Polynomials and Representations XXXIX

Some Invariant Theory

We continue the previous discussion. Recall that for \mu = \overline\lambda we have a GL_n\mathbb{C}-equivariant map

\displaystyle \bigotimes_j \text{Alt}^{\mu_j} \mathbb{C}^n \to \bigotimes_i \text{Sym}^{\lambda_i} \mathbb{C}^n \subset\mathbb{C}[z_{i,j}]^{(\lambda_i)}, \quad e_T^\circ \mapsto D_T

which induces an isomorphism between the unique copies of V(\lambda) in both spaces. The kernel Q of this map is spanned by e_T^\circ - \sum_S e_S^\circ for various fillings T with shape \lambda and entries in [n].

E.g. suppose \lambda = (4, 3, 1) and n=5; then \mu = (3, 2, 2,1) and the map induces:

\displaystyle \text{Alt}^3 \mathbb{C}^5 \otimes \text{Alt}^2 \mathbb{C}^5 \otimes \text{Alt}^2 \mathbb{C}^5\otimes \mathbb{C}^5 \longrightarrow \mathbb{C}[z_{1,j}]^{(4)} \otimes \mathbb{C}[z_{2,j}]^{(3)}\otimes \mathbb{C}[z_{3,j}]^{(1)}, \ 1\le j \le 5,

with kernel Q. For the following filling T, we have the correspondence:

example_determinant_and_alt

Lemma. If \mu_j = \mu_{j+1} = \ldots = \mu_{j+m-1}, then the above map factors through

\displaystyle \text{Alt}^{\mu_j}\mathbb{C}^n \otimes \ldots \otimes \text{Alt}^{\mu_{j+m-1}}\mathbb{C}^n\longrightarrow \text{Sym}^m \left(\text{Alt}^{\mu_j} \mathbb{C}^n\right).

E.g. in our example above, the map factors through \text{Alt}^3 \mathbb{C}^5 \otimes \text{Sym}^2 \left(\text{Alt}^2 \mathbb{C}^5 \right)\otimes \mathbb{C}^5.

Proof

Indeed if \mu_j = \mu_{j+1}, then swapping columns j and j+1 of T gives us the same D_T. ♦

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Now suppose 0\le a\le n and m\ge 0; pick \lambda = (m, \ldots, m) with length a. Now \mu comprises of m copies of a so by the above lemma, we have a map:

\displaystyle \text{Sym}^m \left(\text{Alt}^a \mathbb{C}^n\right) \longrightarrow \mathbb{C}[z_{1,j}]^{(m)} \otimes \ldots \otimes \mathbb{C}[z_{a,j}]^{(m)} \cong \mathbb{C}[z_{i,j}]^{(m,\ldots,m)}

where 1 \le i \le a and 1 \le j \le n. Taking the direct sum over all m we have:

\displaystyle \text{Sym}^* \left(\text{Alt}^a \mathbb{C}^n\right) := \bigoplus_{m\ge 0} \text{Sym}^m \left(\text{Alt}^a \mathbb{C}^n\right)\longrightarrow \bigoplus_{m\ge 0}\mathbb{C}[z_{i,j}]^{(m,\ldots,m)} \subset \mathbb{C}[z_{i,j}]

which is a homomorphism of GL_n\mathbb{C}-representations. Furthermore, \text{Sym}^* V for any vector space V has an algebra structure via \text{Sym}^m V \times \text{Sym}^n V \to \text{Sym}^{m+n}V. The above map clearly preserves multiplication since multiplying e_T^\circ e_{T'}^\circ and D_T D_{T'} both correspond to concatenation of T and T’. So it is also a homomorphism of \mathbb{C}-algebras.

Question. A basis of \text{Alt}^a\mathbb{C}^n is given by

\displaystyle e_{i_1, \ldots, i_a} := e_{i_1} \wedge \ldots \wedge e_{i_a}

for 1 \le i_1 < \ldots < i_a \le n. Hence \text{Sym}^*\left(\text{Alt}^a \mathbb{C}^n\right) \cong \mathbb{C}[e_{i_1,\ldots,i_a}], the ring of polynomials in n\choose a variables. What is the kernel P of the induced map:

\mathbb{C}[e_{i_1, \ldots, i_a}] \longrightarrow \mathbb{C}[z_{1,1}, \ldots, z_{a,n}]?

