## Determinant Modules

We will describe another construction for the Schur module.

Introduce variables $z_{i,j}$ for $i\ge 1, j\ge 1$. For each sequence $i_1, \ldots, i_p\ge 1$ we define the following polynomials in $z_{i,j}$:

$D_{i_1, \ldots, i_p} := \det{\small \begin{pmatrix} z_{1, i_1} & z_{1, i_2} & \ldots & z_{1, i_p} \\ z_{2, i_1} & z_{2, i_2} & \ldots & z_{2, i_p} \\ \vdots & \vdots & \ddots & \vdots \\ z_{p, i_1} & z_{p, i_2} & \ldots & z_{p, i_p}\end{pmatrix}}.$

Now given a filling T of shape λ, we define:

$D_T := D_{\text{col}(T, 1)} D_{\text{col}(T, 2)} \ldots$

where $\text{col}(T, i)$ is the sequence of entries from the i-th column of T. E.g.

Let $\mathbb{C}[z_{i,j}]$ be the ring of polynomials in $z_{ij}$ with complex coefficients. Since we usually take entries of T from [n], we only need to consider the subring $\mathbb{C}[z_{i,1}, \ldots, z_{i,n}]$.

Let $\mu = \overline \lambda.$ Recall from earlier that any non-zero $GL_n\mathbb{C}$-equivariant map

$\displaystyle \bigotimes_j \left(\text{Alt}^{\mu_j} \mathbb{C}^n\right) \longrightarrow \bigotimes_i \left(\text{Sym}^{\lambda_i} \mathbb{C}^n \right)$

must induce an isomorphism between the unique copies of $V(\lambda)$ in the source and target spaces. Given any filling T of shape $\lambda$, we let $e^\circ_T$ be the element of $\otimes_j \text{Alt}^{\mu_j} \mathbb{C}^n$ obtained by replacing each entry k in T by $e_k$, then taking the wedge of elements in each column, followed by the tensor product across columns:

Note that the image of $e^\circ_T$ in $F(V)$ is precisely $e_T$ as defined in the last article.

Definition. We take the map

$\displaystyle \bigotimes_j \left(\text{Alt}^{\mu_j} \mathbb{C}^n\right) \longrightarrow \bigotimes_i \left(\text{Sym}^{\lambda_i} \mathbb{C}^n \right), \quad e_T^\circ \mapsto D_T$

where $z_{i,j}$ belongs to component $\text{Sym}^{\lambda_i}.$

E.g. in our example above, $D_T$ is homogeneous in $z_{1,j}$ of degree 5, $z_{2,j}$ of degree 4 and $z_{3,j}$ of degree 3. We let $g \in GL_n\mathbb{C}$ act on $\mathbb{C}[z_{i,j}]$ via:

$g = (g_{i,j}) : z_{i,j} \mapsto \sum_k z_{i,k}g_{k,j}.$

Thus if we fix i and consider the variables $\mathbf z_i := \{z_{i,j}\}_j$ as a row vector, then $g: \mathbf z_i \mapsto \mathbf z_i g^t$. From another point of view, if we take $z_{i,1}, z_{i,2},\ldots$ as a basis, then the action is represented by matrix g since it takes the standard basis to the column vectors of g.

Proposition. The map is $GL_n\mathbb{C}$-equivariant.

Proof

The element $g = (g_{i,j})$ takes $e_i \mapsto \sum_j g_{j,i} e_j$ by taking the column vectors of g; so

$\displaystyle e_T \mapsto \sum_{j_1, \ldots, j_d} g_{j_1, i_1} g_{j_2, i_2} \ldots g_{j_d, i_d} e_{T'}$

where T’ is the filling obtained from T by replacing its entries $i_1, \ldots, i_d$ with $j_1, \ldots, j_d$ correspondingly.

On the other hand, the determinant $D_{i_1, \ldots, i_p}$ gets mapped to:

$\det{\small \begin{pmatrix} z_{1, i_1} & z_{1, i_2} & \ldots & z_{1, i_p} \\ z_{2, i_1} & z_{2, i_2} & \ldots & z_{2, i_p} \\ \vdots & \vdots & \ddots & \vdots \\ z_{p, i_1} & z_{p, i_2} & \ldots & z_{p, i_p}\end{pmatrix}} \mapsto \det{\small \begin{pmatrix} \sum_{j_1} z_{1,j_1} g_{j_1, i_1} & \ldots & \sum_{j_p} z_{1, j_p}g_{j_p i_p}\\ \sum_{j_1}z_{2, j_1} g_{j_1, i_1} & \ldots & \sum_{j_p} z_{2, j_p}g_{j_p, i_p} \\ \vdots & \ddots & \vdots \\ \sum_{j_1} z_{p, j_1} g_{j_1, i_1} & \ldots & \sum_{j_p} z_{p, j_p} g_{j_p, i_p} \end{pmatrix}}$

which is $\sum_{j_1, \ldots, j_d} g_{j_1, i_1} \ldots g_{j_d, i_d} D_{T'}$. ♦

Since $\otimes_j \text{Alt}^{\mu_j} \mathbb{C}^n$ contains exactly one copy of $V(\lambda)$, it has a unique $GL_n\mathbb{C}$-submodule Q such that the quotient is isomorphic to $V(\lambda).$ The resulting quotient is thus identical to the Schur module F(V), and the above map factors through

$F(V) \to \otimes_i \text{Sym}^{\lambda_i} \mathbb{C}^n, \quad e_T \mapsto D_T.$

Now we can apply results from the last article:

Corollary 1. The polynomials $D_T$ satisfy the following:

• $D_T = 0$ if T has two identical entries in the same column.
• $D_T + D_{T'} = 0$ if T’ is obtained from T by swapping two entries in the same column.
• $D_T = \sum_S D_S$, where S takes the set of all fillings obtained from T by swapping a fixed set of k entries in column j’ with arbitrary sets of k entries in column j (for fixed j < j’) while preserving the order.

Proof

Indeed, the above hold when we replace $D_T$ by $e_T.$ Now apply the above linear map. ♦

Corollary 2. The set of $D_T$, for all SSYT $T$ with shape λ and entries in [n], is linearly independent over $\mathbb{C}.$

Proof

Indeed, the set of these $e_T$ is linearly independent over $\mathbb{C}$ and the above map is injective. ♦

### Example 1.

Consider any bijective filling T for $\lambda = (2, 1)$. Writing out the third relation in corollary 1 gives:

$\left[\det\begin{pmatrix} a & c \\ b & d\end{pmatrix}\right] x = \left[\det\begin{pmatrix} x & c \\ y & d\end{pmatrix}\right] a + \left[ \det\begin{pmatrix} a & x \\ b & y\end{pmatrix}\right] c.$

More generally, if $\lambda$ satisfies $\lambda_j = 2$ and $\lambda_{j'} = 1$, the corresponding third relation is obtained by multiplying the above by a polynomial on both sides.

### Example 2: Sylvester’s Identity

Take the $2 \times n$ SYT by writing $1,\ldots, n$ in the left column and $n+1, \ldots, 2n$ in the right. Now $D_T = D_{1,\ldots, n}D_{n+1, \ldots, 2n}$ is the product:

$\det \overbrace{\begin{pmatrix} z_{1,1} & z_{1,2} & \ldots & z_{1,n} \\ z_{2,1} & z_{2,2} &\ldots & z_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ z_{n,1} & z_{n,2} & \ldots & z_{n,n} \end{pmatrix}}^M \det \overbrace{\begin{pmatrix} z_{1,n+1} & z_{1,n+2} & \ldots & z_{1,2n} \\ z_{2,n+1} & z_{2,n+2} &\ldots & z_{2,2n} \\ \vdots & \vdots & \ddots & \vdots \\ z_{n,n+1} & z_{n,n+2} & \ldots & z_{n,2n} \end{pmatrix}}^N.$

In the sum $D_T = \sum_S D_S$, each summand is of the form $D_S= \det M' \det N'$, where matrices M’N’ are obtained from MN respectively by swapping a fixed set of k columns in N with arbitrary sets of k columns in M while preserving the column order. E.g. for n=3 and k=2, picking the first two columns of N gives:

\begin{aligned} \det ( M_1 | M_2 | M_3) \det(N_1 | N_2| N_3) &= \det(N_1 | N_2 | M_3) \det(M_1 | M_2 | N_3) \\ +\det(N_1 | M_2 | N_2) \det(M_1 | M_3 | N_3) &+ \det(M_1 | N_1 | N_2) \det(M_2 | M_3 | N_3).\end{aligned}

## Notations and Recollections

For a partition $\lambda\vdash d$, one takes its Young diagram comprising of boxes. A filling is given by a function $T:\lambda \to [m]$ for some positive integer m. When m=d, we will require the filling to be bijective, i.e. T contains {1,…,d} and each element occurs exactly once.

