' gl(1), Proposition 6.1.1 shows that the finite dimensional irreducible representations of so(2) are indexed by C. There is another important difference between the representations of SO(2) and those of so(2): whereas SO(2) is reductive, there are finite dimensional representations of so(2) that are not completely reducible, as explained in Note 3.7.5. Example 7.6.5. Assume n ≥ 4. Consider the partition γ = (1, 1,..., 1) of d, where d 1 ≤ d < `. Like in Example 7.4.5, S[γ]V = Λ V is the irreducible so(n)-module with highest weight λd = ε1 + ··· + εd. Thus, for 1 ≤ d ≤ ` − 2, the exterior power ΛdV is the d-th fundamental module for so(n). As noted above, Weyl’s construction does not provide the last two fundamental modules for so(n), i.e., the irreducible modules with highest weights 1 1 λ`−1 = 2 (ε1 + ··· + ε`−1 − ε`) and λ` = 2 (ε1 + ··· + ε`). Example 7.6.6. Suppose that n ≥ 4. In this example we identify the adjoint module for so(n) with the exterior square of the natural module. We consider first the case n = 4. The Lie algebra so(4) is the direct sum of 140 two simple ideals isomorphic to so(3). But these summands are not isomorphic as modules over so(4). In fact, their highest weights are ε1 + ε2 and ε1 − ε2. Applying part (ii) of Theorem 7.6.1 to the partition γ = (1, 1), we deduce that so(4) ' Λ2C4 as so(4)-modules. Now assume n ≥ 6. Then so(n) is simple as a Lie algebra, which means that its adjoint module is irreducible. The corresponding highest weight is ε1 + ε2. Applying Example 7.6.5, we get so(n) ' Λ2Cn as so(n)-modules. Example 7.6.7. Suppose n ≥ 4. The symmetric power SymdV decomposes over so(n) as V (dε1) ⊕ V ((d − 2)ε1) ⊕ V ((d − 4)ε1) ⊕ · · · , where the last summand is V = V (ε1) if d is odd, and C = V (0) if d is even. The proof is the same as in Example 7.4.8, but now taking v = e1e`+1 +e2e`+2 +···+e`e2`. Note 7.6.8. Theorem 7.6.1 and Proposition 7.3.1 show that S[γ]V 6= 0 when the partition γ is n-admissible. The converse implication is also true. For a proof of this fact, we refer the reader to Theorem 5.7.A of [Wey39], or Exercise 19.20 of [FH91]. The remainder of this section is devoted to classify the irreducible regular rep- resentations of the full orthogonal group O(n). We begin with a couple of auxiliary results. Lemma 7.6.9. Let (ρ, W ) be a regular representation of SO(n). Let (ρ0,W ) be the conjugate representation defined by 0 −1 ρ (x) = ρ(g0 xg0) 141 for all x ∈ SO(n). Then ρ0 is regular and, for all X ∈ so(n), 0 −1 dρ (X) = dρ(g0 Xg0). 0 −1 Proof. Notice that ρ is well defined and regular because conjugation by g0 is a regular automorphism of SO(n). Now we fix a basis of W and we identify each element of End(W ) with the corresponding matrix. Then, for all X ∈ so(n), d d dρ0(X) = ρ(g−1 exp(tX)g )| = ρ(exp(tg−1Xg ))| = dρ(g−1Xg ). dt 0 0 t=0 dt 0 0 t=0 0 0 Lemma 7.6.10. Suppose that (ρ, W ) is an irreducible regular representation of SO(n). 0 Define (ρ ,W ) like in the previous lemma. Let γ1ε1 +···+γ`ε` be the highest weight of the representation dρ. Then dρ0 is the irreducible representation of so(n) with highest 0 weight γ1ε1 + ··· + γ`−1ε`−1 − γ`ε`. In particular, ρ and ρ are equivalent if and only if γ` = 0. Proof. Observe first that ρ0 is irreducible because ρ is and ρ0(SO(n)) = ρ(SO(n)). Then both dρ and dρ0 are irreducible. Let h be the diagonal subalgebra of so(n) and set + n = span{xα : α ∈ Φ } −1 with the notation of Chapter 2. A few routine calculations show that g0 hg0 = h and −1 g0 n = n. Therefore, the maximal vector w for dρ (which is unique up to scaling) is also the maximal vector for dρ0. In order to compute the highest weight, take X = diag[d1, . . . , d`, −d1,..., −d`] ∈ h. 142 Then 0 −1 dρ (X)(v) = dρ(g0 Xg0)(w) = dρ(diag[d1, . . . , d`−1, −d`, −d1,..., −d`−1, d`])(w) = = (γ1d1 + ··· + γ`−1d`−1 − γ`d`)w = (γ1ε1 + ··· + γ`−1ε`−1 − γ`ε`)(X)w. 0 0 Thus the highest weight for dρ is γ1ε1 + ··· + γ`−1ε`−1 − γ`ε`. So dρ and dρ are equivalent if and only if γ` = 0, which implies the conclusion in view of Theorem 3.4.10. Theorem 7.6.11. The map γ 7→ S[γ]V gives a bijection between the set of all n- admissible partitions and the set of isomorphism classes of non-trivial irreducible regular modules for O(n). Precisely, let W be a non-trivial irreducible regular module over O(n). Then: (i) Suppose that W is SO(n)-irreducible. Let w be the maximal vector for the action of so(n) on W , and let λ = γ1ε1 + ··· + γ`ε` be the corresponding weight. Then γ` = 0, γ1 ≥ · · · ≥ γ`−1 ≥ 0, and γi ∈ Z for all i. If λ = 0 then 1 W ' D ' S[˜γ]V over O(n). If λ 6= 0, we take γ = (γ1, . . . , γ`−1) and we have W ' S[γ]V or W ' S[γ\]V depending on whether g0 · w = w or g0 · w = −w, respectively. (ii) Suppose that W is not SO(n)-irreducible. Then W ' U1 ⊕ U2 over SO(n), where U1 and U2 are the irreducible regular SO(n)-modules corresponding to 143 the highest weights γ1ε1 + ··· + γ`−1ε`−1 ± γ`ε`, (7.53) where γ1 ≥ · · · ≥ γ` > 0 and γi ∈ Z for all i. Taking γ = (γ1, . . . , γ`), we have W ' S[γ]V over O(n). Proof. Theorem 7.6.1 and Proposition 7.3.1 show that S[γ]V is an irreducible O(n)- module for each n-admissible partition γ. Let U be W viewed as a module for SO(n). Suppose first that U is irreducible. The highest weight theory of so(n) and Propo- sition 6.2.4 imply that the γi are integers such that γ1 ≥ · · · ≥ γ`−1 ≥ |γ`|. By Propo- O(n) 0 sitions 7.5.3, 7.5.4 and 7.5.5, U is isomorphic to its conjugate and IndSO(n)U = W ⊕W 0 1 O(n) with W 6' W = W ⊗ D . Hence W can be seen as a O(n)-submodule of IndSO(n)U. 0 O(n) Since W 6' W , Lemma 7.5.2 implies that IndSO(n)U has exactly two proper non- trivial O(n)-submodules. Consequently, there are at most two isomorphism types of irreducible regular O(n)-modules whose restriction to SO(n) is isomorphic to U. In addition γ` = 0 by Lemma 7.6.10. If λ = 0 then U is trivial. In this case we must have W ' D1 because W is non-trivial. If λ 6= 0, part (i) of Theorem 7.6.1 shows that W must be isomorphic to either S[γ]V or S[γ\]V depending on whether g0 fixes w or sends it to its opposite, respectively. Assume now that U is not irreducible. By Proposition 7.5.4, we have U = U1 ⊕U2 where U1 and U2 are non-isomorphic conjugate irreducible SO(n)-submodules of U. Observe that U1 and U2 are regular because W is. The highest weight theory of so(n) together with Proposition 6.2.4 and Lemma 7.6.10 imply that the highest weights corresponding to U1 and U2 have the form (7.53) with γ1 ≥ · · · ≥ γ` > 0 and γi ∈ Z 144 for all i. Therefore, letting γ = (γ1, . . . , γ`), Proposition 7.5.4 and part (ii) of Theorem 7.6.1 give O(n) W ' IndSO(n)U1 ' S[γ]V as modules over O(n). Note 7.6.12. We saw in the proof that if W is an irreducible regular O(n)-module whose restriction U to SO(n) is still irreducible, then the isomorphism class of W is completely determined by U and by the action of g0 on the maximal vector of U over so(n). Therefore, if γ is an n-admissible partition different fromγ ˜ and from γ\, then 1 S[γ]V ' S[γ\]V ⊗ D (7.54) \ over O(n). In the case γ = γ , S[γ]V is not irreducible over SO(n) and hence Propo- 1 sition 7.5.4 yields S[γ]V ' S[γ]V ⊗ D . In conclusion, (7.54) holds whenever γ 6=γ ˜. Recall that this is also true in the odd orthogonal case. 145 Bibliography [Car05] R. W. Carter. Lie algebras of finite and affine type, volume 96 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 2005. [Die69] J. Dieudonn´e. Foundations of modern analysis. Academic Press, New York, 1969. Enlarged and corrected printing, Pure and Applied Mathematics, Vol. 10-I. [FH91] William Fulton and Joe Harris. Representation theory, volume 129 of Grad- uate Texts in Mathematics. Springer-Verlag, New York, 1991. A first course, Readings in Mathematics. [GW09] Roe Goodman and Nolan R. Wallach. Symmetry, representations, and in- variants, volume 255 of Graduate Texts in Mathematics. Springer, Dor- drecht, 2009. [HR79] Edwin Hewitt and Kenneth A. Ross. Abstract harmonic analysis. Vol. I, volume 115 of Grundlehren der Mathematischen Wissenschaften [Funda- mental Principles of Mathematical Sciences]. Springer-Verlag, Berlin, sec- ond edition, 1979. Structure of topological groups, integration theory, group representations. 146 [Hum78] James E. Humphreys. Introduction to Lie algebras and representation the- ory, volume 9 of Graduate Texts in Mathematics. Springer-Verlag, New York, 1978. Second printing, revised. [Jac89] Nathan Jacobson. Basic algebra. II. W. H. Freeman and Company, New York, second edition, 1989. [Kna02] Anthony W. Knapp. Lie groups beyond an introduction, volume 140 of Progress in Mathematics. Birkh¨auserBoston Inc., Boston, MA, second edition, 2002. [Wey39] Hermann Weyl. The Classical Groups. Their Invariants and Representa- tions. Princeton University Press, Princeton, N.J., 1939. 147 Appendix A Facts on Invariant Theory In Chapter 7 we used a fact from invariant theory at a crucial step, namely in the proof of Theorem 7.1.7. This appendix aims to provide a brief introduction to invariant theory. We fully characterize the polynomial invariants of the classical groups. In the symplectic and orthogonal cases, this information is applied to prove the result quoted in Theorem 7.1.7. Throughout the chapter, V denotes a finite dimensional vector space over a field F of characteristic 0. A.1 Polynomial Invariants A function f : V → F is called polynomial if it is a polynomial in the coordinates with respect to some basis of V . This definition is easily seen to be independent of the basis. The F -algebra of all polynomial functions on V will be denoted by F [V ]. ∗ If {x1, . . . , xn} is any basis of the dual vector space V , then F [V ] = F [x1, . . . , xn]. Lemma A.1.1. Let F [X1,...,Xn] be the algebra of polynomials in n commuting n variables. If p ∈ F [X1,...,Xn] satisfies p(a) = 0 for all a ∈ F , then f = 0. 148 Proof. Use induction on n. Since F is infinite, the case n = 1 follows from the fact that any nonzero polynomial in F [X1] has a finite number of roots. ∗ d1 dn Corollary A.1.2. If {x1, . . . , xn} is a basis of V , then {x1 ··· xn : di ∈ N0} is a basis of F [V ]. In particular F [V ] ' F [X1,...,Xn] as F -algebras. Definition A.1.3. Let d ∈ N. A function f ∈ F [V ] is called homogeneous of degree d if f(tv) = tdf(v) for all t ∈ F and v ∈ V . The homogeneous functions of degree d form a subspace of F [V ], denoted by F [V ]d. For convenience we set F [V ]0 = F . The subspaces F [V ]d with d ≥ 0 are called the homogeneous components of F [V ], by virtue of the following result. L Lemma A.1.4. F [V ] = d≥0 F [V ]d. Suppose now that V is a module for a group G. We say that a polynomial function f on V is G-invariant if f(g · v) = f(v) for all g ∈ G and v ∈ V . The set of all G- invariant polynomial functions is a subalgebra of F [V ] denoted by F [V ]G. The basic problem of invariant theory is to describe F [V ]G and to find a small set of generators for this algebra. There is another way to define F [V ]G. For g ∈ G and f ∈ F [V ], let g · f : V → F be the function given by (g · f)(v) = f(g−1 · v). Since G acts on V by means of linear transformations, we have g · f ∈ F [V ]. In this way we obtain a left action of G on F [V ], whose set of fixed points is precisely F [V ]G. This explains the notation F [V ]G. Lemma A.1.5. F [V ]d is G-invariant for all d ∈ N0. Proof. The claim is trivial for d = 0. Suppose d ∈ N. Let f ∈ F [V ]d and g ∈ G. 149 Then, for all t ∈ F and v ∈ V , (g · f)(tv) = f(g−1 · (tv)) = f(t(g−1 · v)) = tdf(g−1 · v) = td(g · f)(v). The following result is useful because it reduces many questions on F [V ]G to the study of the invariants in each homogeneous component of F [V ]. G G Corollary A.1.6. For each d ∈ N0, define F [V ]d = F [V ]d ∩ F [V ] . Then G M G F [V ] = F [V ]d . d≥0 Lemma A.1.7. Let f1, f2 ∈ F [V ]. Suppose that f1 is G-invariant and nonzero. Assume that f = f1f2 is G-invariant. Then f2 is G-invariant. Proof. The conclusion follows because f1 6= 0 and f1f2 = f = g · f = (g · f1)(g · f2) = f1(g · f2). Given m ∈ N, the vector space V m = V × · · · × V becomes a G-module via g · (v1, . . . , vm) = (g · v1, . . . , g · vm), g ∈ G, vi ∈ V. Our main goal is to describe F [V m]G, where G is a classical complex group (GL(n), SL(n), Sp(n), O(n), or SO(n)), and V is the natural module for G. For the moment, however, we don’t need any further assumption on G or V : just suppose that V is a finite dimensional module for an abstract group G. m m Definition A.1.8. Let f ∈ F [V ] and d = (d1, . . . , dm) ∈ N0 . We say that f is 150 multi-homogeneous (or simply, homogeneous) of degree d if d1 dm f(t1v1, . . . , tmvm) = t1 ··· tm f(v1, . . . , vm) m m for all t1, . . . , tm ∈ F \{0} and v1, . . . , vm ∈ V . The subspace F [V ]d of F [V ] consisting of all homogeneous polynomial functions of degree d is called a multi- homogeneous component of F [V m], as it becomes apparent from the next result. Lemma A.1.9. m M m F [V ] = F [V ]d. m d∈N0 m m Lemma A.1.10. Let d = (d1, . . . , dm) ∈ N0 . Then F [V ]d is G-invariant. Proof. This is just a calculation, analogous to the proof of Lemma A.1.5 The following result is analogous to Corollary A.1.6 and it will be a valuable tool for us. n m G m m G Corollary A.1.11. For each d ∈ N0 , define F [V ]d = F [V ]d ∩ F [V ] . Then m G M m G F [V ] = F [V ]d . m d∈N0 A.2 Polarization Maps Fix m ∈ N. Let i, j be integers between 1 and m. m m Given f ∈ F [V ] and v = (v1, . . . , vm) ∈ V , we define (Dijf)(v) to be the coefficient of t in the polynomial f(v1, . . . , vj−1, vj + tvi, vj+1, . . . , vm). 