## Classical invariants involving exterior powers of standard representation

### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

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

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: `$$\mathrm{Hom}_{GL(V)}\left(  V_{\mu}, V_{\lambda_1} \otimes \cdots \otimes V_{\lambda_k} \right) \cong \mathrm{Hom}_{GL(W)}\left(  W_{\mu^T}, W_{\lambda^T_1} \otimes \cdots \otimes W_{\lambda^T_k} \right).$$`

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.

-

@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... – Abdelmalek Abdesselam Sep 1 2011 at 20:00
...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}. – Abdelmalek Abdesselam Sep 1 2011 at 20:01

### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

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.

-

Quite clean and simple, excellent! – Noah Giansiracusa Aug 30 2011 at 8:02
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 91, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9179966449737549, "perplexity_flag": "head"}