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Let $V$ be a vector space of dimension $n$. Let $S^k V$ be a representation of $GL(n)$. I would like to know if there exists some characterization of finite dimensional $GL(n)$ modules $V_1,V_2$ such that exists $GL(n)$ mapping $T:S^k V \rightarrow V_1\otimes V_2$ such that for some $x$ matrix $T(x)$ is invertible .

Added: Just to clarify: The mapping $T$ itself may not be invertible I am asking that $T(x)$ will be invertible in $V_1\otimes V_2$. For example the mapping $S^2 V \rightarrow V\otimes V$ is not invariable, but $T(x)$ is invartable for most $x$'s.

I am interested for a base field $\mathbb{C}$. And of course this may happen only in case when $S^kV$ is a sub-representation of $V_1 \otimes V_2$. In fact I do not know for which irreps $V_1, V_2$ representation $V_1 \otimes V_2$ have $S^kV$ as a component. I will be happy if you can give me a reference for this question.

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If you work over a field of characteristic $0$, then the category of polynomial representations is semisimple (I assume that you are talking about polynomial representations), and there exists a non-zero $T$ if and only if $V_1\otimes V_2$ has a simple summand isomorphic to a simple summand of $S^kV$. Now such a $T$ will never take an invertible value unless $S^kV$ is isomorphic to $V_1\otimes V_2$, so really, you are asking whether $S^kV$ can be isomorphic to a tensor product of two representations. –  Amritanshu Prasad Jan 11 '13 at 6:31
    
Dear Prasad, Thanks for your comment, but I am not asking for $T$, but for $T(x)$ to be invertible in $V_1\otimes V_2$. –  Klim Efremenko Jan 11 '13 at 9:11
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@Klim: what does it mean for an element of $V_1\otimes V_2$ to be invertible? Are $V_1$ and $V_2$ of the same dimension? What do you mean by invariable in the example $S^2V\to V\otimes V$? –  Dragos Fratila Jan 11 '13 at 9:32
    
@Dragos "invariable" should be a misprint for invertible. –  Anton Fonarev Jan 11 '13 at 10:59
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1 Answer 1

up vote 5 down vote accepted

Regarding the question of pairs, here is the answer (hope, I'm not mistaken).

Let me first change the notation a little bit. Suppose, we are looking for irreps $V_1=\Sigma^{\lambda}V$ and $V_2=\Sigma^{\mu}V$ where $\lambda = (\lambda_1,\ldots,\lambda_n)$ and $\mu = (\mu_1,\ldots,\mu_n)$ resp., such that there exists an embedding $S^kV\to V_1^*\otimes V_2$.

It's immediate that such an embedding exists iff $V_2=\Sigma^{\mu}V$ is an irreducible factor in $V_1\otimes S^kV=\Sigma^{\lambda}V\otimes S^kV$. The Littlewood-Richardson rule gives us the answer. Firstly, $k=|\mu|-|\lambda|$, where $|\lambda| = \sum_i\lambda_i$. Secondly, one has the following inequalities:

  • $\mu_1\geq \lambda_1$,
  • $\lambda_1\geq \mu_2 \geq \lambda_2$,
  • $\ldots$
  • $\lambda_{n-1}\geq \mu_n \geq \lambda_n$.

Finally, if such an embedding exists, it's unique by the very same L-R rule.

UPD: Forgot to mention that the main question still seems to be very hard as even checking that $\dim V_1=\dim V_2$ is a huge problem.

UPD 2: Let me answer your main question (despite UPD). Let's forget about representation theory and do some linear algebra. What we have is a surjective map $V_1\otimes U\to V_2$, where $\dim V_1 = \dim V_2$. Let the corresponding bilinear map be $h:V_1\times U\to V_2$. We want to find such $u\in U$ that $h_u = h(-, u)$ is invertible, equivalently, of maximal rank.

Well, this seems to be quite obvious: let $r$ be the maximal rank of $h_u$ among all $u\in U$. Suppose $r<\dim V_2$. Let $u_0\in U$ be such that $\mathrm{rk}\ h_{u_0}=r$. Take any $v\in V_2\setminus\mathrm{Im}\ h_{u_0}$. Then there exists such $u_1\in U$ that $v\in \mathrm{Im}\ h_{u_1}$. Now it's easy to see that the rank of a general linear combination $\alpha h_{u_0} + \beta h_{u_1} = h_{\alpha u_0+\beta u_1}$ will be greater than $r$.

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Thanks for answer. Why do you say that $k=|\mu|-|\lambda|$? For example take $k$ even then definitely $S^k$ is subreps of $S^{k/2}\otimes S^{k/2}$ –  Klim Efremenko Jan 11 '13 at 13:20
    
@Klim Due to dualization, in my notation $V_1=S^{k/2}V^*$. In particular, $\lambda = (0,\ldots,0,-k/2)$. –  Anton Fonarev Jan 11 '13 at 13:34
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