About the Serre-Tate theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T18:57:32Z http://mathoverflow.net/feeds/question/58265 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58265/about-the-serre-tate-theorem About the Serre-Tate theorem genshin 2011-03-12T13:06:40Z 2011-03-12T18:07:20Z <p>It is somehow a general principle that the (infinitesimal) local behavior of a representable moduli functor $X$ at some point $x$ is closely related to the deformation problem of the structure parameterized by $x$: the deformation functor should be pro-represented by the formal completion of $X$ at $x$ ($X$ representable as assumed). In particular, if $x$ is a smooth geometric point, then the associated deformation functor is pro-represented by a ring formal power series of $d$ variables, $d=\dim_xX$.</p> <p>What does one find when linking this principle with the Serre-Tate theorem? The theorem affirms that the deformation functor of an ordinary abelian variety of dimension $g$ over an algebraically closed field of char.$p$ is pro-represented by a formal torus of dimension $g^2$. does this functor corresponds to the formal completion of some muduli functor of abelian scheme? One cannot expect the Siegel moduli (with level structures) of genus $g>1$ to work as the latter is of rel.dimension $g(g+1)/2$. Does the dimension jump to $g^2$ because one is looking "purely locally" through the $p$-divisible group forgetting the polarization?</p> <p>On the other hand, let $\mathcal{M}$ be the moduli functor of principally polarized $g$-dimensional abelian scheme with $\Gamma_0(p)$ level structure, $p$ being a rational prime (say large enough). It admits a coarse moduli scheme $M$, and according to a theorem of Chai and Norman ("Bad reduction of Siegel moduli scheme of genus two ...") the formal completion of $M$ at a geometric point $x$ corresponding to an ordinary abelian variety of char.$p$ is a formal torus of dimension $g(g+1)/2$, which agrees with the principle mentioned in the beginning.</p> http://mathoverflow.net/questions/58265/about-the-serre-tate-theorem/58274#58274 Answer by Holger Partsch for About the Serre-Tate theorem Holger Partsch 2011-03-12T18:07:20Z 2011-03-12T18:07:20Z <p>Concerning the first question, the answer is yes.</p> <p>Given an abelian variety $A$ over $k$ of genus $g$, we have that the first cohomology of its tangent space is of dimension $g^2$. Moreover, by a theorem of Grothendieck, we know that the deformations of $A$ are unobstructed. This yields that the base ring of the universal deformation is isomorphic to $k[t_1, \dots, t_{g^2}]$.</p> <p>Using the theory of p-divisble groups, one can show that in the case where $A$ is ordinary, the formal scheme representing the deformation functor can be equipped with the structure of a formal group scheme. This corresponds to a more canonical choice of parameters compared to the above one. In any case, the dimensions are the same.</p> <p>In this deformations theoretic arguments, polarizations are not involved. If you add them as additional data to the moduli functor, dimensions can change.</p>