# Determinant of finite group representation

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 $$\rho$$?

• 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. Commented Sep 4 at 22:52
• Oh thanks, the exterior power is a great hint. I think I can work it out now. Commented Sep 4 at 23:07
• 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. Commented Sep 5 at 0:46

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.