Let $K$ be a number field, $L$ a finite abelian extension and $\chi \in \widehat{Gal(L/K)}$ a (non-trivial) character. If we multiply out the associated Artin L-function $L(\chi,s)$ we can write this as a finite sum of "partial" $\zeta$-functions $\zeta_\sigma (s)=\zeta_{\chi,\sigma}(s)$, namely we have $$L(\chi,s) = \sum_{\sigma \in Gal(L/K)} \chi(\sigma) \zeta_\sigma (s)$$ My first question is now if the partial zetas $\zeta_\sigma(s)$ are motivic?

(EDIT: this doesn't seem likely, see the comments)

By finite Fourier inversion we can express every partial zeta as linear combination of motivic L-functions: $$\zeta_\sigma(s) = \frac 1 {[L:K]}\sum_{\chi \in \widehat{Gal(L/K)}}\overline\chi (\sigma) L(\chi,s)$$ but I don't know if this answers my question.


Even if the partical zetas are not motivic it is pretty clear that they are not very far from being motivic. It seems reasonable to ask if known procedures in order to compute special values of motivic L functions carry over in a modified version to partial zetas.

Let us for example consider the case of the Dedekind zeta function $\zeta_K(s)$ by the analytic class number formula we know that the leading coefficient $\zeta_K^*(0)$ at $s=0$ is given by $$\zeta_K^*(0) = \frac {h_K R_K} {\omega_K} $$ where $h_K$ is the class number, $R_K$ the regulator and $\omega_K$ the number of roots of unity.

At present, my "favourite method" to prove the above formula is by means of calculating the volume of the Arakelov-Picard group of degree zero divisors $Pic^{(0)}(K)$. This is for example explained beautifully in the article of van der Geer and Schoof (http://arxiv.org/abs/math/9802121).

One question is now the following: Is it known if the volume of certain subspaces of the Arakelov-Picard group Pic(K) or certain quotients of the latter can be related to the leading coefficients at $s=0$ of partial zeta functions?

If not the Arakelov-Picard group, are their attempts to my question using volumina of different spaces?

(Unfortuanetly, I do not even know if the Arakelov-Picard group can be used to compute the leading coefficients of the Artin L functions $L(\chi,s)$ from above, but at least we have conjectures (Beilinson, ...) at hand which describe the leading coefficient...)


The motivation for my question is the following:

(Roughly speaking) the partial zetas $\zeta_{\sigma}$ appear prominently in the work of Stark on explicit class field theory, their first coefficients of the Taylor expansion at $s=0$ are expected to be logarithms of units in $L$, thereby producing explicit abelian extensions of $K$.

On the other hand Beilinson's program provides conjectural information about special values of motivic L-functions and I would like to understand if Stark's conjectures fit into this program or if they provide a refinement.

  • 4
    $\begingroup$ My gut feeling is that the partial zetas themselves aren't motivic---but as you know they're linear combinations of motivic L-functions. I see your point though: even if I knew all conjectures about motivic L-functions (order of vanishing, special values) then if I add them up there might be miraculous cancellation. The reason I think partial zeta functions can't be motivic is that the general formalism for motivic L-functions implies that they all have Euler product decompositions, and partial zetas don't in general. $\endgroup$ – Kevin Buzzard Nov 30 '10 at 20:36
  • $\begingroup$ Dear Kevin, thank you very much for your comment. I share your feeling (and of course the Euler product decomposition argument pins it down in a very convincing manner). I will reformulate my question in order to ask more precise questions about the relationship between partial zetas and motivic L functions. $\endgroup$ – user5831 Dec 1 '10 at 11:37
  • $\begingroup$ Dear Bora, while it is true that for Artin $L$-functions we have conjectures that explain their special values, none of them seem to explicitly interpret them in terms of volumes of natural sublattices of the units, say, or (which should come down to the same thing) of Arakelov Picard groups. I asked a question here a while ago about precisely this kind of conjectural interpretation, but the MO-community seems to be as unaware of such a conjecture as everyone I have personally spoken to. $\endgroup$ – Alex B. Dec 1 '10 at 13:41
  • $\begingroup$ The question I was referring to is mathoverflow.net/questions/41358/…. Should you have any comments on this or hear anything in the future, I would greatly appreciate if you could let me know. $\endgroup$ – Alex B. Dec 1 '10 at 13:48

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