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Let $X$ be a variety defined over $\mathbb{Q}$. One has the usual Hasse-Weil zeta function.

Now, let's do a different construction. Base change $X$ to $\mathbb{C}$: $X_{\mathbb{C}}$. Now look at its pure numerical motive $hX_{\mathbb{C}}$ (living in $\mathcal{M}_{num}(\mathbb{C})$). I am given to believe that there is a way to define zeta functions on pure numerical motives (although I can't say I understand the construction).

Are these two the same? Could this be? It seems like one would lose a lot of (arithmetic) information when base changing to $\mathbb{C}$. For example, doesn't this imply that one would get the same function if one were to take the Hasse-Weil zeta function for any other $\mathbb{Q}$-model of $X_{\mathbb{C}}$?

Am I misunderstanding, or is this truly an uncanny phenomena? Is the construction conjectural? If not -- then it must mean that $X_{\mathbb{C}}$, a geometric object, contains all of the arithmetic information of a $\mathbb{Q}$-model of it. Why should that be true?

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    $\begingroup$ Half of your instincts are correct: you can't recover the zeta function from what happens at $\mathbb{C}$. You can see this already by just considering Dedekind zeta functions: certainly they carry more information than the degree of the extension. The zeta function is rather something of an amalgamation of what happens at every prime, or a kind of generating function for the number of solutions to some equation or what have you. $\endgroup$
    – Moosbrugger
    Aug 9, 2011 at 2:01
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    $\begingroup$ I think Dipendra Prasad and C.S. Rajan conjecture something close to what you write, that is, that the zeta function for smooth projective curves should almost be determined by $X_{\mathbb{C}}$. Maybe something like: If $X$ and $Y$ over a number field $F$ are isospectral for every complex embedding of $F$, then there is a finite extenstion $K$ of $F$ such that the zeta functions of $X_K$ and $Y_K$ are the same. Here, the spectrum refers to that of the Laplacian, when computed using a constant curvature metric. Obviously, such a statement, if true, is meant to be difficult and deep. $\endgroup$
    – Minhyong Kim
    Aug 9, 2011 at 2:41
  • $\begingroup$ I'm beginning to think that I did misunderstand. I guess the two things that should be equal are the Hasse-Weil zeta function of $X$ and the zeta function of the motive $hX$ living in $\mathcal{M}_{num}(\mathbb{Q})$, rather than $\mathcal{M}_{num}(\mathbb{C})$... $\endgroup$
    – James D. Taylor
    Aug 9, 2011 at 2:58

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