## Return to Answer

4 misprint corrected

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 then than $r$.

3 added 832 characters in body

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 then $r$.

2 added 139 characters in body

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.

1

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.
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