# On the pole of local L-function

Let $F$ be a number field and $v$ a finite place of $F$.

Let $\chi_v$ be a unramified unitary character of $F_v$.

Then we define local L-function $L_v(s,\chi_v):=\frac{1}{1-\chi_v(\omega)q^{-s}}$ where $\omega$ is a uniformizer of $F_v$ for $s>1$.

Then I know that it has a unique meromorphic continuation to $\mathbb{C}$ and has simple a pole at s=$0$ and $1$ if $\chi_v=1$ and no poles if $\chi_v \ne 1$. Am I right?

Then I am confusing beacause I knew that meromorphic contimution is unique and $\frac{1}{1-\chi_v(\omega)q^{-s}}$ has meromorphic continuation to $\mathbb{C}$ by itself.(Furthermore, $\frac{1}{1-\chi_v(\omega)q^{-s}}$ has no pole at $s=1$ and may have a simple pole at $s=0$ depending $\chi_v=1$ or not.

But this meromorphic continution does not fit into the above argument. Because in the above meromorphic continuation definition of local $L$-function, $L_v(s,\chi_v)$ may have a pole at $s=1$.

Why people defined local L-function in this way despite the above natural explcit definition? Is it for the convergence of global L-funtion?

I am also wondering why the global $L$-function $L(s,\chi)$ for non trivial character would have at most simple pole although many local $L$-functions have simple pole there.

Is there a point I am misunderstanding?

Since I am jest a beginner in this area, any tiny comments will be appreciated.

No, you are not right. The local $L$-function $L(s,\chi_v)=(1-\chi_v(\omega)q^{-s})^{-1}$ has infinitely many poles, namely the solutions of the equation $q^s=\chi_v(\omega)$. All these poles are simple of course. Also, it has no zeros. The local $L$-function can be more complicated, but it is usually a product of the above kind of factors. Similarly, a normalized gamma function like $\pi^{-s/2}\Gamma(s/2)$, or a product of shifted versions of it, which occur as local $L$-functions at archimedean places, has infinitely many poles but no zeros.

The local $L$-function has nice properties and tell a lot about the underlying local object (e.g. an admissible representation of $GL_n(\mathbb{Q_p})$). The $L$-function of a global object (e.g. a Dirichlet character) is the product of the local $L$-functions associated with the various local factors of the global object (e.g. various local characters coming from the same Dirichlet character). It has even nicer properties, and properties that are even harder to access. More precisely, the product formula is only valid in a half-plane (usually $\Re s>1$), and elsewhere it is defined by analytic continuation. This also explains why the poles of the local factors are not poles of the global $L$-function.

You have to learn the theory to get familiar with these things. I recommend the following books (already the first one answers your questions):

Davenport: Multiplicative number theory

Neukirch: Algebraic number theory

Weil: Basic number theory

Bump: Automorphic forms and representations

Goldfeld-Hundley: Automorphic representations and $L$-Functions for the general linear group

• Another, similar, common misconception is that "local functional equations" should combine to give the "global functional equation", and this is not the case. – paul garrett Aug 21 '14 at 12:47
• @paulgarrett: I agree, but of course the epsilon factors combine nicely! – GH from MO Aug 21 '14 at 12:50
• Thanks for your clear answer. As you recommeded, I have read Tate's thesis contatined in Bump's book. And I learned that the above local L-function is the same form obtained from the meromorphic continuation of local zeta integral when the test function is the characteristic function. And so I think if $\chi$ is a Hecke character, the $L_v(\chi_v,x)$ would have no pole for $Re(s) >0$ and it may have pole only at $\Re(s)=0$. Your remark that the pole of global L-function do not pertains to that of local L-function is striking to me. I confused it for a long time. Thanks for your illumination. – Monty Aug 23 '14 at 5:57
• On the other hand, may I ask one another question? When $\chi$ is a nontrivial Hecke character of $\mathbb{A}^{\times}/F$ , then how do we know that there is no complex number $s$ such that $\chi \ne |\cdot|^s$ ? Because in the bump's book, it says that every quasi character $\chi$ of $\mathbb{A}^{\times}/F$ is a product of a unitary Hecke character $\chi_1$ and $|\cdot|^s$ for some complex $s$. – Monty Aug 23 '14 at 5:58
• @classnumb: The statement in your second sentence is false. Indeed, for every nonzero complex number $s$, the function $\chi=|\cdot|^s$ on $\mathbb{A}^{\times}/F$ is a nontrivial Hecke character. On the other hand, the statement in Bump's book is true, it is a special case of Corollary 2 in Section VII.3 of Weil: Basic number theory. I recommend studying Sections VII.3 and VII.4 in Weil's book. Also, for new questions you might have, open a new question at MathOverflow (i.e. do not ask further questions as comments here). – GH from MO Aug 23 '14 at 8:36