Tensor product of simple modules over semisimple Lie algebras

lie-algebrasrepresentation-theorysemisimple-lie-algebras

$
\newcommand{\g}{\mathfrak{g}}
\newcommand{\h}{\mathfrak{h}}
\newcommand{\k}{\mathfrak{k}}
$

Let $\g$ and $\h$ be two semisimple Lie algebras, and let $\k := \g \oplus \h$. Show that a $\k$-module is simple iff it is the tensor product of simple $\g$-module and $\h$-module.

We may assume that all Lie algebras/vector spaces involved are finite-dimensional, and the involved field is both algebraically closed and of characteristic $0$.

We first note that if $V$ is a $\g$-module and $W$ is a $\h$-module, then $\k$ acts on $V \otimes W$ by the following:
$$
q = g \oplus h \implies q \cdot (v \otimes w) := (g \cdot v) \otimes w + v \otimes (h \cdot w)
$$
It's easy to check that this action indeed makes $V \otimes W$ a $\k$-module. Unfortunately, I do not have much progress on this from here. For $\impliedby$, initial idea is to assume $V \otimes W$ is not simple, and let $U \subseteq V \otimes W$ be a non-trivial proper submodule. Then, perhaps we can impose some kind of projection of $U$ onto $V$ and $W$ respectively, and contradict their simplicity by showing that this induces a non-trivial proper submodule in $V$ or $W$, contradicting their simplicity. However, as far as I know, there isn't such a notion for tensor product.

For $\implies$, my idea is also similar, but it's clear that $V$ and $W$ are not submodules of $V \otimes W$, so I'm not sure how to proceed with the construction of the $V$ and $W$ given an arbitrary $\k$-module either.

Any help is appreciated.

Note: The following questions are related.

  1. Tensor product of irreducible representations of semisimple Lie algebras: The solution uses Jacobson Density Theorem, which is probably overpowered in the case of finite-dimensional Lie algebras/vector spaces. I am looking for a proof without using this theorem.
  2. Tensor product of irreducible representations: This also seems related, but the question assumes that the field involved is $\Bbb{C}$. Furthermore, the two (unaccepted) answers utilise Clebsch-Gordon formula, which I'm not familiar with.

EDIT: Following Torsten Schoeneberg's hint, I was able to almost construct a proof of the $\implies$ direction. We consider $U$, a $\k$-submodule, as a $\g$-module via the action:
$$
g \cdot u := \underbrace{(g \oplus 0)}_{\in \g \oplus \h} \cdot u
$$
Then since $\g$ is semisimple, $U$ is semisimple as a $\g$-module, so we can write $U = \bigoplus_{i=1}^n X_i$ for simple $X_i$. Similarly, we can write $U = \bigoplus_{i=1}^m Y_i$ as with $Y_i$ being simple $\h$-modules. If all of $X_i$ are isomorphic (and similarly all of $Y_i$), then we can construct an explicit isomorphism from $X_1 \otimes Y_1 \to U$, which completes the proof. However, I have not been able to prove the claim thus far.

Best Answer

To show the reverse implication, consider the associative algebra $A\subseteq End(V)$ generated by the image $\rho: \mathfrak{g} \rightarrow End(V)$. Since $\mathfrak{g}$ is semi-simple, the image $\rho(\mathfrak{g})$ is also semi-simple. In particular, one can find the Casimir operator on $\rho(\mathfrak{g})$, and $A$ contains identity. Note that $V$ is also a simple $A$-module. Lastly, we can use some result from representation theory of associative algebra to show that (check Etingof Introduction to Representation Theory Corollary 3.2.1) to see that $A = End(V)$ and if $U\subseteq V \otimes W$, we can find some $a \in End(V)$ and $b \in End(W)$ such that $b(a(u)) \in U$ is a pure tensor, and finding appropriate $a\in End(V)$ and $b \in End(W)$ one can obtain all basis vectors of $V\otimes W$.

To prove the forward implication, first regard $U$ as a $\mathfrak{g}$ module. Note that $U = \bigoplus_{i=1}^n V_i $ for some simple $V_i$ by Weyl. Choose any one of them and say $V= V_1$. Now consider $\hom_{\mathfrak g}(V,U) \otimes V$ as a $\mathfrak h \oplus \mathfrak g$ module, where $h \in \mathfrak h$ acts on $\phi\in \hom_{\mathfrak g}(V,U)$ by $h \phi$ (and $\mathfrak g$ acts trivially). Then consider the map $f_V: \hom_{\mathfrak g}(V,U) \otimes V \rightarrow U$ given by $$f_V(\phi \otimes v) = \phi(v).$$ One can show that this map is a $\mathfrak{g \oplus h}$ module isomorphism by showing that this linear map is injective and $f_V$ commutes with the action of $\mathfrak{g \oplus h}$ hence it must be surjective. We conclude that it must be the case that $\hom_{\mathfrak g}(V,U)$ is a simple $\mathfrak{h}$ module, for otherwise, $\hom_{\mathfrak{g}}(V,U) \otimes V$ splits into direct sum of $\mathfrak{g} \oplus \mathfrak{h}$ modules.

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