In the book The Symmetric Group the author says:
Let $\chi$ and $\psi$ be characters of the $G$-module $V$.
By picking an orthonormal basis for $V$, we obtain a matrix representation $Y$ for $\psi$, where each $Y(g)$ is unitary; i.e.,
\begin{equation} Y(g^{-1}) = Y(g)^{-1} = \overline{Y(g)}^{t}. \end{equation}
How do we know that $Y$ is unitary with respect to every orthonormal basis?
I have also had a look at the these two posts:
However, I cannot see why having an orthonormal basis for $V$ automatically implies that $Y$ is unitary. Clearly we cannot just write the columns of $Y$ by taking the basis vectors, as suggested in post 1 because this will probably not correspond to the transformation of $g \in G$ in $V$.
Thank you very much for your help!
$\endgroup$1 Answer
$\begingroup$If you have a representation $\varphi:G\rightarrow GL(V)$ of a finite group on a finite-dimensional complex vector space, then there is a procedure for building an inner product on $V$ with respect to which $G$ acts unitarily. Choose any Hermitian inner product $\langle,\rangle$ on $V$ and consider $\langle,\rangle_2$ defined by $$\langle v,w\rangle_2=\frac{1}{\vert G\vert}\sum_{g\in G}\langle gv,gw\rangle.$$ This is a $G$-invariant Hermitian inner product on $V$. So, if you take an orthonormal basis with respect to $\langle,\rangle_2$, then your matrix representatives will be unitary.
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