151 In other words, the polynomial h(t) = f(v1, . . . , vj−1, vj + tvi, vj+1, . . . , vm) − f(v) is divisible by t and hence we can define h(t) (D f)(v) = | . ij t t=0 m m As v varies over V , we obtain a polynomial function Dijf ∈ F [V ]. In this way we m m get a map Dij : F [V ] → F [V ], called a polarization map, or just a polarization. m Lemma A.2.1. Dij is a derivation of F [V ]. m G Lemma A.2.2. Suppose that V is a module for a group G. Then Dij maps F [V ] into F [V m]G. m G m Proof. Let f ∈ F [V ] , g ∈ G and v = (v1, . . . , vm) ∈ V . Then f(v1, . . . , vj + tvi . . . , vm) = f(gv1, . . . , gvj + tgvi, . . . , gvm). By definition, (Dijf)(v) (resp. (Dijf)(g · v)) is the coefficient of t in the left (resp. right) hand side. It follows that (Dijf)(v) = (Dijf)(g · v). Now we introduce coordinates in order to obtain a practical formula for Dij. Given ∗ a basis {u1, . . . , un} of V , let {x1, . . . , xn} be the dual basis of V . For 1 ≤ r ≤ m and 1 ≤ s ≤ n, we define (r) ⊕m us = (0,..., 0, us, 0,..., 0) ∈ V , 152 where the only nonzero component is in the r-th position. Thus (r) {us : 1 ≤ r ≤ m, 1 ≤ s ≤ n} is a basis of V m. Let (r) {xs : 1 ≤ r ≤ m, 1 ≤ s ≤ n} (A.55) m ∗ m be the corresponding dual basis of (V ) . Note that for each v = (v1, . . . , vm) ∈ V we have (r) xs (v) = xs(vr) (r) m ∗ for 1 ≤ r ≤ m , 1 ≤ s ≤ n. For brevity we will say that {xs } is the basis of (V ) associated to {ui}. Lemma A.2.3. Keep the notation above. Then n X (i) ∂ Dij = xk (j) . (A.56) k=1 ∂xk m Proof. Since Dij ∈ Der(F [V ]) it suffices to check (A.56) on the set (A.55), which generates F [V m] as an algebra. This verification is trivial. A.3 General Strategy As we mentioned before, our main goal is to characterize the polynomial invariants for the complex classical groups when their action on the column space Cn is given by right multiplication. However, it will not be always necessary to work over the complex numbers. For the moment, we only require that char(F ) = 0. 153 We begin by introducing some polynomial functions that will play a central role in all that follows. Then we work out a few examples that, besides their intrinsic interest, will illustrate a fruitful method to attack more general situations. We conclude with some remarks on this general strategy. Throughout this section we take V = F n and we fix m ∈ N. Definition A.3.1. Let i1, i2, . . . , in be integers between 1 and m. We define m [i1, i2, . . . , in]: V → F to be the function sending each (v1, v2, . . . , vm) to det(vi1 , vi2 , . . . , vin ), the determi- nant of the n×n matrix whose columns are the vectors vik . It is clear that [i1, i2, . . . , in] is a polynomial function. We call it a bracket. Lemma A.3.2. Keep the notation of the definition and suppose that V is the natural module for GL(n, F ) = {g ∈ Mn(F ) : det g 6= 0}. Let g ∈ GL(n, F ). Then −1 g · [i1, i2, . . . , in] = (det g) [i1, i2, . . . , in]. In particular, any bracket is SL(n, F )-invariant, where SL(n, F ) = {g ∈ Mn(F ) : det g = 1}. Definition A.3.3. Let ω be a nondegenerate bilinear form on V . Given integers i, j between 1 and m, we define a function ω(i j): V m → F by ω(i j)(v1, . . . , vm) = ω(vi, vj). 154 Then ω(i j) ∈ F [V m]. The group of all those matrices that preserve ω will be denoted by G(ω), i.e., 0 0 0 G(ω) = {g ∈ Mn(F ): ω(gu, gu ) = ω(u, u ) for all u, u ∈ V }. Lemma A.3.4. With the notation of the definition, ω(i j) is G(ω)-invariant. Example A.3.5. Suppose F = R, V = R2 and m = 1. Let ω be the bilinear form on V whose Gram matrix in the canonical basis is the identity. Let G = G(ω) ∩ SL(2, R). Thus G is just the group of rotations around the origin. We claim that F [V ]G is generated as an algebra by the function ω(1 1). It is clear that any polynomial in ω(1 1) is G-invariant. Conversely, suppose f ∈ F [V ]G. Let v = x1 ∈ V . Observe x2 first that 2 2 2 ω(1 1)(v) = ω(v, v) = x1 + x2 = ||v|| . ||v|| Since F = R, we can find g ∈ G such that gv = 0 . Geometrically, g is the rotation that takes v onto the positive horizontal axis. Keeping in mind that −I2 ∈ G, ||v|| ||v|| −||v|| f(v) = f g−1 = f = f . 0 0 0 So there exist real numbers c0, c1, c2, . . . , ck such that 2 2 2 k 2 f(v) = c0 + c1||v|| + c2||v|| + ··· + ck||v|| = c0 − c1||v|| + c2||v|| − · · · + (−1) ck||v|| . 155 It follows that ci = 0 for i odd, and hence f is a polynomial in Q(1 1). 2 Example A.3.6. Suppose m = 1 and V = F . Let B = {e1, e2} be the canonical basis of V and let ω be the bilinear form on V with Gram matrix 0 1 (A.57) 1 0 relative to B. Set G = G(ω) ∩ SL(2,F ). Then G consists of the matrices t 0 g(t) = 0 t−1 × with t ∈ F . Let {x1, x2} be the dual basis of B, so that F [V ] = F [x1, x2]. Let G a b f ∈ F [V ] . Suppose that the monomial x1x2 appears with nonzero coefficient c in × a b b−a F . Given t ∈ F , a calculation shows that the coefficient of x1x2 in g(t) · f is ct . So we must have c = ctb−a, whence 1 = tb−a. Since F is infinite, we deduce b = a. Consequently f is a polynomial in x1x2. Conversely, it is clear that any polynomial in x1x2 is G-invariant. Therefore G F [V ] = F [x1x2]. G Since ω(1 1) = 2x1x2 and char(F ) 6= 2, it follows that ω(1 1) generates F [V ] . Suppose that V is a module for a group G. According to Corollary A.1.11, the problem of characterizing F [V m]G reduces to finding the multihomogeneous compo- m G m nents F [V ]d with d ∈ N0 . In order to apply induction on d, we introduce a suitable m order relation in N0 . 156 0 0 0 m Definition A.3.7. Given d = (d1, . . . , dm) and d = (d1, . . . , dm) in N0 , we write d < d0 if P P 0 • di < di, or P P 0 0 0 • di = di and the largest index i for which di and di differ has di < di. m This defines a well-ordering in the set N0 . 2 (1) (1) (2) (2) Example A.3.8. We now suppose V = F and m = 2. Let {x1 , x2 , x1 , x2 } be the basis of (V 2)∗ associated to the canonical basis B of F 2, as defined in Section A.2. Let Ω : F [V 2] → F [V 2] be the map defined by ∂ ∂ 2 2 ∂x(1) ∂x(2) ∂ f ∂ f Ω(f) = 1 1 (f) = − . (1) (2) (1) (2) ∂ ∂ ∂x ∂x ∂x ∂x (1) (2) 1 2 2 1 ∂x2 ∂x2 2 2 Then, for all d = (d1, d2) ∈ N0 and f ∈ F [V ]d, (d1 + 1)d2f − D21 ◦ D12(f) = [1, 2]Ω(f). (A.58) This identity is obtained by direct calculation using (A.56) and keeping in mind that, 2 for i ∈ {1, 2}, the polarization Dii acts on F [V ]d by multiplication by di. Let ω be a symmetric nondegenerate bilinear form on V . Let Γ denote the Gram matrix of ω relative to B. We suppose that Γ is (A.57), or F = R and Γ = I2. Set G = G(ω) ∩ SL(2,F ). We claim that F [V 2]G is generated by the functions [1, 2], ω(1 1), ω(1 2), and ω(2 2). By Lemmas A.3.2 and A.3.4, these functions are G-invariant. We call them 2 G the basic invariants. As observed above, it will suffice to find each F [V ]d . Let 157 2 d = (d1, d2) ∈ N0. If d2 = 0, the conclusion follows from the case n = 1, which was considered in the previous two examples. So suppose d1 > 0 and assume that 0 2 0 2 G the result is true for all those d ∈ N0 such that d < d. Let f ∈ F [V ]d . Using the identity (A.58) together with Lemmas A.1.7 and A.2.2, we see that Ω(f) is G- invariant. By induction, it is a polynomial in the basic invariants. As D12(f) is 2 G-invariant and belongs to F [V ](d1+1,d2−1), the induction hypothesis implies that also D12(f) is a polynomial in the basic invariants. Now we need to show that D21 preserves polynomials in the basic invariants. Since D21 is a derivation, it will suffice to compute its effect on the basic invariants. By direct calculation, we see that D21 maps [1, 2] 7→ 0, ω(1 1) 7→ 2ω(1 2), ω(1 2) 7→ ω(2 2), ω(2 2) 7→ 0. Therefore, we deduce from (A.58) that (d1 + 1)d2 · f is a polynomial in the basic invariants. The result follows because (d1 + 1)d2 6= 0 in F . Let’s meditate a bit on the method used in the example. After finding the poly- nomial invariants in a low dimensional case, we applied induction with