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Irreducible representations of a product of two groups

Let $V$ be an irreducible finite dimensional complex representation of the product of groups $G\times H$ . Is it necessarily isomorphic to a tensor product of irreducible representation of $G$ and $H$ ? If not what is a counter-example, and under what extra assumptions this is known to be true?

Remark. I think for continuous representations of compact groups this is true.

  • gr.group-theory
  • rt.representation-theory

asv's user avatar

  • 2 $\begingroup$ For algebraically irreducible representations this is true and can be found in many sources like Curtis and Reiner. For any finite dimensional algebras the irreducible representations of the tensor product are the tensor products of irreducibles. You can replace the group algebra by the tensor product of the images of CG and CH to reduce to the finite dimensional case. $\endgroup$ –  Benjamin Steinberg Dec 11, 2022 at 12:38
  • 2 $\begingroup$ There is some discussion in comments to this question mathoverflow.net/questions/397592/… which might be useful (I see that @BenjaminSteinberg made many of the same points over there) $\endgroup$ –  Yemon Choi Dec 11, 2022 at 15:41

2 Answers 2

If your groups are finite, then Andy’s answer is perfectly fine. If you want to do topological groups you must be slightly (but not much) more careful. First since we are dealing with finite dimensional representations, algebraic and topological reducibility are the same.

It is enough to prove that your irreducible representation decomposes as a tensor product in the case of discrete groups. For if it is isomorphic to one of the form $U\otimes V$ , then since the linear isomorphism is continuous, it is enough to observe that restricting the original representation to $G$ gives $\dim V$ copies of $U$ and restricting to $H$ gives $\dim U$ copies of $V$ and so $U$ and $V$ must give continuous representations of $G$ , $H$ .

So we may assume that $G$ , $H$ are discrete and we are dealing with no topology. Then replacing $G$ , $H$ with their group algebras and $\mathbb C$ by algebras over an algebraically closed field, it suffices to show that if $A$ , $B$ are $K$ -algebras with $K$ algebraically closed, then any finite dimensional irreducible representation of $A\otimes B$ is equivalent to a tensor product of irreducible representations of $A$ and $B$ .

But the image of $A$ and $B$ under any finite dimensional representation is a finite dimensional algebra and so without loss of generality we may assume that $A$ , $B$ are finite dimensional.

The crux of the matter, which is basically Andy’s proof in different words, is the special case that $A,B$ are matrix algebras over $K$ . Then $M_n(K)\otimes M_m(K)\cong M_{nm}(K)$ and the isomorphism intertwines the actions on $K^n\otimes K^m\cong K^{nm}$ . Thus in the case $A$ , $B$ are matrix algebras then the unique $A\otimes B$ irreducible rep is the tensor product of the unique irreducible $A$ and $B$ reps. Since arbitrary semisimple algebras over an algebraically closed field are finite direct products of matrix algebras and tensor product distributes over direct product, this handles the case $A$ , $B$ are semisimple and also shows that $A\otimes B$ is semisimple in this case.

$\DeclareMathOperator\rad{rad}$ If $A$ , $B$ are general finite dimensional $K$ -algebras with $K$ -algebraically closed, then $A/{\rad(A)}\otimes B/{\rad(B)}$ is semisimple by the above. Moreover, the kernel of the natural map $A\otimes B\to A/{\rad(A)}\otimes B/{\rad(B)}$ is easily checked to be $A\otimes \rad(B)+\rad(A)\otimes B$ , which is a nilpotent ideal. Thus $(A\otimes B)/{\rad(A\otimes B)}\cong A/{\rad(A)}\otimes B/{\rad(B)}$ and so the desired result that irreducible reps are tensor products follows since $C$ and $C/{\rad(C)}$ have the same irreducible reps for any $K$ -algebra $C$ .

Simpler proof Here is an argument along the line of Andy's that works for arbitrary groups or even algebras that avoids the radical and just uses Burnside's theorem that a finite dimensional representation of a $K$ -algebra over an algebraically closed field $K$ is irreducible if and only if it is surjective.

Let $A,B$ be $K$ -algebras (not necessarily finite dimensional) with $K$ -algebraically closed (they could be group algebras) and let $W$ be a finite dimensional irreducible $A\otimes B$ -module. Let $U$ be an irreducible $A$ -subrepresentation of $W$ . Then the sum $S$ of all irreducible $A$ -submodules of $W$ isomorphic to $U$ is invariant under any $A$ -enomorphism of $W$ and hence under $B$ as the $B$ -action commutes with $A$ . Hence $S$ is $A\otimes B$ -invariant and so $S=W$ by irreducibility. Thus $W\cong U^m$ for some $m$ as an $A$ -module (this is a standard argument). Thus, up to isomorphism, we may assume that $W=U\otimes V$ with $V$ a vector space of dimension $m$ and where $A$ acts via matrices of the form $\rho(a)\otimes 1$ where $\rho$ is the irreducible representation associated to $U$ . By Burnside, $\rho$ is onto $End_k(U)$ . But since $End_K(U\otimes V)= End_k(U)\otimes End_k(V)$ , it follows that the centralizer of $End_K(U)\otimes 1$ in $End_K(W)$ is $1\otimes End_k(V)$ . Since the action of $B$ commutes with that of $A$ , as a representation of $B$ , we have $W$ is of the form $b\mapsto 1\otimes \psi(b)$ for some representation $\psi$ of $B$ on $V$ . Clearly, if $V$ is not irreducible, then neither is $U\otimes V$ and so we are done.

Benjamin Steinberg's user avatar

Most books on the representation theory of finite groups prove this using character theory. This is an efficient way of proving it, but I think it doesn’t give much insight into why it is true (and what other settings it generalizes to). I wrote up a more direct proof here .

Andy Putman's user avatar

  • 2 $\begingroup$ The OP doesn't say that the group here is finite. In fact he may be looking at topological irreducibility which for compact groups is the same and I guess this is true in finite dimensions for all topological groups since all subspaces are closed but maybe one has to check continuity. Anyway for reps of discrete infinite groups you need to argue differently because you don't get for free complete reducibility but it isn’t bad. Nonetheless is true as I argue in the comments. The argument for finite dimensional algebras is basically the same as your note after you factor out the radicals. $\endgroup$ –  Benjamin Steinberg Dec 11, 2022 at 14:58
  • 2 $\begingroup$ Actually I think your proof is fine for infinite groups with a tiny modification. You can argue that the restriction of an irreducible to a direct factor (or any normal subgroup) is completely reducible by noting the sum of all the translates of an irreducible subrep of the normal subgroup is the whole representation by irreduciblity and then proceed as in your proof. I'll keep mine since it works for monoids or for any algebras $\endgroup$ –  Benjamin Steinberg Dec 11, 2022 at 15:41

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irreducible representation of product group

Irreducible Representation

1. The dimensionality theorem :

3. Orthogonality of different representations

4. In a given representation, reducible or irreducible, the group characters of all matrices belonging to operations in the same class are identical (but differ from those in other representations).

6. A one-dimensional representation with all 1s (totally symmetric) will always exist for any group .

7. A one-dimensional representation for a group with elements expressed as matrices can be found by taking the group characters of the matrices .

Irreducible representations can be indicated using Mulliken symbols .

Portions of this entry contributed by Todd Rowland

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Rowland, Todd and Weisstein, Eric W. "Irreducible Representation." From MathWorld --A Wolfram Web Resource. https://mathworld.wolfram.com/IrreducibleRepresentation.html

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