# 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!

-
 Tensoring the restrictions together will not work. Restricting a representation from $\mathfrak{g \oplus h}$ to $\mathfrak g$ or to $\mathfrak h$ does not change the dimension, so when you tensor those together you will have squared the dimension of your representation. – Jim Feb 20 at 23:23 @Jim: that's what I thought originally, and have now edited the question to reflect. Can you think of an appropriate way of defining the $\frak g$ and $\frak h$ reps then? I'm stumped at present! Cheers! – Edward Hughes Feb 20 at 23:40

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.