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Suppose $G_1, \ldots, G_k$ are unitary, Hermitian, and anti-commuting, as well as $F_1, \ldots, F_k$. If they are similar, i.e., there exists $T \in GL_n(\mathbb{C})$ such that $$ G_i = T^{-1} F_i T $$ for all $i \in [k]$, does there exist $V \in U_n$, where $U_n$ is the group of unitary matrices, such that $$ G_i = V^{-1} F_i V $$ for all $i \in [k]$.

Added later.

Generally, let $\pi, \sigma : G \to U_n(\mathbb{C})$ be two unitary representations of finite group. If $\pi, \sigma$ are equivalent, i.e., there exists $T \in GL_n(\mathbb{C})$ such that $T \pi(g) = \sigma(g) T$ for all $g \in G$, is $\pi$ and $\sigma$ unitarily equivalent, that is, there exists $T' \in U_n$ such that $T' \pi = \sigma T'$?

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Motivation? Ideas already tried? Context? –  Yemon Choi Sep 20 '12 at 19:55
    
More generally, if we have two equivalent (in the sense of similar) unitary representations (not necessarily irreducible) of a group G, is it true that they are unitarily similar? I am not sure whether it is a well-known conclusion in representation theory. –  jsliyuan Sep 20 '12 at 19:59
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I can not see a solution even when $k = 1$ .... –  jsliyuan Sep 20 '12 at 20:00
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It seems to be a simple application of polar decomposition. –  Mateusz Wasilewski Sep 20 '12 at 20:25
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Thank you. I see. –  jsliyuan Sep 21 '12 at 0:24
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up vote 1 down vote accepted

Maybe this question deserves an answer which doesn't use decomposition into irreducible representations. Let $\pi$ and $\sigma$ be equivalent unitary representations. If $T$ intertwines, then so does $T^{\ast}$, because $\pi(g^{-1})=\pi(g)^{\ast}$ (same for $\sigma$). Now we can conclude that $|T|=\sqrt{T^{\ast}T}$ is also an intertwiner. Finally, $T|T|^{-1}$ is the unitary we need.

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Nice. I still am partial to an argument that just uses lin algebra at 1st/2nd course level, though –  Yemon Choi Sep 21 '12 at 17:46
    
It's very clear. Thanks –  jsliyuan Sep 21 '12 at 20:36
    
It seems that there is some problem... Using the notations above, $T \pi=\sigma T$, and $\pi T^* = (\pi^{-1})^* T^* = (T \pi^{-1})^* = (\sigma^{-1} T)^* = T^* \sigma$. Thus, both $T$ and $(T^*)^{−1}$ intertwines. $T|T|^{−1}$ is unitary. However, I can't see why $T|T|^{−1}π=\sigma T|T|^{−1}$, which amounts to $\pi|T|=|T|\pi$. Why this is true? –  jsliyuan Sep 21 '12 at 22:06
    
$|T|$ is a norm limit of polynomials in $T^{\ast}T$ and the latter operator can be easily seen to commute with $\pi$. –  Mateusz Wasilewski Sep 21 '12 at 22:28
    
Could you tell me the definition of "norm limit of polynomials"? I haven't seen it before. Thanks :) –  jsliyuan Sep 21 '12 at 22:32
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Two similar unitary representations are always unitarily similar if they are irreducible. An isomorphism $V_1 \to V_2$ pulls back the unitary metric on $V_2$ to a unitary metric on $V_1$. To make this unitary, we need to change that metric into the original metric on $V_1$ through a $G$-equivariant map $V_1\to V_1$. But any two unitary metrics on a representation are related by a $G$-equivariant map, so this is always possible.

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If two irreducible unitary representations are similar, then T would be a scalar multiple of the identity by Schur's lemma. What happens if two representations are not irreducible? Did you mean that through a $G$-equivariant map $V_1 \to V_1$, we can make $T$ unitary? And how to find such a $G$-equivariant map? –  jsliyuan Sep 21 '12 at 0:28
    
We can write a unitary metric as a unitary matrix with one row or column for each irreducible subrepresentation, where all entries that correspond to two non-isomorphic representations are zero. Similarly, a G-equivariant map is a matrix where all entries that correspond to two non-isomorphi representations are zero. G-equivariant maps act by simultaneous left multiplication and right conjugate multiplication. The only invariant preserved by this action is the signature of each block. Unitary metrics are always positive-definite, so this is the same for both metrics. –  Will Sawin Sep 21 '12 at 1:22
    
In other words, you take a square root. –  Will Sawin Sep 21 '12 at 1:22
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Did you mean that the answer is "yes" for irreducible representations, and the answer may be "no" for the others? –  jsliyuan Sep 21 '12 at 1:45
    
No, I mean it's true in general. –  Will Sawin Sep 21 '12 at 4:03
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