## Is the “L-function of the complex cohomology” of a motive equal to the L-function of its l-adic realization?

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Let's say I have a motive in $\mathcal{M}_{num}(K)$ ($K$ a number field). For each prime $l$ there is a realization of this motive in terms of etale cohomology with coefficients in $\mathbb{Q}_l$. This has more structure than being a mere vector space: it is a representation of $Gal(K)$! This representation has an $L$-function. As I understand it, the $L$ function doesn't depend on $l$.

What I wonder is whether there is something special about etale cohomology with coefficients in an $l$-adic field, or whether every Weil cohomology has an $L$-function attached to it that would equal the $L$-function of any other realization.

The usual (singular) cohomology with the complex topology, is also a representation of $Gal(K)$ (factoring through the motivic Galois group, if this means anything to you). Is it true, then, that the $L$-function attached to this representation is the same as the $L$-function of the realizations via etale cohomology with coefficients in $\mathbb{Q}_l$?

How does one go about proving equalities between $L$-functions of different realizations of the same motive?

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Although I'm not a number theorist, I'm a bit skeptical as to whether $Gal(K)$ acts on singular cohomology with $\mathbb{Z}$ or $\mathbb{Q}$-coefficients in an interesting way. Or do I misunderstand the set up? Perhaps the work of Denef, Loeser and Kapranov on motivic zeta functions, might be what you are looking for. – Donu Arapura Aug 10 2011 at 10:40

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I think that the problem is that maps are going the wrong way: the (conjectural) motivic Galois group should map to $Gal(K)$. – Donu Arapura Aug 10 2011 at 10:59
How would you propose to construct such an $L$-function for, e.g., an elliptic curve? The Betti realization seems to be insensitive to moduli, while the usual $L$ function is very sensitive. – S. Carnahan♦ Aug 11 2011 at 3:55

## 1 Answer

As far as I can tell, the Galois action that you assert exists doesn't in fact exist.
First of all, to define the motivic Galois group, you have to choose a realization; it seems that you are choosing the Betti realization, so the motivic Galois group is a pro-algebraic group $G$ over $\mathbb Q$ which acts on $H^i(X(\mathbb C),\mathbb Q)$ for every smooth projective variety $X$ over $\mathbb K$.

If we consider the $\mathbb Q_{\ell}$ points of $G$, then we get a map $\rho_{\ell}:Gal(\overline{K}/K) \to G(\mathbb Q_{\ell})$, corresponding precisely to the Galois action on $\ell$-adic cohomology (and using the canonical isomorphism of $\ell$-adic cohomology with $H^i(X(\mathbb C),\mathbb Q_{\ell})$). This just encodes Galois actions on $\ell$-adic cohomology, by its construction.

There is also a map of pro-algebraic groups $G \to Gal(\overline{K}/K)$, given by the action of $G$ on $H^0$s. (This is the map that Donu refers to in his comment.) The maps $\rho_{\ell}$ are sections to this map (after passing to $\mathbb Q_{\ell}$-valued points), again by construction.

There is nothing in this formalism (and it is just formalism) which suggests a natural action of the Galois group on $H^i(X(\mathbb C),\mathbb Q)$ for $i > 0$.

As for proving the equality of $L$-functions for different realizations (say different $\ell$-adic realizations), this is a very difficult problem which remains open in general, as far as I know. (If one wants to compare the $L$-functions built from the $\ell$-adic and $\ell'$-adic realizations, then the equality of Euler factors at primes of good reduction different from $\ell$ and $\ell'$ is due to Deligne as part of his proof of Weil's Riemann hypothesis. If $\ell$ or $\ell'$ is a prime of good reduction, then the equality of Euler factors at $\ell$ or $\ell'$ is known but is much more involved (since even defining the Euler factors in this case is much harder). At primes of bad reduction, my understanding is that equality is not know in general.)

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So as things stand, there is no way to define the L-function of a motive without the notion of l-adic cohomology? – James D. Taylor Aug 11 2011 at 16:20
Well, you could try to directly "count points" on the reduction mod $p$, but you have the problem of non-uniqueness of models, especially at primes of bad reduction, and this is related to the whole difficulty of working at those points.    – Emerton Aug 11 2011 at 18:15
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