If $w\in S_m$ and $T:\lambda \to [m]$ is a filling, then $w(T) = w\circ T$ is obtained by replacing each i in the filling with w(i). For a filling T, the corresponding row (resp. column) tabloid is denoted by {T} (resp. [T]).

Recall from an earlier discussion that we can express the $S_d$-irrep $V_\lambda$ as a quotient of $\mathbb{C}[S_d]b_{T_0}$ from the surjection:

$\mathbb{C}[S_d] b_{T_0} \to \mathbb{C}[S_d] b_{T_0} a_{T_0}, \quad v \mapsto v a_{T_0}.$

Here $T_0$ is any fixed bijective filling $\lambda \to [d]$.

Concretely, a C-basis for $\mathbb{C}[S_d]b_{T_0}$ is given by column tabloids [T] and the quotient is given by relations: $[T] = \sum_{T'} [T']$ where T’ runs through all column tabloids obtained from T as follows:

• fix columns jj’ and a set B of k boxes in column j’ of T; then T’ is obtained by switching B with a set of k boxes in column j of T, while preserving the order. E.g.

## For Representations of GLn

From the previous article we have $V(\lambda) = V^{\otimes d} \otimes_{\mathbb{C}[S_d]} V_\lambda$, where $V_\lambda$ is the quotient of the space of column tabloids described above. We let $V^{\times \lambda}$ be the set of all functions $\lambda \to V$, i.e. the set of all fillings of λ with elements of V. We define the map:

$\Psi : V^{\times\lambda} \to V^{\otimes d}\otimes_{\mathbb{C}[S_d]} V_\lambda, \quad (v_s)_{s\in\lambda} \mapsto \overbrace{\left[v_{T^{-1}(1)} \otimes \ldots \otimes v_{T^{-1}(d)}\right]}^{\in V^{\otimes d}} \otimes [T]$

for any bijective filling $T:\lambda \to [d].$ This is independent of the T we pick; indeed if we replace T by $w(T) = w\circ T$  for $w\in S_d$, the resulting RHS would be:

\begin{aligned}\left[v_{T^{-1}w^{-1}(1)} \otimes \ldots \otimes v_{T^{-1}w^{-1}(d)}\right] \otimes [w(T)] &= \left[v_{T^{-1}w^{-1}(1)}\otimes \ldots \otimes v_{T^{-1} w^{-1}(d)}\right]w \otimes [T]\\ &= \left[v_{T^{-1}(1)} \otimes \ldots \otimes v_{T^{-1}(d)}\right] \otimes [T]\end{aligned}

where the first equality holds since the outer tensor product is over $\mathbb{C}[S_d]$ and the second equality follows from our definition $(v_1' \otimes \ldots \otimes v_d')w = v_{w(1)}' \otimes \ldots \otimes v_{w(d)}'$. Hence $\Psi$ is well-defined. It satisfies the following three properties.

Property C1. $\Psi$ is multilinear in each component V.

In other words, if we fix $s\in \lambda$ and consider $\Psi$ as a function on V in component s of $V^{\times\lambda}$, then the resulting map is C-linear. E.g. if $w'' = 2w + 3w'$, then:

This is clear.

Property C2. Suppose $(v_s), (v'_s)\in V^{\times\lambda}$ are identical except $v'_s = v_t$ and $v'_t = v_s$, where $s,t\in \lambda$ are in the same column. Then $\Psi((v'_s)) = -\Psi((v_s)).$

Proof

Let $w\in S_d$ be the transposition swapping s and t. Then $w([T]) = -[T]$ by alternating property of the column tabloid and $w^2 = e$. Thus:

\begin{aligned}\left[v'_{T^{-1}(1)} \otimes \ldots \otimes v'_{T^{-1}(d)}\right] \otimes [T] &= \left[ v_{T^{-1}w^{-1}(1)} \otimes \ldots \otimes v_{T^{-1}w^{-1}(d)}\right] \otimes -w([T])\\ &= -\left[v_{T^{-1}(1)} \otimes\ldots \otimes v_{T^{-1}(d)}\right]\otimes [T]. \end{aligned} ♦

Finally, we have:

Property C3. Let $(v_s)\in V^{\times\lambda}.$ Fix two columns $j in the Young diagram for λ, and a set B of k boxes in column j’. As A runs through all sets  of k boxes in column j, let $(v_s^A) \in V^{\times\lambda}$ be obtained by swapping entries in A with entries in B while preserving the order. Then:

$\displaystyle \Psi((v_s)) = \sum_{|A| = |B|} \Psi((v_s^A)).$

E.g. for any $u,v,w,x,y,z\in V$ we have:

Proof

Fix a bijective filling $T:\lambda \to [d].$ Then:

\begin{aligned}\Psi((v_s^A)) &= \left[v_{T^{-1}(1)}^A \otimes \ldots \otimes v_{T^{-1}(d)}^A\right] \otimes [T ]\\ &= \left[v_{T^{-1}w^{-1}(1)} \otimes \ldots \otimes v_{T^{-1}w^{-1}(d)}\right] \otimes [T] \\ &= \left[v_{T^{-1}(1)} \otimes \ldots \otimes v_{T^{-1}(d)}\right] \otimes w([T])\end{aligned}

where $w\in S_d$ swaps the entries in A with those in B while preserving the order (note that $w^2 =e$). But the sum of all such $w([T])$ vanishes in $V_\lambda.$ Hence $\sum_A \Psi((v_s^A)) = 0.$ ♦

## Universality

Definition. Let V, W be complex vector spaces. A map $\Psi : V^{\times \lambda} \to W$ is said to be λ-alternating if properties C1, C2 and C3 hold.

The universal λ-alternating space (or the Schur module) for V is a pair $(F(V), \Phi_V)$ where

• $F(V)$ is a complex vector space;
• $\Phi_V : V^{\times\lambda} \to F(V)$ is a λ-alternating map,

satisfying the following universal property: for any λ-alternating map $\Psi : V^{\times\lambda} \to W$ to a complex vector space W, there is a unique linear map $\alpha : F(V) \to W$ such that $\alpha\circ \Phi_V = \Psi.$

F(V) is not hard to construct: the universal space which satisfies C1 and C2 is the alternating space:

$\displaystyle \left(\text{Alt}^{\mu_1} V\right) \otimes \ldots \otimes \left(\text{Alt}^{\mu_e}V\right), \quad \mu := \overline\lambda.$

So the desired F(V) is obtained by taking the quotient of this space with all relations obtained by swapping a fixed set B of coordinates in $\text{Alt}^{j'}$ with a set A of coordinates in $\text{Alt}^j$, and letting A vary over all |A| = |B|. E.g. the relation corresponding to our above example for C3 is:

\begin{aligned} &\left[ (u\wedge x\wedge z) \otimes (v\wedge y) \otimes w\right] -\left[ (u\wedge y\wedge z) \otimes (u\wedge x) \otimes w\right] \\ - &\left[ (v\wedge x\wedge y)\otimes (u\wedge z)\otimes w\right] - \left[ (u\wedge x\wedge w) \otimes (v\wedge z) \otimes w\right]\end{aligned}

over all $u,v,w,x,y,z\in V.$

By universality, the λ-alternating map $\Psi: V^{\times\lambda} \to V^{\otimes d} \otimes_{\mathbb{C}[S_d]} V_\lambda$ thus induces a linear:

$\alpha: F(V) \longrightarrow V^{\otimes d} \otimes_{\mathbb{C}[S_d]} V_\lambda.$

You can probably guess what’s coming next.

Main Theorem. The above $\alpha$ is an isomorphism.

## Proof of Main Theorem

First observe that $\alpha$ is surjective by the explicit construction of F(V) so it remains to show injectivity via dim(LHS) ≤ dim(RHS).

Now $V^{\otimes d}\otimes_{\mathbb{C}[S_d]}V_\lambda \cong V(\lambda)$, and we saw earlier that its dimension is the number of SSYT with shape λ and entries in [n].

On the other hand, let $e_1, \ldots, e_n$ be the standard basis of $V= \mathbb{C}^n.$ If T is any filling with shape λ and entries in [n], we let $e_T$ be the element of F(V) obtained by replacing each i in T by $e_i \in V$; then running through the map $\Phi_V: V^{\times \lambda} \to F(V).$

Claim. The set of $e_T$ generates $F(V)$, where T runs through all SSYT with shape λ and entries in [n].

Proof

Note that the set of $e_T$, as T runs through all fillings with shape λ and entries in [n], generates F(V).

Let us order the set of all fillings of T as follows: T’ > T if, in the rightmost column j where T’ and T differ, at the lowest $(i,j)$ in which $T_{ij}' \ne T_{ij}$, we have $T_{ij}' > T_{ij}$.

This gives a total ordering on the set of fillings. We claim that if T is a filling which is not an SSYT, then $e_T$ is a linear combination of $e_S$ for S > T.

• If two entries in a column of T are equal, then $e_T = 0$ by definition.
• If a column j and row i of T satisfy $T_{i,j} > T_{i+1,j}$, assume j is the rightmost column for which this happens, and in that column, i is as large as possible. Swapping entries $(i,j)$ and $(i+1, j)$ of T gives us T’T and $e_T = -e_{T'}.$
• Now suppose all the columns are strictly ascending. Assume we have $T_{i,j} > T_{i, j+1}$, where j is the largest for which this happens, and $T_{k,j} \le T_{k,j+1}$, for $k=1,\ldots, i-1$. Swapping the topmost i entries of column j+1, with various  i entries of column j, all the resulting fillings are strictly greater than T. Hence $e_T = -\sum_S e_S$, where each S > T.

Thus, if T is not an SSYT we can replace $e_T$ with a linear combination of $e_S$ where S > T. Since there are finitely many fillings T (with entries in [n]), this process must eventually terminate so each $e_T$ can be written as a linear sum of $e_S$ for SSYT S. ♦

Thus $\dim F(V)$ ≤ number of SSYT with shape λ and entries in [n], and the proof for the main theorem is complete. From our proof, we have also obtained:

Lemma. The set of $\{e_T\}$ forms a basis for F(V), where T runs through the set of all SSYT with shape λ and entries in [n].

## V(λ) as Schur Functor

Again, we will denote $V := \mathbb{C}^n$ throughout this article. In the previous article, we saw that the Schur-Weyl duality can be described as a functor:

• given a $\mathbb{C}[S_d]$-module M, the corresponding $GL_n\mathbb{C}$-module is set as $\text{Hom}_{S_d}(M, V^{\otimes d}).$

Definition. The construction

$F_M(V) := \text{Hom}_{S_d}(M, V^{\otimes d})$

is functorial in $V$ and is called the Schur functor when M is fixed.

Here, functoriality means that any linear map $V\to W$ induces a linear $F_M(V) \to F_M(W)$.

For example, when $M = \mathbb{C}[S_d]$, the functor $F_M$ is the identity functor. By Schur-Weyl duality, when M is irreducible as an $S_d$-module, the resulting $F_M(V)$ is either 0 or irreducible. We will see the Schur functor cropping up in two other instances.

Following the reasoning as in $S_d$-modules, we have for partitions $\lambda\vdash d$ and $\mu := \overline\lambda$,

\begin{aligned}\text{Sym}^\lambda V &:= \bigotimes_i \text{Sym}^{\lambda_i} V = V(\lambda) \oplus\left( \bigoplus_{\nu\trianglerighteq \lambda, \nu\ne\lambda} V(\nu)^{\oplus K_{\nu\lambda}}\right),\\ \text{Alt}^{\mu} V &:= \bigotimes_j \text{Alt}^{\mu_j}V = V(\lambda) \oplus \left(\bigoplus_{\nu\trianglelefteq \lambda, \nu\ne\lambda} V(\nu)^{\oplus K_{\overline\nu\mu}} \right).\end{aligned}

Since the only common irrep between the two representations is $V(\lambda)$, any non-zero G-equivariant $f: \text{Sym}^\lambda V\to \text{Alt}^\mu V$ must induce an isomorphism between those two components. We proceed to construct such a map.

For illustration, take $\lambda = (3, 1)$ and pick the following filling:

To construct the map, we will take $\text{Sym}^d V$ and $\text{Alt}^d$ as subspaces of $V^{\otimes d}.$ Thus:

\begin{aligned}\text{Sym}^d V \subseteq V^{\otimes d},\quad& v_1 \ldots v_d \mapsto \sum_{w\in S_d} v_{w(1)} \otimes \ldots \otimes v_{w(d)},\\ \text{Alt}^d V\subseteq V^{\otimes d},\quad &v_1 \wedge \ldots \wedge v_d \mapsto \sum_{w\in S_d} \chi(w) v_{w(1)} \otimes \ldots \otimes v_{w(d)},\\ V^{\otimes d}\twoheadrightarrow \text{Sym}^d V, \quad &v_1 \otimes \ldots\otimes v_d \mapsto v_1 \ldots v_d, \\ V^{\otimes d} \twoheadrightarrow \text{Alt}^d V, \quad &v_1 \otimes \ldots \otimes v_d \mapsto v_1 \wedge \ldots \wedge v_d.\end{aligned}

Let us map $\text{Sym}^\lambda V \to V^{\otimes 4}$ according to the above filling, i.e. $\text{Sym}^3 V$ goes into components 1, 4, 2 of $V^{\otimes 4}$ while $V$ goes into component 3. Similarly, we map $V^{\otimes 4} \to \text{Alt}^\mu V$ by mapping components 1, 3 to $\text{Alt}^2 V$, components 4 and 2 to the other two copies of V. In diagram, we have:

This construction is clearly functorial in V. Hence, if $f:V\to W$ is a linear map of vector spaces, then this induces a linear map $f(\lambda) : V(\lambda) \to W(\lambda).$

## Young Symmetrizer Revisited

Another means of defining the Schur functor is by the Young symmetrizer. Here we shall let $GL_n\mathbb{C}$ act on $V^{\otimes d}$ on the left and $S_d$ act on it on the right via:

$w\in S_d \implies(v_1 \otimes \ldots \otimes v_d)w := v_{w(1)} \otimes \ldots \otimes v_{w(d)}.$

Now given any (left) $\mathbb{C}[S_d]$-module M, consider:

$V(M) := V^{\otimes d} \otimes_{\mathbb{C}[G]} M,$

a left $GL_n\mathbb{C}$-module. We shall prove that $V(M)$ corresponds to the Schur-Weyl duality, i.e. $M = V_\lambda \implies V(M) \cong V(\lambda).$ Once again, by additivity, we only need to consider the case $M = \mathbb{C}[X_\lambda]$. This gives $M \cong \mathbb{C}[G]a_T$ where T is any filling of shape λ and thus:

$V(M) = V^{\otimes d} \otimes_{\mathbb{C}[G]} \mathbb{C}[G]a_T \cong V^{\otimes d}a_T.$

From here, it is clear that $V(M) \cong \text{Sym}^\lambda V$ and so $V\mapsto V(M)$ is yet another expression of the Schur functor.

