Explicit formula for invariant inner product of the standard representation of $S_3$

Question

Let $V$ be a representation of a group $G$ over $\mathbb{C}$. Given the standard Hermmitian inner product $\langle\cdot,\cdot\rangle$ on $V$ we can always define a $G$-invariant inner product by $$\langle x,y\rangle_\text{new} := \frac{1}{|G|} \sum_{g \in G} \langle g x, g y\rangle$$ Can anyone calculate this explicitly for the standard representation of $S_3$? Is there any easy way to do this in general (for other reps of $S_n$) aside from the above method?

Answer 1

Let $V = \Bbb C^2$ with the standard basis, and let $\rho: S_3 \to GL_2(\Bbb C)$ be given by:

$\rho(e) = I$, $\rho((1\ 2\ 3)) = \begin{bmatrix}-\frac{1}{2}&-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&-\frac{1}{2} \end{bmatrix}$, $\rho((1\ 3\ 2)) = \begin{bmatrix}-\frac{1}{2}&\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&-\frac{1}{2} \end{bmatrix}$

$\rho((2\ 3)) = \begin{bmatrix}1&0\\0&-1\end{bmatrix}$, $\rho((1\ 2)) = \begin{bmatrix}-\frac{1}{2}&\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&\frac{1}{2} \end{bmatrix}$, $\rho((1\ 3)) = \begin{bmatrix}-\frac{1}{2}&-\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&\frac{1}{2} \end{bmatrix}$

As one can readily verify, these are isometries under the standard Hermetian inner product, so for each $g \in S_3$ we have:

$\langle gu,gv\rangle = \langle u,v\rangle$, and our $G$-invariant inner product is the same as the standard one.