Extending representations of Lie subalgebras to the whole Lie algebra 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 $\frak{g}$?
 A: 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.
