Let $\frak{g}$ be a complex simple Lie algebra and let $\frak{k}$ be a non-zero semisimple Lie subalgebra of $\frak{g}$. Is it possible to realize every simple $\frak{k}$-module $W$ as a $\frak{k}$-submodule of a some $\frak{g}$-module $V$. (Of course we are thinking about $V$ as a $\frak{k}$-module by restriction). I guess semsimplicity is necessary here, since if I start to look at solvable subalgebras I can imagine problems arising . . .

Edit: I would like to add that I am really interested in the finite dimensional case: If $V$ is f.d. will it extend to a f.d. representation of $\frak{g}$?


1 Answer 1


Suppose you have an inclusion of algebraic objects $A \subset B$. In this case, $A = \mathfrak{k}$ and $B = \mathfrak{g}$ are Lie algebras, but it doesn't make a big difference — you can read "algebraic object" as "group" or "Lie group" or "algebraic group" or "Hopf algebra" or many other things. What's important is that algebraic objects have a natural notion of "module".

Now, whatever "modules" are, surely there will be a way to restrict modules: you will have a restriction functor $\mathrm{Res} : \mathrm{Mod}(B) \to \mathrm{Mod}(A)$.

This functor might or might not have an adjoint, and it depends on what "algebraic object" and "module" mean in your case. The simplest case is where a $B$-module is a vector space $V$ and a homomorphism $B \to \mathrm{End}(V)$ without any "size" constraints. In this case, abstract nonsense will promise you that $\mathrm{Res}$ has both left and right adjoints (which might or might not agree). If there are size constraints built into the meaning of "module" (for example, if you are talking about integrable modules of algebraic groups), then you might have only one adjoint. Anyway, in the case you care about of Lie algebras, there are no real issues, and $\mathrm{Res}$ has a left adjoint $$\mathrm{Ind} : \mathrm{Mod}(A) \to \mathrm{Mod}(B),$$ called "induction". (The right adjoint to $\mathrm{Res}$, if it exists, is called "coinduction".)

Just because you have an adjoint pair, you in particular find, for every $V \in \mathrm{Mod}(A)$, a canonical natural-in-$V$ homomorphism $$ V \to \mathrm{Res}(\mathrm{Ind}(V)).$$

Most of the time, this homomorphism will be an injection. This is in particular true when $A = \mathfrak{k}$ and $B = \mathfrak{g}$ are semisimple Lie algebras, as in your question.

  • $\begingroup$ What is the adjoint explicitly? I guess it involves the universal enveloping algebra. But I fear that $\mathrm{Res \circ Ind}(V)$ may not be finite dimensional . . . which is something I sould have required in my question. $\endgroup$ Oct 31, 2021 at 12:21
  • $\begingroup$ I think everything has now been answered in this attached question: mathoverflow.net/questions/407464/… $\endgroup$ Oct 31, 2021 at 19:22
  • 1
    $\begingroup$ @TimMontegue Yes, if you had insisted that you find $W$ inside a finite-dimensional module $V$, then you would have needed less-abstract nonsense. I think I would have built the (infinite-dimensional) module Res(Ind(V)), and then used some semisimplicity to look inside it and find a finite solution. $\endgroup$ Nov 1, 2021 at 0:37
  • $\begingroup$ Great, thanks a lot for the explanation. $\endgroup$ Nov 1, 2021 at 10:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .