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Let $\mathfrak{g}$ be a complex semisimple Lie algebra.

Suppose $M,N,L$ are $\mathfrak{g}$-modules, $N$ is a $\mathfrak{g}$-submodule of $M$.

Does this implies $\text{Hom}_{\mathfrak{g}}(N,L)\le \text{Hom}_{\mathfrak{g}}(M,L)$ as a vector subspace?

If yes, how to prove it?

James Cheung
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1 Answers1

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Firstly, since the functor Hom is contravariant in the first variable, what you expect is a map going the other way. For a general abelian category, the only situation in which you expect to have an inclusion $$\mathrm{Hom}(A,B) \subseteq \mathrm{Hom}(C,B)$$ is when $A$ is a quotient of $C$. For a semi-simple Lie algebra $\mathfrak{g}$ over $\mathbf{C}$, the category of finite-dimensional modules is semi-simple, so there are (in general, many) ways to represent a given submodule $N \subseteq M$ as a quotient. Choosing one of them gives you the desired inclusion (which is not canonical in general!).

However, the category of all $\mathfrak{g}$-modules is far from semi-simple, so there is no such inclusion in the generality you are asking for. For instance, if $\mathfrak{g}=\mathfrak{sl}_2(\mathbf{C})$ and you take $N=L=\mathbf{C}$ to be the trivial representation and $M$ to be the co-Verma module containing it as a submodule, then $$\mathrm{Hom}_\mathfrak{g}(M,\mathbf{C})=0,$$ so there can be no inclusion as in your question.

Stephen
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  • But how to represent a given submodule $N\subseteq M$ as a quotient of $M$ in the semisimple Lie algebra case? – James Cheung Nov 27 '18 at 14:21
  • @JamesCheung Are your representations finite dimensional $\mathbf{C}$-vector spaces? If so, you must choose a complementary submodule. There are many ways to do so, in general. It's impossible for me to specify one of them canonically, as I noted in my answer. One procedure would be to start with a positive definite Hermitian form on $M$ and to average it over the special unitary group to get an invariant positive definite form, which you could then use by taking orthogonal complements. – Stephen Nov 27 '18 at 14:23
  • Thank you for your detailed answer. – James Cheung Nov 27 '18 at 15:50
  • By the way, do you have any reference book talking about the fact: For a general abelian category, Hom(A,B)⊆Hom(C,B) is when A is a quotient of C? – James Cheung Mar 23 '19 at 03:24
  • @JamesCheung I don't know a reference off the top of my head, but it's a quick consquence of the axioms for an abelian category: the inclusion is given by precomposition with the quotient map. – Stephen Mar 23 '19 at 16:47