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I'm working on a problem with unitary representations. Basically the irreducible representations of a group $G$ have been shown to be of the form of $\chi(z) \otimes \beta$ where $\chi(z)$ maps the elements of the group to a complex number $z$ and $\beta$ is a irreducible representation of $G$. I've shown that this is unitary if and only if the $\beta(g) = U$ have the property that $\langle Uv | Uw \rangle_A = c\langle v | w \rangle_A$ for some Hermitian inner product $\langle \cdot | \cdot \rangle_A$ which I have taken to calling near unitarizable.

It would be very helpful if near unitarizable implied unitarizable (ie. $\langle Uv | Uw \rangle_B = \langle v | w \rangle_B$ for some inner product). I have been trying to find either a proof or a counter example with no luck. If you have any imput on either it would be very helpful.

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  • $\begingroup$ in short: you have an operator $U$ and an inner product such that $\langle Uv|Uw\rangle_A=c\langle u|v\rangle_A$ for some constant $c$ and you want to know if this imply the existence of another inner product $\langle\cdot|\cdot\rangle_B$ that is unitary, right? $\endgroup$
    – Masacroso
    Jul 11, 2017 at 0:37
  • $\begingroup$ Yeah that's right, I just thought I would give some motivation. $\endgroup$
    – Paul Vienhage
    Jul 11, 2017 at 1:23
  • $\begingroup$ You have $0\leq \langle Uv|Uv\rangle_{A}=c\langle v|v\rangle_{A},$ and $0\leq \langle v|v\rangle_{A},$ so $c\geq 0.$ Then define $\langle u|v\rangle_{B}=c\langle u|v\rangle_{A},$ and this is a perfectly valid inner product so long as $c>0$ (i.e., nonzero). $\endgroup$
    – RideTheWavelet
    Jul 11, 2017 at 7:24
  • $\begingroup$ Hey I'm not sure I understand, if $\langle u | v \rangle _B = c \langle u | v \rangle _A $ then $ \langle U u | U v \rangle _B =c \langle U u | U v \rangle _A = c^2 \langle u | v \rangle _A = c \langle u | v \rangle _B $ which makes U it not unitary. $\endgroup$
    – Paul Vienhage
    Jul 11, 2017 at 15:08

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No, this is not possible. For instance, if $U=dI$ is a scalar multiple of the identity, then it is near unitarizable with $c=|d|^2$ for any inner product. In particular, $U$ is unitary for one inner product iff it is unitary for every inner product iff $|d|^2=1$.

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