Polynomials and Representations XXXVIII

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.


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.


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}.


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}


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Polynomials and Representations XXXVII

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)).



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<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:



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. ♦



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].


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].


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Polynomials and Representations XXXVI

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<k since if i=k we have \alpha_{i,j,k} = 0 and if k <i\le j 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<k\} 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. ♦


Observe that the set \{(i,j,k) \in [n]^d : i\le j, i<k\} 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.


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Polynomials and Representations XXXV

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\}.


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}).


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.


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. ♦


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).


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\}.


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. ♦


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.

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Polynomials and Representations XXXIV


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)}


  • \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<j \le n}\mathbb{C} \cdot (e_i \wedge e_j).\end{aligned}

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):



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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}.


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.


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. ♦



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!)


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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).


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.


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}

warningWe 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.


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.


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.

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