## K-Representations and G-Representations

As mentioned at the end of the previous article, we shall attempt to construct analytic representations of $G = GL_n\mathbb{C}$ from continuous representations of $K = U_n.$

Let $\rho : K \to GL(V)$. Consider $\rho|_S$, where $S$ is the group of diagonal matrices in K so

$S = \left\{\begin{pmatrix} e^{i\theta_1} & 0 & \ldots & 0 \\ 0 & e^{i\theta_2} & \ldots & 0\\ \vdots & \vdots & \ddots &\vdots\\ 0 & 0 & \ldots & e^{i\theta_n}\end{pmatrix} : \theta_1, \ldots, \theta_n \in \mathbb{R}\right\} \cong (\mathbb R/\mathbb Z)^n$

as a topological group. From our study of representations of the n-torus, we know that $\rho|_S$ is a direct sum of 1-dimensional irreps of the form:

$\rho_{\mathbf a} = \rho_{a_1, \ldots, a_n} : S \to \mathbb C^*, \quad \begin{pmatrix}x_1 & \ldots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \ldots & x_n \end{pmatrix} \mapsto x_1^{a_1} \ldots x_n^{a_n},$

where $\mathbf a = (a_1, \ldots, a_n) \in \mathbb Z^n$. E.g. if $\mathbf a = \mathbf 0$ then the representation is trivial; if $\mathbf a = (1,\ldots, 1),$ the representation is obtained by taking the determinant.

Hence, the character of $\rho|_S$ is a Laurent polynomial with non-negative integer coefficients, i.e.

$\chi(\rho|_S) \in \mathbb{Z}[x_1, \ldots, x_n, x_1^{-1}, \ldots, x_n^{-1}].$

Definition. For a continuous finite-dimensional representation $\rho :K \to GL (V)$, we will write $\psi_V$ for $\chi(\rho|_S)$ expressed as a Laurent polynomial in $x_1, \ldots x_n$.

The same holds for a complex analytic representation of G.

Examples

1. If $\rho: G \to GL (\mathbb C^n)$ is the identity map, its Laurent polynomial is $x_1 + x_2 + \ldots + x_n.$
2. For $\det : G\to \mathbb C^*$, its Laurent polynomial is $x_1 x_2 \ldots x_n.$

The following are clear for any continuous representations VW of K.

\begin{aligned} \psi_{V\oplus W} &= \psi_V + \psi_W \\ \psi_{V\otimes W} &= \psi_V \cdot \psi_W\\ \psi_{V} &= \psi_W + \psi_{V/W} \text{ if } W\subseteq V \\ \psi_{V^\vee}(x_1, \ldots, x_n) &= \psi_V(x_1^{-1}, \ldots, x_n^{-1}).\end{aligned}

In summary, so far we have the following:

## Main Examples: Sym and Alt

For any vector space V, the group $S_d$ acts on $V^{\otimes d}$ by permuting the components. We denote:

\begin{aligned} \text{Sym}^d V &:= \{ v \in V^{\otimes d} : w(v) = v \text{ for each } w\in S_d\},\\ \text{Alt}^d V &:= \{ v \in V^{\otimes d} : w(v) = \chi(w)v \text{ for each } w\in S_d\}\end{aligned}

where $\chi(w)$ is the sign of w. Let $e_1, \ldots, e_n$ be a fixed basis of V. The case d=2 is quite easy to describe for the above spaces, for we can just take the following bases:

• $\text{Sym}^2 V : \{ e_i \otimes e_j + e_j \otimes e_i : 1 \le i \le j \le n\}$;
• $\text{Alt}^2 V : \{ e_i \otimes e_j - e_j \otimes e_i : 1 \le i < j \le n\}$.

Denote these two types of elements by $e_i e_j$ and $e_i \wedge e_j$ respectively, so that $e_i e_j = e_j e_i$ and $e_i \wedge e_j = -e_j \wedge e_i.$ Note that $V\otimes V \cong \text{Sym}^2 V \oplus\text{Alt}^2 V$; this is not true for higher values of d.

Similarly, in general, we can pick the following as bases:

• $\text{Sym}^d V : \{ e_{i_1} \ldots e_{i_d} : 1 \le i_1 \le \ldots \le i_d \le n\}$;
• $\text{Alt}^d V : \{ e_{i_1} \wedge \ldots \wedge e_{i_d}: 1 \le i_1 < \ldots < i_d \le n\}$.

where the components commute in Sym and anticommute in Alt (e.g. $e_1 \wedge e_2 \wedge e_4 = - e_2 \wedge e_1 \wedge e_4 = e_2 \wedge e_4 \wedge e_1$).

Now suppose $V = \mathbb{C}^n$ and $G = GL_n\mathbb{C}$ acts on it; G and $S_d$ act on $V^{\otimes d}$ in the following manner:

\begin{aligned} g\in G &\implies g(v_1 \otimes \ldots \otimes v_d) = g(v_1) \otimes \ldots \otimes g(v_d), \\ w\in S_d &\implies w(v_1 \otimes \ldots \otimes v_d) = v_{w^{-1}(1)} \otimes \ldots \otimes v_{w^{-1}(d)}.\end{aligned}

These two actions commute, from which one easily shows that $\text{Sym}^d V$ and $\text{Alt}^d V$ are G-invariant subspaces of $V^{\otimes d}$.

Example

Suppose we have $g = \begin{pmatrix} 2 & -1\\ 1 & 3\end{pmatrix}$ acting on $\text{Sym}^3 \mathbb{C}^2$. This takes:

$e_1^2 e_2 \mapsto (2e_1 - e_2)^2 (e_1 + 3e_2) = 4e_1^3 + 8e_1^2 e_2 - 11e_1 e_2^2 + 3e_2^3.$

To compute the Laurent polynomials of these spaces, we let the diagonal matrix $D(x_1, \ldots, x_n)$ act on them, giving:

\begin{aligned} e_{i_1} e_{i_2} \ldots e_{i_d} &\mapsto (x_{i_1} x_{i_2} \ldots x_{i_d}) e_{i_1} e_{i_2} \ldots e_{i_d},\\ e_{i_1} \wedge \ldots \wedge e_{i_d}& \mapsto (x_{i_1} x_{i_2}\ldots x_{i_d}) e_{i_1} \wedge \ldots \wedge e_{i_d}.\end{aligned}

Hence we have:

\displaystyle\begin{aligned} \psi_{\text{Sym}^d} &= \sum_{1\le i_1 \le \ldots \le i_d} x_{i_1} x_{i_2} \ldots x_{i_d} = h_d(x_1, \ldots, x_n),\\ \psi_{\text{Alt}^d} &= \sum_{1 \le i_1 < \ldots < i_d} x_{i_1} x_{i_2} \ldots x_{i_d} = e_d(x_1, \ldots, x_n).\end{aligned}

## Some Lemmas

Lemma 1. For a representation V of K, the Laurent polynomial $\psi_V$ is symmetric.

