# Are all representations of a finite group unitary?

Every representation $(D,V)$ of a finite group $G$ is

*equivalent*to a unitary representation.It is often termed as

*Weyl's unitary trick*. This works by simply redefining your inner product by averaging over on the space $V$. This smoothening trick works precisely because of finite number of elements and invariance of sum of finite elements (i.e. $\sum_{g \in G} D(gh) = \sum_{gh \in G} D(gh) = \sum_{g \in G} D(g)$).

$$\langle v| w \rangle = \frac{1}{|G|}\sum_{g \in G} \langle D(g)v| D(g)w \rangle$$

You can verify how $\langle v| w \rangle = \langle D(h)v| D(h)w \rangle$ for some $h \in G$. In fact, this claim is similar to that of existence of a common basis change matrix $S$ such that $D'(g) = SD(g)S^{-1}$ is a unitary representation.

They are all *equivalent* to unitary representations. It's not quite the same thing. Merely checking unitarity might not do the trick.

For instance \begin{align} \Gamma(e)&=\left(\begin{array}{cc} 1&0 \\ 0 & 1\end{array}\right)\, ,\qquad \Gamma(P_{12})=\left(\begin{array}{cc} 1&-1 \\ 0 & -1\end{array}\right)\, ,\qquad \Gamma(P_{13})=\left(\begin{array}{cc} 0&1 \\ 1 & 0\end{array}\right)\\ \Gamma(P_{23})&=\left(\begin{array}{cc} -1&0 \\ -1 & 1\end{array}\right)\, ,\qquad \Gamma(P_{123})=\left(\begin{array}{cc} 0&-1 \\ 1 & -1\end{array}\right)\, ,\qquad \Gamma(P_{132})=\left(\begin{array}{cc}-1&1 \\ -1 & 0\end{array}\right) \end{align} is an irreducible representation of $S_3$ but clearly not unitary. It can be conjugated to a unitary rep, i.e. there is a common $V$ so that $V\Gamma V^{-1}$ is unitary.