Representation Theory – Isotypical Component of the Tensor Product of Irreducible Representations of $\mathrm{U}(n)$

Question

The following question is closely related to this one.

Let $\mathrm{U}(n)$ be the group of all (complex) unitary matrices $n\times n$. It is known that all irreducible representations of $\mathrm{U}(n)$ are parameterized by weights $\lambda=(\lambda_1,\ldots,\lambda_n)\in\mathbb{Z}^n$, where $\lambda_1\ge \lambda_1\ge\ldots\ge\lambda_n$; denote this representation by $V_{\lambda}$.
It also worths mentioning that if all $\lambda_i\ge 0$, then the corresponding irreducible representation is just $\mathbb{S}^{\lambda}(V)$, where $V\simeq\mathbb{C}^n$ is the standard $n$-dimensional representation. Similarly, if all $\lambda_i\le 0$, then $V_{\lambda}$ is isomorphic to $\mathbb{S}^{\mu}(V)^*$, where $\mu=(-\lambda_n,\ldots,-\lambda_1)$. Here $\mathbb{S}^{\mu}$ is the Schur functor associated to a partition $\mu$.

Now suppose that we have two decreasing sequences of integers: $\lambda=(\lambda_1,\ldots,\lambda_l)$ and $\mu=(\mu_1,\ldots,\mu_k)$, where $\lambda_1\ge\ldots\ge\lambda_l\ge 0\ge\mu_1\ge\ldots\ge\mu_k$ and $k+l\le n$. We are interested in $V_{(\lambda,\mu)}$-isotypic component of the tensor product $V_{\lambda}\otimes V_{\mu}$ which can be regarded as a submodule of $V^{\otimes|\lambda|}\otimes (V^*)^{\otimes|\mu|}$. Here $|\lambda|=\sum_{j=1}^{l}\lambda_j$, $|\mu|=\sum_{j=1}^{k}(-\mu_j)$ and weight $(\lambda,\mu)\in\mathbb{Z}^n$ is defined as
$$
(\lambda,\mu)=(\lambda_1,\ldots,\lambda_l,\underbrace{0,\ldots,0}_{n-k-l~\text{zeroes}},\mu_1,\ldots,\mu_k).
$$

I think that I saw somewhere the following statement (probably it was in The Classical Groups by H. Weyl but I am not sure and now I can't find this statement there).

Claim. The $V_{(\lambda,\mu)}$-isotypic component of $V_{\lambda}\otimes V_{\mu}$ is the intersection of kernels of all tensor contractions $\psi_{p,q}$, where $p=\overline{1,|\lambda|}$ and $q=\overline{1,|\mu|}$, that correspond to the $p$-th contravariant component and to the $q$-th covariant component.

Does anyone know whether this statement is correct or not? Of course I could have forgotten some details, so any similar facts (or corrections to the original statement) would be appreciated.

Update. It seems that a similar fact is claimed in A. Kirillov's Elements of the theory of representations (see 17.2. Representations of classical compact Lie groups) but the author also refers to H. Weyl's Classical groups without any additional details. Hence, any other references would be appreciated.

Answer 1

This is true. Let $r=|\lambda|,s=|\mu|$. The important observations are this:

• $V_{(\lambda,\mu)}$ appears with multiplicity 1. It's the "Cartan component" of the tensor product.
• $V_{(\lambda,\mu)}$ is not a summand of $V^{\otimes r-1}\otimes (V^*)^{\otimes s-1}$. This can be seen by some basic convexity results (or just the Pieri rule). Thus, it lies in the kernel of any map to this space.
• Any other summand of $V_{\lambda}\otimes V_{\mu}$ is a summand of $V^{\otimes r-1}\otimes (V^*)^{\otimes s-1}$. This is actually probably the trickiest part. There should be some nice combinatorial way to see this, but at the moment, the only way I can see is the Schur-Weyl duality with the walled Brauer algebra (See here, for example). By eq. (2) in that paper, any endomorphism of $V_{\lambda}\otimes V_{\mu}$ must be a scalar plus a homomorphism factoring through $V^{\otimes r-1}\otimes (V^*)^{\otimes s-1}$. Applying this to projection to any summand other than $V_{\lambda}\otimes V_{\mu}$ shows that summand is isomorphic to one of $V^{\otimes r-1}\otimes (V^*)^{\otimes s-1}$.
• Any homomorphism $ V_{\lambda}\otimes V_{\mu}\to V^{\otimes r-1}\otimes (V^*)^{\otimes s-1}$ factors through one of the vector contractions. Again, I would show this using the walled Brauer algebra (or maybe better, the oriented Brauer category of this paper), since $\mathrm{Hom}(V^{\otimes r}\otimes (V^*)^{\otimes s},V^{\otimes r-1}\otimes (V^*)^{\otimes s-1})$ is an easily understood bimodule of diagrams, and you just have to shorten one of the cups (in the usual graphical calculus, a contraction is a cup joining two inputs with nothing else happening).