Recall that the irreducible $S_d$-module $V_\lambda$ can be written as $\mathbb{C}[S_d]c_T$ where $c_T$ is the Young symmetrizer for a fixed filling of shape λ. Hence, the irrep $V(\lambda)$ can be written as:

$V^{\otimes d} \otimes_{\mathbb{C}[G]} V_\lambda \cong V^{\otimes d}\otimes_{\mathbb{C}[G]} \mathbb{C}[G]c_T \cong V^{\otimes d}c_T.$

## Example: d=3

For d=3, and $\lambda = (2,1)$, let us take the Young symmetrizer:

$c_T = a_T b_T = (e + (1,2))(e - (1,3)) = e + (1,2) - (1,3) - (1,3,2).$

If $e_1, \ldots, e_n$ is the standard basis for $V= \mathbb{C}^n$, then $V^{\otimes d}c_T$ is spanned by elements of the form:

$\alpha_{i,j,k} := e_i \otimes e_j \otimes e_k + e_j \otimes e_i \otimes e_k - e_k \otimes e_j \otimes e_i - e_k \otimes e_i \otimes e_j, \ 1 \le i, j, k\le n.$

These satisfy the following:

$\alpha_{j,i,k} = \alpha_{i,j,k},\quad \alpha_{i,j,k} + \alpha_{j,k,i} + \alpha_{k,i,j} = 0.$

By the first relation, we only include those $\alpha_{i,j,k}$ with $i \le j$. By the second relation, we may further restrict to the case $i since if $i=k$ we have $\alpha_{i,j,k} = 0$ and if $k we replace $\alpha_{i,j,k} = \alpha_{k,j,i}+ \alpha_{k,i,j}.$ We claim that the resulting spanning set $\{\alpha_{i,j,k} : i\le j, i forms a basis. Indeed the number of such triplets (ijk) is:

$d = \sum_{i=1}^n (n-i+1)(n-i) = \frac{n(n+1)(n-1)}3.$

On the other hand, we know that $V^{\otimes 3}$ has one copy of $\text{Sym}^3$, one copy of $\text{Alt}^3$ and two copies of $V(\lambda)$ so

$2\dim V = n^3 - \frac{(n+2)(n+1)n}6 - \frac{n(n-1)(n-2)}6 =\frac{2n(n+1)(n-1)}3.$

Thus $\dim V$ is the cardinality of the set and we are done. ♦

Note

Observe that the set $\{(i,j,k) \in [n]^d : i\le j, i corresponds to the set of all SSYT with shape (2, 1) and entries in [n] (by writing ij in the first row and k below i). This is an example of our earlier claim that a basis of $V(\lambda)$ can be indexed by SSYT’s with shape $\lambda$ and entries in [n]. For that, we will explore $V(\lambda)$ as a quotient module of $\otimes_j \text{Alt}^{\mu_j} V$ in the next article. This corresponds to an earlier article, which expressed $S_d$-irrep $V_\lambda$ as a quotient of $\mathbb{C}[S_d]b_T$.

## Schur-Weyl Duality

Throughout the article, we denote $V = \mathbb{C}^n$ for convenience.

So far we have seen:

• the Frobenius map gives a correspondence between symmetric polynomials in $x_1, x_2, \ldots$ of degree d and representations of $S_d$;
• there is a correspondence between symmetric polynomials in $x_1, \ldots, x_n$ and polynomial representations of $GL_n\mathbb C$.

Here we will describe a more direct relationship between representations of $S_d$ and polynomial representations of $GL_n\mathbb{C}.$ Recall from earlier, that $S_d$ and $GL_n\mathbb C$ act on $V^{\otimes d}$ as follows:

\begin{aligned} w\in S_d &\implies v_1 \otimes \ldots \otimes v_d \mapsto v_{w^{-1}(1)} \otimes \ldots \otimes v_{w^{-1}(d)},\\ g\in GL_n(\mathbb C) &\implies v_1 \otimes \ldots \otimes v_n \mapsto g(v_1) \otimes \ldots \otimes g(v_n),\end{aligned}

and the two actions commute, so $w\circ g = g\circ w$ as endomorphisms of $V^{\otimes d}.$

Lemma. The subspace $\text{Sym}^d V \subset V^{\otimes d}$ of all elements fixed by every $w\in S_d$ is spanned by $\{v^{\otimes d}: v\in V\}.$

Proof

Use induction on d; the case d=1 is trivial so suppose d>1. For integers $k\ge 0$, consider the binomial expansion in $\text{Sym}^d V$:

$\displaystyle(v + kw)^d = v^d + \left(\sum_{i=1}^{d-1} k^i {d\choose i} v^{d-i} w^{i}\right) + k^d w^d.$

We claim: for large k, the $(d+1)\times k$ matrix with (i, j)-entry $j^i {d\choose i}$ (where $0\le i \le d$) has rank d+1.

• Indeed, otherwise there are $\alpha_0, \ldots, \alpha_d \in \mathbb{C}$, not all zero, such that $\alpha_0 {d\choose 0} + \alpha_1 {d\choose 1} k + \ldots + \alpha_d {d\choose d} k^d = 0$ for all large k, which is absurd since this is a polynomial in k.

Hence, we can find a linear combination summing up to:

$\alpha_0 v^d + \alpha_1 (v+w)^d + \ldots + \alpha_k (v+kw)^d = vw^{d-1}, \qquad \text{ for all }v, w \in V.$

Thus $vw^{d-1}$ lies in the subspace spanned by all $v^d$. By induction hypothesis, the set of all $w^{d-1} \in \text{Sym}^{d-1} V$ spans the whole space. Hence, the set of all $v^d$ spans $\text{Sym}^d V$. ♦

This gives:

Proposition. If $f: V^{\otimes d} \to V^{\otimes d}$ is an $S_d$-equivariant map, then it is a linear combination of the image of $GL_n\mathbb{C} \hookrightarrow \text{End}(V^{\otimes d})$.

Proof

Note that since $\text{End}(V) \cong V\otimes V^\vee$ we have $\text{End}(V^{\otimes d}) \cong \text{End}(V)^{\otimes d}.$ Hence from the given condition

$f \in \text{End}(V^{\otimes d})^{S_d} = (\text{End}(V)^{\otimes d})^{S_d}.$

By the above lemma, f is a linear combination of $u^{\otimes d}$ for all $u\in\text{End}(V).$ Since $\text{GL}_n\mathbb{C} \subset \text{End}(V)$ is dense, f is also a linear combination of $u^{\otimes d}$ for $u\in \text{GL}_n\mathbb{C}$. ♦

## Main Statement

Now let U be any complex vector space and consider the complex algebra $\text{End}(U).$ Recall: if $A\subseteq \text{End}(U)$ is any subset,

$C(A) = \{a \in \text{End}_{\mathbb C}(U) : ab = ba \text{ for all }b \in A\}$

is called the centralizer of A. Clearly $C(A) \subseteq \text{End}(U)$ is a subalgebra and we have $A\subseteq C(C(A)).$

Theorem (Schur-Weyl Duality). Let $A\subseteq \text{End}(U)$ be a subalgebra which is semisimple. Then:

• $B:=C(A)$ is semisimple;
• $C(B) = A$; (double centralizer theorem)
• U decomposes as $\oplus_{\lambda} (U_\lambda \otimes W_\lambda)$, where $U_\lambda, W_\lambda$ are respectively complete lists of irreducible A-modules and B-modules.