respect to the 2 order relation in N0 using (A.58) as our main tool. This strategy generalizes to many other situations. We only need to replace (A.58) by a suitable generalization. In the late nineteenth century, the Italian mathematician Alfredo Capelli discovered such a formula, which is known as Capelli’s identity. Observe that (A.58) can be written in the more suggestive form D11 + 1 D12 (f) = [1, 2]Ω(f). D D 21 22 158 (i) Suppose that n = dim V = m. Let {xj : 1 ≤ i ≤ m, 1 ≤ j ≤ m} be the basis of (V m)∗ associated to the canonical basis of V . Capelli’s identity says that ∂ ∂ ∂ D11 + m − 1 D12 ··· D1m (1) (2) ··· (m) ∂x1 ∂x1 ∂x1 ∂ ∂ ∂ D21 D22 + m − 2 ··· D2m (1) (2) ··· (m) ∂x ∂x ∂x (f)=[1, 2, . . . , m] 2 2 2 (f) ...... D D ··· D ∂ ∂ ∂ m1 m2 mm (1) (2) ··· (m) ∂xm ∂xm ∂xm for all f ∈ F [V m]. For a proof of this identity, we refer the reader to [FH91]. The determinants in this expressions are expanded as usual with compositions of operators reading from left to right. This remark is important because Der(F [V m]) is not commutative in general. For example, the determinant on the right equals X ∂m (sgσ) (1) (2) (m) . ∂xσ(1)∂xσ(2)···∂xσ(m) σ∈Sm This operator is denoted by Ω and called the Cayley’s omega operator. m Let K be the determinant in the left hand side. Let d = (d1, . . . , dm) ∈ N0 m and f ∈ F [V ]d. The operators in the main diagonal of K act as scalars on each multi-homogeneous component of F [V m]. Hence their composition acts on f by multiplication by ρ = (d1 + m − 1)(d2 + m − 2) ··· (dm−1 + 1)dm. Note that in any other term of the expansion of K, the rightmost off-diagonal operator 159 has the form Dij with i < j. Therefore X K(f) = ρf − PijDij(f), i where each Pij is a linear combination of compositions of various Dab with a 6= b. So Capelli’s identity yields X ρf = PijDij(f) + [1, 2, . . . , m]Ω(f), n = m. (A.59) i Suppose now that n < m. We can consider f as a function on V m which is independent of the last m − n coordinates of each argument. Then Ω(f) = 0 and hence X ρf = PijDij(f), n < m. (A.60) i In the oncoming sections, identities (A.59) and (A.60) will play the role of (A.58) in Example A.3.8. To conclude the section, we mention that in some occasions it will be necessary to prove that a certain polynomial function is constant. To this end, it is useful to introduce the concept of Zariski-dense sets. Definition A.3.9. Let A ⊂ V . We say that A is Zariski-dense in V if the only polynomial function f ∈ F [V ] that vanishes on A is the zero function. Lemma A.3.10. Given h ∈ F [V ], we define Vh = {v ∈ V : h(v) 6= 0}. If h 6= 0 then Vh is Zariski-dense in V . Proof. If f ∈ F [V ] vanishes on Vh, then fh = 0. The result follows because F [V ] 160 does not have divisors of zero. A.4 Polynomial Invariants of SL(n, F ) and GL(n, F ) Here we compute the polynomial invariants of SL(n, F ) and GL(n, F ) acting on V m, where V = F n is the natural module for these groups. The conclusions for the general linear group will follow almost immediately from those for the special linear group. In the latter case, we proceed by induction on the number of variables, and, m for each fixed m, we make induction with respect to the order relation < in N0 . The following technical lemma is useful to deal with the base case. Lemma A.4.1. Suppose 1 ≤ m ≤ n and let m A(m) = {(v1, . . . , vm) ∈ V : {v1, . . . , vm} is linearly independent in V }. Then A(m) is Zariski-dense in V m. Proof. Suppose first that m = n. Then, with the notation of Lemma A.3.10, we have m m A(m) = (V )h with h = [1, 2, . . . , m]. Since h 6= 0, A(m) is Zariski dense in V . m Assume now that m < n and let f ∈ F [V ] such that f|A(m) = 0. We can consider f as a function on V n which is independent of the last n − m vector arguments. So f vanishes on A(n). Hence f = 0 by the previous case. Theorem A.4.2. Suppose n ≥ 2. Then the polynomial invariants of SL(n, F ) acting on V m are the polynomials in the brackets [i1, i2, . . . , in] (A.61) 161 with 1 ≤ i1 < i2 < ··· < in ≤ m. In particular, if m < n then the only polynomial invariants of SL(n, F ) are the constant functions. Proof. Write G = SL(n, F ). We proceed by induction on m. The functions (A.61) will be called the basic invariants. Suppose first that 1 ≤ m < n. Define A(m) as above and let f ∈ F [V m]G. We will show that f is constant on A(m) and Lemma A.4.1 will imply that f is constant. Let v = (v1, . . . , vm) ∈ A(m) and w = (w1, . . . , wm) ∈ A(m). Since m < n, there exists g ∈ G such that g · v = w. (Just complete {v1, . . . , vm} and {w1, . . . , wm} to obtain bases {v1, . . . , vn} and {w1, . . . , wn} of V . Then map vi 7→ wi and scale the image of vn so that we obtain an isomorphism with determinant 1.) Since f is G-invariant, f(w) = f(g · v) = f(v). Therefore f is constant. Now suppose n ≤ m and assume that the theorem holds for polynomial functions m G with less than m variables. As explained before, it will suffice to find F [V ]d for m each d ∈ N0 . At this point we regard m as fixed and we make induction on d. Let m m G d = (d1, . . . , dm) ∈ N0 . If dm = 0, the elements of F [V ]d can be seen as functions in m − 1 variables and so they are polynomials in the basic invariants by induction. Henceforth suppose dm 6= 0 and assume that all the elements of M m G F [V ]d0 d0 m G are polynomials in the basic invariants. Let f ∈ F [V ]d . Recall from the previous 162 section that X 0 if m > n, ρ · f − PijDij(f) = i ρ = (d1 + m − 1)(d2 + m − 2) ··· (dm), and each Pij is a linear combination of compositions of various Dab with a 6= b. Notice that ρ 6= 0 because dm 6= 0 and char(F ) = 0. For i < j, Dij(f) is G-invariant and hence a polynomial in the basic invariants by induction. We need to show that the operators Dab preserve polynomials in the basic invariants. Since the Dab are derivations, it will suffice to compute their action on the basic invariants. This is an obvious calculation: Dab takes a bracket [i1, i2, . . . , in] to zero if b does not appear as one of the indices, or to the bracket with the index b replaced by a if b does occur. The latter is zero if a also occurs. Finally we can reorder the variables in the bracket, so that the indices appear in increasing order. This may change the sign of the bracket. If m > n, the conclusion follows because ρ 6= 0. Now suppose m = n. Since X ρ · f − PijDij(f) and [1, 2, . . . , m] i Corollary A.4.3. The only polynomial invariants of GL(n, F ) acting on V m are the 163 constant functions. Proof. If m < n, the result follows immediately from Theorem A.4.2. Assume n ≤ m. By Corollary A.1.6, we have m G M m G F [V ] = F [V ]k , k≥0 m G where G = GL(n, F ). Suppose, if possible, that F [V ]k 6= 0 for some k ∈ N. By Theorem A.4.2, k must be a multiple of n, say k = tn with t ∈ N, and the elements m G of F [V ]k are linear combinations of homogeneous functions of total degree k, each of which is a product of t brackets (A.61). Applying Lemma A.3.2, each g ∈ G m G −t t acts on F [V ]k by multiplication by (det g) . So we must have (det g) = 1 for all g ∈ G. Since F is infinite and the polynomial Xt − 1 has a finite number of roots × t in F , there exists a ∈ F such that a 6= 1. Set g0 = diag[a, 1,..., 1] ∈ G. Then t t (det g0) = a 6= 1, a contradiction. A.5 Polynomial Invariants of Sp(n, F ) In this section we take V = F n with n even, say n = 2`. Let ω be the bilinear form on V whose Gram matrix relative to the canonical basis is 0 I` J − = . ` −I` 0 We define Sp(n, F ) = G(ω) and we call it a symplectic group. Our goal is to char- acterize the polynomial invariants of Sp(n, F ) acting on V m, where V is the natural module for Sp(n, F ). Once we do this, a simple change of coordinates will provide 164 a description of the polynomial invariants of G(Q), where Q is any nondegenerate skew-symmetric bilinear form on V . We begin by showing that the brackets are not needed, for they are polynomials in the functions ω(i j). Lemma A.5.1. For all v1, . . . , vn ∈ V we have (−1)`(`−1)/2 X det(v , . . . , v ) = sg(σ)ω(v , v ) ··· ω(v , v ). 