Classical invariants involving exterior powers of standard representation - MathOverflow

Classical invariants involving exterior powers of standard representation

2011-08-29T16:23:12Z

While investigating certain conformal blocks line bundles on $\overline{M}_{0,n}$, I was led to what seems to be an identification between two spaces of invariants, and I am curious if there is a direct way to see this identification.

Statement: for any integers $n\ge 4$ and $r\ge 2$, and any integers $i_1,\ldots,i_n$ such that $1 \le i_j \le r-1$ and $r=\frac{1}{2}\sum_{j=1}^n i_j$ , I believe there is a vector space isomorphism \[(\wedge^{i_1}\mathbb{C}^r\otimes\cdots\otimes \wedge^{i_n}\mathbb{C}^r)^{SL(r)} \cong (S^{i_1}\mathbb{C}^2\otimes\cdots\otimes S^{i_n}\mathbb{C}^2)^{SL(2)},\] where $\mathbb{C}^m$ denotes the standard representation of $SL(m)$. The invariants on the RHS are classical and well-known: a basis is given by all $2\times r$ semi-standard tableaux with entries in ${1,\ldots,n}$ such that $j$ occurs exactly $i_j$ times. I wonder if the invariants on the LHS are also known, and if there's a conceptual reason why they might be in bijection with those on the RHS.

Background: This is not relevant for the question itself, but I am including it in case you are curious how this purported identity arose. It seems likely that for conformal blocks bundles on $\overline{M}_{0,n}$ of level 1 and Lie algebra $\mathfrak{sl}(r)$, the global sections are naturally identified with a space of covariants. Specifically, the conformal blocks line bundle with weights $(\omega_{i_1},\ldots,\omega_{i_n})$ , where $\omega_i$ are fundamental weights, should have global sections $(\wedge^{i_1}\mathbb{C}^r\otimes\cdots\otimes\wedge^{i_n}\mathbb{C}^r)_{\mathfrak{sl}(r)}$ , since the irreducible representation associated to $\omega_i$ is $\wedge^i\mathbb{C}^r$. This space of $\mathfrak{sl}(r)$-covariants is isomorphic to the corresponding space of $\mathfrak{sl}(r)$-invariants, which in turn is the same as the space of $SL(r)$-invariants for this representation. On the other hand, it is known (by a result of Fakhruddin) that when $\sum_{j=1}^n i_j = 2r$ then this conformal blocks line bundle induces the GIT morphism $\overline{M}_{0,n} \rightarrow (\mathbb{P}^1)^n//_{(i_1,\ldots,i_n)}SL(2)$ , so we know that its space of global sections is $H^0((\mathbb{P}^1)^n,\mathcal{O}(i_1,\ldots,i_n))^{SL(2)}$.

Answer by Sasha for Classical invariants involving exterior powers of standard representation

2011-08-29T20:00:42Z

The dimension of the spaces of invariants are given by the number of semistandard Young tableaux of a specific form. If one applies the Littlewood-Richardson rule for tensoring wedge powers, one sees that for the first space one needs to count the tableaux of the rectangle form with $r$ rows and $2$ columns which a filled with $i_1$ of $1$'s, $i_2$ of $2$'s, ..., $i_n$ of $n$'s such that the numbers increase strictly in rows and non-strictly in columns. And if one applies the Littlewood-Richardson rule for tensoring symmetric products, one sees that For the second space one has to count the tableaux of the rectangle form with $2$ rows and $r$ columns which a filled with $i_1$ of $1$'s, $i_2$ of $2$'s, ..., $i_n$ of $n$'s such that the numbers increase strictly in columns and non-strictly in rows. Now you can see that the transposition turns the tableaux of the first type into the tableaux of the second type. Which give the equality you want.

Answer by David Speyer for Classical invariants involving exterior powers of standard representation

2011-08-30T00:54:08Z

To follow up on Sasha's answer, yes there is a natural isomorphism of vector spaces which lifts the combinatorial equality. All isomorphisms in this answer will be natural.

Schur-Weyl duality:

Let $\lambda$ be a partion; set $d = |\lambda|$ and let $m$ be greater than or equal to the number of parts of $\lambda$. Let $V$ be a vector space of dimension $m$, let $V_{\lambda}$ be the irrep of $GL(V)$ with highest weight $\lambda$ an let $M_{\lambda}$ be the $S_d$-irrep (aka Specht module) indexed by $\lambda$. Schur-Weyl duality is the isomorphism of $GL(V)$ representations: $$\mathrm{Hom}_{S_d}(M_{\lambda}, V^{\otimes d}) \cong V_{\lambda}.$$ I'll take this to be the definition of $V_{\lambda}$.