Proof

Since A is semisimple, we can write it as a finite product $\prod_\lambda \text{End}(\mathbb{C}^{m_\lambda})$. Each simple A-module is of the form $U_\lambda := \mathbb{C}^{m_\lambda}$ for some $m_\lambda >0.$ As an A-module, we can decompose: $\displaystyle U \cong \oplus_{\lambda} U_\lambda^{n_\lambda}.$ Here $n_\lambda > 0$ since as A-modules we have:

$U_\lambda \subseteq A \subseteq \text{End}(U) \cong U^{\dim U}.$

By Schur’s lemma $\text{End}_A(U_\lambda, U_\mu) \cong \mathbb{C}$ if $\lambda = \mu$ and 0 otherwise. This gives:

$\displaystyle B = C(A) = \text{End}_A(U) = \text{End}_A\left(\prod_\lambda U_\lambda^{n_\lambda} \right) \cong \prod_\lambda \text{End}(\mathbb{C}^{n_\lambda})$

which is also semisimple. Now each simple B-module $W_\lambda$ has dimension $n_\lambda$. From the action of B on U, we can write $U \cong \oplus_\lambda U_\lambda ^{n_\lambda} \cong \oplus_\lambda (U_\lambda \otimes W_\lambda)$ where A acts on the $U_\lambda$ and B acts on the $W_\lambda$. Expressed as a sum of simple B-modules, we have $U \cong \oplus_\lambda W_\lambda^{m_\lambda}$; thus repeating the above with A replaced by B gives:

$C(B) \cong \prod_\lambda \text{End}(\mathbb C^{m_\lambda})\cong A.$

From $A\subseteq C(C(A))$ we thus have $A= C(B).$ This proves all three properties. ♦

Note

From the proof, we see that

• $U = \oplus_\lambda (U_\lambda \otimes W_\lambda)$ as complex vector spaces,
• $A \cong \prod_\lambda \text{End}_{\mathbb{C}}U_\lambda$ acts on the $U_\lambda$, and
• $B\cong \prod_\lambda \text{End}_{\mathbb{C}} W_\lambda$ acts on the $W_\lambda$.

Thus the correspondence between $U_\lambda$ and $W_\lambda$ works as follows:

\begin{aligned}\text{Hom}_A(U_\lambda, U) &= \text{Hom}_{\prod \text{End}(U_\mu)}(U_\lambda, \oplus_\mu (U_\mu \otimes W_\mu))\\ &\cong \text{Hom}_{\text{End}(U_\lambda)} (U_\lambda, U_\lambda^{n_\lambda})\\ &\cong W_\lambda.\end{aligned}

The nice thing about this point-of-view is that the construction is now functorial, i.e. for any A-module M, we can define the corresponding: $F: M \mapsto\text{Hom}_A(M, U).$ This functor is additive, i.e. $F(M_1 \oplus M_2) \cong F(M_1) \oplus F(M_2)$, since the Hom functor is bi-additive.

## The Case of Sd and GLnC

Now for our main application.

Consider $S_d$ and $GL_n\mathbb C$ acting on $V^{\otimes d}$; their actions span subalgebras $A, B\subseteq \text{End}_{\mathbb C}(V)$. Now A is semisimple since it is a quotient of $\mathbb{C}[S_d]$. From the lemma, we have B = C(A) so Schur-Weyl duality says A = C(B), B is semisimple and

$V^{\otimes d} \cong \oplus_\lambda (U_\lambda \otimes W_\lambda)$

where $U_\lambda, W_\lambda$ are complete lists of simple A– and B-modules respectively. Since A is a quotient of $\mathbb{C}[S_d]$, the $U_\lambda$ are also irreps of $S_d$ so they can be parametrized by $\lambda \vdash d$.

Proposition. If $U_\lambda$ is the irrep for $S_d$ isomorphic to $V_\lambda$, then $W_\lambda$ is the irrep for $GL_n\mathbb{C}$ corresponding to $V(\lambda).$

Proof

It suffices to show: if $\mathbb{C}[X_\lambda]$ corresponds to $W'$ via the functor in the above note, then

$W'\cong \text{Sym}^\lambda V = \text{Sym}^{\lambda_1} V \otimes \ldots \otimes \text{Sym}^{\lambda_l} V.$

By definition $W' = \text{Hom}_{S_d}(\mathbb{C}[X_\lambda], V^{\otimes d}).$ Recall that $X_\lambda$ is a transitive $S_d$-set; picking a point $A=(A_i) \in X_\lambda$, any map $f:\mathbb{C}[X_\lambda] \to V^{\otimes d}$ which is $S_d$-equivariant is uniquely defined by the element $f(A)\in V^{\otimes d}$, as long as this element is invariant under the stabilizer group:

$H := \{w\in S_d : w(A) = A\} \cong S_{\lambda_1} \times S_{\lambda_2} \times \ldots \times S_{\lambda_l}.$

Thus, the coefficients $c_{i_1\ldots i_d}$ of $e_{i_1}\otimes\ldots\otimes e_{i_d}$ in $f(A)$ remain invariant when acted upon by $\prod_i S_{\lambda_i}$. So we have an element of $\text{Sym}^\lambda V.$ ♦

Theorem. The set of irreps $V_\lambda$ of $S_d$ occurring in $V^{\otimes d}$ is:

$\{ V_\lambda : \lambda \vdash d, l(\lambda) \le n\}.$

Proof

The following is the complete set of $GL_n\mathbb{C}$-irreps of degree d:

$\{V(\lambda) : \lambda\vdash d, l(\lambda) \le n\}$

We claim that this is also the set of all irreps in $V^{\otimes d}.$ Clearly, each irrep in $V^{\otimes d}$ is of degree d; conversely, $V^{\otimes d}$ has

$\psi = (x_1 + \ldots + x_n)^d = h_\mu(x_1, \ldots, x_n), \ \mu = (1,1,\ldots, 1).$

Clearly $K_{\lambda\mu} > 0$ so $V^{\otimes d}$ contains all $V(\lambda)$ of degree d. Now apply the above proposition. ♦

Example

The simplest non-trivial example follows from the decomposition

$V^{\otimes 2} = (\text{Sym}^2 V) \oplus (\text{Alt}^2 V).$

The action of $S_2$ is trivial on the first component and alternating on the second.

## Twisting

From the previous article, any irreducible polynomial representation of $G= GL_n\mathbb{C}$ is of the form $V(\lambda)$ for some $\lambda \vdash d, l(\lambda) \le n$ such that $\psi_{V(\lambda)}$ is the Schur polynomial $s_\lambda(x_1, \ldots, x_n)$.

Now given any analytic representation V of G, we can twist it by taking $V\otimes \det^k$ for an integer k. Then:

$\displaystyle\psi_{V \otimes \det^k}= \psi_V \cdot \psi_{\det}^k = (x_1 \ldots x_n)^k \psi_V.$

Twisting the irrep $V(\lambda)$ with $k\ge 0$ gives us another irrep, necessarily of the form $V(\mu)$. What is this $\mu$? Note that from $\psi_{V(\lambda)}$ we can recover the partition $\lambda$ by taking the minimal partition (with respect to $\trianglelefteq$). Hence from $\psi_{V(\mu)} = (x_1\ldots x_n)^k\psi_{V(\lambda)}$ we must have $\mu = \lambda + (k, \ldots, k).$ Thus:

Proposition. Any irreducible analytic representation of G can be uniquely written as:

$\{\psi_{V(\lambda)} \otimes \det^k : l(\lambda) \le n-1, k\in\mathbb{Z}\}$

where $V(\lambda)$ is the polynomial representation satisfying:

$\psi_{V(\lambda)} = s_\lambda(x_1, \ldots, x_n).$

Since $l(\lambda) \le n-1$, $s_\lambda(x_1, \ldots, x_n)$ is not divisible by $x_n$ so the representation $V(\lambda) \otimes \det^k$ is polynomial if and only if $k\ge 0.$ Its degree is $|\lambda| + kn.$

## Dual of Irrep

The dual of the irrep $V(\lambda)$ is also a rational irrep, so it is of the form $V(\mu) \otimes \det^k$ for some partition $\mu$ with $l(\mu) \le n-1$ and integer k. From:

$\psi_{V(\lambda)^\vee}(x_1, \ldots, x_n) = \psi_{V(\lambda)}(x_1^{-1}, \ldots, x_n^{-1})$

we take the term with the smallest exponent for $x^\mu$ in lexicographical order. For large N, denoting $\text{rev}(\alpha)$ for the reverse of a sequence $\alpha$, we have:

$\lambda, \mu \vdash d, \lambda \trianglerighteq \mu \implies \text{rev}((N,\ldots, N) - \lambda) \trianglerighteq \text{rev}((N,\ldots, N) - \mu).$

Hence $\mu = (\lambda_1, \lambda_1 - \lambda_{n-1}, \lambda_1 - \lambda_{n-2}, \ldots, \lambda_1 - \lambda_2)$ and $k = -\lambda_1.$ Pictorially we have:

## Weight Space Decomposition

By definition $\psi_{V(\lambda)}$ is the character of $V(\lambda)$ when acted upon by the torus group S. Since this polynomial is $s_\lambda = \sum_{\mu} K_{\lambda\mu} m_\mu$, as vector spaces we have:

$\displaystyle V(\lambda) = \bigoplus_{\mu} \bigoplus_{\sigma} V(\lambda)_{\sigma(\mu)}$

where:

• $\mu$ runs through all partitions with $|\mu| = |\lambda|$ and $l(\mu) \le n$;
• $\sigma(\mu)$ runs through all permutations of $\mu$ without repetition, e.g. if $\mu = (5, 3, 3)$ we get 3 terms: (5, 3, 3), (3, 5, 3) and (3, 3, 5);
• $V(\lambda)_{\nu}$ is the space of all $v\in V(\lambda)$ for which S acts with character $x^\nu$, i.e.