Answer

We have seen that for any 1 \le i_1 < \ldots < i_a \le n and 1 \le j_1 < \ldots < j_a \le n we have:

\displaystyle e_{i_1, \ldots, i_a} e_{j_1, \ldots, j_a} = \sum e_{i_1', \ldots, i_a'} e_{j_1', \ldots, j_k', j_{k+1},\ldots, j_a}

where we swap j_1, \ldots, j_k with various sets of k indices in i_1, \ldots, i_a while preserving the order to give (i_1', \ldots, i_a') and (j_1', \ldots, j_k'). Hence, P contains the ideal generated by all such quadratic relations.

On the other hand, any relation e_T = \sum_S e_S is a multiple of such a quadratic equation with a polynomial. This is clear by taking the two columns used in swapping; the remaining columns simply multiply the quadratic relation with a polynomial. Hence P is the ideal generated by these quadratic equations. ♦

Since the quotient of \mathbb{C}[e_{i_1, \ldots, i_a}] by P is a subring of \mathbb{C}[z_{i,j}], we have:

Corollary. The ideal generated by the above quadratic equations is prime.

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Fundamental Theorems of Invariant Theory

Recall that g = (g_{i,j}) \in GL_n\mathbb{C} takes z_{i,j} \mapsto \sum_k z_{i,k} g_{k,j}, which is a left action; here 1\le i\le a, 1\le j\le n. We also let GL_a\mathbb{C} act on the right via:

\displaystyle g=(g_{i,j}) \in GL_a\mathbb{C} :z_{i,j} \mapsto \sum_k g_{i,k}z_{k,j}

so that \mathbb{C}[z_{i,j}] becomes a (GL_n\mathbb{C}, GL_a\mathbb{C})-bimodule. A basic problem in invariant theory is to describe the ring \mathbb{C}[z_{i,j}]^{SL_a\mathbb{C}} comprising of all f such that g\cdot f = f for all g\in SL_a\mathbb{C}.

Theorem. The ring \mathbb{C}[z_{i,j}]^{SL_a\mathbb{C}} is the image of:

\displaystyle \mathbb{C}[D_{i_1, \ldots, i_a}] \cong \mathbb{C}[e_{i_1, \ldots, i_a}]/P \hookrightarrow \mathbb{C}[z_{i,j}]

where (i_1, \ldots, i_a) runs over all 1 \le i_1 < \ldots < i_a \le n.

In other words, we have:

  • First Fundamental Theorem : the ring of SL_a\mathbb{C}-invariants in \mathbb{C}[z_{i,j}] is generated by \{D_{i_1, \ldots, i_a}: 1 \le i_1 < \ldots < i_a \le n\}.
  • Second Fundamental Theorem : the relations satisfied by these polynomials are generated by the above quadratic relations.

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Proof of Fundamental Theorems

Note that g takes D_{i_1, \ldots, i_a} to:

\displaystyle \det{\small \begin{pmatrix} z_{1, i_1} & z_{1, i_2} & \ldots & z_{1, i_a} \\ z_{2, i_1} & z_{2, i_2} & \ldots & z_{2, i_a} \\ \vdots & \vdots & \ddots & \vdots \\ z_{a, i_1} & z_{a, i_2} & \ldots & z_{a, i_a}\end{pmatrix}} \mapsto \det {\small \begin{pmatrix}\sum_{j_1} g_{1,j_1} z_{j_1, i_1} & \ldots & \sum_{j_a} g_{1,j_a} z_{j_a, i_a} \\ \sum_{j_1} g_{2,j_1} z_{j_1, i_1} & \ldots & \sum_{j_a} g_{2,j_a} z_{j_a, i_a} \\ \vdots & \ddots & \vdots \\ \sum_{j_1} g_{a,j_1} z_{j_1, i_1} & \ldots & \sum_{j_a} g_{a,j_a} z_{j_a, i_a} \end{pmatrix}} = \det(g) D_{i_1, \ldots, i_a}

