I am trying to prove that for $G$ and $H$ two compact groups, it's true that any irreducible $G \times H$-representation $U$ is isomorphic to a tensor product of an irreducible $G$-representation $V$ and an irreducible $H$-representation $W$, ie. that $U \simeq V \otimes W$.

After thinking a lot about this, I found the following argument in some online lecture notes (Lemma 22.6 in https://math.berkeley.edu/~teleman/math/RepThry.pdf):

"From the properties of the tensor product of matrices, it follows that the character of $V \otimes W$ at the element $g \times h$ is $\chi_{V \otimes W} (g \times h) = \chi_V (g) \cdot \chi_W (h)$. Now, a conjugacy class in $G \times H$ is a Cartesian product of conjugacy classes in $G$ and $H$, and character theory ensures that the $\chi_V$ and $\chi_W$ form a Hilbert space basis of the $L^2$ class functions on the two groups. It follows that the $\chi_V (g) \cdot \chi_W (h)$ form a Hilbert space basis of the class functions on $G\times H$, so this is a complete list of irreducible characters. $\blacksquare$"

I understand the claim about the conjugacy classes of the direct product and everything else, except how the conclusion that $\chi_V \cdot \chi_W$ are a full set of irreducible characters for the product group follows from this.

I just don't get why a class function $f : G \times H \rightarrow \mathbb{C}$ can be written as a linear combination of the various $\chi_V \cdot \chi_W$, as $V$ and $W$ vary through all irreps of $G$ and $H$. I tried to convince myself of this by looking at, say, \begin{align*} f_G : G \times \lbrace e \rbrace &\longrightarrow \mathbb{C}, \text{ and}\\ f_H : \lbrace e \rbrace \times H &\longrightarrow \mathbb{C}, \end{align*} and use that since $f_G$ and $f_H$ are class functions, they can each be written as a linear combination of $\lbrace \chi_V \rbrace_V$ and $\lbrace \chi_W \rbrace_W$ respectively. But I can't seem to connect this fact to the conclusion I'm trying to reach.

Any tips on why the specific argument works or how the result might be proved in a different way are most welcome!


Take $U$ a finite dimensional representation of $G\times H$. Consider only the $G$ action. Then every element from $H$ is an intertwining operator. Decompose $U$ into $G$-isotypic components. Then $H$ invariates each of them. Note that for every irreducible representation of $G$ the intertwining operators are scalers. Therefore, each $G$ isotypic of type $\rho$ of $U$ can be written as $\rho\otimes \eta$, where $\eta$ is a representation of $H$. Therefore, each finite dimensional (continuous) representation of $G\times H$ can be written as a direct sum of $\rho_1\otimes \rho_2$, where $\rho_1$ is an irrep of $G$ and $\rho_2$ is an irrep of $H$. From here it is easy also to show that $\rho_1\otimes \rho_2$ are irreducible.

This argument avoids most of the analysis. It only uses Schur's lemma on intertwining operators. One needs to argue that $\eta$ above is continuous, if we started with a continuous representation of $G\times H$, which is straightforward, looking at matrix coefficients.


This result requires (in general) that the underlying field is algebraically closed. The problem is that the intertwining operators are not (in general) scalars.

A simple counterexample over the field $\mathbb{R}$ is to consider the natural 2-dimensional irreducible representation of the cyclic group $C_{12}$. However, one can write $C_{12}\simeq C_3\times C_4$, but it's easy to see that this representation is not (isomorphic to) the tensor product of irreducible reps of $C_3$ and $C_4$.

Presumably (I've not checked the details) the OP's statement is true if either of the groups $G$ or $H$ has the property that every irreducible rep of that group is absolutely irreducible (over the field in question). For example, if $G=S_n$, the symmetric group. Would anyone like to confirm?


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