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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!

Edward Hughes
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  • 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 '13 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 '13 at 23:40
  • Cf. https://math.stackexchange.com/q/3826725/96384 and https://math.stackexchange.com/q/3350849/96384 with their answers and discussion in comments. – Torsten Schoeneberg Apr 08 '23 at 23:43

3 Answers3

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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.

Eric O. Korman
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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.

Jim
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  • 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! – Edward Hughes Feb 21 '13 at 08:22
  • By the way, here is the reference I've got the statement from, if that sheds any light on the required conditions! – Edward Hughes Feb 21 '13 at 08:27
  • I don't know if semisimple would fix it. My intuition says no but an immediate counterexample doesn't come to mind. – Jim Feb 21 '13 at 16:34
  • I would have thought something would fix it, else that's a pretty glaring error in the linked book! – Edward Hughes Feb 21 '13 at 20:17
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    This counterexample is hardly ever irreducible. For instance, if $k$ is algebraic closed then it follows from Schur's lemma that every finite-dimensional irreducible representation of an Abelian Lie algebra over $k$ is 1-dimensional. This goes to shows that perhaps you can fix your statement by requiring irreducibility. In fact this answer claims this requirement is sufficient regardless of wether of not $\mathfrak{g}$ and $\mathfrak{h}$ are semisimple. – Thiago Brevidelli Garcia Jan 02 '23 at 20:56
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    To be clear, I'm not claiming the counterexample is wrong, I'm just claiming it is not a counter example to the corrected statement (at least for algebraically closed $k$). – Thiago Brevidelli Garcia Jan 02 '23 at 20:58
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    I've added an answer containing a proof of the corrected statement. – Thiago Brevidelli Garcia Apr 07 '23 at 22:42
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I would like to add to Eric's answer by showing that the result is true for finite-dimensional complex Lie algebras regardless of whether or not the Lie algebras in question are semisimple. Namely, I'll show that if $\mathfrak{g}$ and $\mathfrak{h}$ are finite-dimensional complex Lie algebras (not necessarily semisimple) then every finite-dimensional irreducible representation $M$ of $\mathfrak{g} \oplus \mathfrak{h}$ is isomorphic to the tensor product of representations $M_1$ of $\mathfrak{g}$ and $M_2$ of $\mathfrak{h}$.

This is a consequence of the following corollary of Lie's theorem, whose proof can be found in the 9th chapter of Fulton-Haris (see Proposition 9.17).

Proposition. Let $\mathfrak{g}$ be a complex Lie algebra. Then every irreducible representation $M$ is the tensor product of a 1-dimensional representation of $\mathfrak{rad(g)}$ and an irreducible representation of $\mathfrak{g/rad(g)}$, where $\mathfrak{rad(g)}$ is the radical of $\mathfrak{g}$.

In our case, this implies that $M = Z \otimes N$ where $Z$ is a 1-dimensional representation of $\mathfrak{rad(g \oplus h)}$ and N is an irreducible representation of $\frac{\mathfrak{g \oplus h}}{\mathfrak{rad(g \oplus h)}}$. Now recall that $\mathfrak{rad(g \oplus h)} = \mathfrak{rad(g)} \oplus \mathfrak{rad(h)}$ and $\frac{\mathfrak{g \oplus h}}{\mathfrak{rad(g \oplus h)}} = \mathfrak{g / rad(g)} \oplus \mathfrak{h / rad(h)}$.

As Eric indicated, since $\mathfrak{g/rad(g)}$ and $\mathfrak{h/rad(h)}$ are finite-dimensional semisimple complex Lie algebras, it follows from the character theory of compact Lie groups and the existence of the compact forms that $N = N_1 \otimes N_2$, where $N_1$ is an irreducible representation of $\mathfrak{g/rad(g)}$ and $N_2$ is an irreducible representation of $\mathfrak{h/rad(h)}$. In addition, since $Z$ is 1-dimensional, $Z = Z_1 \otimes Z_2$, where $Z_1 = \operatorname{Res}_{\mathfrak{rad(g)}}^{\mathfrak{rad(g) \oplus rad(h)}} Z$ and $Z_2 = \operatorname{Res}_{\mathfrak{rad(h)}}^{\mathfrak{rad(g) \oplus rad(h)}} Z$.

If we then take $M_1 = Z_1 \otimes N_1$ and $M_2 = Z_2 \otimes N_2$ we find $M = Z \otimes N = (Z_1 \otimes Z_2) \otimes (N_1 \otimes N_2) = (Z_1 \otimes N_1) \otimes (Z_2 \otimes N_2) = M_1 \otimes M_2$. It is not hard to verify that $M_1$ and $M_2$ are indeed irreducible representations of $\mathfrak{g} = \mathfrak{rad(g) \oplus g / rad(g)}$ and $\mathfrak{h} = \mathfrak{rad(h) \oplus h / rad(h)}$ respectively, which concludes our proof.

I would like to conclude the answer with some comments on potential generalizations of this result. First of all, I am fairly certain that the proposition of Fulton-Harris holds for Lie algebras over any algebraicly closed field of characteristic zero, given that it is a pretty straightforward consequence of Lie's theorem. I'm also confident that if $\mathfrak{g}$ and $\mathfrak{h}$ are finite-dimensional semisimple Lie algebras over an algebraic closed field $k$ of characteristic $0$ then the finite-dimensional irreducible representations of $\mathfrak{g \oplus h}$ are all given by tensor products, so that our proof applies for Lie algebras over $k$ -- even though Eric's proof doesn't work in this setting.

Furtheremore, as far as I can tell the proposition in Fulton-Harris holds for infinite-dimensional representations too. Our argument may thus be generalized to the infinite-dimensional setting.

Thiago Brevidelli Garcia
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  • To be honest, I think there should be a cleaner proof of this state, one which proves that $M$ splits as $M_1 \otimes M_2$ without relying on the fact the theorem holds for semisimple algebras a-priory. Nevertheless, at least we know the result is true (unless someone can find a mistake on my proof ). – Thiago Brevidelli Garcia Apr 08 '23 at 14:02