Why does the determinant come in for Artin L-Functions?

Question

Let $L/K$ be a Galois extension of number fields. For $\mathfrak p$ a prime of $K$, unramified in $L$, the Frobenius elements $\sigma_{\mathfrak P}$ for $\mathfrak P \mid \mathfrak p$ are conjugate, so if $\chi$ is a class function on $G = \operatorname{Gal}(L/K)$, $\chi(\sigma_{\mathfrak p}) := \chi(\sigma_{\mathfrak P})$ is well defined.

Assume that $L/K$ is abelian, and $\chi$ is a character of $\operatorname{Id}(c)/P_c \mathfrak N(c)$ (which is isomorphic to $G$ via the Artin map). The Weber L-function is defined for $\operatorname{Re}(s) > 1$ by

$$L(s,\chi) = \prod\limits_{\mathfrak p} L_{\mathfrak p}(s,\chi) = \prod\limits_{\mathfrak p} (1 - \chi(\mathfrak p) N(\mathfrak p)^{-s})^{-1}$$

Identify $\chi$ as a character of $G$ via the Artin map. When $\mathfrak p$ is unramified, it identifies with $\sigma_{\mathfrak p}$. Now instead of a character of $G$, take a finite dimensional representation $\rho: G \rightarrow \operatorname{GL}_n(\mathbb C)$ of $G$ with character $\chi$. Consider the formal logarithm of the local factor:

$$\log L_{\mathfrak p}(s,\chi) = \log (1 - \chi(\sigma_{\mathfrak p}) N(\mathfrak p)^{-s})^{-1} = \sum\limits_{k=1}^{\infty} \frac{\chi(\sigma_{\mathfrak p}^k)}{kN(\mathfrak p)^{sk}}$$

The notes I'm reading say "This exponentiates to $L_{\mathfrak p}(s,\chi) = \operatorname{Det}(I_n - N(\mathfrak p)^{-s} \rho(\sigma_{\mathfrak p}))^{-1}$." I don't understand how this is done. Where does the determinant arise from the trace $\chi(g) := \operatorname{tr}(\rho(g))$ and the exponential map?

Answer 1

Where does the determinant arise from the trace $χ(g):=tr(ρ(g))$ and the exponential map?

In general, we have the identity $\exp(\mbox{tr } A) = \det(\exp A)$. This is obvious for diagonal matrices, then it follows easily for diagonalizable matrices, then by continuity for all matrices since diagonalizable matrices are dense in the space of all $n \times n$ matrices.