Classical invariants involving exterior powers of standard representation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T16:40:52Z http://mathoverflow.net/feeds/question/73984 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73984/classical-invariants-involving-exterior-powers-of-standard-representation Classical invariants involving exterior powers of standard representation Noah Giansiracusa 2011-08-29T16:23:12Z 2011-08-30T00:54:08Z <p>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.</p> <p><strong>Statement</strong>: for any integers $n\ge 4$ and $r\ge 2$, and any integers <code>$i_1,\ldots,i_n$</code> such that <code>$1 \le i_j \le r-1$</code> and <code>$r=\frac{1}{2}\sum_{j=1}^n i_j$</code>, I believe there is a vector space isomorphism <code>$(\wedge^{i_1}\mathbb{C}^r\otimes\cdots\otimes \wedge^{i_n}\mathbb{C}^r)^{SL(r)} \cong (S^{i_1}\mathbb{C}^2\otimes\cdots\otimes S^{i_n}\mathbb{C}^2)^{SL(2)},$</code> 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 <code>$i_j$</code> 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. </p> <p><strong>Background</strong>: 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 <code>$\overline{M}_{0,n}$</code> 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 <code>$(\omega_{i_1},\ldots,\omega_{i_n})$</code>, where <code>$\omega_i$</code> are fundamental weights, should have global sections <code>$(\wedge^{i_1}\mathbb{C}^r\otimes\cdots\otimes\wedge^{i_n}\mathbb{C}^r)_{\mathfrak{sl}(r)}$</code>, 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 <code>$\sum_{j=1}^n i_j = 2r$</code> then this conformal blocks line bundle induces the GIT morphism <code>$\overline{M}_{0,n} \rightarrow (\mathbb{P}^1)^n//_{(i_1,\ldots,i_n)}SL(2)$</code>, so we know that its space of global sections is $H^0((\mathbb{P}^1)^n,\mathcal{O}(i_1,\ldots,i_n))^{SL(2)}$.</p> http://mathoverflow.net/questions/73984/classical-invariants-involving-exterior-powers-of-standard-representation/74002#74002 Answer by Sasha for Classical invariants involving exterior powers of standard representation Sasha 2011-08-29T20:00:42Z 2011-08-29T20:00:42Z <p>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.</p> http://mathoverflow.net/questions/73984/classical-invariants-involving-exterior-powers-of-standard-representation/74024#74024 Answer by David Speyer for Classical invariants involving exterior powers of standard representation David Speyer 2011-08-30T00:54:08Z 2011-08-30T00:54:08Z <p>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.</p> <p><strong>Schur-Weyl duality:</strong> </p> <p>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 $\lambda$. <a href="http://en.wikipedia.org/wiki/Schur%25E2%2580%2593Weyl_duality" rel="nofollow">Schur-Weyl duality</a> is the isomorphism of $GL(V)$ representations: <code>$$\mathrm{Hom}_{S_d}(M_{\lambda}, V^{\otimes d}) \cong V_{\lambda}.$$</code> I'll take this to be the definition of $V_{\lambda}$.</p> <p><strong>Tensor product and restriction:</strong> </p> <p>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 <code>$$\begin{array}{rl} \mathrm{Hom}_{GL(V)}\left( V_{\mu}, V_{\lambda_1} \otimes \cdots \otimes V_{\lambda_k} \right) &amp;\cong \mathrm{Hom}_{GL(V)} \left( \mathrm{Hom}_{S_{\sum d_i}} \left( M_{\mu}, V^{\otimes \sum d_i} \right), \mathrm{Hom}_{S_{d_1} \times \cdots S_{d_k}}\left( M_{\lambda_1} \boxtimes \cdots \boxtimes M_{\lambda_k}, V^{\otimes \sum d_i} \right) \right) \\ &amp; \cong \mathrm{Hom}_{S_{d_1} \times \cdots \times S_{d_k}} \left( M_{\lambda_1} \boxtimes \cdots \boxtimes M_{\lambda_k}, \left( M_{\mu} \right)|_{S_{d_1} \times \cdots \times S_{d_k}} \right) \end{array}.$$</code> 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.</p> <p><strong>Relation to transpose:</strong> <p>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 <a href="http://arxiv.org/abs/math.RT/0503040" rel="nofollow">Vershik-Okounkov</a> approach, the bases they construct for $M_{\lambda}$ and $M_{\lambda^T}$ correspond to each other.</p></p> <p>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.</p> <p>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: <code>$$\mathrm{Hom}_{GL(V)}\left( V_{\mu}, V_{\lambda_1} \otimes \cdots \otimes V_{\lambda_k} \right) \cong \mathrm{Hom}_{GL(W)}\left( W_{\mu^T}, W_{\lambda^T_1} \otimes \cdots \otimes W_{\lambda^T_k} \right).$$</code></p> <p><strong>Your question:</strong></p> <p>Take $\lambda_j = 1^{i_j}$ and $\mu = 2^r$. Let $\dim V = r$ and $\dim W=2$. </p> <p>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 $W_{\lambda_j^T} \cong \mathrm{Sym}^{i_j}(W)$.</p> <p>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.</p>