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 $H^0((\mathbb{P}^1)^n,\mathcal{O}(i_1,\ldots,i_n))^{SL(2)}$.


2 Answers 2


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{align*} &\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 \times 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{align*} 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.

  • $\begingroup$ @David: great post. Do you see a way of doing this by hand? One can use the FFT of invariant theory to describe a linearly generating set for the SL(V) side as follows: take 2r column vectors v_1,...,v_2r in C^r. For a partition of 1,...,2r in two sets of r indices take the product of the two r by r determinants corresponding to this partition. Then antisymmetrize the blocks {1,...,i1} etc. A subset corresponds to the 2 by r SSYT. The rows give the selection of indices for each of the two determinants. Can one prove by hand using Grassmann-Plucker that the subset is a basis. I wonder also... $\endgroup$ Sep 1, 2011 at 20:00
  • $\begingroup$ ...if one can make a direct relation to the binary domain (the SL2 side) by specializing to v's of the form 1,x,...,x^{r-1}. $\endgroup$ Sep 1, 2011 at 20:01

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.

  • $\begingroup$ Quite clean and simple, excellent! $\endgroup$ Aug 30, 2011 at 8:02

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