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Let $\rho:G\to GL_n(\mathbb{C})$ be a finite-dimensional representation of a finite group $G$ over $\mathbb{C}$, and $C_\rho\subset M_n(\mathbb{C})$ its centralizer, i.e. $m\in C$ iff $m$ commutes with each $\rho(g)$, $g\in G$.

In some cases, e.g. if $\rho$ is irreducible, or $\rho$ is unitary, $C_\rho$ is closed under conjugate transpose $*$, where $m^*:=\overline{m}^\top$.

What are reducible non-unitary $\rho$ for which $C_\rho$ is closed under conjugate transpose?

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I do not have a complete answer, but it might be helpful anyway.

Since every representation of a finite group is unitarisable, there is an invertible operator $T$ and a unitary representation $\sigma$ such that $\rho(g) = T \sigma(g) T^{-1}$. If $C_{\sigma}$ denotes the centraliser (or commutant) of $\sigma$, then $C_{\rho} = T C_{\sigma} T^{-1}$. Now $C_{\sigma}$ may be an arbitrary unital, $\ast$-subalgebra. The condition $C_{\rho}^{\ast}=C_{\rho}$ is equivalent to $T^{\ast}T C_{\sigma} (T^{\ast}T)^{-1} = C_{\sigma}$. Since $T^{\ast}T$ is positive, there exists a unitary $U$ such that $T^{\ast}T = UD U^{\ast}$, where $D$ is a diagonal matrix. The condition we get now is $U^{\ast}C_{\sigma} U = D U^{\ast}C_{\sigma} U D^{-1}$. Now we would like to force $U^{\ast} C_{\sigma} U$ be a subalgebra of block diagonal matrices, where the structure of blocks is determined by the representation. Note that any unital, $\ast$-subalgebra of $M_n$ is conjugate by a unitary to such a thing, so we may achieve that by replacing $T$ by a slightly different operator. Namely, fix a unitary $V$ such that $VC_{\sigma} V^{\ast} = B$, where $B$ is a block diagonal subalgebra. If we replace $T$ by $TUV$ (i.e. we consider new $\sigma'(g):= (UV)^{\ast} \sigma(g) UV$), then we get what we want. And in the block diagonal case the conjugation by a diagonal matrix preserves the block structure, so everything is fine.

So, it seems that the main point is to find an equivalent unitary representation and then tweak it a little bit, so that the centraliser is of a particularly nice form. The procedure in this answer does not exhaust all the possibilities but at least it shows the following: if $\sigma$ is a unitary representation and $V$ is a unitary such that $VC_{\sigma}V^{\ast}$ is block diagonal then any $T=UDV$, with $D$ diagonal and $U$ unitary will provide an example of a representation $\rho = T\sigma T^{-1}$, whose centraliser is self-adjoint.

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  • $\begingroup$ So, basically, multiplicity-free $\rho$ has a self-adjoint (the correct term, thanks!) $C_\rho$, right? $\endgroup$ – Dima Pasechnik Feb 21 at 19:39
  • $\begingroup$ I suspect that's not true in general. Actually, I realised now that my answer is incorrect; the block structure is more complicated than what I suggested. Actually each block consists of a number of copies of the same matrix, so conjugation by an arbitrary diagonal matrix will destroy the structure. I will try to fix it tomorrow (or retract the answer). I am sorry for that. $\endgroup$ – Mateusz Wasilewski Feb 21 at 21:19
  • $\begingroup$ The only thing I can do is to impose additional conditions on the diagonal matrix. Say, if the first irreducible representation has dimension $n$ and multiplicity $m$ then in the commutant we would get $m\times m$ matrices, but repeated $n$ times, so the pattern in the diagonal matrix would also have to be repeated: first $m$ entries are arbitrary but then you have to repeat it $n$ times. Is that in any way interesting for you? For me it does not sound very satisfactory. $\endgroup$ – Mateusz Wasilewski Feb 22 at 9:47

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