# If $\Pi$ and $\Sigma$ agree at almost all places, then the central character of $\Pi$ corresponds to $\operatorname{Det} \Sigma$

Let $$\Sigma$$ be an $$n$$-dimensional representation of the global Weil group $$W_F$$ for a number field $$F$$, and $$\Pi$$ an automorphic representation of $$\operatorname{GL}_n(\mathbb A_F)$$. Suppose that at every place $$v$$ of $$F$$ where $$\Sigma_v$$ and $$\Pi_v$$ are both unramified, we have $$\Pi_v \leftrightarrow \Sigma_v$$ under the local Langlands correspondence.

In Guy Henniart's proof on the local Langlands correspondence, he claims that $$\operatorname{Det} \Sigma$$ then corresponds to the central character of $$\Pi$$ under the isomorphism $$W_F^{\operatorname{ab}} \cong \mathbb A_F^{\ast}/F^{\ast}$$ of global class field theory.

Why should this be the case? It is true that at almost all places $$v$$, $$\operatorname{Det} \Sigma_v$$ identifies with the central character of $$\Pi_v$$ under the isomorphism $$F_v^{\ast} \cong W_{F_v}^{\operatorname{ab}}$$. We have for $$a = (a_v) \in \mathbb A_F^{\ast}$$,

$$\varpi_{\Pi}(a) = \prod\limits_v \varpi_{\Pi_v}(a_v)$$

so it seems like something could go wrong with $$\varpi_{\Pi}$$ at the ramified places.

Now that I have thought about it a little more, I think this just comes down to the following fact: If $$\chi = \otimes \chi_v$$ is a character of $$\mathbb A_F^{\ast}/F^{\ast}$$, and $$S$$ is a finite set of places of $$F$$, then $$\chi$$ is completely determined by the characters $$\chi_v : v \not\in S$$. This is because if $$\eta = \otimes_v \eta_v$$ is another character of $$\mathbb A_F^{\ast}/F^{\ast}$$ with $$\eta_v = \chi_v$$ for all $$v \not\in S$$, then $$\chi \eta^{-1}$$ defines a character of $$\prod\limits_{v \in S} k_v^{\ast}$$ which is trivial on the diagonal embedding $$\Delta$$ of $$k^{\ast}$$. But $$\Delta$$ is dense in this product by weak approximation, so in fact $$\chi = \eta$$ on $$\mathbb A_F^{\ast}$$.

Now $$\operatorname{Det} \circ \Sigma$$ corresponds to a character of $$\mathbb A_F^{\ast}/F^{\ast}$$ which agrees at almost all places with the central character of $$\Pi$$ (also trivial on $$F^{\ast}$$), so they must be the same character.