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Suppose G_i are finite groups for i=1,2 and G is the direct product of G_i. If V is a finite dimensional irreducible representation of G, then it is well known that V is a tensor product of V_i,i=1,2 and each V_i is an irreducible representation of G_i.

The question I have is when V is given, is there a canonical way to construct V_i from V?

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it there any reference for it? – Omar Shehab Jun 19 at 14:13
For example, Theorem 10 in section 3.2 of Serre's book 'linear representations of finite groups.' – user1832 Jun 21 at 8:53
up vote 14 down vote accepted

I agree with David Speyer's answer, and furthermore there is no canonical way to construct $V_i$ from $V$. This is a subtle and oft-overlooked point in representation theory, in my opinion. Many texts prove that an irrep of $G_1 \times G_2$ is isomorphic to a tensor product of an irrep of $G_1$ with an irrep of $G_2$. The typical slick proof relies on character theory -- kind of a cheat, in my view, since it only says something about isomorphism classes.

Here's a categorical explanation of the theorem: Let $G_1$ and $G_2$ be finite groups, and let $\pi$ be an irrep of $G_1 \times G_2$ on a complex vector space $V$.

Then, for every pair $(\rho_1, W_1)$, $(\rho_2, W_2)$ of representations of $G_1, G_2$, one gets a complex vector space: $$H_\pi(\rho_1, \rho_2) := Hom_G(\rho_1 \boxtimes \rho_2, \pi).$$ In fact, this extends to a contravariant functor: $$H_\pi: Rep_{G_1} \times Rep_{G_2} \rightarrow Vec.$$ Here we use categories of finite-dimensional complex representations and vector spaces.

This is also functorial in $\pi$, yielding a functor: $$H: Rep_G \rightarrow [ Rep_{G_1} \times Rep_{G_2}, Vec ],$$ where the right side of the arrow denotes the category of functors (for categories enriched in $Vec$). What this demonstrates is that the canonical thing is to take representations of $G$ to objects of an appropriate functor category related to $Rep_{G_1}$ and $Rep_{G_2}$. By Yoneda's lemma (for categories enriched in $Vec$), there is an embedding of categories: $$Rep_{G_1} \times Rep_{G_2} \hookrightarrow [ Rep_{G_1} \times Rep_{G_2}, Vec ].$$

It turns out -- and this is where some finiteness is important, and a proof necessarily uses some counting, character theory, or the like -- that for any irrep $\pi$ of $Rep_G$, the functor $H_\pi \in [Rep_{G_1} \times Rep_{G_2}, Vec]$ is representable. It is not uniquely representable, but it is uniquely representable up to natural isomorphism.

Practically, what this means is that given an irrep $\pi$ of $G$, there exists an isomorphism $\iota: \pi \rightarrow \pi_1 \boxtimes \pi_2$ for some irreps $\pi_1, \pi_2$ of $G_1, G_2$, respectively. The pair $(\pi_1, \pi_2)$ is not unique, but the triple $(\iota, \pi_1, \pi_2)$ is unique up to unique isomorphism. This is usually good enough.

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If you choose a representation $W$ of $G_1$ which is isomorphic to $V_1$, then you can construct $V_2$ as $\mathrm{Hom}_{G_1}(W, V)$. But I don't think there can be a fully choice-free construction of $(V_1, V_2)$.

Here is a hazy argument; if you give me a rigorous definition of choice-free then I may able to do better. Suppose that I find $V_1$, $V_2$ and an isomorphism $V_1 \boxtimes V_2 \to V$. And suppose that you likewise find $V'_1$, $V'_2$ and $V'_1 \boxtimes V'_2 \to V$. If your construction is canonical, you should be able to give canonical isomorphisms $a_1: V_1 \to V'_1$ and $a_2: V_2 \to V'_2$, making the obvious diagram commute.

But the obvious diagram also commutes when $(a_1, a_2)$ is replaced by $(-a_1, -a_2)$. I don't see how you can possibly single out which of $a_1$ and $- a_1$ is better.

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