Tensor product and restriction:

Let $\mu$, $\lambda_1$, ..., $\lambda_k$ be partitions, with $|\mu| = \sum |\lambda_i|$. Let $m$ be greater than or equal to the number of parts of $\mu$. Set $d_i = |\lambda_i|$. Then the above shows that $$\begin{array}{rl} \mathrm{Hom}_{GL(V)}\left( V_{\mu}, V_{\lambda_1} \otimes \cdots \otimes V_{\lambda_k} \right) &\cong \mathrm{Hom}_{GL(V)} \left( \mathrm{Hom}_{S_{\sum d_i}} \left( M_{\mu}, V^{\otimes \sum d_i} \right), \mathrm{Hom}_{S_{d_1} \times \cdots S_{d_k}}\left( M_{\lambda_1} \boxtimes \cdots \boxtimes M_{\lambda_k}, V^{\otimes \sum d_i} \right) \right) \\ & \cong \mathrm{Hom}_{S_{d_1} \times \cdots \times S_{d_k}} \left( M_{\lambda_1} \boxtimes \cdots \boxtimes M_{\lambda_k}, \left( M_{\mu} \right)|_{S_{d_1} \times \cdots \times S_{d_k}} \right) \end{array}.$$ In other words, every Hom in the second expression is induced from composition with one of the Hom's in the third expression. Here, if $U$ and $V$ are $G$ and $H$-representations, then $U \boxtimes V$ denotes $U \otimes V$ with $G$ and $H$ acting on the first and second factors respectively.

Relation to transpose:

Now let $\lambda_i$ and $\mu$ be as above. Let $\epsilon(d)$ be the sign representation of $S_d$, and let $\lambda^T$ be the transpose of $\lambda$. Then $M_{\lambda} \cong M_{\lambda^T} \otimes \epsilon(|\lambda|)$. How to make this natural depends on exactly how you define $M_{\lambda}$. For example, if you use the Vershik-Okounkov approach, the bases they construct for $M_{\lambda}$ and $M_{\lambda^T}$ correspond to each other.

Now, $\epsilon(d_1) \boxtimes \cdots \boxtimes \epsilon(d_k) \cong \epsilon(\sum d_i)|_{S_{d_1} \times \cdots \times S_{d_k}}$, and is irreducible. So we deduce that $$\mathrm{Hom}_{S_{d_1} \times \cdots \times S_{d_k}} \left( M_{\lambda_1} \boxtimes \cdots \boxtimes M_{\lambda_k}, \left( M_{\mu} \right)|_{S_{d_1} \times \cdots \times S_{d_k}} \right) \cong \mathrm{Hom}_{S_{d_1} \times \cdots \times S_{d_k}} \left( M_{\lambda_1}^T \boxtimes \cdots \boxtimes M_{\lambda_k}^T, \left( M_{\mu}^T \right)|_{S_{d_1} \times \cdots \times S_{d_k}} \right)$$ and this isomorphism is natural if we define $M_{\lambda} \cong M_{\lambda^T} \otimes \epsilon(|\lambda|)$ in a natural way.

Putting it all together, if $\dim V$ is greater than or equal to the number of parts in $\mu$, and $\dim W$ is greater than or equal to the number of parts in $\mu^T$, then we have a natural isomorphism: $$\mathrm{Hom}_{GL(V)}\left( V_{\mu}, V_{\lambda_1} \otimes \cdots \otimes V_{\lambda_k} \right) \cong \mathrm{Hom}_{GL(W)}\left( W_{\mu^T}, W_{\lambda^T_1} \otimes \cdots \otimes W_{\lambda^T_k} \right).$$

Your question:

Take $\lambda_j = 1^{i_j}$ and $\mu = 2^r$. Let $\dim V = r$ and $\dim W=2$.

Since $M_{\lambda_j}$ is the sign rep, we have $V_{\lambda_j} \cong \bigwedge^{i_j} V$ naturally. Similarly, since $M_{\lambda_j^T}$ is the trivial rep, we have $W_{\lambda_j^T} \cong \mathrm{Sym}^{i_j}(W)$.

Finally, one needs to use the fact that $V_{\mu}$ and $W_{\mu^T}$ are trivial one-dimensional $SL(V)$ and $SL(W)$ reps. (More precisely, they are $\det^2$ and $\det^r$, as $GL$-reps.) I'm not sure what the easiest proof of this is, but it can only introduce a scalar factor to your isomorphisms.