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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?

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Concerning question (2), Shrawan Kumar assures me that there is a natural embedding here (implicit in the PRV paper). Thus works in any characteristic if you take the modules to be the spaces of global sections of suitable line bundles on the flag variety, which in prime characteristic are dual Weyl modules. This takes some care to write down, but a key point is that $V_\nu^* \otimes V_\nu \cong \text{ End}(V_\nu)$ has a unique $G$-invariant line. –  Jim Humphreys Apr 23 at 16:30

3 Answers 3

up vote 6 down vote accepted

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.

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Here is the answer to question (1). Let $V$ be a simple module; then the multiplicity of $V$ in $V_\lambda \otimes V^*_\mu$ equals the dimension of the space of vectors in $V$ of weight $\lambda -\mu$ and annihilated by $e_i^{\langle \mu,\alpha_i^\vee \rangle +1}$ (I believe that this formula is due to Parthasarathy, Ranga Rao and Varadarajan, see e.g. paper by Khare Theorem 3.1). This gives an answer to (1). I think that with little more work one should get a canonical map for (2).

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This reference to PRV's Theorem 2.1 eluded me at first, but it's definitely the most standard way to see that each irreducible summand of the smaller tensor product occurs at least as often in the larger one. It does take some work to translate things into their notation or Khare's. The proof doesn't seem to involve a "natural" or "canonical" embedding (however those terms are defined), but maybe there is a more conceptual method lurking in later work by Shrawan Kumar or others. –  Jim Humphreys Apr 13 at 13:35

EDITED (my arithmetic being too hasty):

I assume you are working over a field like $\mathbb{C}$ of characteristic 0, where it's equivalent to work with the Lie algebra (and in any case $G$ might as well be taken to be semisimple).

At first sight I don't see an immediate reason for your proposed inclusion to be correct, though such an inclusion would presumably be "natural" when it exists. Since the irreducible summands are uniquely determined in each tensor product, an inclusion would be the same as having the first list of summands included in the second list (multiplicity counted).

In any case it's helpful to clarify the role of dual representations here. First, recall that $V_\mu^*$ is itself irreducible of highest weight $\mu^*:= -w_\circ \mu$ (where $w_\circ$ is the longest element of the Weyl group). Moreover, on the right side you have $(\mu + \nu)^* = \mu^* + \nu^*$. Putting this together, your inclusion would be equivalent to a statement that $V_\lambda \otimes V_\mu$ is included in $V_{\lambda+\nu} \otimes V_{\mu + \nu^*}$ for any dominant weight $\nu$.

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I don't think that's quite accurate; it should be that $V_\lambda \otimes V_{\mu'}$ is included in $V_{\lambda + \nu} \otimes V_{\mu' + (-w_0 \nu)}$, if you are making the substitution $\mu' = -w_0 \mu$. Consider, for example, the case of $SL_3$ with $\lambda = \mu = 0, \nu = \omega$; clearly, whenever $\lambda = \mu = 0$, you get the natural inclusion of $\mathbb{C}$ into $V_\nu \otimes V_\nu^*$, but there is no inclusion of $\mathbb{C}$ into $V_\nu \otimes V_\nu$. –  user44191 Jul 11 at 17:33
My last paragraph is accurate, but I was simplifying the original formulation by replacing $\mu^*$ by $\mu$. Anyway, Victor's answer and the fuller version I got from Kumar extract the answers to (1) and (2) mainly from PRV. For (2) it's probably not strictly necessary to interpret the modules as global sections of line bundles, but for this version the best convention is to use $B^-$ rather than $B$. Algebraic group people like Andersen and Jantzen write things this way. Mixing $B, B^-$ gets confusing. –  Jim Humphreys Jul 12 at 15:00
Ah, sorry; I think I missed the tiny asterisk in the second subscript. –  user44191 Jul 12 at 15:19

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