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

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 $$V$$?

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?)

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.