# representations with centralizer stable under conjugate transpose

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?

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.
• So, basically, multiplicity-free $\rho$ has a self-adjoint (the correct term, thanks!) $C_\rho$, right? – Dima Pasechnik Feb 21 at 19:39
• 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. – Mateusz Wasilewski Feb 22 at 9:47