Proof

For any $w\in S_n$, take the corresponding permutation matrix $M\in K$; we have:

$\displaystyle M\cdot \begin{pmatrix} x_1 & \ldots & 0 \\ \vdots & \ddots & \vdots \\0 & \ldots & x_n \end{pmatrix}\cdot M^{-1} = \begin{pmatrix}x_{w(1)} & \ldots & 0 \\ \vdots & \ddots & \vdots \\0 & \ldots & x_{w(n)}\end{pmatrix}.$

Thus $\psi_V(x_1, \ldots, x_n) = \psi_V(x_{w(1)}, \ldots, x_{w(n)})$ for any $w\in S_n.$ ♦

Lemma 2. Given K-representations V, W, if $\psi_{V} = \psi_{W}$, then $V\cong W.$

Hence by the previous article, the same holds for analytic G-representations V, W.

Proof

Any $M\in K$, being unitary, is diagonalizable by a unitary matrix, i.e. there exists $Q\in K$ such that $QMQ^{-1} \in S$. Hence the given condition implies:

$\chi_{V}(M) = \chi_{V}(\overbrace{QMQ^{-1}}^{\in S})= \chi_{W}(QMQ^{-1}) = \chi_{W}(M).$.

By character theory of compact groups, $V \cong W$ as K-reps. ♦

Now for the final piece of the puzzle.

Lemma 3. Let $f \in \mathbb{Z}[x_1, \ldots, x_n]$ be a symmetric polynomial. There are polynomial representations:

$\rho_1 : G\to GL(V_1),\quad \rho_2 : G\to GL(V_2),$

such that $\psi_{V_1} - \psi_{V_2} = f.$

Proof

Taking homogeneous parts, let us assume $f \in \Lambda_n^{(d)}$ for some degree $d\ge 0.$ Write f as a linear sum of elementary symmetric polynomials $\{h_\lambda\}_{\lambda\vdash d, \lambda_1 \le n}$ with integer coefficients; separating terms we have $f = g - h$, where gh are both linear combinations of $\{h_\lambda\}$ with non-negative integer coefficients. Hence, it suffices to show that $h_\lambda = \psi_V$ for some polynomial representation $G\to GL(V).$

Since $h_\lambda = h_{\lambda_1} \ldots h_{\lambda_l}$, we can just pick:

$V = \left(\text{Sym}^{\lambda_1} \mathbb{C}^n\right) \otimes \left(\text{Sym}^{\lambda_2} \mathbb{C}^n\right) \otimes \ldots \otimes \left(\text{Sym}^{\lambda_d} \mathbb{C}^n\right),$

from the above. ♦

## Consequences

Immediately we have:

Corollary 1. For any symmetric Laurent polynomial $f\in \mathbb{Z}[x_1^{\pm 1}, \ldots, x_n^{\pm 1}]$, there exist rational representations:

$\rho_1 : G\to GL(V_1), \quad \rho_2 : G\to GL(V_2)$

such that $f = \psi_{V_1} - \psi_{V_2}.$

Proof

Indeed, $f = (x_1\ldots x_n)^{-M}g$ for some symmetric polynomial g in $x_1, \ldots x_n.$ Applying lemma 3, we can find $\rho_1, \rho_2$ such that $g = \psi_{V_1} - \psi_{V_2}$; now we take $\rho_1 \otimes \det^{-M}$ and $\rho_2\otimes \det^{-M}.$ ♦

Finally we have:

Corollary 2. Any irreducible K-module V can be lifted to a rational irreducible G-module.

Proof

By corollary 1, $\psi_V = \psi_{W} - \psi_{W'}$ where WW’ are rational G-modules. Then $\psi_W = \psi_V + \psi_{W'} = \psi_{V\oplus W'}$ so $W\cong V\oplus W'$ as K-modules by lemma 2 above. By corollary 3 here, $V\subseteq W$ is a rational irreducible G-module. ♦

## Summary of Results

Thus we obtain:

Note the following.

• When we consider virtual representations (recall that these are formal differences of two representations), this corresponds to all symmetric Laurent polynomials with integer coefficients.
• Any rational G-representation is of the form $\rho \otimes \det^m$ where $\rho$ is a polynomial G-representation.

Moving Ahead

Our next task is to identify the irreducible rational G-modules V. Tensoring by some power of det, we may assume V is polynomial, so $\psi_V(x_1, \ldots, x_n)$ is a symmetric polynomial in $x_1, \ldots, x_n.$ Studying these will lead us back to Schur polynomials.

## Representations of GLn and Un

Note: all representations of topological groups are assumed to be continuous and finite-dimensional.

Here, we will look at representations of the general linear group $GL_n \mathbb{C}.$ We fix the following notations:

• $G$ denotes $GL_n\mathbb{C}$ for some fixed $n$;
• $K$ is the subgroup $U_n \subset GL_n(\mathbb{C})$ of unitary matrices; this group is compact.

Now, a representation

$\rho: G = GL_n\mathbb{C}\to GL_N\mathbb{C}$

is said to be continuousanalyticrationalpolynomial if it is so in each entry of $GL_N\mathbb{C}.$ For example:

• $G \to \mathbb{C}^*, \ g \mapsto \det(g)^{m}$ is rational for each integer n but polynomial only for non-negative values of m.
• $G \to \mathbb{C}^*,\ g \mapsto |\det(g)|$ is continuous but not analytic.

Note that the above definition makes sense for G since it is an open subset of $\mathbb{C}^{n^2}$, the space of all $n\times n$ matrices.

Definition. For a coordinate-free variant, we say that the representation $\rho : G = GL_n\mathbb{C} \to GL(V)$ is continuous / analytic / rational / polynomial if:

• for any $v\in V$ and $\alpha \in V^\vee$, the function $G\to \mathbb{C},$ $g\mapsto \alpha(\rho(g)(v))$ is so.

Our focus is in analytic representations of G. For K, we will only look at continuous representations since it does not have a natural analytic structure. It turns out there is a correspondence between these two classes of representations, so the former can be analyzed using our earlier tools.

## Main Theorem

We begin with a basic result in complex analysis.