1 n `!2` σ(1) σ(2) σ(n−1) σ(n) σ∈Sn n times Proof. Let f : V × · · · × V → F be the function defined by X f(v1, . . . , vn) = sg(σ)ω(vσ(1), vσ(2)) ··· ω(vσ(n−1), vσ(n)). σ∈Sn It is clear that f is linear in each variable. We claim that f is alternating. Indeed, let n v = (v1, . . . , vn) ∈ V such that vi = vj for some i 6= j. Let τ = (i j) ∈ Sn. Then ω(vσ(1), vσ(2)) ··· ω(vσ(n−1), vσ(n)) = ω(vτσ(1), vτσ(2)) ··· ω(vτσ(n−1), vτσ(n)) for all σ ∈ Sn. Classifying the elements of Sn according to their parity and keeping in mind that Sn is the disjoint union of the alternating group An and its coset τAn, we deduce f(v) = 0. Thus f is an alternating n-linear form, which implies that f is a scalar multiple of det. Therefore it will suffice to compute f(e1, . . . , en). For each σ ∈ Sn, we set ξ(σ) = sg(σ)ω(eσ(1), eσ(2)) ··· ω(eσ(n−1), eσ(n)). 165 So f(e1, . . . , en) is the sum of all the ξ(σ). Looking at the Gram matrix of ω, we see that most of the terms ξ(σ) are 0. In fact a term ξ(σ) is non-zero if and only if |σ(2i − 1) − σ(2i)| = ` for 1 ≤ i ≤ `, (A.62) in which case ξ(σ) = ±1. Let Ξ be the set of all permutations σ ∈ Sn satisfying the condition (A.62). This set has size `!2` and is closed under composition on the left by (a) τi = (i i + `) with 1 ≤ i ≤ `, and (b) τij = (i j)(i + ` j + `) where i, j are between 1 and `. Since ω is skew-symmetric, τi is odd, and τij is even, we easily see that ξ(τiσ) = ξ(σ) = ξ(τijσ). for each σ ∈ Ξ. In addition, if we start with any element of Ξ and we repeatedly apply operations (a) and (b), we can reach any other element of Ξ. Therefore X ` f(e1, . . . , en) = ξ(σ) = `!2 ξ(σ0), σ∈Ξ where σ0 is any element of Ξ. The result follows taking 1 2 3 4 ······ n − 1 n σ0 = . 1 ` + 1 2 ` + 2 ······ ` 2` = n 166 Corollary A.5.2. Let i1, i2, . . . , in be integers between 1 and m. Then (−1)`(`−1)/2 X [i , i , . . . , i ] = sg(σ)ω(i i ) ··· ω(i i ). 1 2 n `!2` σ(1) σ(2) σ(n−1) σ(n) σ∈Sn Note A.5.3. Lemmas A.3.2 and A.5.1 imply Sp(n, F ) ⊂ SL(n, F ), which is not obvious from the definitions. Theorem A.5.4. The polynomial invariants of Sp(n, F ) acting on V m are the poly- nomials in the basic invariants ω(i j) with 1 ≤ i < j ≤ m. m m Let Tn be the statement that the theorem holds for Sp(n, F ) acting on V where m dim V = n. Since Sp(2,F ) = SL(2,F ), Theorem A.4.2 and Lemma A.5.1 imply T2 for all m ∈ N. Thus it suffices to prove the following implications: m m0 0 (i) Tn ⇒ Tn for 1 ≤ m < m; n−1 n m (ii) Tn ⇒ Tn ⇒ Tn for 2 ≤ n < m; n−1 n−1 (iii) Tn−2 ⇒ Tn for n ≥ 4. m m0 0 Lemma A.5.5. Tn ⇒ Tn for 1 ≤ m < m. Proof. The result follows regarding a function in m0 vector arguments as a function in m vector arguments. n−1 n m Lemma A.5.6. Tn ⇒ Tn ⇒ Tn for 2 ≤ n < m. m−1 Proof. Let n, m be integers such that 2 ≤ n ≤ m. We assume Tn and we want to m deduce Tn . The argument is analogous to the induction step in the proof of Theorem A.4.2, now keeping in mind Lemma A.5.1. We only need to compute the action of the 167 operators Dab with a 6= b on the basic invariants. This is an obvious calculation: Dab takes Q(i j) to zero if b does not appear as one of the indices, or replaces the index b by a if b does occur. The latter is zero if a also occurs. Finally we can reorder the variables, so that the indices appear in increasing order. This operation only changes signs. Therefore, the operators Dab preserve polynomials in the basic invariants. Before we prove (iii), we need a couple of facts from linear algebra. Lemma A.5.7. Sp(n, F ) acts transitively on V − {0}. Proof. Given v ∈ V − {0}, it will suffice to show that gv = e1 for some g ∈ Sp(n, F ). By basic linear algebra, we can complete {v} to a basis {v = v1, v2, . . . , vn} of V − relative to which ω has Gram matrix J` . Let g be the only element of GL(n, F ) such that gvi = ei for 1 ≤ i ≤ n. Since ω(gvi, gvj) = ω(ei, ej) = ω(vi, vj) for all i and j, we deduce g ∈ Sp(n, F ). Lemma A.5.8. Sp(n, F ) acts transitively on the hyperplanes of V . Proof. Since ω is non-degenerate, the map ψ : V → V ∗ sending u 7→ ω(u, −) is a linear isomorphism. Let H and H0 be hyperplanes of V . Let ϕ : V → F be a linear functional with kernel H. Then ϕ = ψ(u) for some non-zero u ∈ V . Hence H = ker(ψ(u)). Likewise, there exists v ∈ V − {0} such that H0 = ker(ψ(v)). By Lemma A.5.7, gu = v for some g ∈ Sp(n, F ). A calculation shows that g(H) ⊂ H0. Equality holds because both sides have the same dimension. n−1 n−1 Lemma A.5.9. Tn−2 ⇒ Tn for n ≥ 4. 168 m Proof. Let n be an integer such that n ≥ 4 and let G = Sp(n, F ). Suppose Tn−2, 0 where m = n − 1. Let B be the canonical basis of V and B = B − {e`, en}. Set 0 0 0 0 V = spanB and G = {g ∈ G : ge` = e`, gen = en}. Then G is a subgroup of G isomorphic to Sp(n − 2,F ). The action of G0 on V 0 is equivalent to the action of Sp(n − 2,F ) on F n−2, in the sense that the diagram Sp(n − 2,F ) −−−→' G0 y y GL(F n−2) −−−→' GL(V 0) commutes. Thus the action of G0 on F [(V 0)m] is equivalent to the action of Sp(n−2,F ) on F [(F n−2)m]. The basic invariants in F [(F n−2)m] correspond in F [(V 0)m] to the restrictions to (V 0)m of the functions ω(i j) with 1 ≤ i < j ≤ m. Let f be an arbitrary element of F [V m] and let f 0 denote its restriction to (V 0)m. 0 0 0 Since f is G invariant, the induction hypothesis implies f = p|(V 0)m , where p is a polynomial in the basic invariants ω(i j), 1 ≤ i < j ≤ m. We claim that f = p. Since f − p is 0 on (V 0)m, it will suffice to show that any G-invariant polynomial function that vanishes on (V 0)m is necessarily zero. So we assume f 0 = 0 and we aim to deduce f = 0. 