which is D_{i_1, \ldots, i_a} if g \in SL_a\mathbb{C}. Hence we have \mathbb{C}[D_{i_1, \ldots, i_a}] \subseteq \mathbb{C}[z_{i,j}]^{SL_a\mathbb{C}}. To prove equality, we show that their dimensions in degree d agree. By the previous article, the degree-d component of \mathbb{C}[e_{i_1, \ldots, i_a}]/P has a basis indexed by SSYT of type \lambda = (\frac d a, \ldots, \frac d a) and entries in [n]; if d is not a multiple of a, the component is 0.

Next we check the degree-d component of \mathbb{C}[z_{i,j}]^{SL_a\mathbb{C} }. As GL_a\mathbb{C}-representations, we have

\displaystyle \mathbb{C}[z_{i,j}] \cong \text{Sym}^*\left( (\mathbb{C}^a)^{\oplus n}\right)

where GL_a\mathbb{C} acts on \mathbb{C}^a canonically. Taking the degree-d component, once again this component is 0 if d is not a multiple of a. If a|d, it is the direct sum of \text{Sym}^{d_1} \mathbb{C}^a \otimes \ldots \otimes \text{Sym}^{d_n}\mathbb{C}^a over all d_1 + \ldots + d_n = \frac d a. The \psi of this submodule is h_{\lambda'} = \sum_{\mu'\vdash \frac d a} K_{\mu'\lambda'} s_{\mu'} where \lambda' is the sequence (d_1, \ldots, d_n). Hence

\displaystyle \text{Sym}^{d_1} \mathbb{C}^a \otimes \ldots \otimes \text{Sym}^{d_n}\mathbb{C}^a \cong \bigoplus_{\mu'\vdash \frac d a} V(\mu')^{\oplus K_{\mu'\lambda'}}.

Fix \mu' and sum over all (d_1, \ldots, d_n); we see that the number of copies of V(\mu') is the number of SSYT with shape \mu' and entries in [n]. The key observation is that each V(\mu') is an SL_a\mathbb{C}-irrep.

  • Indeed, \mathbb{C}^* \subset GL_a\mathbb{C} acts as a constant scalar on the whole of V(\mu') since \psi_{V(\mu')} = s_{\mu'} is homogeneous. Hence any SL_a\mathbb{C}-invariant subspace of V(\mu') is also GL_a\mathbb{C}-invariant.

Hence V(\mu')^{SL_a\mathbb{C}} is either the whole space or 0. From the proposition here, it is the whole space if and only if \mu' = (\frac d a, \ldots, \frac d a) with a terms (which corresponds to \det^{d/a}). Hence, the required dimension is the number of SSYT with shape (\frac d a, \ldots) and entries in [n]. ♦

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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.

determinant_products_and_filling

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}].

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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:

basis_element_of_alt_tensor_module

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'}. ♦

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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}

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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.

modulo_relations_for_specht

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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:

c1_multilinear

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)).

c2_alternating

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<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:

c3_column_swaps

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. ♦

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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.

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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).

basis_element_of_schur_module

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}.

comparison_of_two_fillings

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.

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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:

young_diagram_label_1_to_n_v2

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:

schur_functor_vector_space_constr

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).

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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.

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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. ♦

Note

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\}.

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}. ♦

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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.

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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.

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Polynomials and Representations XXXIV

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.

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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:

partition_of_dual_representation

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).

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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:

gl2_example_ssyt

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.

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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:

triangular_simplex

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:

simplex_lambda_and_mu

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

geometric_kostka_v2

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

geometric_kostka_v1

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):

weight_spaces_and_ssyt

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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}.

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. ♦

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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\}.

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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. ♦

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Summary

We have a correspondence between:

representation_gln_corr

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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