Representations of Direct Sum of Lie Algebras

I’m trying to prove the following. Let $\frak{g}$ and $\frak{h}$ be (semisimple) Lie algebras. Then every representation $d$ of $\frak{g}\oplus\frak{h}$ is the tensor product of representations $d^1$ on $\frak{g}$ and $d^2$ on $\frak{h}$; that is $d=d^1\otimes d^2$.

$d^1$ and $d^2$ are cannot be defined by restriction (see comment below). I don’t know how to define them so that their tensor product must give back $d$. In particular how do I know that the vector space $V$ of $d$ decomposes appropriately for this to work?

Any hints would be much appreciated!

Solutions Collecting From Web of "Representations of Direct Sum of Lie Algebras"

I think you meant to add that your representation is irreducible.

This statement is true for representations of compact Lie groups– see theorem 3.9 of Sepanski’s Compact Lie groups (unfortunately the relevant pages are not on google books but the book is on springerlink if you have institutional access). The proof of this relies on the use of characters. Your result then follows for semi-simple Lie algebras since they all have compact real forms.

The following example should show that this statement is actually false.

Let $\mathfrak g$ be a $1$-dimensional abelian Lie algebra. Then $\mathfrak{g \oplus g}$ is a $2$-dimensional abelian Lie algebra. The universal enveloping algebra of $\mathfrak{g \oplus g}$ is $k[x, y]$ where $k$ is whatever field you’d like to work over.

Now consider a $3$-dimensional $k[x, y]$-module where we let both $x$ and $y$ act via a nilpotent Jordan block. For this to be the tensor product of two representations they would have to be dimensions $1$ and $3$ but neither $x$ nor $y$ act diagonally which is what we would get after tensoring with a $1$-dimensional representation.