What are supersingular varieties? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T06:25:45Z http://mathoverflow.net/feeds/question/95729 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95729/what-are-supersingular-varieties What are supersingular varieties? temp 2012-05-02T06:50:51Z 2012-05-03T11:29:14Z <p>For varieties over a field of characteristic $p$, I saw people talking about supersingular varieties.</p> <p>I wanted to ask "why are supersingular varieties interesting". However, as I don't want to ask an MO question with no background/context I thought I'd better define what a supersingular variety is. Unfortunately I can't. (And search engine doesn't help...) Can anyone here help me with this and explain why they are interesting?</p> <p>A little bit more context: I have seen the definition of supersingular elliptic curves on textbooks by Hartshorne and Silverman. When I read about abelian varieties I saw "for abelian varieties of $\text{dim}>2$ being supersingular $\neq p$ rank being 0".</p> <p>Illusie mentioned (in the "Motives" volume) that for $X$ a variety over a perfect field of characteristic $p$, $X$ is said to be ordinary if "$H_\text{cris}^*(X/W(k))$ has no torsion and $\text{Newt}_m(X)=\text{Hdg}_m(X)$ for all $m$". My guess is being supersingular should correspond to the other extreme, but what precisely is it? ($\text{Newt}_m(X)$ being a straight line for all $m$?)</p> http://mathoverflow.net/questions/95729/what-are-supersingular-varieties/95784#95784 Answer by Ariyan Javanpeykar for What are supersingular varieties? Ariyan Javanpeykar 2012-05-02T16:26:07Z 2012-05-02T17:48:45Z <p>This doesn't really answer your question. But I think you might find these comments interesting (even though I might not be saying anything you don't know).</p> <p>By a theorem of Mazur-Ogus (Katz' conjecture) the $m$-dimensional Newton polygon of a variety lies above or is equal to its $m$-dimensional Hodge polygon.</p> <p>A variety is ordinary if these polygons are equal (for all $m$).</p> <p>For abelian varieties the $m=1$ case suffices and you see that an abelian variety is ordinary iff it is ordinary in the usual sense.</p> <p>By a theorem of Grothendieck-Katz most varieties are ordinary. This is stated more precisely also in Illusie's paper you mention. </p> <p>You should take a look at Mazur's beautiful paper on Katz' conjecture.</p> <p><a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183533965" rel="nofollow">http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183533965</a></p> <p>Let me also talk about another nice subject which has to do with Newton polygons and supersingular varieties. Namely, "constructing" varieties with given Newton polygon.</p> <p>If you stick to the case of curves there are many open questions (to my knowledge). For example, Mazur asks in <em>loc. cit.</em> (page 659) if all five different possible Newton polygons arising from a smooth projective curve of genus $3$ allowed by the restraint of Poincaré duality really arise from some curve or not. I don't know if this question has been answered by now (and let me add that it might actually be answered by now). </p> <p>I can't really tell you anything else on supersingular varieties. It does seem to be a nice sport to look at "strata" in the moduli spaces of abelian varieties (=Shimura varieties). For example, every "symmetric" Newton polygon arises from an abelian variety (and the Newton polygon of an abelian variety is symmetric). See <a href="http://arxiv.org/abs/math/0007201" rel="nofollow">http://arxiv.org/abs/math/0007201</a> for even more beautiful statements.</p> <p>For "strata" of Shimura varieties see </p> <p><a href="http://arxiv.org/abs/1011.3230" rel="nofollow">http://arxiv.org/abs/1011.3230</a> (Wedhorn-Viehmann) <a href="http://arxiv.org/abs/1111.6830" rel="nofollow">http://arxiv.org/abs/1111.6830</a> (Kret)</p> <p>These have to do with showing existence of abelian varieties with certain Newton polygons.</p> <p>So the moral of the story is that it's already pretty difficult to prove that certain polygons arise from geometric objects, e.g., the case of the genus $3$ curves and the Shimura variety business.</p> http://mathoverflow.net/questions/95729/what-are-supersingular-varieties/95828#95828 Answer by Xavier Roulleau for What are supersingular varieties? Xavier Roulleau 2012-05-03T06:53:09Z 2012-05-03T06:53:09Z <p>For a surface $S$, supersingular means that the étale cohomology group $H^{2}(S,\mathbb{Q}_\ell)$ ($\ell$ a prime, prime to the characteristic $p$) is generated by divisors on $S$ (thus the Picard number equals to the second Betti number). Supersingularity is useful if one wishes to compute the Zeta function of $S$.</p> <p>Shioda worked on supersingular surfaces, see e.g. "An Example of Unirational Surfaces in Characteristic $p$" or "On Unirationality of Supersingular Surfaces". Supersingularity is a necessary condition for a surface to be unirational (be it is not sufficient). He gives examples of Fermat surfaces of degree $>4$ that are unirational.</p> http://mathoverflow.net/questions/95729/what-are-supersingular-varieties/95847#95847 Answer by Joe Silverman for What are supersingular varieties? Joe Silverman 2012-05-03T11:29:14Z 2012-05-03T11:29:14Z <p>Not an answer, but some historical context (which I think is correct). An elliptic curve over $\mathbb{C}$ used to be called "singular" if its endomorphism ring was larger than $\mathbb{Z}$, i.e., what we now call having complex multiplication. Presumably this use of the word singular was to indicate that the curve was unusual. Then, when people looked at elliptic curves over finite fields, the found that some of them had endomorphism rings that were even larger than an order in a quadratic imaginary field, so those curves were "supersingular" in the sense of being even more unusual. Of course, it turns out that an alternative way to characterize those curves is as having no $p$-torsion (over the algebraic closure of their base field). </p>