$V(\lambda)_{\nu} = \{v \in V(\lambda) : D(x_1, \ldots, x_n) \in S \text{ takes } v\mapsto x^\nu v\}$

and the dimension of $V(\lambda)_{\nu}$ is $K_{\lambda\nu}$. This is called the weight space decomposition of $V(\lambda).$ We will go through some explicit examples later.

### Foreshadowing: SSYTs as a Basis

As noted above, the dimension of $V(\lambda)_{\sigma(\mu)}$ is exactly the number of SSYT with shape $\lambda$ and type $\sigma(\mu)$. Thus in a somewhat ambiguous way, we can take, as a basis of $V(\lambda)$, elements of the form $\{v_T\}$ over all SSYT T of shape $\lambda$ and entries from [n]={1,2,…,n}; each $v_T$ lies in the space $V(\lambda)_{\text{shape}(T)}.$

However, such a description does not distinguish between distinct SSYT of the same type. For that, one needs a construction like the determinant modules (to be described later).

## Example: n=2

Consider $G = GL_2\mathbb C.$ By the above proposition, each irreducible representation is given by $V(m) \otimes \det^k$ where mk are integers and $m\ge 0.$ To compute V(m), we need to find a polynomial representation of G such that

$\psi_{V(m)} = s_m(x, y)= x^m + x^{m-1}y + \ldots + y^m$

corresponding to the SSYT with shape (m) and entries comprising of only 1’s and 2’s. E.g. $s_4(x,y) = x^4 + x^3 y + x^2 y^2 + xy^3 + y^4$ from:

Such a V(m) is easy to construct: take $\text{Sym}^m V$; if {ef} is a basis of V, then a corresponding basis of $\text{Sym}^m V$ is given by $\{e^i f^{4-i}\}_{i=0,\ldots,4}$. If $v_i := e^{2+i} f^{2-i}$, then the diagonal matrix D(ab) takes $v_i \mapsto a^{2+i} b^{2-i} v_i$ so its character is $a^4 + a^3 b + a^2 b^2 + ab^3 + b^4$ as desired.

The weight space decomposition thus gives:

$V(m) = V(m)_{4,0} \oplus V(m)_{3,1} \oplus V(m)_{2,2} \oplus V(m)_{1,3} \oplus V(m)_{0,4}$

where each $V(m)_{i,j}$ is 1-dimensional and spanned by $e^i f^{4-i}.$

## Example: d=2

Consider $G = GL_n\mathbb{C}$. We have:

$\mathbb{C}^n \otimes_{\mathbb C} \mathbb{C}^n \cong \text{Sym}^2 \mathbb{C}^n \oplus \text{Alt}^2 \mathbb{C}^n,$

where each component is G-invariant. As shown earlier, we have:

\begin{aligned}\psi_{\text{Sym}^2} &= \sum_{1\le i\le j \le n} x_i x_j = h_2(x_1, \ldots, x_n),\\ \psi_{\text{Alt}^2} &= \sum_{1 \le i < j \le n} x_i x_j = e_2(x_1, \ldots, x_n).\end{aligned}

Since the Schur polynomials are $s_2 = h_2$ and $s_{11} = e_2$, both $\text{Alt}^2$ and $\text{Sym}^2$ are irreps of G. The weight space decomposition of the two spaces are:

\displaystyle \begin{aligned}\text{Sym}^2\mathbb{C}^n &= \bigoplus_{1\le i\le j\le n}\mathbb{C}\cdot e_i e_j, \\ \text{Alt}^2\mathbb{C}^n &= \bigoplus_{1\le i

Hence in their weight space decompositions, all components have dimension 1.

## Example: n=3

Now let us take $G = GL_3\mathbb{C}$ and compute $V(\lambda)$ where $\lambda = (\lambda_1, \lambda_2)$ and $\lambda_1 + \lambda_2 = d.$ To find $s_\lambda(x,y,z)$ we need to compute $K_{\lambda\mu}$ for all $\mu\vdash d$ and $l(\mu) \le 3.$ We will work in the plane $X_1+ X_2+ X_3 = d;$ since partitions lie in the region $X_1 \ge X_2 \ge X_3$, we only consider the coloured region:

The point $\lambda$ has $\lambda_3 = 0$. Assuming $|\mu| = d$, the condition $\mu \trianglelefteq\lambda$ then reduces to a single inequality $\mu_1 \le \lambda_1.$ Hence, $\mu$ lies in the brown region below:

To fix ideas, consider the case $\lambda = (5,3).$ Calculating the Kostka coefficients gives us:

Taking into account all coefficients then gives us a rather nice diagram for the weight space decomposition.

E.g. we have $\dim V(\lambda)_{2,3,3} =3$ and $\dim V(\lambda)_{4,3,1} = 2$. These correspond to the following SSYT of shape (5,3):

## Polynomials and Representations XXXIII

We are back to the convention $G = GL_n\mathbb{C}$ and $K = U_n.$ We wish to focus on irreducible polynomial representations of G.

The weak Peter-Weyl theorem gives:

$\displaystyle\mathcal{O}(K) \cong \bigoplus_{K-\text{irrep } V} \text{End}(V)^\vee = \bigoplus_{\text{rat. } G-\text{irrep } V} \text{End}(V)^\vee.$

Theorem. Restricting the RHS to only polynomial irreducible V gives us $\mathbb{C}[z_{ij}]_{1\le i, j\le n}$ on the LHS, where each polynomial $f$ in $z_{ij}$ restricts to a function $K \to \mathbb{C}.$

Proof

Since $z_{ij} \in \mathcal{O}(K)$ is a matrix coefficient, we have $\mathbb{C}[z_{ij}] \subseteq \mathcal{O}(K)$ and this is clearly a $G\times G$-submodule. Furthermore, as functions $K\to\mathbb{C}$, the $z_{ij}$ are algebraically independent over $\mathbb{C}$ by the main lemma here.

Since each $\text{End}(V)^\vee$ is irreducible as a $G\times G$-module, $\mathbb{C}[z_{ij}]$ corresponds to a direct sum $\oplus_S \text{End}(V)^\vee$ over some set S of G-irreps V. It remains to show that S is precisely the set of polynomial irreps.

Next, as $G\times G$-representations, we have a decomposition

$\displaystyle\mathbb{C}[z_{ij}] = \bigoplus_{d\ge 0} \mathbb{C}[z_{ij}]^{(d)}$

into homogeneous components; each is a finite-dimensional representations of $G\times G.$ By considering the action of $1\times G$, each component is a polynomial representation of G. Hence every irrep in S must be polynomial.

Conversely, if W is any polynomial irrep of G of dimension m, upon taking a basis the action of every $g\in G$ can be written as an $m\times m$ matrix with entries in $\mathbb{C}[z_{ij}].$ Hence $\mathbb{C}[z_{ij}]^m$ contains W; since W is irreducible, $\mathbb{C}[z_{ij}]$ contains W. ♦

## Decomposing By Degrees

Thus we have as isomorphism of $G\times G$-reps (*):

$\displaystyle\bigoplus_{\text{poly irrep } V \text{of }G} \text{End}(V)^\vee \cong \mathbb{C}[z_{ij}] = \bigoplus_{d\ge 0} \mathbb{C}[z_{ij}]^{(d)}.$

Definition. The degree of a polynomial irrep V is the unique $d\ge 0$ for which $\mathbb{C}[z_{ij}]^{(d)}$ contains it.

