Timeline for Is the following the right definition of $L$-functions (on the Galois side)?
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| Feb 3, 2013 at 22:12 | comment | added | Vesselin Dimitrov | ... where the latter, by the Lefschetz formula, is also equal, simply, to $\prod_{x}' (1-|k(x)^{-s}|)^{-1}$, where the product is over the closed points $x$ of $X$ of residue characteristic not in $S$. | |
| Feb 3, 2013 at 22:04 | comment | added | Vesselin Dimitrov | If you take a smooth model $X$ over the localization $\mathbb{Z}_S$ at a finite set of primes $S$ (which is always possible by clearing denominators), then your partial zeta function is $\prod_i \prod_{p \notin S} \det \Big( 1 - p^{-s}\mathrm{Frob}_p | H^i(\overline{X}_{\mathbb{Z}/p},\mathbb{Q}_l) \Big)^{(-1)^i}$. | |
| Feb 3, 2013 at 22:00 | comment | added | Vesselin Dimitrov | Yes, I did not intend to write that there is no $\mathbb{Z}$-model in general, only that there is no canonical choice. | |
| Feb 3, 2013 at 21:57 | comment | added | Vesselin Dimitrov | Yes, there is no preferred integral model in general (with the notable exception of curves and abelian varieties). If you take an arbitrary integral model, you would only get the zeta function up an undetermined rational factor, which for many purposes is fine. But for obtaining the precise zeta function, the way out is to look at the Frobenius action on $H^i(\overline{X},\mathbb{Q}_{l})^{I_p}$ (which is well-defined precisely because we take inertial invariants). | |
| Feb 3, 2013 at 21:54 | comment | added | Vesselin Dimitrov | With the above definition, $\prod_i L_i$ is the Hasse-Weil zeta function only up to a rational factor. In the above, $L_{i,p}(s) := P_{i,p}(p^{-s})^{(-1)^i}$, and $L_i := \prod_p L_{i,p}$. | |
| Feb 3, 2013 at 21:53 | comment | added | Nicole | What do you mean that there is no integral model in general? Can't you just clear denominators in all of the polynomials that define $X_{\mathbb{Q}}$? Did you mean to say that there's no integral model with good reduction at every prime in general? | |
| Feb 3, 2013 at 21:43 | comment | added | Vesselin Dimitrov | (I meant to write: the product over the closed points; that is to say, those of finite residue field $k(x)$). | |
| Feb 3, 2013 at 21:42 | comment | added | Vesselin Dimitrov | By the way, the Hasse-Weil zeta function of a finite type scheme $X_{\mathbb{Z}}$ over $\mathbb{Z}$ (the integral model, that is), is simply the product $\prod_x (1 - |k(x)|^{-s})^{-1}$ over the closed space. But since there is no $\mathbb{Z}$-model in general, what you do instead is to take your $P_{i,p}$ to be the char. poly. of $\mathrm{Frob}_p$ acting on $H^i(\overline{X},\mathbb{Q}_l)^{I_p}$, the invariants of inertia. This does not refer to a $\mathbb{Z}$-model, and works to define the L-function of an abstract Galois representation. See this: ucl.ac.uk/~ucahmki/ihes3.pdf | |
| Feb 3, 2013 at 21:31 | comment | added | Nicole | By the way, with your definition of $L_i$ is it true that $\prod_i L_i$ is the Hasse Weil zeta function? Or is it an alternating series? And is $L_{i,p}$ (for good $p$'s) the characteristic polynomial itself, or is it $P_{i,p}^{(-1)^i}$ where $P_{i,p}$ is the characteristic polynomial? | |
| Feb 3, 2013 at 21:27 | comment | added | Nicole | $L_i$ is the product $\prod_{p \in Spec(\mathbb{Z})} L_{i,p}$? | |
| Feb 3, 2013 at 21:27 | comment | added | Nicole | Yeah, that was silly of me that I took the product over the primes instead of the $i$'s... It was just a typo. So essentially what I'm taking from this is the following: Let $X_{\mathbb{Z}}$ be any model over $\mathbb{Z}$. Then what you call the "bad primes" are the primes of bad reduction of this model? (Where the fibers are not geometrically irreducible and smooth?) And that for the good primes, you define the $L_{i,p}$ in the way I defined it above, but for the bad primes nobody quite knows what to do. But somehow conjecturally there should be a definition at the bad primes so that | |
| Feb 3, 2013 at 20:48 | history | answered | Felipe Voloch | CC BY-SA 3.0 |