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!

  • $\begingroup$ 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. $\endgroup$ – Jim Feb 20 '13 at 23:23
  • $\begingroup$ @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! $\endgroup$ – Edward Hughes Feb 20 '13 at 23:40

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.

  • $\begingroup$ That does seem like a convincing counterexample. And that's an irrep as well, isn't it? So we can't fix it by requiring irreps only. Is the result true if we only consider semisimple Lie algebras? Thanks for your help! $\endgroup$ – Edward Hughes Feb 21 '13 at 8:22
  • $\begingroup$ By the way, here is the reference I've got the statement from, if that sheds any light on the required conditions! $\endgroup$ – Edward Hughes Feb 21 '13 at 8:27
  • $\begingroup$ I don't know if semisimple would fix it. My intuition says no but an immediate counterexample doesn't come to mind. $\endgroup$ – Jim Feb 21 '13 at 16:34
  • $\begingroup$ I would have thought something would fix it, else that's a pretty glaring error in the linked book! $\endgroup$ – Edward Hughes Feb 21 '13 at 20:17

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