Theorem. Let $f(z_1, \ldots, z_n)$ be an analytic function on $\mathbb{C}^n.$ If $f=0$ on $\mathbb{R}^n$, then $f=0.$

Proof

When n=0, $\mathbb{R}^n = \mathbb{C}^n$ has only one point so this is trivial. Suppose n>0 and $f\ne 0.$ Treat f as a function in $z_n$ and we get:

$f(z_1, \ldots, z_n) = z_n^d g(z_1, \ldots, z_{n-1}) + z_n^{d+1} h(z_1, \ldots, z_n)$

where $g\ne 0$ and h are analytic. Suppose $f=0$ on $\mathbb{R}^n.$ Then $f z_n^{-d} = g + z_n h$ is analytic and on $\mathbb{R}^n - (\mathbb{R}^{n-1} \times \{0\})$ it is zero; since this set is dense in $\mathbb{R}^n,$ we have $g + z_n h=0$ on $\mathbb{R}^n.$ Thus $g=0$ on $\mathbb{R}^{n-1} \times \{0\}$. By induction hypothesis, this gives g=0, a contradiction. ♦

From this, we obtain the following result, which is the main case of concern.

Main Lemma. If $f : G\to\mathbb{C}$ is an analytic function and $f|_K = 0$, then $f=0$.

Proof

Let $h : M_{n\times n}(\mathbb{C}) \cong \mathbb{C}^{n^2} \to \mathbb{C}$ be defined by $h(A) := f(\exp(2\pi i A))$, which is analytic. We have:

$A \in M_{n\times n}(\mathbb{R})\implies \exp(2\pi i A) \in U_n= K$

so h=0 when restricted to $M_{n\times n}(\mathbb{R})$. By the above theorem, h=0. Note that any diagonalizable matrix can be written as $\exp(A)$ for some matrix A, so f(B) = 0 if B is diagonalizable. Since the set of diagonalizable matrices in $GL_n\mathbb{C}$ is dense, f = 0. ♦

## Consequences

From the main lemma we get, in turn:

Corollary 1. Let $V, W$ be finite-dimensional analytic representations of G. Then:

$\text{Hom}_K(V, W) = \text{Hom}_G(V, W).$

Proof

A linear map $f:V\to W$ is G-equivariant if and only if for each $g\in G$$\rho_W(g)\circ f = f\circ \rho_V(g)$. This is equivalent to:

$\lambda \in \text{Hom}_\mathbb{C}(V, W)^\vee, g\in G \implies\lambda(\rho_W(g) \circ f - f\circ \rho_V(g))= 0.$

Since $f, \rho_V(g), \rho_W(g), \lambda$ are all analytic, by the main lemma, this condition remains equivalent when we replace G by K, in which case the condition says that $f\in \text{Hom}_K(V, W).$ ♦

Corollary 2. Let $V, W$ be finite-dimensional analytic G-modules. They are isomorphic as G-modules ⇔ they are isomorphic as K-modules.

Proof

Indeed, saying $V\cong W$ (as G– or K-modules) just means there is an invertible $f\in \text{Hom}_G(V, W)$ or $\text{Hom}_K(V, W)$. Apply corollary 1. ♦

Corollary 3. Let V be a finite-dimensional analytic G-module. A subspace $W\subseteq V$ is G-invariant ⇔ it is K-invariant. Hence V is irreducible as a G-module ⇔ it is irreducible as a K-module.

Proof

W is G-invariant if and only if for all $g\in G$, $\rho_V(g)(W) \subseteq W$; this is equivalent to:

• if $w\in W$ and $\lambda \in V^\vee$ satisfies $\lambda(W) =0$, then the function $G\to \mathbb{C}, g \mapsto \lambda(\rho_V(g)(w))$ vanishes.

By corollary 1, this condition is equivalent when we replace G by K, but now the condition says $\rho_V(g)(W)\subseteq W$ for all $g\in K.$ ♦

Summary

Let V be a finite-dimensional analytic representation of G. Since K is compact, V as a K-module can be written as a direct sum of irreps:

$V = W_1^{\oplus m_1} \oplus W_2^{\oplus m_2} \oplus \ldots \oplus W_r^{\oplus m_r},$

where the $W_i$ are pairwise non-isomorphic K-irreps. By corollary 3, each $W_i$ is a G-irrep; the $W_i$ are also pairwise non-isomorphic G-modules by corollary 2.

Question

Given a K-module, can we necessarily lift it to an analytic G-module? We will explore this in the next article.

## Characters

Definition. The character of a continuous G-module V is defined as:

$\chi_V : G\to\mathbb{C}, \quad g \mapsto \text{tr}(g: V\to V).$

This is a continuous map since it is an example of a matrix coefficient.

Clearly $\chi_V(hgh^{-1}) = \chi_V(g)$ for any $g,h\in G$. The following are quite easy to show:

\displaystyle \begin{aligned}\chi_{V\oplus W}(g) &= \chi_V(g) + \chi_W(g); \\ \chi_{V\otimes_{\mathbb C} W}(g) &= \chi_V(g) \chi_W(g); \\ W\subseteq V\implies \chi_V(g) &= \chi_W(g) + \chi_{V/W}(g); \\ \chi_{V^\vee}(g) &= \chi_V(g^{-1}) = \overline{\chi_V(g)}.\end{aligned}

The last equality, that $\chi_V(g^{-1})$ is the complex conjugate of $\chi_V(g)$, follows from the following:

Lemma 1. For $g\in G$, the linear map $g:V\to V$ has all eigenvalues on the unit circle T.

Proof

Suppose the eigenvalues are $\lambda_1, \ldots, \lambda_N$ with the right multiplicity. Then $g^m$ has trace $\chi_V(g^m)= \sum_i \lambda_i^m.$ But $\chi_V : G\to \mathbb{C}$ is continuous; since G is compact the image is bounded. It remains to show: if $\sum_i \lambda_i^m$ is bounded for all integers m we must have all $|\lambda_i| = 1$.

Suppose $|\lambda_1| \ne 1$; replacing g by its inverse, we assume $|\lambda_1| > 1$ and $|\lambda_1| \ge |\lambda_2| \ge \ldots.$ If $|\lambda_1| > |\lambda_2|$, $|\lambda_1|^m$ increases much faster than the other $|\lambda_i|^m$ and we get a contradiction. On the other hand, if $|\lambda_1| = \ldots = |\lambda_k|$, write $\lambda_j = Re^{i\theta_j}$ for $1\le j \le k$ where R>1 and $\theta_j$ are real. Consider the continuous homomorphism

$\phi:\mathbb{Z} \mapsto (\mathbb{C}^*)^k, \ m \mapsto (e^{im\theta_1}, \ldots, e^{im\theta_k}).$

Its image lies in the torus group $T_k$ which is compact. Hence, there are arbitrarily large m such that $\text{Re}(e^{im\theta_1}), \ldots, \text{Re}(e^{im\theta_k}) > \frac 1 2$ (see exercise). So the real part of $\lambda_1^m + \ldots + \lambda_k^m$ exceeds $\frac 1 2 R^m$ and increases much faster than the remaining $|\lambda_i|^m.$ ♦

Exercise

Prove that if $f:\mathbb{Z} \to G$ is a continuous homomorphism to a compact Hausdorff group, then $f^{-1}(U)$ is infinite for any open neighbourhood U of e.

