Tensoring $\frak{g}$-modules by fundamental representations

Question

Given a fundamental representation $V(\nu_k)$ of a semisimple Lie algebra $\frak{g}$, and a general irreducible finite-dimensional representation $V$, is it ever possible that the tensor product $V \otimes V(\nu_k)$ can contain a copy of $V$?

Answer 1

Yes. First of all, if we take a fixed irrep $ V(\mu) $, then for big enough $ \lambda$, we know that $ V(\lambda + \nu) $ will appear in $ V(\lambda) \otimes V(\mu) $ with multiplicity equal to the weight multiplicity of $ \nu $ in $V(\mu) $.

So your question is equivalent to: is $ 0 $ a weight of a fundamental representation? For $ \mathfrak{sl}_n $, this is not possible, but for other simple Lie algebras, this is possible. (Is it possible for every simple Lie algebra?)

Answer 2

Joel Kamnitzer explained that $\omega_k$ is such a fundamental representation iff $0$ is a weight of the corresponding representation. In this answer I want to explain that such a weight exists for an irreducible root system exactly when the root system in not of Type A.

Note that the highest short root $\widehat{\theta}$ is the minimal (in root order) dominant weight greater than $0$ in root order. In particular, $0$ is a weight of $V(\widehat{\theta})$. (This representation is called a "quasi-minuscule representation"; see https://en.wikipedia.org/wiki/Minuscule_representation.) Thus, if the highest short root happens to coincide with a fundamental weight $\omega_k$, then it will satisfy your requirement that there is some $V$ for which $V$ belongs to $V\otimes V(\omega_k)$.

I claim that for every irreducible root system other than Type A, we indeed have that $\widehat{\theta}$ is a fundamental weight.

For simply laced root systems, $\widehat{\theta}$ is the same as the highest root $\theta$, and there is a nice combinatorial rule to write the coefficients of $\theta$ in the basis of fundamental weights: writing $\theta = \sum_{i=1}^{n}c_i\omega_i$, the coefficient $c_i$ is the number of edges between the node numbered $i$ and the special affine node in the extended Dynkin diagram. (This is in Bourbaki somewhere.) Then we can just check the extended Dynkin diagrams (https://en.wikipedia.org/wiki/Dynkin_diagram#Affine_Dynkin_diagrams) and see that the simply laced extended Dynkin diagrams have a single edge adjacent to their affine nodes, except in Type A.

I think there's an extension of this combinatorial rule to non-simply laced cases and for the highest short root using some variant of extended Dynkin diagrams, but I don't remember exactly. So let me just say that it's not hard to check, using say the standard realizations, that for $B_n$ the highest short root is $\omega_1$, and for $C_n$ the highest short root is $\omega_2$. And similarly it can be checked that for $G_2$ the highest short root is $\omega_1$, and for $F_4$ the highest short root is $\omega_4$.

A table of these quasi-minuscule weights appears in https://books.google.com/books?id=Np7y-LVcwSwC&pg=PA221#v=onepage&q&f=false.