Let $G$ be a group and $\pi$ be a finite-dimensional (not necessarily unitary) representation of $G$ on a complex Hilbert space $H$. We shall say that $\pi$ is completely reducible if there exists a decomposition of $H$ into *orthogonal* irreducible sub-representations of $\pi$.



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> **Question 1.** Suppose $(\pi\_1, H\_1)$, $(\pi\_2, H\_2)$ are completely reducible representations of $G$. Then is the tensor product $(\pi\_1 \otimes \pi\_2, H\_1 \otimes H\_2)$ also completely reducible?
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Note, this question is different from the below question because in that question it does not require the direct sum decomposition to be orthogonal since the group is simply acting on a vector space.


[Semisimple representations of discrete groups](https://mathoverflow.net/questions/257583/semisimple-representations-of-discrete-groups)


Though, I had a read of the wonderful answer by nfdc23 to the above question which explains that the crux of proof uses the fact that all finite-dimensional linear representations of a reductive smooth affine group over a complex vector space are completely reducible (where this notion is weaker than the one that I have defined in this question). This leads me to my second question.



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> **Question 2.** Are there any "large" family of groups such that all their finite-dimensional (*insert here any "mild" adjectives*) representations on a complex Hilbert space are completely reducible in the sense defined in this question?
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When are finite-dimensional representations on Hilbert spaces completely reducible?
I don't know about Q1 off the top of my head, but I think that for Q2 you are unlikely to get a good answer. Of course I may have a different view from you as to what "mild" adjectives are reasonable to impose.


The reason I say this is that in your definition of completely reducible you are requiring the summands of the decomposition of $\pi$ to be orthogonal, yet you are allowing $\pi$ to be non-unitary. Therefore I can always take a reducible unitary representation $\sigma: G \to {\mathcal U}(H)$ and then conjugate it with a non-unitary invertible operator on $H$ which will almost surely mess up the the orthogonality inside the decomposition of $\sigma$.


For a concrete example: let $G=\{\pm 1\}$, fix $\varepsilon>0$, and let $\pi$ be the representation of $G$ on ${\bf C}^2$ which sends $-1$ to $\begin{pmatrix} 1 & \varepsilon \\ 0 & -1 \end{pmatrix}$. Since the eigenvectors of this matrix are not orthogonal to each other, $\pi$ does not decompose into orthogonal summands.


Another example showing that "nice" groups can fail to have the property you seek: fix $\varepsilon>0$ and let $\pi: {\bf Z} \to {\rm GL}\_2({\bf C})$ be the representation which sends $1$ to $\begin{pmatrix} 1 & \varepsilon \\ 0 & 1 \end{pmatrix}$. This representation is reducible but not decomposable; however, I am not sure if you are ruling out such examples with one of your unspecified "mild" adjectives.


By the way, if there are no examples other than the trivial representation, which satisfy the conditions in Q2 then your Q1 will have a positive answer. This is why I think Q1 is actually not such a good question unless you pin down what your intended definition of "completely reducible" should be.


