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Let $G$ be a finite group with $n$ elements. Let $\rho\colon G \to \operatorname{GL}(V)$ be a linear representation of $G$, where $V$ is a finite dimensional $\Bbb C$-vector space.

The action extends to a ring homomorphism from the group ring $\Bbb C[G]$ to $\operatorname{End}(V)$, again denoted by $\rho$.

Given a general element $\gamma = \sum x_i g_i \in \Bbb C[G]$, I'm interested in the determinant of the linear map $\rho(\gamma)\colon V \to V$, as a homogeneous polynomial in $x_1, \dots, x_n$ of (homogeneous) degree $\dim V$.

In particular, is there an easy way to compute this polynomial from the character of the representation $\rho$?

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    $\begingroup$ The determinant of an $m$ dimensional representation is the exterior $m$th power. To compute exterior powers, you need the power maps on conjugacy classes, and then there's a formula. For example, $\chi(g,\Lambda^2V)=(\chi(g,V)^2-\chi(g^2,V))/2$, and so on. $\endgroup$ Commented Sep 4 at 22:52
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    $\begingroup$ Oh thanks, the exterior power is a great hint. I think I can work it out now. $\endgroup$
    – WhatsUp
    Commented Sep 4 at 23:07
  • $\begingroup$ I think Frobenius’s method for computing the group determinant works more or less here. Maybe one needs to first cut down to a subset of G which is a basis for the image of the group algebra under the representation. Given a semisimple algebra and a basis I think Noether gave the formula for the determinant of the generic element. $\endgroup$ Commented Sep 5 at 0:46

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Thanks to the hint of @DaveBenson, I now realize that it's simply the usual relation between determinant and trace.

For example, if $V$ has dimension $2$, then for $\gamma = \sum x_i g_i \in \Bbb C[G]$ we have \begin{align} \det(\rho(\gamma)) &= \frac 1 2 (\operatorname{tr}(\rho(\gamma))^2 - \operatorname{tr}(\rho(\gamma^2)))\\ &= \frac 1 2 ((\sum_i \chi(g_i)x_i)^2 - \sum_{i, j}\chi(g_ig_j)x_ix_j). \end{align} When the dimension of $V$ increases, the formula gets complicated very fast.

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