## Notations and Recollections

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

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

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

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

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

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

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

## For Representations of GLn

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

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

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

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

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

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

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

This is clear.

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

Proof

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

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

Finally, we have:

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

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

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

Proof

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

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

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

## Universality

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

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

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

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

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

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

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

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

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

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

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

You can probably guess what’s coming next.

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

## Proof of Main Theorem

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

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

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

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

Proof

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

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

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

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

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

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

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

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