Question on irreducible representation of tensor products

Question

Question:

Suppose that $V_1$ and $V_2$ are two finite dimensional representations of lie algebra $\mathfrak{g}$ generated by highest weight vectors $v_1^*$ and $v_2^*$ of weights $\mu_1$ and $\mu_2$ respectively.

Now, suppose that $V_3$ is an irreducible representation of $V_1 \otimes V_2$ such that the highest weight of $V_3$ has weight $\mu$. I want to show that $\mu = \mu_1 + \lambda $ for some weight $\lambda$ of $V_2$.

Approach: I know that the weight spaces of the tensor $V_1 \otimes V_2$ are spanned by tensor products of weight vectors of $V_1$ and $V_2$. Then $v_1 \otimes v_2$ is a highest weight vector of $V_1 \otimes V_2$ of highest weight $\mu_1+ \mu_2$. If $w$ is a highest weight vector of $V_3$, then it is a sum of the form $v_{1,k} \otimes v_{2,k}$ with $v_{1,k}$ of weight $\lambda_{1,k}$ and $v_{2,k}$ of weight $\lambda_{2,k}$ such that $\lambda_{1,k} + \lambda_{1,k} =\mu.$

The goal is to show that for one of these tensors $\lambda_{1,k} = \mu_1.$ Considering a simpler case when $w = v_1 \otimes v_2$ with $v_1$ of weight $\lambda_1$, $v_2$ of weight $\lambda_2$ and $\lambda_1 +\lambda_2 = \mu$. Suppose that $\lambda_1 \neq \mu_1$, how can I get a contradiction here? I believe if this part is cleared, we can then next suppose that $w = \sum_{k}v_{1,k} \otimes v_{2,k}$ and also that $\lambda_{1,k} \neq \mu_1$ for all $k$ ( the sum being over some indexing set) and obtain a contradiction ( I believe this approach answers this question).

P.S: I apologize this is not a research question but I think its an interesting question. I have been struggling to come up with a proof and that is why I am seeking help. I will be glad if anyone interested in this question to please provide me a proof/explanation to this question. Thanks

Answer 1

The question is definitely not trivial, but is not currently at research level because the answer has been known for a long time. A good modern source available online here is the 2010 ICM write-up by Shrawan Kumar. He attributes the result to the late Bert Kostant's seminal (though in retrospect overcomplicated) 1959 paper, with free access online here, deriving a closed formula for weight multiplicities in irreducible finite dimensional representations. See Proposition (3.2) in Kumar's recent survey, where there is also a natural further refinement giving an upper bound on the multiplicity of each irreducible summand of such a tensor product.

[By the way, I included a version of this result as Exercise 24.12 in my 1972 graduate text on Lie algebras, though for some reason I can't recall now the exercise stated that it should be deduced from Steinberg's tensor product formula. That seems misleading.]

Answer 2

It is probably complete overkill but for $G=GL(n)$ your assertion follows immediately from the Littlewood-Richardson rule while for general $G$ one can use Littelmann's path model.

Recall that Littelmann represents every weight vector of an irreducible representation by a piecewise linear path $p(t)$ starting in $0$ and ending in a weight. There is exactly one path staying in the dominant Weyl chamber and the endpoint of that path is the highest weight.

The weights of the tensor product $U:=V_1\otimes V_2$ of two representations are then represented by the concatentions of a path for $V_1$ with a path for $V_2$. Now Littelmann shows that the highest weights $\mu$ of $U$ correspond to those concatenations which stay entirely in the dominant Weyl chamber. But then the first part has to be the unique path connecting $0$ with $\mu_1$ while the second connects $0$ with some weight $\lambda$ of $V_2$. So the concatenated path connects $0$ with $\mu=\mu_1+\lambda$. qed

Answer 3

Your question is about the vectors in $V_1\otimes V_2$ that provide $1$-dimensional $B_\Delta$-subrepresentations of weight $\mu$, where $B_\Delta$ is the diagonal in $B\times B \leq G\times G$. Let's call this subspace $J$.

Let $S \hookrightarrow T$ be a regular dominant coweight, and let $B_z := (S(z)\times 1)\ B_\Delta\ (S(z^{-1})\times 1)$. Then $B_1 = B_\Delta$, and $\lim_{z\to\infty} B_z = (B' \times 1) T_\Delta$, where $B'\leq B$ is the unipotent radical; call this limit subgroup $B_\infty$. We can also consider $J_z := (S(z)\times 1)\cdot J$, and its limit $J_\infty$ (computed inside the Grassmannian of $(\dim J)$-dimensional subspaces of $V_1\otimes V_2$).

Now I claim that $J_z$ is $B_z$-invariant for each $z$, and hence, $J_\infty$ is invariant under $(B'\times 1) T_\Delta$. Your assumption was that $J$ was not $0$; hence there are vectors of $(B'\times 1) T_\Delta$-weight $\mu$. Call that potentially larger space $J'$.

Unlike $B_\Delta$, the limit group $B_\infty$ is normalized by $T\times T$, which will act on $J'$ and break it into weight spaces, i.e. weight vectors of $V_1$ tensor weight vectors of $V_2$. The $(B'\times 1)$-invariance means that the only $V_1$ weight space met this way is the high weight space.

Summing up, the assumption that $\mu$ is a high weight in $V_1 \otimes V_2$ implies that $\mu$ can be written as the high weight of $V_1$ plus a weight of $V_2$.

One way to rephrase your question was that you were looking for asymmetry. I get it here with this $S(z)\times 1$, whereas Friedrich Knop's crystal-based answer implicitly derives it from the quantum group (for which $V_1\otimes V_2$ is not obviously isomorphic to $V_2\otimes V_1$, especially as $q\to 0$).