Lemma 2. For $g\in G$, the linear map $g:V\to V$ is diagonalizable.

Proof

Write $g:V\to V$ in block Jordan form. Suppose the matrix is not diagonal, say entry (1, 2) is 1 with corresponding eigenvalue $\lambda$. The matrix for $g^n : V\to V$ has entry (1, 2) equal to $n \lambda^{n-1}$ which is unbounded since $|\lambda| = 1.$ However, G is compact so taking entry (1, 2) of the representation $G\to GL(V)$ has bounded image, which is a contradiction. ♦

## Orthogonality of Characters

The space of G-invariant elements of a G-module V is denoted by:

$V^G := \{v\in V: gv = v \, \forall\, g\in G\}.$

We have:

Proposition. The dimension of $V^G$ is:

$\displaystyle \int_{x\in G} \chi_V(x) dx.$

Proof

Take $\pi: V\to V$ via $v\mapsto \int_{x\in G} xv\cdot dx.$

Left-invariance then implies that the image of π lies in $V^G.$ Also, if $v\in V^G$ then $\pi(v) = v.$ Putting these two facts together gives us $\pi^2 = \pi$ so  π is the projection map onto $V^G.$ Hence:

$\displaystyle \dim V^G = \text{tr}(\pi) = \int_{x\in G} \text{tr}(x:V\to V)dx = \int_{x\in G} \chi_V(x)\cdot dx.$

as desired. ♦

In particular, replacing V by $\text{Hom}_{\mathbb C}(V, W)$, note that:

$\text{Hom}_{\mathbb C}(V, W)^G = \text{Hom}_{\mathbb C[G]}(V, W).$

So the dimension of this space is the integral of:

$\chi_{\text{Hom}(V, W)}(g) = \chi_{V^\vee \otimes W}(g) = \overline{\chi_V(g)} \chi_W(g).$

In other words:

$\dim \text{Hom}_{\mathbb{C}[G]}(V, W) = \int_{x\in G} \overline{\chi_V(x)} \chi_W(x)\cdot dx$

which we denote by $\left.$ Schur’s lemma then gives us:

Orthonormality of Irreducible Characters.

If V, W are irreps of G, then

$\left< V, W\right> = \int_{x\in G}\overline{\chi_V(x)} \chi_W(x)\cdot dx= \begin{cases} 1, \ &\text{if } V\cong W, \\ 0 \ &\text{else.}\end{cases}$

## Abelian Case

Suppose now G is compact Hausdorff and abelian.

Theorem. Each irreducible representation V of G is 1-dimensional.

Proof

First we show that every $g\in G$ acts as a scalar. By lemma 2 above, V decomposes as a direct sum of eigenspaces $W_\lambda$ for $g:V\to V.$ Each $W_\lambda$ is G-invariant since:

$g' \in G, v\in W_\lambda \implies g(g'v) = g'(gv) = g'(\lambda v) = \lambda (g'v) \implies g'v \in W_\lambda.$

Since V is irreducible, we have $W_\lambda = V$ so g acts as a scalar.

Since every element of G acts as a scalar, every vector subspace is G-invariant; so dim(V) = 1. ♦

Hence by lemma 1, every irrep of G is a continuous homomorphism $G\to T$, where $T\subset \mathbb{C}^*$ is the unit circle. The following special case is of interest.

Definition. The n-torus is the topological group $T^n \cong (\mathbb{R}/\mathbb{Z})^n.$

[In fact, any compact Hausdorff abelian G must be of the form (finite group) $\times T^n$ but we won’t be needing this fact.]

For i = 1,…,n, let $\chi_i : T^n \to T$ be the homomorphism taken by projecting to the i-th coordinate. More generally, for any $\mathbf m := (m_1, \ldots, m_n) \in \mathbb{Z}^n$, we define:

$\chi^{\mathbf m} := \chi_1^{m_1} \cdot \ldots \cdot \chi_n^{m_n} : T^n \to T.$

Thus it takes $\mathbf v =(v_1, \ldots, v_n)$ to $v_1^{m_1} v_2^{m_2} \ldots v_n^{m_n}.$

Theorem. Any irrep of $T^n$ is of the form $\chi^{\mathbf m}$ for some $\mathbf m \in \mathbb{Z}^n.$

Proof

Suppose n=1: for a continuous $\chi:T\to T$ write T as $\mathbb{R} / \mathbb{Z}$ and compose $\mathbb{R} \to \mathbb{R}/\mathbb{Z} \stackrel\chi\to \mathbb{R}/\mathbb{Z}.$ The subgroup $\mathbb{Z}$ lies in the kernel; since the kernel is a closed subgroup it is either $\mathbb{R}$ or $\frac 1 m \mathbb{Z}$ for some m>0. So $\chi$ is of the form $v\mapsto v^m$ for some $m\in\mathbb{Z}.$

For n>1 consider the homomorphism $\phi_i : T \to T^n$ for = 1,…,n, by taking $v\mapsto (1,\ldots, 1,v,1\ldots,1),$ where the i-th component is v. By the previous case, $\chi\circ\phi_i : T\to T$ is of the form $v\mapsto v^{m_i}$ for some $m_i\in\mathbb{Z}.$ Since $\mathbf v = \prod_i \phi_i\chi_i(\mathbf v)$ for all $\mathbf v\in T^n,$ we get:

$\chi(\mathbf v) = \prod_i \chi \phi_i\chi_i(\mathbf v) = \prod_i \chi_i(\mathbf v)^{m_i} = \chi^{\mathbf m}(\mathbf v).$ ♦

Corollary. A character of $T^n$ is a finite linear combination of $\chi^{\mathbf m}$ whose coefficients are positive integers.

Replacing $\chi_i$ with variables $x_i$, this gives us a polynomial in $x_1, \ldots, x_n, x_1^{-1},\ldots, x_n^{-1}$, also known as a Laurent polynomial in $x_1, \ldots, x_n.$ Hence we have a correspondence:

## Polynomials and Representations XXVIII

Starting from this article, we will look at representations of $G := GL_n\mathbb{C}$. Now, $GL_n\mathbb{C}$ itself is extremely complicated so we will only focus on representations of particular types. Generally, for any topological group G, we want:

$\rho : G\to GL_N \mathbb{C}$

to be a continuous homomorphism of groups.

## Continuous Representations of Topological Groups

Let G be a topological group.

Our main objects of study are continuous representations $\rho: G \to GL_N\mathbb{C}.$ These correspond to $\mathbb{C}[G]$-modules (also called G-modules) V of finite complex dimension, with an additional condition for continuity:

• for any $v\in V$ and $\alpha \in V^\vee$ (dual), the map $G\to \mathbb{C}$ given by $g\mapsto \alpha(g(v))$ is continuous.

