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I was reading the question: Symmetric and exterior power of representation regarding how to determine the character of an exterior power of a representation from the original representation. One of the responses mentioned that the formula for $\wedge^2$ is \begin{equation} \chi_{\wedge^2V}(g)=1/2 (\chi_V(g)^2-\chi_V(g^2))\end{equation}

The user 'draks' observed that this looks like the formula for variance and asked if there was any connection. There were no responses and I thought it was an interesting question so I'm reposting it as a full question.

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goatman2743
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  • Pick a basis on $V$. I think the probability space should be $\Omega=S(V)$ the elements in $V$ of norm one, and the random variable should be $X(v) = v^\vee \rho(g) v $, where $\rho(g)$ is the value of the representation on $g$ and $v^\vee\in V$ is the dual vector to $v\in V$. In particular, $E[X] = \int v^\vee \rho(g) v \ dV $ is the trace of $\rho(g)$, i.e. $E[X]=\chi_V(g)$, and similarly $E[X^2]= \int (v^\vee \rho(g) v) (v^\vee \rho(g) v) \ dV= \int (v^\vee \rho(g) \rho(g) v) dV = \int (v^\vee \rho(g^2) v) dV$. I don't know how to interpret the LHS in terms of $E[(X-\mu)^2]$ though. – Pulcinella May 06 '20 at 18:48

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