0 m Let W = V ⊕ F u`. We show first that f is zero on W . We fix an element 0 0 0 m m (v1, . . . , vm) in (V ) and we define a function h : F → F by 0 0 h(t1, . . . , tm) = f(v1 + t1ur, . . . , vm + tmur). × −1 Given a ∈ F , let g be the only element of GL(n, F ) that gu` = au`, gun = a un, and gui = ui for i∈ / {`, n}. It is straightforward to check that g ∈ G. Using the fact 169 that f is G-invariant, for all t1, . . . , tm ∈ F we have 0 0 h(t1, . . . , tm) = f(g(v1 + t1ur), . . . , g(vm + tmur)) = h(at1, . . . , atm). Since h is a polynomial in t1, . . . , tm and a is arbitrary, it follows that h is constant. But h(0,..., 0) = 0. Thus h = 0, which means that f is zero on W m. m Finally, let (v1, . . . , vm) ∈ V . As m = n − 1, there is a hyperplane H of V containing v1, . . . , vm. By Lemma A.5.8, W = h · H for some h ∈ G. Then f(v1, . . . , vm) = f(hv1, . . . , hvm) = 0, whence f = 0. Now the proof of Theorem A.5.4 is complete. The following result will enable us to generalize Theorem A.5.4. The proof is just a routine verification and will be omitted. Lemma A.5.10. Let Q be a nondegenerate bilinear form on V and let p ∈ GL(n, F ). Let Q0 be the nondegenerate bilinear form on V defined by Q0(v, w) = Q(pv, pw), v, w ∈ V. Then p−1G(Q)p = G(Q0). In particular G(Q) ' G(Q0). The map f 7→ p−1 · f is an algebra automorphism of F [V m] that sends F [V m]G(Q) onto F [V m]G(Q0). In addition −1 0 −1 p · Q(i1 i2) = Q (i1 i2) and p · [i1, i2, . . . , in] = (det p)[i1, i2, . . . , in] 170 for all i1, i2, . . . , in between 1 and m. Corollary A.5.11. If Q is a nondegenerate skew-symmetric bilinear form on V , then the polynomial invariants of G(Q) acting on V m are the polynomials in the functions Q(i j) with i < j. Proof. This follows from Theorem A.5.4 and Lemma A.5.10 because any nondegen- erate skew-symmetric bilinear form on V is equivalent to ω. A.6 Polynomial Invariants of the Orthogonal Groups In this section we take V = F n, where F = R or F contains the square roots of all its elements. The latter condition holds, for instance, if F is algebraically closed. Although we are only interested in the situation F = C, the real case will be considered as well because of its geometric relevance. Lemma A.6.1. Let ω be the bilinear form on V with Gram matrix In relative to the canonical basis. Let i1, i2, . . . , in, j1, j2, . . . , jn be integers between 1 and m. Then [i1, i2, . . . , in][j1, j2, . . . , jn] = det (ω(ir js))1≤r,s≤n . Theorem A.6.2. Suppose n > 1. Let ωn be a nondegenerate symmetric bilinear from on V . If F = R, we assume that ωn has Gram matrix In relative to the canonical basis of V . Let Gn = G(ωn) ∩ SL(n, F ). Then the polynomial invariants of Gn acting on V m are the polynomials in the functions ωn(i j) and [i1, i2, . . . , in] 171 with 1 ≤ i ≤ j ≤ m and 1 ≤ i1 < i2 < ··· < in ≤ m. m m As before, let Tn be the assertion that the theorem holds for G acting on V 1 and dim V = n. Example A.3.5 implies T2 when F = R. If F contains the square roots of all its elements, then any two nondegenerate symmetric bilinear forms on V 1 are equivalent. In this case T2 follows from Example A.3.6 and Lemma A.5.10. It suffices to prove the following implications. m m0 0 (i) Tn ⇒ Tn for 1 ≤ m < m; n−1 n m (ii) Tn ⇒ Tn ⇒ Tn for 2 ≤ n < m; n−1 n−1 (iii) Tn−1 ⇒ Tn for n ≥ 3. The proofs of (i) and (ii) are exactly as in the previous cases and require no further comment. n−1 n−1 Lemma A.6.3. Tn−1 ⇒ Tn for n ≥ 3. Proof. By Lemma A.5.10, we may assume that ωn has Gram matrix In relative to m the canonical basis of V . Let n ≥ 3 and write m = n − 1. Suppose Tm . We consider the following subgroups of Gn: 0 00 Gn = {g ∈ Gn : gen = ±en} ,Gn = {g ∈ Gn : gen = en}. 0 Notice that Gn is the set of those matrices with block form A 0 (A.63) 0 ε 172 00 0 where A ∈ G(ωm) and ε = det A. In addition, Gn is the subgroup of Gn consisting of the matrices (A.63) with ε = 1 and hence A ∈ Gm. m Gn 0 0 m 0 Let f ∈ F [V ] . Let f be f restricted to (V ) , where V = span{e1, . . . , em}. 0 0 00 Then f is invariant under Gn and Gn. Like in the proof of Lemma A.5.9, the action 00 0 m of Gn on V is equivalent to the action of Gm on F . Thus, applying the induction hypothesis and Lemma A.6.1, f 0 has the form C + D[1, 2, . . . , m] where 0 m • C and D are restrictions to (V ) of polynomials in the functions ωn(i j), i ≤ j; • the bracket [1, 2, . . . , m] ignores the last component of each vector argument. 0 Consider be the element g of Gn such that gei = ei for 1 ≤ i ≤ n − 2 and gei = −ei for i ∈ {n − 1, n}. Observe that g has the form (A.63) with det A = ε = −1. Then, applying Lemmas A.3.2 and A.3.4, we have C + D[1, . . . , m] = f 0 = g · f 0 = C − D[1, . . . , m], 0 0 which implies f = C because char(F ) 6= 2. Hence f = p|(V 0)m where p is a polynomial in the functions ωn(i j) with i ≤ j. We claim that f equals p. Since their difference is zero on (V 0)m, it suffices to show that f 0 = 0 implies f = 0. Assume f 0 = 0. 0 m Since f is Gn invariant, it is zero on the set Y = Gn · (V ) . If F = R then Gn acts transitively on the hyperplanes of V and hence Y = V m. Suppose now that F contains the square roots of all its elements. We will prove that Y is Zariski dense 173 in V m and it will follow that f is identically 0. Let d : V m → F be the polynomial function defined by d(v1, . . . , vm) = det (ωn(vi, vj))1≤i,j≤m . Since d 6= 0, the set Z = {v ∈ V m : d(v) 6= 0} is Zariski-dense in V m by Lemma A.3.10. We claim that Z ⊂ Y . Indeed, let v = (v1, . . . , vm) ∈ Z. The vectors v1, . . . , vm are linearly independent for if one of them, say vk, were a linear combination of the others, then the k-th column of the matrix (ωn(vi, vj))i,j would be a linear combination of the other columns, and hence d(v) = 0, a con- tradiction. Thus W = span{v1, . . . , vm} has dimension m. By definition of Z, the 0 restriction ωn of ωn to W × W is non-degenerate. By the assumption on F , there is a basis {w1, . . . , wm} of W relative to which ωn has Gram matrix Im. Let wn ∈ V be a nonzero vector orthogonal to w1, . . . , wm. (The existence of such a vector follows from the fact that any homogeneous linear system with less equations than variables 0 has a nontrivial solution.) Since ωn is nondegenerate, wn ∈/ W and ωn(wn, wn) 6= 0. So we can assume ωn(wn, wn) = 1 by scaling wn. Thus {w1, . . . , wn} is a basis of V relative to which ωn has Gram matrix In. Let g be the only element of GL(n, F ) such that gei = wi for all i. Clearly g ∈ G(ωn). Moreover, we can assume det g = 1 because otherwise we could define 0 0 m gen = −wn. Thus g ∈ Gn. Since g maps V onto W , we have v ∈ Gn · (V ) = Y . Consequently Z ⊂ Y . Then Y is Zariski-dense in V m, which implies f = 0. 174 The proof of Theorem A.6.2 is now complete. As a corollary, we are able to characterize the polynomial invariants of the full orthogonal group. Corollary A.6.4. Keep the hypothesis of Theorem A.6.2. Then the polynomial in- m variants of G(ωn) acting on V are the polynomials in the basic invariants ωn(i j), 1 ≤ i ≤ j ≤ m. m G(ωn) Proof. Let f ∈ F [V ] . In particular, f is invariant under G(ωn) ∩ SL(n, F ). Applying Theorem A.6.2 and Lemma A.6.1, f has the form X C + CiDi, where the C’s are polynomials in the basic invariants and the D’s are brackets. If g is any element of G(ωn) with negative determinant, we have X X C + CiDi = f = g · f = C − CiDi, which implies f = C because char(F ) 6= 2. For future use, we state the following particular case of Corollary A.6.4. Corollary A.6.5. Suppose n > 1. Let V = Cn. Let ω be the bilinear form on V + whose Gram matrix relative to the canonical basis is J` if n = 2` is even, or S` if n = 2` + 1 is odd, where 1 0 0 0 I + ` J = ,S` = 0 0 I . ` ` I` 0 0 I` 0 175 Notice that G(ω) = O(n, C). Then the polynomial invariants of