[Note the above condition condition is independent of coordinates and does not require a choice of basis. When we replace G with $GL_n\mathbb{C}$ later, we can also replace the word continuous with “analytic” to obtain a narrower class of representations, but let’s not get ahead of ourselves.]

A function $G\to\mathbb{C}$ of the form $g\mapsto \alpha(g(v))$ as above is called a matrix coefficient. E.g. if we fix a basis for V which gives us $G \to GL_N\mathbb{C},$ picking entry (ij) in the matrix gives us a matrix coefficient $G\to\mathbb{C}.$

Henceforth, all representations of topological groups are assumed to be continuous and finite-dimensional.

### Constructing New Representations

These constructions are identical to the case for representations of finite groups.

• Given any complex vector space V, the trivial action takes gvv for any $g\in G, v\in V$; it is clearly continuous.
• Given a G-module V, a vector space $W\subseteq V$ is said to be Ginvariant if $g\in G, w\in W \implies g(w) \in W.$ Clearly, the resulting $G\to GL(W)$ is continuous.
• We say that V is irreducible if $V\ne 0$ and it has no non-trivial G-invariant submodules. An irreducible representation V is often called irrep for short.
• If $W\subseteq V$ is G-invariant, then V/W is a G-module; clearly it is continuous.
• If V is a G-module, the dual $V^\vee$ is also continuous, where G acts via

$G\times V^\vee \to V^\vee, \qquad (g, \alpha) \mapsto \alpha\circ g^{-1} \in V^\vee.$

• Given G-modules V, W, the action of G on $V\oplus W$ and $V \otimes_{\mathbb C} W$ are continuous as well.
• Finally we have $\text{Hom}_{\mathbb C}(V, W) \cong V^\vee \otimes_{\mathbb C} W$ naturally. Using the above constructions, we get a continuous G-representation for this space via:

$g\in G, f:V\to W \mapsto \rho_W(g)\circ f \circ \rho_V(g^{-1}) : V\to W.$

How do we know that the above constructions are all continuous?

Here’s one example. To show that $V\otimes_{\mathbb{C}} W$ is a continuous representation, pick a basis $e_i$ (resp. $f_j$) of V (resp. W) so that $e_i \otimes f_j$ form a basis of $V\otimes W.$ The matrix for $g: V\otimes W \to V\otimes W$ is obtained from the matrices of $g:V\to V$ and $g:W\to W$ by multiplying entries; thus the map $G\to \mathbb{C}$ taking g to a fixed entry of the matrix for $g:V\otimes W\to V\otimes W$ is continuous.

Finally, a linear map of G-modules $f:V\to W$ is said to be Gequivariant if

$g\in G, v\in V \implies f(gv) = g\cdot f(v).$

Note that this is equivalent to saying G acts trivially on $f\in \text{Hom}_{\mathbb C}(V, W).$

## Compact Groups

From now till the end of the next article, G is assumed to be compact and Hausdorff; this case will be instrumental to the general theory even though $GL_n\mathbb{C}$ is not compact. Here are some common examples of compact topological groups in representation theory:

\begin{aligned} U_n &= \{ M \in GL_n \mathbb{C} : M \overline{M}^t = I\},\\ SU_n &= \{ M \in GL_n\mathbb{C} : M\overline{M}^t = I, \det(M) = 1\},\\ O_n &=\{ M\in GL_n\mathbb{R} : M M^t = I\}, \\ SO_n &= \{M \in GL_n\mathbb{R} : MM^t = I, \det(M) = 1\}, \\ T &= \{ z \in \mathbb{C} : |z| = 1\}.\end{aligned}

These groups are compact because they are closed and bounded subsets of $\mathbb{C}^{n\times n}.$ E.g. the condition for $M \in U_n$ can be written out as a sequence of quadratic equations in $\text{Re}(z_{ij})$ and $\text{Im}(z_{ij})$ for $1\le i,j\le n.$ Thus the set $U_n$ is closed; it is bounded since the condition gives:

$|z_{i1}|^2 + |z_{i2}|^2 + \ldots + |z_{in}|^2 = 1, \quad\text{for } 1\le i\le n$

so each $|z_{ij}| \le 1.$

## Haar Measure

To proceed, we will borrow a result from harmonic analysis:

Theorem. If G is a compact Hausdorff group, then there is a measure m on G (called the Haar measure) satisfying:

• m(G) = 1;
• m is “left-invariant”, i.e. for any continuous $f:G\to \mathbb{C}$, we have:

$\displaystyle y\in G\implies \int_{x\in G} f(yx)\cdot dm(x) = \int_{x\in G} f(x)\cdot dm(x),$

The Haar measure is unique and in fact right-invariant, i.e. $\int_{x\in G} f(xy)\cdot dm(x) = \int_{x\in G} f(x) \cdot dm(x).$ We will implicitly assume the Haar measure is used when integrating over G.

Note

The Haar measure is usually defined for locally compact Hausdorff groups in general, in which case we have to modify the statements a little. In particular, left-invariant measures are not necessarily right-invariant.

## Semisimplicity of Continuous Representations

The following result is crucial.

Theorem. Let V be a G-module. If $W\subseteq V$ is a G-invariant subspace, then we can find a G-invariant subspace $W' \subseteq V$ such that $V = W\oplus W'.$

Proof

Let $\pi : V\to V$ be any projection map onto W, so $\pi^2 = \pi$ and $\text{im}(\pi) = W$. Now define:

$\pi' : V\to V, \quad \pi'(v) := \int_{x\in G} x\pi(x^{-1}v)\cdot dx.$

[Explicitly, one may take an isomorphism $V\cong \mathbb{C}^N$ so each $v\in V$ is a tuple of complex numbers; the above integral can be carried out component-wise, as N integrals of continuous functions.]

We then have:

• The image of $\pi'$ is in W, since $\pi(V)=W$ and $x(W)\subseteq W$ for $x\in G.$
• If $v\in W$ then $x^{-1}v \in W\implies x\pi(x^{-1}v) = xx^{-1}v = v$ and $\pi'(v) = v.$
• $\pi'^2 = \pi'$: indeed if $v\in V$ then $\pi'(v) \in W$ by the first property and so $\pi'(\pi'(v)) = \pi'(v)$ by the second.
• The map $\pi'$ is G-equivariant: for $y\in G$ we have $\pi'(yv) = y\pi'(v).$ Indeed by left-invariance we have: $\pi'(v) = \int_{x\in G} yx \pi(x^{-1}y^{-1}v) dx.$ Replacing v with yv gives us the desired outcome.

