Let $G$ be a reductive group and $\lambda$, $\mu$ and $\nu$ be dominant weights of $G$. Denote by $V_\lambda$ the irreducible representation of $G$ of highest weight $\lambda$. It seems to be true that
$$
V_\lambda \otimes V_\mu^* \subseteq V_{\lambda+\nu}\otimes V_{\mu+\nu}^*,
$$
where $W^*$ as usual is the dual representation for $W$.
This is obvious in the case $\lambda=0$ or $\mu=0$.
Here are two questions:
(1) about multiplicities, i.e. how to prove that for any $\theta$ the multiplicity of $V_\theta$ in right hand side is more or equal than in the left hand side?
(2) is there a natural inclusion?
Best Answer
There is a natural inclusion, defined as follows:
The space of sections of weight $\mu$ on $\mathcal{F} = G/B$ can naturally be identified with $V_\mu^*$ (via the map $w \rightarrow [g \rightarrow \langle w, g v_\mu \rangle ]$ where $v_\mu$ is a highest-weight vector). Similarly, if we consider $\mathcal{F} = G/B^-$, we can identify $V_\lambda$ with the space of sections of weight $-\lambda$ (which is dominant with respect to $B^-$). We can therefore identify $V_\lambda \otimes V_\mu^*$ with sections on $\mathcal{F} \times \mathcal{F}$.
Consider the $G$-invariant vector $\iota_\nu \in V_\nu \otimes V_\nu^*$ (we can choose this naturally by choosing the identity on $V_\nu$); by the above, we can identify it with a section on $\mathcal{F} \times \mathcal{F}$. Then the inclusion of $V_\lambda \otimes V_\mu^*$ into $V_{\lambda + \nu} \otimes V_{\mu + \nu}^*$ is given by taking the product of the relevant section with $\iota_\nu$; because $\mathcal{F} \times \mathcal{F}$ is irreducible, there are no zero-divisors, which makes this an inclusion. It is $G$-equivariant because $\iota_\nu$ is $G$-invariant and the product map is $G \times G$-equivariant, and so $G$-diagonally equivariant.