Elkies' supersingularity theorem in higher dimension - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T01:27:24Z http://mathoverflow.net/feeds/question/8003 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8003/elkies-supersingularity-theorem-in-higher-dimension Elkies' supersingularity theorem in higher dimension David Speyer 2009-12-06T15:55:01Z 2011-02-09T15:54:14Z <p>The following is a theorem of <a href="http://www.ams.org/mathscinet-getitem?mr=903384" rel="nofollow">Elkies</a>:</p> <p>Let $X$ be an elliptic curve over $\mathbb{Q}$. Then there are infinitely many primes $p$ such that the action of Frobenius on $H^1(\mathcal{O}, X)$ is zero.</p> <p>Allen Knutson and I would like a similar theorem for higher dimensional Calabi-Yau varieties. Unfortunately, we've been told that this is probably open. (References for the fact that it is open are appreciated.) But we don't need the full strength of Elkies' result. It would be enough for us to know the following:</p> <p>Let $X$ be an $n$-dimensional, smooth, complete Calabi-Yau variety over $\mathbb{Q}$, for $n>0$. Write $X/p$ for the fiber of $X$ over $p$. Let $T(p)$ be the action of Frobenius on $H^n(\mathcal{O}, X/p)$. Are there infinitely many $p$ for which $T(p) \neq 1$? Also, in the same generality, are there infinitely many primes for which $T(p) \neq 0$?</p> http://mathoverflow.net/questions/8003/elkies-supersingularity-theorem-in-higher-dimension/8006#8006 Answer by Pete L. Clark for Elkies' supersingularity theorem in higher dimension Pete L. Clark 2009-12-06T16:21:09Z 2009-12-06T16:35:12Z <p>First, I am not an expert on this. Let me give you the names of three experts: Noam Elkies, Abhinav Kumar, Matthias Schuett.</p> <p>The first result that you want is so strong that it must be open. I am pretty sure that it is open even in dimension 2 -- K3 surfaces -- so you should probably start with that. (I don't see how a reference could establish that a problem is open. At best, it <em>was</em> open as of such and such a time. Better to ask the people above.) </p> <p>Maybe the right question to ask is what is the most general class of K3 surfaces which can be shown to have infinitely many primes of supersingular (resp. ordinary) reduction? I think the case of singular K3's (i.e., with Picard number 20) should be the easiest due to connections with the theory of complex multiplication. I expect that the result is probably known in this case, at least for certain CM types. You should also look at Kummer surfaces because of the connection to abelian surfaces (is it true that supersingularity/ordinary passes from the abelian surface to its Kummer surface? it seems plausible). There are a lot of results on primes of ordinary reduction for abelian varieties: a conjecture of Serre is that, after a finite base change, the density of primes of ordinary reduction is always equal to one, and a lot of special cases of that conjecture are now known (e.g., possibly for all abelian surfaces).</p> <p>Some of the above should generalize to Calabi-Yau's with complex multiplication, I think.</p> <p>It's a very interesting question: please let us know what you find out.</p> <p>ADDENDUM: Here's an off the cuff idea to show that the problem must be open: start with an elliptic curve E over an imaginary quadratic field K. Let A be the Weil restriction from K down to Q, an abelian surface. Let X be the Kummer surface. As above, I am guessing that X is ordinary/supersingular iff E/K is, and this is well known to be an open problem in general: there are some examples due to Elkies and Jao where infinitude of supersingular primes can be proven, but very few.</p> http://mathoverflow.net/questions/8003/elkies-supersingularity-theorem-in-higher-dimension/8035#8035 Answer by Lavender Honey for Elkies' supersingularity theorem in higher dimension Lavender Honey 2009-12-06T21:16:47Z 2009-12-06T21:16:47Z <p>If I understand your question correctly, the "easy" version of the question you ask is unknown in dimension $\ge 3$, and is basically easy for surfaces of interest.