Hence, $\pi'$ is projection onto W and its kernel W’ satisfies $V = W \oplus W'.$ Since $\pi'$ is G-equivariant, W’ is a G-invariant subspace. ♦

Thus, every continuous representation of G can be written as a direct sum of irreps. Next, we have:

Schur’s Lemma. If $V, W$ are irreps of G, then:

$\text{Hom}_{\mathbb{C}[G]}(V, W) \cong \begin{cases} \mathbb{C}, \qquad &\text{if }V \cong W,\\ 0, \qquad &\text{otherwise.} \end{cases}$

The proof is identical to that of the usual Schur’s lemma.

Proof

Suppose $V\not\cong W$ and $f:V\to W$ is G-equivariant. Then $\text{ker}(f) \subseteq V$ and $\text{im}(f)\subseteq W$ are subrepresentations, so $\text{ker}(f) = 0, V$ and $\text{im}(f) = 0, W$. By considering various cases we have either

1. $\text{ker}(f) = 0, \text{im}(f) = W$, in which case f is an isomorphism or
2. $\text{ker}(f) = V, \text{im}(f) = 0$, in which case f = 0.

It remains to show that $\text{Hom}_{\mathbb{C}[G]}(V,W) \cong \mathbb{C}$ when $V\cong W$.

It suffices to show that any isomorphism $f:V\to V$ is a scalar: let $v\in V$ be an eigenvector of with eigenvalue $\lambda\in\mathbb{C}$ so $f(v) = \lambda v.$ Then $f - \lambda\cdot I : V\to V$ is G-equivariant and its kernel is non-zero; hence $f - \lambda\cdot I = 0 \implies f = \lambda\cdot I.$ ♦

## Polynomials and Representations XXVII

From the previous article, we have columns jj’  in the column tabloid U, and given a set A (resp. B) of boxes in column j (resp. j’), we get:

$U-\beta_B(U) = U- \sum_{A,\, |A| = |B|} \alpha_{A,B}(U) \in \text{ker}(\pi)$

where $\alpha_{A,B}(U)$ is the column tabloid obtained by swapping contents of A with B while preserving the order. For example, we have the following modulo ker(π):

Here is our main result.

Main Theorem. Each of the following spans $\ker\pi$ as a vector space:

• the collection of $U - \beta_B(U)$ over all sets B in column j’ and (j, j’) such that j < j’;
• as above, except B comprises of the top k squares (for all k) of column j’ and j’ = j+1.

We have shown that the first set spans a subspace of ker(π); since the second set is a subset of the first, it remains to show that Q = ker(π), where Q is the vector space spanned by the second set.

Lemma. Consider $[T]$ as $T$ varies through all SYT with shape $\lambda$. These span the vector space $\mathbb{C}[G]b_{T_0} / Q.$

Proof

Create a new total ordering on the set of all fillings T of shape λ. Given T’T, we consider the rightmost column in which T’ and T differ; in that column, take the lowest square in which T’ and T differ; we write T’T if the entry in that square of T’ is larger than that of T:

Suppose T is a filling which is not an SYT. We claim that modulo Q, we can express [T] as a linear sum of [T’] for T’T.

• Let T’ be obtained from T by sorting each column in ascending order; then [T’] = ±[T] and T’ ≥ T, so we assume each column of T is increasing.
• If T is not an SYT, then we can find row i and column j of T such that $T_{i, j} > T_{i,j+1}.$ Taking j’j+1 and the top i squares B of column j’ of T, we can replace $[T]$ (modulo Q) by $\beta_B([T]) = \sum_{|A|=|B|} \alpha_{A,B}([T])$ where each term is [T’] for some T’>T.

Applying this iteratively, since there are finitely many fillings we eventually obtain any [T] as a linear combination of [T’] where T’ is SYT. ♦

Proof of Main Theorem

We know that $Q\subseteq \text{ker}(\pi)$ so we get a surjection

$\overline\pi : \mathbb{C}[G]b_{T_0}/Q \to \mathbb{C}[G]b_{T_0}a_{T_0} \cong V_\lambda.$

Hence the dimension of $\mathbb{C}[G]b_{T_0}/Q$ is at least $\dim V_\lambda = f^\lambda.$ On the other hand, by our lemma above, the space is spanned by [T] for all SYT T so its dimension is at most the number of SYT of shape λ, i.e. $f^\lambda.$ Thus $\overline\pi$ is a bijection so $Q = \text{ker}(\pi)$ as desired. ♦

As a side consequence, we have also proved that the set of [T] for all SYT T gives a basis for $\mathbb{C}[G]b_{T_0}a_{T_0}.$

This concludes our discussion of representations of the symmetric group $S_d$. Our next target is the representation theory of the general linear group $GL_n \mathbb{C}.$

## Polynomials and Representations XXVI

Let us fix a filling $T_0$ of shape $\lambda$ and consider the surjective homomorphism of $\mathbb{C}[G]$-modules

$\pi : \mathbb{C}[G]b_{T_0} \to \mathbb{C}[G]b_{T_0} a_{T_0} \subseteq \mathbb{C}[G]a_{T_0},$

given by right-multiplying by $a_{T_0}.$ Specifically, we will describe its kernel, which will have interesting consequences when we examine representations of $GL_n\mathbb{C}$ later.

## Row and Column Tabloids

By the lemma here, $\mathbb{C}[G]a_{T_0}$ and $\mathbb{C}[G]b_{T_0}$ can be described as follows.

For the former, take a basis comprising of “row tabloids“, i.e. for each filling T, take each row as a set, thus giving a partitioning $[d] = A_1 \coprod A_2 \coprod \ldots.$ E.g. in the example below, the first two row tabloids are identical since they give rise to the same partitioning.

For $\mathbb{C}[G]b_{T_0}$, take a basis of “column tabloids“, by taking a filling T and taking its columns as sets, with the condition that swapping two elements in the same column flips the sign. E.g.

Note

• Given a filling T, the corresponding row tabloid is denoted by {T} and the row tabloid is [T].
• If $g\in S_d$ we have $g(\{T\}) = \{g(T)\}$ and $g([T]) = [g(T)]$. E.g. (1,2) takes the above column tabloid to its negative while (1,2)(3,4) leaves it invariant.
• Let us describe explicitly the correspondence between a column tabloid and an element of $\mathbb{C}[G]b_{T_0}.$ For any column tabloid U = [T], where T is a filling, write $T = g(T_0)$ for a unique $g\in S_d.$ The corresponding element is $gb_{T_0} \in \mathbb{C}[G]b_{T_0}.$
• The reader may worry that we left out the twist $\chi(g),$ but you may check that this is consistent when we swap two elements in the same column.
• For a row tabloid, we take $\{g(T_0)\}$ to the element $ga_{T_0} \in \mathbb{C}[G]a_{T_0}$.

