most recent 30 from http://mathoverflow.net 2013-05-24T01:59:33Z http://mathoverflow.net/feeds/user/31172 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120749/what-is-an-algebro-geometric-family-of-modular-forms/120835#120835 Michigan J Frog 2013-02-05T05:35:43Z 2013-02-05T05:35:43Z

<p>$\newcommand\Q{\mathbf{Q}}$ $\newcommand\A{\mathbf{A}}$ $\newcommand\AQ{\A_{\Q}}$</p> <p>How about instead of considering lisse sheaves, derived pushforwards, étale cohomology, modular forms, blah blah blah, you consider instead the following situation: over the affine line $\Q[t]$, you can consider the equation $x^2 - t$; it's a family of quadratic extensions. If you like, you can turn this into a smooth map of curves (eliminating the bad fibre at $0$) $\pi: X \rightarrow Y$, and you can consider the lisse sheaf $R^0 \pi_* \mathbf{Z}_l$, the stalks of which are Galois representations corresponding to the quadratic character of the associated quadratic extension (along with the trivial character, which I'll suppress). All the Galois representations corresponding to rational points are automorphic/modular for $\mathrm{GL}_1(\AQ)$ - that's a theorem of Gauss called quadratic reciprocity. What does this family look like? Well, it is what it is; the global data is related to the factorization of $t$ in the expected way, and it's not really a "family" is any analytic sense. There is, however, one useful fact to observe about how this family behaves as one varies $t$. Namely, the characters $\chi_t$ are <em>locally</em> constant. That is, if $t$ and $s$ are close in $\Q_v$ for any place $v$, then the local characters $\chi_{t,v}$ and $\chi_{s,v}$ are equal. For example, if $t$ and $s$ are both the same sign, then $\chi_{t,v}(-1) = \chi_{s,v}(-1)$. This is Krasner's lemma (we use smoothness here). It turns out that this "local constancy" of Galois representations is true more generally, I think Kisin proved something along these lines (although you should think of that result as also being Krasner's Lemma). </p>