For a Dirichlet character $\chi: \left(\mathbb{Z}/p\mathbb{Z}\right)^{\times} \to \mathbb{C}$, the Dirichlet L function is $$\prod_{q \neq p} (1 - \chi(q)q^{-s})^{-1}$$ If we lift this character to $\mathbb{A}_{\mathbb{Q}}^{\times}$, we would get a Hecke character. The lifting comes from $$\mathbb{A}_{\mathbb{Q}}^{\times} = \mathbb{R}_{>0} \times \mathbb{Q}^{\times} \times \prod_{q} \mathbb{Z_q}^{\times}$$ and the projection $$\prod_{q} \mathbb{Z_q}^{\times} \to \mathbb{Z_p}^{\times} \to \frac{\mathbb{Z}_p^{\times}}{1+p\mathbb{Z}_p} \cong \left(\mathbb{Z}/p\mathbb{Z}\right)^{\times} \stackrel{\chi}{\longrightarrow} \mathbb{C}$$ (See e.g. the answers here)

However, the local $L$-factor for the lifted character at finite place other than $p$ seems to be $(1 - \chi(q)^{-1} q^{-s})^{-1}$, since if we restrict the lifted character to $\mathbb{Q}_q^{\times}$, we see that $$(1,1,\cdots,q,\cdots,1) = (1/q, 1,\cdots,1,\cdots,1) \cdot q \cdots (1, 1/q, \cdots, 1, \cdots, 1/q)$$ where the first entry is for real place, second entry is for the first prime, and so forth. $q$ is at the $q$-th entry. So by the lifting process, we would be considering $\chi(1/q)$ rather than $\chi(q)$.

Question: Where did I make mistake so that $\chi(1/q)$ arises instead of $\chi(q)$ in the local $L$-factor for $q \neq p$?



There is a question of conventions here. In lifting a Dirichlet character $\chi$ mod $N$ to a character $\chi_{\mathbf{A}}$ of $\mathbf{A}^\times / \mathbf{Q}^\times$, you can either

(1) arrange that the restriction of $\chi_{\mathbf{A}}$ to $\widehat{\mathbf{Z}}^\times \subset \mathbf{A}^\times$ coincides with $\chi$ under the natural map $\widehat{\mathbf{Z}}^\times \to (\mathbf{Z}/N\mathbf{Z})^\times$;

(2) arrange that $\chi_{\mathbf{A}}(\varpi_q)$, for $q \nmid N$ prime and $\varpi_q$ any uniformizer at $q$, agrees with $\chi(q)$.

Both of these would seem like pretty reasonable things to ask for, but you can't have both at once -- as you've noticed, one is the inverse of the other -- and the usual convention is (2). If you use convention (2) not convention (1), then the adelic and non-adelic definitions of the L-functions coincide.

  • $\begingroup$ Thanks! Is there any situation where we would prefer convention 1 other than its naturality? $\endgroup$ – user27126 Jul 14 '13 at 18:01

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