What does "supersingular" mean? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T22:16:37Z http://mathoverflow.net/feeds/question/1269 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1269/what-does-supersingular-mean What does "supersingular" mean? Ilya Nikokoshev 2009-10-19T19:39:37Z 2010-02-15T16:49:35Z <p>Are <strong>supersingular primes</strong> and <strong>supersingular elliptic curves</strong> related?</p> <p>(this was essentially a subquestion in <a href="http://mathoverflow.net/questions/1249/ways-to-characterize-supersingular-primes" rel="nofollow">my earlier question</a>, but still looks sufficiently different to me to deserve a separate post)</p> http://mathoverflow.net/questions/1269/what-does-supersingular-mean/1324#1324 Answer by Jonah Sinick for What does "supersingular" mean? Jonah Sinick 2009-10-20T00:27:49Z 2010-02-15T16:49:35Z <p>Let F be the finite field with p elements.</p> <p>A supersingular elliptic curve is an elliptic curve E/F with the property that the endomorphism ring (ring of homomorphisms from E to E) of E over the algebraic closure of F_p is has rank 4 as a Z-module. </p> <p>It is a theorem that End(E/F) has rank 2 or rank 4 (and that in the former case, End(E/F) is isomorphic to an order in an imaginary quadratic number ring whereas in the latter case, End(E/F) is isomorphic to an order in a quaternion algebra over Q). Hence, a supersingular elliptic curve over F_p can be thought of as an elliptic curve over F with a "big" endomorphism ring.</p> <p>It is a theorem that another characterization supersingular elliptic curves is that E/F is supersingular is if and only if the number of points on E/F is exactly p + 1 (edit: for p > 3, see Voloch's comment below).</p> <p>I believe that the etymology of the term "supersingular" is as follows: if you start with an elliptic curve E/Q over Q, for all but finitely many primes p, reduction (mod p) gives an elliptic curve E/F. For a generic E/Q (specifically, one without "complex multiplication") then the set of primes such that reduction (mod p) turns E/Q into a supersingular E/F has asymptotic density 0. Such primes are called "supersingular primes for E/Q" - supersingular refers to "really unusual." The reductions for such primes are then called supersingular elliptic curves over F. I'm pretty sure that every elliptic curve over F is a (mod p) reduction of an elliptic curve over Q so that all supersingular elliptic curves arise in this way.</p> <p>I'll remark (following Silverman) that a supersingular elliptic curve over F is not "singular" in the sense of algebraic geometry - by definition all elliptic curves are nonsingular.</p> <p>I do not know much about supersingular primes in the context of monstrous moonshine. According to Wikipedia, a supersingular prime is a prime that divides the order of the monster group; and there are 15 such primes. Given E/Q, by a theorem of Elkies there will be infinitely many super singular primes p for E/Q. So it's difficult to imagine how the list of 15 supersingular (with respect to moonshine) primes could emerge from the notion of "supersingular elliptic curve." I imagine that the etymology of the term "supersingular" in the context of moonshine is again that that supersingular primes are special - but that they special in a completely different way from the supersingular primes for a elliptic curve over Q.</p> http://mathoverflow.net/questions/1269/what-does-supersingular-mean/1339#1339 Answer by S. Carnahan for What does "supersingular" mean? S. Carnahan 2009-10-20T01:50:08Z 2010-02-10T03:44:56Z <p>There are many equivalent ways to define supersingularity for an elliptic curve over a characteristic $p$ field. One of them is that the $p$-torsion of the curve is connected, i.e., it is a purely infinitesimal group scheme of order $p^2$. As Jonah mentioned, supersingular means very special, and is not a statement about smoothness. There is a theorem of Deuring that implies the j-invariant of a supersingular elliptic curve always lies in $\mathbb{F}_{p^2}$, and as a consequence, all such curves are defined over a finite degree extension of $\mathbb{F}_p$. </p> <p>There are two notions of supersingular prime: one is relative to a fixed elliptic curve over $\mathbb{Q}$, and one is absolute. For any elliptic curve $E/\mathbb{Q}$, a prime $p$ is supersingular for $E$ if $E$ has good supersingular reduction at $p$. Such primes are known to be asymptotically density zero, but infinite in number (by a theorem of Elkies). Lang and Trotter conjectured that the number of supersingular primes less than $N$ limits to a constant times $\sqrt{N}/\log(N)$ as $N$ gets large.