O(n, C) acting on V m are the polynomials in the functions ω(i j) with i ≤ j. A.7 Useful Consequences In this section we take F = C and V = Cn. Let ω be the bilinear form on V − + whose Gram matrix relative to the canonical basis {e1, . . . , en} is J` , J` , or S`. Let G = G(ω). Thus G is O(n, C) or Sp(n, C), depending on whether ω is symmetric or skew-symmetric. We regard V as the natural module for G. Let ρ : G → GL(V ) be the associated representation. The action of G on the tensor power V ⊗d is given by the representation d times g 7→ ρ(g)⊗ · · · ⊗ρ(g). ∗ Let {x1, . . . , xn} be the basis of V that is dual with respect to {e1, . . . , en}. Since ω is nondegenerate, the map V → V ∗ sending v 7→ ω(−, v) is an isomorphism of G-modules. So we have a basis {u1, . . . , un} of V such that xi = ω(−, ui) for all i. Thus ω(ej, ui) = δij for all i, j. Given a pair J = (p, q) of integers such that 1 ≤ p < q ≤ d, recall the linear ⊗d ⊗d operator ϑJ : V → V sending each elementary tensor v1 ⊗ · · · ⊗ vd to n X ω(vp, vq)v1 ⊗ · · · ⊗ ei ⊗ · · · ⊗ ui ⊗ · · · ⊗ vd. |{z} |{z} i=1 p-th pos. q-th pos. ⊗d Also recall that for each σ ∈ Sd we have a linear automorphism πσ of V given 176 by v1 ⊗ · · · ⊗ vd 7→ vσ(1) ⊗ · · · ⊗ vσ(d). ⊗d The formula v · σ = πσ(v) defines a right action of the symmetric group Sd on V that commutes with the left action of G. The objective of this section is to prove the following result. Theorem A.7.1. The elements of End(V ⊗d) that commute with all the operators ϑ(p,q) (1 ≤ p < q ≤ d) and with all the operators πσ (σ ∈ Sd) are the linear combina- tions of operators of the form ρ(g) ⊗ · · · ⊗ ρ(g) with g ∈ G. One of the implications is almost trivial. We have already observed that the ⊗d ⊗d actions of G and Sd on V commute. To prove that the action of G on V commutes with the operators ϑJ , it suffices to check that the element n X ψ = ei ⊗ ui. i=1 is fixed by G. Although this can be achieved by direct calculation, it will follow later from a more general argument. The main ingredients in the proof of Theorem A.7.1 will be some results on poly- nomial invariants from previous sections, namely Theorem A.5.4 and Corollary A.6.5, together with the following fact from the theory of associative algebras. Theorem A.7.2. (Double Commutant Theorem) Let W be finite dimensional vector space over an arbitrary field. For each subset C of End(W ) we define Comm(C) = {x ∈ End(W ): xy = yx for all y ∈ C}. 177 Suppose that A is a subalgebra of End(W ) with identity element IW . Assume that W is a completely reducible module over A. Let B = Comm(A). Then A = Comm(B). Proof. By definition we have A ⊂ Comm(B). Let T ∈ Comm(B) and fix a basis m m {v1, . . . , vm} of W . Let v0 = (v1, . . . , vm) ∈ W . The A-module W is completely reducible because W is. So the cyclic submodule M = A· v0 has an A-invariant complement N. Let P : W m = M ⊕ N → M be the projection along N. 2 Since P is linear, there exist m maps pij ∈ End(W ) (1 ≤ i ≤ m, 1 ≤ j ≤ m) such that m m ! X X P (w) = p1j(wj),..., pmj(wj) j=1 j=1 m for all w = (w1, . . . , wm) ∈ W . The fact that A preserves both M and N implies m that P commutes with the action of A on W . Consequently pij ∈ B for all i, j. We have v0 ∈ M because IW ∈ A. So P (v0) = v0, which means m X pij(vj) = vi, j=1 for 1 ≤ i ≤ m. Then, keeping in mind that pij ∈ B and T ∈ Comm(B), m ! m ! X X P (T · v0) = pij(T (vj)) = T pij(vj) = (T (v1),...,T (vm)). j=1 1≤i≤m j=1 1≤i≤m It follows that (T (v1),...,T (vm)) ∈ M = A· v0. So there exists S ∈ A such that (T (v1),...,T (vm)) = S · v0 = (S(v1),...,S(vm)). 178 Therefore T = S ∈ A. By Theorem 3.7.4, G is reductive as a linear algebraic group. Thus V ⊗d is a completely reducible G-module. Set A = span{ρ(g) ⊗ · · · ⊗ ρ(g): g ∈ G} ⊂ End(V ⊗d). Then A is a subalgebra of End(V ⊗d). Notice that the identity map is in A. Since each g ∈ G acts on V ⊗d by means of ρ(g) ⊗ · · · ⊗ ρ(g), the actions of G and A on V ⊗d have the same invariant subspaces. It follows that V ⊗d is completely reducible as a module over A. Let B be the subalgebra of End(V ⊗d) consisting of all polynomials in the operators ϑ(p,q) (1 ≤ p < q ≤ d) and πσ (σ ∈ Sd). With this notation, Theorem A.7.1 states that A = Comm(B). By the Double Commutant Theorem, the proof of Theorem A.7.1 reduces to show that B = Comm(A). In other words, all we need to do is to prove the following result. Theorem A.7.3. The elements of End(V ⊗d) that commute with the operators ρ(g) ⊗ · · · ⊗ ρ(g) with g ∈ G are the polynomials in the operators ϑ(p,q) (1 ≤ p < q ≤ d) and πσ (σ ∈ Sd). We can view End(V ⊗d) as a module for G via (g · φ)(v) = g · φ(g−1 · v), g ∈ G, φ ∈ End(V ⊗d), v ∈ V ⊗d. Consequently, the elements of Comm(A) are precisely the endomorphisms of V ⊗d 179 that are fixed by G, i.e., ⊗d G ⊗d Comm(A) = [End(V )] = EndG(V ). Next we consider a series of isomorphisms of G-modules in order to reduce the com- putation of Comm(A) to a question on polynomial invariants in C[V 2d]. In first place, recall that the map V → V ∗ given by v 7→ ω(−, v) is a G- isomorphism. Secondly, if W is any finite dimensional module over G, then we have an isomorphism of G-modules W ∗ ⊗ W → End(W ) sending each h ⊗ w to the map u 7→ f(u)w. Thirdly, there is a G-isomorphism (V ∗)⊗d → (V ⊗d)∗ mapping each elementary tensor h1 ⊗ · · · ⊗ hd to the linear function ⊗d h1⊗¯ · · · ⊗¯ hd : V → C such that (h1⊗¯ · · · ⊗¯ hd)(v1 ⊗ · · · ⊗ vd) = h1(v1) ··· hd(vd). 2d 2d Finally, consider the multi-homogeneous component C[V ]1 of C[V ], where 1 stands for the 2d-tuple (1,..., 1). The linear map (V ∗)⊗(2d) → C[V 2d] defined by h1 ⊗ · · · ⊗ h2d 7→ [(v1, . . . , v2d) 7→ h1(v1) ··· h2d(v2d)] is an isomorphism of G-modules. By combining all these identifications, we obtain the following chain of G-isomorphisms: 2d ∗ ⊗(2d) ⊗(2d) ∗ ⊗d ⊗d ∗ ⊗d ⊗d C[V ]1 ' (V ) ' V ' (V ⊗ V ) ' (V ) ⊗ V ' End(V ). (A.64) 180 2d G As a consequence of Theorem A.5.4 and Corollary A.6.5, C[V ]1 is linearly spanned by the polynomial functions ω(σ(1) σ(2))ω(σ(3) σ(4)) ··· ω(σ(2d − 1) σ(2d)) (A.65) with σ ∈ S2d. Our task consists of passing this information through the isomorphisms ⊗d 2d ⊗(2d) (A.64) in order to characterize EndG(V ). We will go first from C[V ]1 to V , and then from here to End(V ⊗d). Explicitly, our G-isomorphism T : V ⊗(2d) → End(V ⊗d) is given on elementary tensors by the formula T (v1 ⊗ · · · ⊗ v2d)(w1 ⊗ · · · ⊗ wd) = ω(w1, v2)ω(w2, v4) ··· ω(wd, v2d)v1 ⊗ v3 ⊗ · · · ⊗ v2d−1. The preimage under T of the identity map IV ⊗d is the element d times ψd = ψ⊗ · · · ⊗ψ, that is, n X ψd = ei1 ⊗ ui1 ⊗ ei2 ⊗ ui2 ⊗ · · · ⊗ ein ⊗ uin . i1,i2,...,id=1 Since IV ⊗d is G-invariant, we deduce that ψd and hence ψ are G-invariant. As men- tioned before, this implies that all the operators ϑJ commute with the action of G on V ⊗d. In the case d = 1, we have n ! X T ui ⊗ ei = εIV i=1 181 where 1 if ω is symmetric, ε = −1 if ω is skew-symmetric. Thus n X ψ = ε ui ⊗ ei. (A.66) i=1 ⊗(2d) 2d Now we consider the isomorphism S : V → C[V ]1 of G-modules sending each v1 ⊗ · · · ⊗ v2d to the polynomial function (w1, . . . , w2d) 7→ ω(w1, v1)ω(w2, v2) ··· ω(w2d, v2d). −1 It is readily checked that the preimage under S of the function (A.65) is ψd · σ . 