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?

enter image description here


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.

  • $\begingroup$ Thanks. I don't know how I have never seen this before. $\endgroup$ – D_S Apr 1 '18 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.