## Column Swaps

Let us define some operations on column tabloids. Consider a column tabloid U := [T].

Suppose, in U, we have a (possibly empty) set B of boxes in column j’ and a set A of squares in column j such that |A| = |B|, with j’j. Let $\alpha_{A,B}(U)$ denote the resulting column tabloid obtained by swapping squares A and B in order.

Now given a set B of squares in column j’ of U, define a linear map

$\displaystyle \beta_B : \mathbb{C}[G]b_{T_0} \to \mathbb{C}[G]b_{T_0},\qquad \beta_B(U) := \sum_A \alpha_{A,B}(U),$

where A runs through all sets of squares in column j of U such that |A| = |B|. Next, given a set Y of squares in column j’ of U, let:

$\displaystyle\gamma_Y(U) := \sum_{B\subseteq Y} (-1)^{|B|}\beta_B(U) = \sum_{\substack{A,B\\ B\subseteq Y,\, |A| = |B|}}(-1)^{|B|}\alpha_{A,B}(U).$

Clearly $\gamma_Y(U)$ is an integer linear combination of the $\beta_B(U)$; the following shows that the converse is true.

Lemma 1. For each set $Y'$ of squares in column $j'$, we have:

$\displaystyle\sum_{Y \subseteq Y'} (-1)^{|Y|}\gamma_{Y}(U) = \beta_{Y'}(U).$

Proof

Fix A, B satisfying $B\subseteq Y'$ and |A| = |B|; we keep track of the coefficient of $\alpha_{A,B}(U)$ in the expansion of LHS. The term is included in $\gamma_Y(U)$ for all Y satisfying $B\subseteq Y \subseteq Y'.$ Its coefficient is thus:

$\displaystyle(-1)^{|B|}\cdot \sum_{Y,\, B\subseteq Y\subseteq Y'} (-1)^{|Y|}.$

If B = Y’, this is 1; otherwise, fixing some $y\in Y' - B$, there is a 1-1 correspondence between Y containing y and Y not containing y; these cancel each other out so the sum is 0. Hence the overall sum is:

$\displaystyle \sum_{A,\, |A| = |Y'|} \alpha_{A, Y'}(U) = \beta_{Y'}(U)$

as desired. ♦

Lemma 2. We have $\gamma_Y(U) \in \text{ker}\pi$ if Y is a non-empty set of squares in the j’-th column of U.

Proof

Let $K\le H\le S_d$ be subgroups defined as follows:

• H is the set of $g \in S_d$ acting as the identity outside Y or the j-th column of U;
• K is the set of $g\in H$ taking $Y\to Y$ and the j-th column of U back to itself.

For each $g \in K$ we have $\chi(g) g(U) = U$ so $\sum_{g\in K} \chi(g) g(U) = |K|U.$

Let $h\in S_d$ run through all swaps of a subset A of the j-th column with $B\subseteq Y$ preserving the order; this gives a complete set of coset representatives for $H/K.$ On the other hand, the sum of these $h(U)$ gives us $\gamma_Y(U)$ so:

$\displaystyle \sum_{g'\in H} \chi(g')g'(U) = \text{positive multiple of } \gamma_Y(U).$

It remains to show that the LHS is in ker(π).

Fix filling T such that U = [T]. Unwinding the definition for map π, we first take $g\in S_d$ such that $T = g(T_0)$ then take the row tabloid for $gb_{T_0}$. Now, $gb_{T_0} = gb_{T_0} g^{-1} g = b_T g$ so

$\pi:[T] \ \mapsto\ b_{T}\{T\} = \sum_{g''\in C(T)} \chi(g'') g''\{T\}.$

Since $\pi$ is $\mathbb{C}[G]$-linear, we now need to show:

$\displaystyle \sum_{g'\in H} \chi(g') g' \sum_{g''\in C(T)} \chi(g'')g''\{T\} \overset{?}=0.\quad (*)$

Since $Y\ne\emptyset,$ pick a square in it and a square to its left in the j-th column (since jj’). Fix the element $g''\in C(T)$ and let $h\in H$ be the transposition that swaps those two squares in $g''T$ so $h(g''\{T\}) = \{g''T\}$. Taking $g', g'h$ only in the outer sum of (*) gives:

$(\chi(g')g' + \chi(g'h)g'h) \chi(g'') \{g''T\} = \chi(g')\chi(g'')g'(e - h) \{g''T\} = 0.$

Summing over all cosets of $\{e,h\} \le H$ and $g'' \in C(T)$ we see that (*) sums to zero. ♦

From lemmas 1 and 2, we have:

Corollary. We have:

$\beta_{Y'}(U) - \gamma_\emptyset(U) = \beta_{Y'}(U) - U\in \text{ker}(\pi).$

E.g. modulo ker(π), we have:

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## Properties of the Young Symmetrizer

Recall that for a filling $T$, we have $R(T), C(T) \le S_d,$ the subgroup of elements which take an element of the i-th row (resp. column) of T to the i-th row (resp. column) of T. Then:

$a_T = \sum_{g\in R(T)} g,\quad b_T = \sum_{g\in C(T)} \chi(g)g,\quad c_T = a_T b_T,$

where $c_T$ is the Young symmetrizer. Recall the following results from earlier:

$g\in R(T) \implies a_T g = g a_T = a_T, \\ g\in C(T) \implies b_T g = g b_T = \chi(g)b_T.$

The following is obvious.

Lemma 1. If $g\in S_d$, then $a_{g(T)} = g a_T g^{-1}$ and $b_{g(T)} = g b_T g^{-1}.$ Thus $c_{g(T)} = g c_T g^{-1}.$

Proof

This follows from $R(g(T)) = g R(T) g^{-1}$ and $C(g(T)) = g C(T)g^{-1}.$ E.g. the latter gives:

$b_{g(T)} = \sum_{x \in C(g(T))} \chi(x)x = \sum_{x\in C(T)} \chi(gxg^{-1}) gxg^{-1} = g b_T g^{-1}.$

The following generalizes the Young symmetrizer.

Proposition 1. Consider the element $c = a_T b_{T'}$ for fillings $T, T'$ of shape $\lambda.$ Note that since $\mathbb{C}[G]c$ is contained in both $\mathbb{C}[G]a_T$ and $\mathbb{C}[G]b_{T'}$ it is either 0 or isomorphic to $V_\lambda.$

The following are equivalent.

• $c \ne 0$;
• no two distinct $i,j$ lie in the same row of $T$ and same column of $T'$;
• there exist $g \in R(T)$ and $g' \in C(T')$ such that $g(T) = g'(T').$

Note

The third condition says: we can change T to T’ by permuting elements in each row, then elements in each column.