</p> <p>Supersingular primes in the absolute sense are those primes $p$ for which all supersingular elliptic curves over an algebraic closure of <code>$\mathbb{F}_p$</code> have $j$-invariant in <code>$\mathbb{F}_p$</code> instead of just <code>$\mathbb{F}_{p^2}$</code>. These happen to be the primes that divide the order of the monster simple group, and they are also the primes for which the normalizer of <code>$\Gamma_0(p) \in SL_2(\mathbb{R})$</code> acts on the complex upper half plane with a genus zero quotient. For general $p$, this normalizer contains $\Gamma_0(p)$ as an index 2 subgroup, with the nontrivial coset called the "Fricke involution" (a special case of Atkin-Lehner involultion). There is a standard order 2 representative, taking $z \mapsto -1/pz$. The quotient curve classifies unordered pairs of elliptic curves with dual degree p isogenies between them. <strike>I do not know any canonical relations between these characterizations of supersingularity.</strike></p> <p><b>Edit:</b> Thanks to Emerton for pointing out the connection. I'll try to expand on it a bit. The moduli problem of generalized elliptic curves with $\Gamma_0(p)$-structure has a coarse moduli space that is a smooth irreducible curve away from p, but has mod p fiber given by taking a disjoint union of two copies of $X(1)$ (a genus zero curve) and gluing along supersingular points (this description is more or less in Katz-Mazur, chapter 13). A geometric point describing an elliptic curve with j-invariant <code>$\alpha \in \mathbb{F}_{p^2}$</code> is glued to a geometric point on the other irreducible component describing an elliptic curve with j-invariant $\alpha^p$. The Fricke involution switches the components, so the quotient of <code>$X_0(p)$</code> by this involution is a genus zero curve glued to itself at finitely many supersingular points. The quotient has arithmetic genus zero if and only if all supersingular geometric points are glued to themselves - otherwise, the flat modular deformation to characteristic zero yields a smooth curve of higher genus. In other words, it is necessary and sufficient that all supersingular geometric points have no Frobenius conjugates, i.e., that the j-invariants of all supersingular curves lie in <code>$\mathbb{F}_p$</code>.</p> <p><b>More Edit:</b> I should give a more honest reply to Mariano's question, which was originally raised by Ogg in the mid 1970s (and he famously offered a bottle of Jack Daniels to anyone who could solve it). Half of the question has an answer. If we combine the results of Borcherds's paper <a href="http://math.berkeley.edu/~reb/papers/index.html" rel="nofollow">Monstrous moonshine and monstrous Lie superalgebras</a> with the results of the paper <i>Modular equations and the genus zero property of moonshine functions</i> by Cummins and Gannon, we get the following fact:</p> <blockquote> <p>Let G be a finite group acting faithfully on a conformal vertex algebra V by conformal symmetries, and suppose V has central charge 24 and character <code>$\operatorname{Tr}(q^{L_0-1}|V) = j(\tau)-744$</code>. Then for any element $g \in G$, the series <code>$\operatorname{Tr}(gq^{L_0-1}|V)$</code> is the q-expansion of a modular function that is holomorphic on the upper half plane, invariant under a discrete group <code>$\Gamma \subset PSL_2(\mathbb{R})$</code> satisfying <code>$\Gamma \supset \Gamma_0(N)$</code> for some $N$, and generates the function field of the quotient curve $\Gamma \backslash \mathbf{H}$. In particular, the quotient curve is genus zero.</p> </blockquote> <p>The monster simple group arises in this context because I. Frenkel, Lepowsky, and Meurman constructed a conformal vertex algebra satisfying the above hypotheses, whose group of conformal automorphisms is the monster. The fact given above implies that for each prime p dividing the order of the monster, the quotient <code>$X_0^+(p) = X_0(p)/\langle w_p \rangle$</code> is genus zero, where <code>$w_p$</code> is the Fricke involution. In particular, each prime dividing the order of the monster is necessarily supersingular.</p> <p>I know some people who would like there to be a conceptual (read: non-enumerative) explanation for why <b>all</b> supersingular primes divide the order of the monster. So far, the best I've heard is that the monster is really big, while there aren't that many supersingular primes.</p>