Let us pick one particular component $\mathbb{C}[z_{ij}]^{(d)}.$ We get a G-rep by taking the action of $1\times G.$ Hence its Laurent polynomial can be computed by considering the action of $1\times S$ on it. Since the space is spanned by monomials of degree d, we have:

Property 1. For a polynomial G-irrep V, we have $\deg V = \deg \psi_V.$

Recall that if G acts on V, then $G\times G$ acts on $\text{End}(V)$ via:

$(x,y) \in G\times G, f\in \text{End}(V) \ \mapsto \ (x,y)f = \rho_V(x)\circ f\circ \rho_V(y^{-1}).$

Hence, taking the dual gives:

Property 2. The action of $S\times S \subset G\times G$ on $\text{End}(V)^\vee$ gives the character

$\displaystyle \psi_V(x_1^{-1}, \ldots, x_n^{-1}) \psi_V(y_1, \ldots, y_n)$

The pair of diagonal matrices $(D(x_1, \ldots, x_n), D(y_1, \ldots, y_n)) \in S\times S$ takes $z_{ij} \mapsto x_i^{-1} z_{ij} y_j$. Hence, taking the basis of monomials of degree d, we have:

Property 3. The action of $S\times S\subset G\times G$ on $\mathbb{C}[z_{ij}]^{(d)}$ has character:

$\displaystyle \sum_{\substack{m_{11}, m_{12},\ldots, m_{nn} \ge 0\\ m_{11} + m_{12} + \ldots + m_{nn} = d}} \left( \prod_{i=1}^n \prod_{j=1}^n x_i^{-m_{ij}} y_j^{m_{ij}}\right).$

Finally, from lemma 3 here and property 1 above, we see that:

Property 4. For each d, the number of polynomial G-irreps of degree d is exactly the cardinality of $\{\lambda \vdash d: \lambda_1 \le n\}.$

## Main Theorem

Now we are ready to prove:

Theorem. A polynomial representation V of G of degree d is irreducible if and only if $\psi_V$ is a Schur polynomial in $x_1, \ldots, x_n$ of of degree d.

Proof

For each d, let $V_{d1}, \ldots, V_{de}$ be the polynomial irreps of degree d; let $p_{dj} := \psi_{V_{dj}},$ a homogeneous symmetric polynomial in $x_1, \ldots, x_n$ of degree d. By property 2, the character of $S\times S$ on $\mathbb{C}[z_{ij}]^{(d)}$ is:

$\displaystyle\sum_{j} p_{dj}(x_1^{-1}, \ldots, x_n^{-1}) p_{dj}(y_1, \ldots, y_n).$

By property 3 and (*), summing this over all d and j gives the power series:

$\displaystyle\sum_{m_{11}, \ldots, m_{nn}\ge 0} \left( \prod_{i=1}^n \prod_{j=1}^n x_i^{-m_{ij}} y_j^{m_{ij}}\right) = \prod_{i=1}^n \prod_{j=1}^n \sum_{m\ge 0} (x_i^{-1} y_j)^m = \prod_{1\le i, j\le n} \frac 1 {1 - x_i^{-1}y_j}.$

Finally by property 4, for each d, the number of $p_{dj}$ is exactly the size of $\{\lambda \vdash d : \lambda_1 \le n\}.$ Thus by the criterion for orthonormal basis proven (much) earlier, the $\{p_{dj}\}_j$ forms an orthonormal basis of $\Lambda_n^{(d)}.$ Hence, each $p_{dj}$ is, up to sign, a Schur polynomial of degree d. Since the coefficients of $p_{dj}$ are non-negative, they are the Schur polynomials. ♦

## Summary

We have a correspondence between:

which takes $V\mapsto \psi_V.$ Under this correspondence, for partition $\lambda = (\lambda_1, \ldots, \lambda_l)$,

\begin{aligned}\text{Sym}^{\lambda_1} \mathbb{C}^n \otimes \ldots \otimes \text{Sym}^{\lambda_l} \mathbb{C}^n &\leftrightarrow h_\lambda(x_1, x_2, \ldots, x_n), \\\text{Alt}^{\lambda_1} \mathbb{C}^n \otimes \ldots \otimes \text{Alt}^{\lambda_l} \mathbb{C}^n &\leftrightarrow e_\lambda(x_1, x_2, \ldots, x_n), \\ \text{poly. irrep } &\leftrightarrow s_\lambda(x_1, x_2, \ldots, x_n), \\ \otimes \text{ of reps } &\leftrightarrow \text{ multiplication of polynomials},\\ \text{Hom}_G(V, W) &\leftrightarrow \text{ Hall inner product}. \end{aligned}

Indeed, the first two correspondences are obvious; the third is what we just proved. The fourth is immediate from the definition of $\psi_V$. The final correspondence follows from the third one. Denote $V(\lambda)$ for the corresponding irrep of $GL_n\mathbb{C}$; we can now port over everything we know about symmetric polynomials, such as:

\displaystyle \begin{aligned} h_\lambda = \sum_{\mu\vdash d} K_{\mu\lambda} s_\mu \ &\implies\ \bigotimes_i \text{Sym}^{\lambda_i} \mathbb{C}^n = \bigoplus_{\mu\vdash d} V(\mu)^{\oplus K_{\mu\lambda}},\\ e_\lambda = \sum_{\mu\vdash d} K_{\overline\mu\lambda} s_\mu\ &\implies\ \bigotimes_i \text{Alt}^{\lambda_i} \mathbb{C}^n = \bigoplus_{\mu\vdash d} V(\mu)^{\oplus K_{\overline\mu\lambda}},\\ s_\lambda s_\mu = \sum_{\nu\vdash d} c_{\lambda\mu}^\nu s_\nu \ &\implies\ V(\lambda) \otimes V(\mu) \cong \bigoplus_{\nu\vdash d} V(\nu)^{\oplus c_{\lambda\mu}^\nu}.\end{aligned}

Setting $\lambda = (1,\ldots, 1)$ for $h_\lambda$ gives:

$\displaystyle(\mathbb{C}^n)^{\otimes d} \cong \bigoplus_{\mu\vdash d} V(\mu) ^{d_\mu}$

where $d_\mu$ is the number of SYT of shape $\mu.$

Unfortunately, the involution map $\omega : \Lambda_n \to \Lambda_n$ does not have a nice interpretation in our context. (No it does not take the polynomial irrep to its dual!)

## Polynomials and Representations XXXII

We attempt to identify the irreducible rational representations of $G = GL_n\mathbb C.$ From the last article, we may tensor it with a suitable power of det and assume it is polynomial.

One key ingredient is the following rather ambiguous statement.

• Peter-Weyl Principle: any irrep can be embedded inside the ring of functions on G.

To make sense of this statement, we need to define a group action on the latter. This article is rather general; throughout, G is allowed to be any topological group and all representations are assumed to be continuous and finite-dimensional.

## C(G) as a Representation

Let C(G) be the ring of continuous functions $G\to \mathbb{C}.$

Definition. The action of $G\times G$ on $C(G)$ is defined as follows:

$(x, y)\in G\times G, f \in C(G)\ \implies\ (x,y)\cdot f : G\to\mathbb{C}, g\mapsto f(x^{-1}gy).$

One checks that $(x',y')\cdot((x,y)\cdot f) = (x'x, y'y)\cdot f$.

Some examples of functions in C(G) are the matrix coefficients; recall that these are functions of the form $f : G \to \mathbb{C}, g\mapsto \alpha (\rho(v))$ for some representation $\rho : G\to GL(V)$, $v\in V$ and $\alpha \in V^\vee.$ A more economical way of writing this is $f = \beta\circ \rho$ for some $\beta \in \text{End}_{\mathbb C}(V)^\vee.$

Proposition. Let $\mathcal{O}(G)$ be the set of matrix coefficients. Then $\mathcal{O}(G)$ is a (unital) $\mathbb{C}$-subalgebra of $C(G)$.

