MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Ok, so this might be a really naive question (and clearly related to Special values of Artin L-functions).

The Stark conjecture postulates that all Artin L-functions has a transcendental (over $\mathbb{Q}$) leading coefficient at $s=1$ (or $s=0$) in the Taylor expansion (well, it implies this).

My question: are all such leading coefficients, at various values of $s$, transcendental over $\mathbb{Q}$, or can there be algebraic ones?

Edit: I exclude the Dirichlet L-functions for which there is a positive answer to my question.

So the question becomes: given an Artin L-function (corresponding to some representation of some Galois group), are there always $s\in\mathbb{C}$ such that the leading term is algebraic?

My gut feeling tells me that there can indeed be algebraic ones.

As I said, this is in all probability an extremely naive question, but a fleeting google search led to nothing (at least nothing I could understand).

share|cite|improve this question
If you look at Dirichlet L-functions, there are in fact algebraicity results. For example if $\chi$ is a non trivial Dirichlet character then $L(\chi,m) \in \mathbf{Q}(\chi)$ for any $m \in \mathbf{Z}_{\leq 0}$ (it can be expressed in terms of Bernoulli numbers) and such values can indeed be nonzero (in fact $L(\chi,m) \neq 0$ iff $\chi(-1)=(-1)^{m+1}$). – François Brunault Jun 14 '11 at 10:56
There are other examples with Artin L-functions of degree >1, for example the Klingen-Siegel theorem tells you that $\zeta_K(-1) \in \mathbf{Q}^{\times}$ for any totally real number field $K$. – François Brunault Jun 14 '11 at 11:02
So I think you want to consider the Artin motive $M_\rho$ associated to the Artin representation $\rho$. Your question is then for which integers $m \geq 0$ the motive $M_\rho(m)$ is critical AND the associated period $c^+(M_\rho(m))$ (defined by Deligne) is trivial. I'm pretty sure this can be worked out but don't know a precise reference. You could try to read Deligne's \emph{Périodes d'intégrales} (if you don't know it already) and make everything explicit in the case of an Artin motive. – François Brunault Jun 14 '11 at 11:41
For Artin motives things are worked out more explicitly in the following reference : Ramakrishnan, Regulators, algebraic cycles and values of L-functions, Section 3 (see Conj. 3.3.9). Using the notations of loc. cit. you want to determine for which integers $m$ one has $d_m = 0$ ($m$ is critical) AND $R_m=1$ (the period is trivial). This depends only on the behaviour of $M_\rho$ at infinity (= the action of complex conjugation). – François Brunault Jun 14 '11 at 12:06
A quick comment on why the period is trivial: an Artin motive is just a bunch of points (geometrically speaking), so there is no non-trivial geometry present that could give rise to an interesting period. Regards, Matthew – Emerton Jun 15 '11 at 2:08
up vote 7 down vote accepted

Ok so I think I have worked out which integers are critical. Let $F$ be a finite Galois extension of $\mathbf{Q}$ and $\rho:\operatorname{Gal}(F/\mathbf{Q}) \to GL(V)$ be an irreducible Artin representation, with $\rho \neq 1$.

First I should mention that there is a functional equation relating $L(\rho,s)$ and $L(\overline{\rho},1-s)$ but it involves some power of $\pi$, so one should be careful when formulating an algebraicity conjecture. So let me restrict to the case $s=1-m$ where $m$ is an integer $\geq 1$.

As proved in Deligne's article Valeurs de fonctions L et périodes d'intégrales, if $1-m$ is critical then the associated period is trivial and Deligne's conjecture is true, so that $L(\rho,1-m)$ is a nonzero algebraic number. One can even prove, using a theorem of Siegel, that $L(\rho,1-m)$ belongs to the number field generated by the values of the character of $\rho$, and that $L(\rho,1-m)^{\sigma}=L(\rho^\sigma,1-m)$ for any $\sigma \in \operatorname{Aut}(\mathbf{C})$, for a detailed proof see Thm 1.2 in Coates-Lichtenbaum, On $\ell$-adic zeta functions, Annals of Math. 98 n°3 (1973) (I'm grateful to Junkie for pointing out to me this reference). If $1-m$ is not critical then $L(\rho,1-m)=0$ but algebraicity of the leading term is not expected (regulators are expected to be transcendental, but even their irrationality is very difficult to prove).

Thus it remains to find the critical integers. By definition $1-m$ is critical iff the gamma factor $L_\infty(\rho,s)$ has no pole at $s=1-m$. By definition $L_\infty(\rho,s)=\Gamma_{\mathbf{R}}(s)^{\dim V^+} \Gamma_{\mathbf{R}}(s+1)^{\dim V^-}$ where $\Gamma_{\mathbf{R}}(s)=\pi^{-s} \Gamma(s/2)$ and $V^{\pm}$ is the $\pm$-eigenspace for the action of $\rho(c)$, where $c \in \operatorname{Gal}(F/\mathbf{Q})$ is a choice of complex conjugation. A small computation then gives :

$$s=1-m \textrm{ is critical if and only if } V=V^{(-1)^m}.$$

Examples and remarks.

  1. If $\rho$ is $1$-dimensional, this is consistent with the situation for Dirichlet characters.
  2. If $F$ is totally real then $V=V^+$ so the result is consistent with the Klingen-Siegel theorem.
  3. If $\rho$ is an odd irreducible $2$-dimensional Artin representation (correponding to a weight 1 newform thanks to the proof of Serre's conjectures) then $\dim V^+ = \dim V^{-}=1$ so there is no critical integer for $L(\rho,s)$.
  4. On the other hand if $\rho$ is $2$-dimensional and even (corresponding conjecturally to a non-holomorphic Maass cusp form), then half of the integers are critical for $L(\rho,s)$.
  5. It is possible to extend the above analysis to Artin representations associated to arbitrary Galois extensions of number fields.
share|cite|improve this answer
Nice! Thank you very much! – Daniel Larsson Jun 14 '11 at 21:45
Regarding #4, Kevin Buzzard and I did some computations for a specific case, and found numerically that the values were algebraic (cyclotomic of degree 5 here).… – Junkie Jun 14 '11 at 22:15
@Junkie : Very interesting, thanks! A side question : could you actually prove that the L-values were equal to the algebraic numbers you guessed ? Siegel's theorem gives the algebraicity, but maybe doesn't give a recipe to compute the exact value. – François Brunault Jun 15 '11 at 1:17
I do not really know. Looking at Coates/Lichtenbaum (where Siegel's theorem is stated): on page 503, it seems the special $L$-value can be written rather explicitly, involving Bernoulli polynomials and a sum over a ray class group $R(f)$. They say $L(\Phi,-n)=\star\sum_{r\in R(f)} \Phi(r)Q_{n+1}(r)$, where the $\star$ is power of 2 depending on the parity of $n$, and the $Q_{n+1}(r)$ are rational and given fairly explicitly, involving a sum over $E(f)$ equivalence classes where $E(f)$ is units that are 1 modulo $f$. It would be interesting to explicit this. – Junkie Jun 15 '11 at 1:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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