</p> <p>These questions are certainly unknown in dimension $\ge 3$, or else the Sato-Tate conjecture for weight $>3$ would have been known $12$ months ago, rather than $4$ months ago.</p> <p>I take it you are only asking that the action of Frobenius is neither almost always trivial ($T(p) = 0$) or acts as the identity ($T(p) = 1$), which is much weaker than asking that there exist any supersingular primes. Assume otherwise. Fix a prime $\ell$, and Let $V$ be the $\ell$-adic etale cohomology $H^2(X)$ with the usual action of $\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$. Let $F_p(T)$ denote the characteristic polynomial of Frobenius. It has coefficients in $\mathbf{Z}[x]$. Say one wants to show that there exist infinitely many primes $p$ such that $T(p) \ne 0$. You are imposing the condition that $F_p(T) \equiv T^n \mod p$ for all but finitely many $p$, where $n = \mathrm{dim}(V)$. </p> <p>Choose a prime $L > 2n$. There will a positive density of primes such that $F_p(T) \equiv (T-1)^n \mod L$. If $p \equiv 1$ modulo $L$, These conditions together imply that the trace of $F_p(T)$, which is <em>a</em> <em>priori</em> an integer, is $np \mod pL$. By the Weil conjectures, the roots of $F_p(T)$ have absolute value $p$, and thus, since $L > 2n$, they must all equal $p$. By Cebatarev density, it follows that a finite index subgroup of the Galois group acts (on the semisimplification of $V$) via the cylotomic character on $V$. By looking at Hodge-Tate weights this implies that $h^{2,0} = h^{0,2} = 0$, and so (for example) one has a contradiction for a K3-surfaces.</p> <p>If one wants to rule out that $T(p) = 1$ for (almost all) $p$, one is imposing the condition that $F_p(T) = (T-1)^n \mod p$. In the same way one obtains an open subgroup of the Galois group such that the trace of Frobenius is always $n$, and then one deduces that $h^{0,2} > 0$ and $h^{2,0} = 0$, which can never happen.</p> <p>EDIT: Forgot to mention that it was Ogus who proved in ~70's that Abelian surfaces had infinitely primes of ordinary reduction, presumably by a very similar argument. Well, except for the comparison theorem of Faltings I used above...</p> http://mathoverflow.net/questions/8003/elkies-supersingularity-theorem-in-higher-dimension/54897#54897 Answer by Piotr Achinger for Elkies' supersingularity theorem in higher dimension Piotr Achinger 2011-02-09T15:54:14Z 2011-02-09T15:54:14Z <p>Your question is connected with Frobenius splitting. Namely, we have the following fact (Brion-Kumar "Frobenius splitting methods in geometry and representation theory", Remark 1.3.9 (ii)):</p> <blockquote> <p>A complete smooth variety $X$ over an algebraically closed field of characteristic $p>0$ is Frobenius split if and only if the pullback map $F^*: H^n(X, \omega_X)\to H^n(X, \omega_X^p)$ is nonzero.</p> </blockquote> <p>Since in your case $\omega_X = \mathcal{O}_X$, this is exactly the same question. We can be even more general and ask questions like "which characteristic zero varieties are (or are not) Frobenius split for $p$ sufficiently large (or infinitely many $p$)". </p> <p>One result of this kind (Brion-Kumar, exercise in paragraph 1.6) is that if $X$ is Fano, then $X/p$ is Frobenius split for $p$ large enough.</p> <p>Moreover, in their recent work, Mustata and Smith propose a conjecture of this kind in the local case. I don't know enough details to state it precisely, but it says something like "$fpt(X/p) = lct(X)$ for infinitely many $p$", where $X$ is a singularity we reduce mod $p$, $fpt$ is the Frobenius pure threshold and $lct$ is the log-canonical threshold. I heard that this conjecture will imply that all characteristic zero abelian varieties become ordinary when reduced modulo infinitely many $p$.</p>