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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 $\pi$.

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?

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

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.

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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  • $\begingroup$ I don't know the answer to question 1 but just wanted to make the trivial remark that you might as well assume that $\pi_1$ and $\pi_2$ are irreducible. I'd instinctively guess that the answer is no in general. If I wanted to think about it I'd start by considering $\pi_1$ and $\pi_2$ to be the irreducible $2$-dimensional representations of $S_3$ so $\pi_1\otimes \pi_2$ is uniquely decomposed as a direct sum of $3$ irreducible subreps. And then try to put inner products on $\pi_1$ and $\pi_2$ so that these are not mutually orthogonal in the tensor product. $\endgroup$
    – Simon Wadsley
    Commented Jul 14, 2023 at 11:43
  • $\begingroup$ Right, that is a good point about manipulating the inner product to make then not orthogonal. That makes me incline to agree with you that the answer is no. I will have a play around and to try find a specific example. $\endgroup$
    – Nanoputian
    Commented Jul 14, 2023 at 11:58

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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.

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  • $\begingroup$ Thank you for your answer. Regarding what I meant by "mild", I didn't really have anything in mind. I am actually more interested in Q1. Q2 was just me being curious if anything could be said in some general case. But I can see from your explanation that there isn't any hope for Q2 so happy to leave that there. Could you please explain what you mean by "pin down what your intended definition of completely reducible should be"? The definition that I stated in the question is exactly what I intend. $\endgroup$
    – Nanoputian
    Commented Jul 14, 2023 at 11:45
  • $\begingroup$ Well it seems to me that for most groups, your definition of "completely reducible" probably requires the representations to already be unitary, or to be irreducible. Your Q1 has the form "if there are two representations which have a property, does this other representation have the same property". I am pointing out that if your property is restrictive or reduces to an existing one, then we have not gained anything. $\endgroup$
    – Yemon Choi
    Commented Jul 14, 2023 at 11:51
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This is only an attempt at answering question 2.

If $G$ is a group with a non-trivial finite dimensional representation $V$ then it has a finite dimensional representation on a complex Hilbert space $W$ that is not completely irreducible in the sense of this question. In particular one can take $W$ to be the direct sum of the representation $V$ and the trivial representation $\mathbf{1}$ and put a Hilbert space structure on $W$ such that $V$ and $\mathbf{1}$ are not mutually orthogonal. Since the only way to decompose of $W$ as a direct sum of irreducible subrepresentations is $W\oplus \mathbf{1}$, $W$ is not completely irreducible in the sense of the question.

I think it follows that the largest possible family of groups that you can consider is the family of groups all of whose finite dimensional representations are trivial.

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  • $\begingroup$ I should say that this is essentially the same as Yemon Choi's answer which I hadn't seen until I posted but perhaps different enough to be worth leaving here. $\endgroup$
    – Simon Wadsley
    Commented Jul 14, 2023 at 11:36
  • $\begingroup$ Out of interest, do you know any groups apart from the trivial one which satisfy the property in your last paragraph? I am used to thinking only about continuous unitary representations, and there are a good supply of locally compact groups which have no non-trivial continuous unitary representations, but the ones I am aware of are all matrix Lie groups or coverings thereof... $\endgroup$
    – Yemon Choi
    Commented Jul 14, 2023 at 11:48
  • $\begingroup$ I think the (simple) group of even finitely supported permutations of $\mathbb{N}$ (ie the 'union' of all finite alternating groups) may have this property though I don't know a proof of this fact. $\endgroup$
    – Simon Wadsley
    Commented Jul 14, 2023 at 11:59
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    $\begingroup$ Actually maybe it is easy to prove that this infinite alternating group $A$ has no non-trivial finite dimensional representation since any non-trivial representation of the $A$ restricts to a non-trivial representation of $A_n$ for all $n\geq 3$ and the dimension of the smallest non-trivial representation of $A_n$ goes to infinity as $n$ goes to infinity. $\endgroup$
    – Simon Wadsley
    Commented Jul 14, 2023 at 12:43
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    $\begingroup$ A finitely generated group has a nontrivial finite dimensional representation iff it has a nontrivial finite image by Malcev theorem on residual finiteness of finitely generated linear groups. So finitely generated simple groups or Higmans group has no nontrivial finite dim reps $\endgroup$
    – Benjamin Steinberg
    Commented Jul 14, 2023 at 17:25

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