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This is a crosspost from math.SE. Suppose $G$ and $H$ are discrete groups. Is it always the case that any finite dimensional complex representation of $G\times H$ is of the form $$ \bigoplus_i V_i \otimes W_i, $$ where $V_i, W_i$ are reps of $G$ and $H$, respectively?

I know this is true when $G$ and $H$ are finite and when the representation of $G\times H$ is completely reducible, but is there a simple counterexample to the general case?

I'm also curious if it is ``usually true," in some sense, that any rep of $G\times H$ has the above form.

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You mean the $V_{i}$ and $W_{i}$ are irreducibe reps? –  Geoff Robinson Jun 11 '13 at 18:40
    
I'm not assuming $V_i$ and $W_i$ are irreducible. –  Eric O. Korman Jun 11 '13 at 19:44

2 Answers 2

up vote 10 down vote accepted

Let $G = H = \mathbb{Z}$. Now a $G \times H$ representation is a pair of commuting invertible matrices. Let's try $$\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right)\mbox{ and } \left( \begin{array}{cc} -1 & 1 \\ 0 & -1 \\ \end{array} \right).$$ Certainly this rep is indecomposable since it already is after restricting to either group. Why can't it be a tensor product of two representations of $\mathbb{Z}$? One representation would have to be one-dimensional, given by a scalar multiplication. But this is impossible since neither matrix is a scalar.

I think it is true that there is an open set on the representation variety where things decompose into a sum of tensor products as you say but I do not see why exactly.

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Thanks for the nice counterexample! –  Eric O. Korman Jun 11 '13 at 19:56
    
Isn't the set where both matrices are diagonalizable an open set? Then since they commute they're simultaneously diagonalizable, and in this case the representation decomposes as desired. I'm not sure if this generalizes or not, though... –  Peter Samuelson Jun 12 '13 at 2:03

This is not true, in general. For example, take $G = H = \mathbb Z$. Let $G \times H$ act on $\mathbb C^3$ in such a way that a generator of $G$ carries $e_1$ to $e_2$, and $e_2$ and $e_3$ to $0$, while a generator of $H$ carries $e_1$ to $e_3$, and $e_2$ and $e_3$ to $0$. It is easy to see that this representation is indecomposable, and that it is not of the form above.

[Edit:] as was, correctly, pointed out to me, my maps are not invertible; but you can make them invertible by adding the identity to both.

I had in mind the following. A representation of $G \times H$ corresponds to a $\mathbb C[x^{\pm 1}, y^{\pm 1}]$-module. The representation above corresponds to $\mathbb C[x,y]/((x-1)^2, (y-1)^2, (x-1)(y-1))$.

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