Proof

To prove $1\in\mathcal{O}(G)$, let $\rho :G \to \mathbb{C}^*$ be the trivial representation.

Closure under addition and multiplication: suppose $\rho_i:G\to GL(V_i)$, $\beta_i\in \text{End}(V_i)^\vee$ for i = 1,2; let $f_i = \beta_i \circ \rho_i:G\to\mathbb{C}$ be the matrix coefficient. Taking:

\begin{aligned} \rho_1 \oplus \rho_2 : G \to GL(V_1 \oplus V_2), \ \ &\beta_1 \oplus \beta_2 \in \text{End}(V_1)^\vee \oplus \text{End}(V_2)^\vee \subseteq \text{End}(V_1 \oplus V_2)^\vee, \\ \rho_1\otimes \rho_2 : G\to GL(V_1 \otimes V_2), \ \ &\beta_1 \otimes \beta_2 \in \text{End}(V_1)^\vee \otimes \text{End}(V_2)^\vee \cong \text{End}(V_1 \otimes V_2)^\vee,\end{aligned}

we have $f_1 + f_2 = (\beta_1 \oplus \beta_2) \circ (\rho_1 \oplus \rho_2)$ and $f_1 f_2 = (\beta_1 \otimes \beta_2) \circ(\rho_1 \otimes \rho_2).$ ♦

## Identifying Matrix Coefficients in C(G)

Next we have:

Lemma. For any $f\in C(G)$, the following are equivalent.

1. The span of all $(g,g')\cdot f$ over all $(g, g')\in G \times G$ is finite-dimensional.
2. The span of all $(g,1)\cdot f$ over all $g\in G$ is finite-dimensional.
3. The span of all $(1,g)\cdot f$ over all $g\in G$ is finite-dimensional.
4. $f$ is a matrix coefficient.

Proof

That 1 ⇒ 2, 3 is obvious. Let us now prove 4 ⇒  1. Suppose $f = \beta\circ \rho$ for $\rho : G\to GL(V)$ and $\beta : \text{End}(V) \to \mathbb{C}$. Then the function $(x, y)f : G\to \mathbb{C}$ takes

$g \mapsto f(x^{-1}gy) = \beta(\rho(x^{-1}) \rho(g) \rho(y)) = \beta'(\rho(g))$

where $\beta' : \text{End} (V) \to \mathbb{C}$ is a linear map which takes $\phi \mapsto \beta(\rho (x^{-1})\phi\rho(y))$. Hence the span of all $(x,y)f$ lies in $\{\beta' \circ \rho : \beta' \in \text{End}(V)^\vee\}$ and (1) follows since $\text{End}(V)^\vee$ is finite-dimensional.

Finally, we prove 3 ⇒ 4; the case of 2 ⇒ 4 is similar. Let V be the space spanned by all $(1, g)f$, so V is a finite-dimensional subspace of C(G) which is $(1\times G)$-invariant. Let

• $\rho : G \to GL(V)$ be given by the action of $(1\times G)$ on V;
• $\beta : \text{End}(V) \to \mathbb{C}$ be the map $\alpha \mapsto (\alpha(f))(e_G)$.

Now tracing through $\beta\circ \rho : G \to \mathbb{C}$ gives:

$\overbrace{g}^{\in G}\ \mapsto \overbrace{(f_1\in V \mapsto (x\mapsto f_1(xg)))}^{\in GL(V)} \mapsto \left.(x\mapsto f(xg))\right|_{x=e} = f(g).$

Thus $f = \beta\circ \rho$ is a matrix coefficient. [Note: for 2 ⇒ 4, one needs to take the dual of V.]   ♦

Corollary. $\mathcal{O}(G)$ is the sum of all finite-dimensional $G\times G$-subrepresentations of $C(G)$.

## Weak Peter-Weyl Theorem

Let V be a representation of G; the vector space End(V) becomes a $G\times G$-rep via:

$(x,y) : (\alpha : V\to V) \mapsto (\rho_V(x)\circ \alpha \circ\rho_V(y^{-1}) : V\to V).$

Another way of seeing this is: $\text{End}(V) \cong V\otimes V^\vee$ naturally and the action of $G\times G$ on the LHS is obtained from the RHS. Next, we obtain a linear map:

$\text{End}(V)^\vee\to \mathcal{O}(G),\qquad \beta \mapsto (\beta \circ \rho_V : G\to \mathbb{C}).$ (*)

This is $G\times G$-equivariant:

• Indeed (xy) takes $\alpha\in \text{End}(V)$ to $\rho_V(x) \alpha \rho_V(y^{-1})$ so dually $\beta \in \text{End}(V)^\vee$ is taken to the function $(\alpha \in \text{End}(V)) \mapsto \beta(\rho_V(x^{-1})\alpha \rho_V(y)).$
• In $\mathcal{O}(G)$, (xy) takes $f:G\to\mathbb{C}$ to $g\mapsto f(x^{-1}gy).$ Letting $f = \beta\circ \rho_V$ it gets taken to $g \mapsto \beta(\rho_V(x^{-1}) \rho_V(g) \rho_V(y))$ so the following commutes:

$\begin{matrix} \text{End}(V)^\vee &\longrightarrow & \mathcal{O}(G)\\ \downarrow & &\downarrow & \text{action of }(x,y)\\ \text{End}(V)^\vee &\longrightarrow &\mathcal{O}(G)\end{matrix}$

We have $\text{End}(V) \cong V \otimes V^\vee$ so its dual $V^\vee \otimes V$ is naturally isomorphic to End(V) as vector spaces and as G-reps, but not as $G\times G$-reps. This is because the isomorphism $V^\vee \otimes V \cong V\otimes V^\vee$ requires a twist of the two components, which is not $G\times G$-equivariant.

Now suppose G is compact Hausdorff.

Lemma. Let V, W be irreducible representations of G. Then:

• $\text{End}(V)$ is an irreducible representation of $G\times G$;
• $\text{End}(V) \cong \text{End}(W)$ as $G\times G$-reps if and only if $V\cong W$ as G-reps.

Proof

The character of End(V) is: $\chi_{\text{End}(V)}(g_1, g_2) = \chi_{V}(g_1) \chi_{V^\vee}(g_2).$ Taking the inner product:

\begin{aligned}\int_{G\times G} \chi_{\text{End}(V)}(g_1, g_2) \chi_{\text{End}(W)}(g_1, g_2) d(g_1, g_2) &=\int_G \int_G \chi_{V}(g_1) \chi_{V^\vee}(g_2) \chi_{W}(g_1) \chi_{W^\vee}(g_2) dg_1 \, dg_2\\ &= \left(\int_G \chi_{V}(g_1) \chi_{W}(g_1) dg_1\right) \left(\int_G \chi_{V^\vee}(g_2) \chi_{W^\vee}(g_2) dg_2\right)\\&= \begin{cases} 1, \ \ &\text{if } V\cong W; \\ 0,\ \ &\text{otherwise.}\end{cases}\end{aligned}

This proves both claims together. ♦

Taking the direct sum of (*) over all irreps V of G, we obtain:

$\displaystyle\bigoplus_{V \text{irrep of } G} \text{End}(V)^\vee\to \mathcal{O}(G).$

Weak Peter-Weyl Theorem. The above map is an isomorphism.

Proof

To show injectivity: the kernel of the map is a $G\times G$-submodule, so it is a direct sum of $G\times G$-irreps $\text{End}(V)^\vee.$ On the other hand, $\text{End}(V)^\vee \to \mathcal{O}(G)$ clearly cannot map to 0, since that would mean $\rho_V = 0.$

To show surjectivity: each $f\in \mathcal{O}(G)$ is, by definition, of the form $\beta \circ \rho$ for some $\rho : G \to GL(V)$, $\beta \in \text{End}(V)^\vee$. Thus lies in the image. ♦

We remind the reader that this theorem only works when G is compact Hausdorff.