Proof

Suppose $c\ne 0$ but two distinct $i, j$ lie in the same row of T and same column of T’. Let $g\in S_d$ swap those two elements; then $g\in R(T) \cap C(T').$ Letting C run through a set of coset representatives of $R(T)/\left,$ we have $a_T = (\sum_{x\in C} x)(1+g).$ But now we have $gb_{T'} = \chi(g)b_{T'} = -b_{T'}$ so $(1+g)b_{T'} = 0$ and we have c=0.

Suppose the second condition holds. Elements of the first row of T are in different columns of T’. Bringing those in T’ to the first row, there exist $g'_1\in C(T')$ and $g_1\in R(T)$ such that $g_1(T)$ and $g_1'(T')$ have identical first rows.

Likewise, since the second row of $g_1(T)$ are in different columns of $g'_1(T')$ there exist $g'_2 \in C(T')$ and $g_2 \in R(T)$ such that $g_2 g_1(T)$ and $g'_2 g'_1(T')$ have the same first and second rows. Repeating, we eventually get the desired $g\in R(T)$ and $g'\in C(T').$

Finally, suppose the third condition holds; let $T'' = g(T) = g'(T')$. By lemma 1,

$a_{T''} = g a_T g^{-1} = a_T, \quad b_{T''} = g' b_T g'^{-1} = b_{T'},$

So $c=a_T b_{T'} = a_{T''}b_{T''} = c_{T''}\ne 0.$ ♦

Lemma 2. If $T, T'$ are distinct SYT of the same shape, they do not satisfy the conditions of proposition 1.

Proof

Order all the fillings of shape $\lambda$ as follows: given TT’, let k be the largest number occurring in different squares of T and T’; we write T’T if k occurs earlier in the word w(T’) than in w(T). For example, we have:

Note that for any SYT T we have:

$g \in R(T), g' \in C(T) \implies g(T) \ge T, g'(T) \le T$

since the largest entry of T which is moved by g must move to its left in w(T), and the largest entry moved by g’ moves to its right in w(T). Thus if TT’ are SYT and g(T) = g’(T’) for some $g\in R(T), g'\in C(T')$, we have $T \ge g(T) = g'(T') \ge T'$ so TT’. ♦

Lemma 3. For any $v\in \mathbb{C}[G]$, $v$ is a multiple of $c_T$ if and only if:

• for any $g\in R(T)$ and $h \in C(T)$, we have $gvh = \chi(h)v.$

Proof

⇒ is left as an easy exercise, by proving $gv = v\chi(h)h = v.$

For ⇐, write $v= \sum_{x\in G} \alpha_x x$ where $\alpha_x \in \mathbb{C}.$ Then $\alpha_{gxh} = \chi(h)\alpha_x$ for any $x\in G$, $g\in R(T)$ and $h\in C(T).$ Taking x=e, it suffices to show: if $x\in G$ is not of the form $gh$ for $g\in R(T), h\in C(T)$, then $\alpha_x = 0.$

For that, consider the filling $T' := x(T)$. We claim that $T, T'$ satisfy the conditions of proposition 1.

• Indeed, if $g(T) = g'(T')$ for some $g\in R(T)$ and $g'\in C(T'),$ then

$x=g'^{-1}g, \quad g' \in C(T') = C(xT) = x C(T)x^{-1}.$

• So $x^{-1} g' x = g^{-1} g' g \in C(T)$ and we have $x = g(g^{-1}g'g)^{-1} \in R(T)C(T)$ which is a contradiction.

Hence there exist distinct ij in the same row of T and same column of T’. Let t be the transposition (ij); then $t \in R(T)$ so $tv = v$ by the given condition. Since $t \in C(T') = x C(T) x^{-1}$ so $t' := x^{-1}tx$ satisfies $vt'=\chi(t') v = -v.$ Now consider the coefficient of tx in v. We have:

\begin{aligned} tv = v&\implies \alpha_x = \alpha_{tx},\\ vt' = -v &\implies \alpha_{x} = -\alpha_{tx}.\end{aligned}

Thus $\alpha_x = 0$ as desired. ♦

Here is one immediate application of lemma 2.

Proposition 2. For any $w \in \mathbb{C}[G]$, $c_T w c_T$ is a scalar multiple of $c_T.$

In the case $w=e$ we have $c_T^2 = nT$ where $n = \frac {d!}{\dim V_\lambda}.$

Proof

Indeed if $g\in R(T)$ and $h\in C(T)$ we have:

$g(c_T w c_T) h = (gc_T) w(c_T h) = (c_T)w(\chi(h) c_T) = \chi(h) c_T w c_T$

so by lemma 2, $c_T w c_T$ is a scalar multiple of $c_T.$

For $c_T^2$, consider right-multiplication on $\mathbb{C}[G]$ by $c_T$ and let A be its trace. This map takes $g\mapsto gc_T$ where the coefficient of g is 1; taking the basis comprising of elements of G, we have $A = d!.$ On the other hand, the image of this map is $\mathbb{C}[G]c_T \cong V_\lambda$ and takes each $vc_T \mapsto vc_T^2 = nvc_T.$ Thus the map is a scalar map $n$ on $V_\lambda$ and we have $n\cdot \dim V_\lambda = d!.$ ♦

This gives the following.

Theorem. We have, as a direct sum of irreps:

$\displaystyle \mathbb{C}[S_d] = \bigoplus_{\lambda \vdash d} \bigoplus_{\substack{T = SYT \\ \text{of shape }\lambda}} \mathbb{C}[S_d] c_T.$

Proof

Let us show that $\sum_T \mathbb{C}[S_d] c_T$ is a direct sum, where T runs over all SYT of shape $\lambda.$ Indeed, if not we have:

$\overbrace{v_1 c_{T_1}}^{\ne 0} + v_2 c_{T_2} + \ldots + v_m c_{T_m} = 0,\quad (*)$

for some distinct SYT $T_1, \ldots, T_m$ and some $v_1, \ldots, v_m \in \mathbb{C}[G].$ Note that if $T\ne T'$ then by lemma 2, T and T’ do not satisfy the conditions in proposition 1, so we can find distinct ij in the same row of T and same column of T’; using the same technique as the first paragraph of proof of proposition 1, we have $b_{T} a_{T'} =0$ and thus $c_T c_{T'} = 0.$

Right-multiplying (*) by $c_{T_1}$ thus gives $0 =v_1 c_{T_1}^2 = n v_1 c_{T_1}$ which gives $v_1 c_{T_1} = 0$, a contradiction. Hence, the inner sum is a direct sum. The entire sum is a direct sum since there is no common irrep between distinct $\lambda.$

Hence, the RHS is a subspace of the LHS. Now we count the dimension: on LHS we get d!; on RHS we get $\sum_\lambda f_\lambda\dim V_\lambda = \sum_\lambda f_\lambda^2 = d!$ so the two sides are equal. ♦

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