2d G Since the functions (A.65) span C[V ]1 , we obtain the following characterization of [V ⊗(2d)]G, which is interesting in its own right. ⊗(2d) G −1 Theorem A.7.4. [V ] is spanned by the elements ψd · σ with σ ∈ S2d. ⊗(2d−1) G Note A.7.5. We have [V ] = 0 because the matrix −In is in G and acts on V ⊗(2d−1) by multiplication by −1. Next we introduce some useful subgroups of S2d. ˜ Let Sd be the subgroup of S2d that permutes the ordered pairs (1, 2) , (3, 4) ,..., (2d − 1, 2d), Precisely, for each σ ∈ Sd we define an elementσ ˜ of S2d that sends 2i − 1 7→ 2σ(i) − 1 , 2i 7→ 2σ(i) 182 for 1 ≤ i ≤ d. The map Sd → S2d given by σ 7→ σ˜ is a group monomorphism, whose ˜ ˜ image we denote by Sd. In particular |Sd| = d!. Observe that ψd · σ˜ = ψd (A.67) ˜ for allσ ˜ ∈ Sd. Now let Nd be the subgroup of S2d generated by the transpositions (2i − 1 2i) with 1 ≤ i ≤ d. Thus a1 a2 ad d Nd = {(1 2) (3 4) ··· (2d − 1 2d) : ai ∈ {0, 1}} ' Z2. d In particular |Nd| = 2 . Using (A.66) we see that ψd · σ = ±ψd (A.68) for all σ ∈ Nd. −1 ˜ ˜ A calculation shows that σ Ndσ ⊂ Nd for all σ ∈ Sd. Therefore Bd = SdNd is ˜ d a subgroup of S2d. Since Sd ∩ Nd is trivial, we deduce |Bd| = d!2 . It follows from (A.67) and (A.68) that ψd · σ = ±ψd (A.69) for all σ ∈ Bd. Using (A.69) we get a better version of Theorem A.7.4: ⊗(2d) G −1 Theorem A.7.6. [V ] = span{ψd · σ : σ ∈ Ωd}, where Ωd is any set of left coset representatives for Bd in S2d. 183 For each s ∈ Sd we define a permutation τ(s) ∈ S2d by τ(s)(2i − 1) = 2s(i) − 1 , τ(s)(2i) = 2i, for 1 ≤ i ≤ d. So τ(s) permutes {1, 3,..., 2d − 1} and fixes {2, 4,..., 2d} pointwise. The map τ : Sd → S2d is a group monomorphism. Then τ(Sd) is a subgroup of S2d of order d!. Before we state and prove the next result, recall that, given subgroups H1,H2 of a group K, we can define an equivalence relation ∼ in K by k ∼ k0 if and only if 0 k = h1k h2 for some h1 ∈ H1 and h2 ∈ H2. The equivalence class of each k ∈ K is {h1kh2 : h1 ∈ H, h2 ∈ H} = H1kH2. The quotient set is denoted by H1 \K/H2 and its elements are called the double cosets of (H1,H2) in K. Lemma A.7.7. Let Γ ⊂ S2d be a set of representatives of the double cosets of (τ(S)d, Bd) in S2d. Then ⊗d −1 EndG(V ) = span{πs−1 ◦ T (ψd · γ ): s ∈ Sd, γ ∈ Γ}. Proof. Applying Theorem A.7.4, we have ⊗d −1 EndG(V ) = span{T (ψd · γ ): γ ∈ S2d}. (A.70) 184 It follows from the definitions that πs ◦ T (v) = T (v · τ(s)) (A.71) ⊗(2d) for all s ∈ Sd and v ∈ V . Suppose that the permutations γ1, γ2 ∈ S2d belong to the same double coset, i.e., γ1 = τ(s)γ2σ for some s ∈ Sd and σ ∈ Bd. Then, using (A.69) and (A.71), −1 −1 −1 −1 −1 T (ψd · γ1 ) = T (ψd · σ γ2 τ(s )) = ±πs−1 ◦ T (ψd · γ2 ). (A.72) The result follows form (A.70) and (A.72). Now we are quite close to prove Theorem A.7.3. In view of this lemma, it suffices to find a set Γ of representatives for the double cosets in τ(Sd) \ S2d/Bd such that −1 T (ψd · γ ) is a polynomial in the operators ϑJ for every γ ∈ Γ. A 2-partition of {1, 2,..., 2d} is a partition of this set into d classes with 2 elements each. The set of all such partitions will be denoted by Xd. Each element x of Xd can be represented by an undirected graph, called a Brauer diagram. We draw a 2 × d array of dots, with the dots in the top row labeled 1, 3,..., 2d − 1 from left to right, and the dots in the bottom row labeled 2, 4,..., 2d from left to right. In the Brauer diagram for x, the dots i, j are joined by an edge if and only if {i, j} ∈ x. For instance, if d = 4 and x is {{1, 3}, {2, 5}, {4, 6}, {7, 8}}, (A.73) 185 the associated diagram is 1 3 5 7 • • • • • • • • 2 4 6 8 (A.74) Since each element x of Xd is uniquely determined by its Brauer diagram D, we sometimes make no distinction between x and D. In a Brauer diagram D, a top bar is an edge that joins two dots from the top row. Similarly, a bottom bar is an edge that connects two dots from the bottom row. Clearly, the number of top bars in D coincides the number of bottom bars. If this number is r, we say that D is an r-bar diagram. For example, (A.74) is a 1-bar diagram. There is a natural action of S2d on Xd: for σ ∈ S2d and x = {{i1, j1},..., {id, jd}} ∈ Xd, we define σ · x = {{σ(i1), σ(j1)},..., {σ(id), σ(jd)}}. For example, if σ = (3 5 4) ∈ S8 and x is (A.73), then σ · x is represented by 1 3 5 7 • • • • • • • • 2 4 6 8 Let x0 = {{1, 2}, {3, 4},..., {2d − 1, 2d}}. The corresponding picture is 186 1 3 2d − 1 • • • ··· • • • 2 4 2d The action of S2d on Xd is clearly transitive. The stabilizer group of x0 is precisely Bd. Thus we can identify Xd with the quotient S2d/Bd by means of the map σ·x0 7→ σBd. Consequently, the action of τ(Sd) on Xd can be seen as an action on S2d/Bd by left multiplication. The orbits for the latter action coincide with the double cosets of (τ(Sd), Bd) in S2d. Each orbit of τ(Sd) on Xd has a unique representative z with the following prop- erties: (a) Whenever {2i, 2j} ∈ z, we have {2i − 1, 2j − 1} ∈ z, in which case we say that z contains the (i, j)-bar. Geometrically, this means that each top bar in z lays on top of a bottom bar. (b) Whenever {2i − 1, 2j} ∈ z, we have i = j. Graphically, the diagram has no diagonal edges, i.e., the edges that join a dot from the top row with a dot from the bottom row are vertical. A Brauer diagram that meets conditions (a) and (b) is called normalized. The set of all such diagrams is denoted by Zd. For example, (A.74) is not normalized. The normalized representative in the τ(S4)-orbit of (A.74) is 1 3 5 7 • • • • • • • • 2 4 6 8 This is a 1-bar diagram containing the (2, 3)-bar. 187 ⊗d Now we associate an element of End(V ) to each element z of Zd. We begin with the 1-bar diagram containing the (1, 2)-bar, i.e., 1 3 5 7 2d − 1 • • • • • z = ··· • • • • • 2 4 6 8 2d Observe that z = (2 3) · x0. In addition, for all w1, . . . , wd ∈ V , T (ψd · (2 3))(w1 ⊗ · · · ⊗ wd) = n ! X = T ei1 ⊗ ei2 ⊗ ui1 ⊗ ui2 ⊗ ei3 ⊗ ui3 ⊗ · · · ⊗ ein ⊗ uin (w1 ⊗ · · · ⊗ wd) i1,...,id=1 n X = ω(w1, ei2 )ω(w2, ui2 )ω(w3, ui3 ) ··· ω(wd, uid )ei1 ⊗ ui1 ⊗ ei3 ⊗ · · · ⊗ eid i1,...,id=1 n X = ω(w1, ei2 )xi2 (w2)xi3 (w3) ··· xid (wd)ei1 ⊗ ui1 ⊗ ei3 ⊗ · · · ⊗ eid i1,...,id=1 n X = ω(w1, w2) ei1 ⊗ ui1 ⊗ w3 ⊗ · · · ⊗ wd i1=1 = ϑ(1,2)(w1 ⊗ · · · ⊗ wd). Thus T (ψd · (2 3)) = ϑ(1,2). (A.75) More generally, suppose that z ∈ Zd is a 1-bar diagram containing the (p, q)-bar, with 1 ≤ p < q ≤ d. Then z = (2p 2q − 1) · x0. Computing just as before, we obtain T (ψd · (2p 2q − 1)) = ϑ(p,q). (A.76) 188 Suppose now that z ∈ Zd is an r-bar diagram containing the (i1, j1),..., (ir, jr)- bars. Let r Y γz = (2ik 2jk − 1). k=1 Then z = γz · x0. Since the transpositions (2ik 2jk − 1) act on different tensor positions, we obtain −1 T (ψ · γz ) = T (ψ · γz) = ϑ(i1,j1) ◦ · · · ◦ ϑ(ir,jr) (A.77) by the same calculations that led to (A.75) and (A.76). Since Zd is a set of representa- tives for the orbits of τ(Sd) on Xd, Γ = {γz : z ∈ Zd} is a set of representatives for the double cosets of (τ(Sd), Bd) in S2d. Therefore, Lemma A.7.7 and equation (A.77) imply Theorem A.7.3. As explained before, Theorem A.7.1 follows from Theorem A.7.3. 189