User stankewicz - MathOverflow

Answer by stankewicz for Can one bound the Quadratic Points on Curves?

2013-05-23T20:05:17Z

Hi Barinder!

As far as I know there is not an algorithm to do so. See for instance the following paper of Harris and Silverman:

http://www.ams.org/journals/proc/1991-112-02/S0002-9939-1991-1055774-0/

Especially in the proof of Corollary 3 you can really get a clear picture of the argument. Basically if your curve is not hyperelliptic or bielliptic then you find that the symmetric product of $C$ with itself has only finitely many rational points by Faltings' Theorem. Then you have a finite-degree map of sets between the rational points of the symmetric product and your $\Gamma_C$.

If you had a way to compute $D_C$ for any curve, then you'd have something close to an effective version of Faltings for these symmetric product surfaces. Given that Faltings theorem isn't effective for curves I think we're quite far away from that.

For modular curves in particular, well that question is actually on my upcoming agenda. If you'd like to talk about this some time I would probably not require much persuasion.

Answer by stankewicz for Field of definition of canonical morphism between (congruence) modular curves

2013-05-21T18:55:14Z

Yes. Please see Theorem 7.1.3 of Katz-Mazur.

Answer by stankewicz for ramified quaternion algebras

2013-03-20T08:26:07Z

Near as I can tell, you're confusing things because you're trying to consider a general number field as an analogue to a quaternion algebra when the right analogue is a quadratic field.

In a quadratic field $F$, a prime $p$ of $\mathbf{Z}$ is ramified if and only if there is a prime $\mathfrak{p}$ of $\mathbf{Z}_F$ such that $\mathfrak{p}^2 = p\mathbf{Z}_F$.

Similarly in a quaternion algebra $B$ over $\mathbf{Q}$, you will have many orders $\mathcal{O}$. We will say that a prime $p$ of $\mathbf{Z}$ is ramified in $\mathcal{O}$ if there is a two-sided ideal $\mathfrak{P}$ of norm $p$ in $\mathcal{O}$ such that $\mathfrak{P}^2 = p\mathcal{O}$. If $\mathcal{O}$ is maximal, then the ramified primes of $\mathcal{O}$ are exactly the primes at which $B$ is not division.

Of course historically, these two notions probably had no more connection than "these are the finite set of primes at which something funny happens." In general, just because two things in mathematics have the same name does not necessarily mean there is a deep connection between the two. See for instance http://mathoverflow.net/questions/7389/what-are-the-most-overloaded-words-in-mathematics

Answer by stankewicz for Deducing BSD from Gross-Zagier and Kolyvagin

2013-03-11T06:45:16Z

No papers because it's not proven for elliptic curves of rank zero or one.

The work of Kolyvagin, together with the work of Gross and Zagier almost proves that if $E_/{\mathbf{Q}}$ is an elliptic curve of analytic rank zero or one then the rank is also zero or one. Depending on the form of BSD you use, this may or may not prove it.

To take care of the almost, you need to keep in mind the following. Kolyvagin's work proves that if an elliptic curve over $\mathbf{Q}$ has analytic rank zero, then its rank is zero. Gross and Zagier's work proves that if an elliptic curve over an imaginary quadratic field has rank one, then its rank over that field is one. To bridge the gap, you need to say that if $E$ is an elliptic curve over $\mathbf{Q}$ whose analytic rank is one, then there is an imaginary quadratic field $K$ such that the twist of $E$ by $K$ has analytic rank zero, and thus the base change of $E$ to $K$ has analytic rank one.

This last bit was proven independently by either

the brothers Murty : http://www.mast.queensu.ca/~murty/murty-murty-annals.pdf

Bump, Friedberg, and Hoffstein: http://wintrac.sagemath.org/sage_summer/bsd_comp/Bump-Friedberg-Hoffstein-Nonvanishing_theorems_of_L-functions_of_modular_forms_and_their_derivates.pdf

Answer by stankewicz for Fuchsian groups and automorphisms of Riemann surfaces

2013-02-25T22:37:42Z

How about $\Gamma = PSL_2(\mathbf{Z})$, where $N(\Gamma)/\Gamma)$ is the trivial group but the automorphism group of $\Gamma \setminus \mathcal{H}^*$ is infinite?

In general the best you can do is that you have a map $N(\Gamma) \to \mathrm{Aut}(\Gamma \setminus \mathcal{H}^*)$ whose kernel is $\Gamma$. If you have no cusps, then you can say something else.

Answer by stankewicz for the dual abelian scheme

2013-01-15T17:00:40Z

Alternately, see Faltings, Chai, "Degeneration of Abelian Varieties" Chapter I, especially Theorem 1.9

The general idea is: it can be shown that the Picard functor of a scheme $X/S$ is represented by a group algebraic space over $S$. If $X$ is an abelian scheme, then it can be shown that $Pic^0(X/S)$ is an abelian algebraic space. A Theorem of Raynaud shows that any such algebraic space is automatically a scheme. I think I recall that BLR only proves this fact for certain types of schemes $S$.

Answer by stankewicz for How do you find the genus of a Fuchsian group derived from a quaternion algebra?

2013-01-02T14:34:58Z

You are essentially asking for a fundamental domain for your Fuchsian group $G$. This is an area of specialty for John Voight, who has produced an algorithm for finding such a fundamental domain and thus a presentation for $G$: http://jtnb.cedram.org/jtnb-bin/item?id=JTNB_2009__21_2_467_0

In particular, he pays special attention to the case of quaternion algebras. I'm a little bit confused by a few things in your question: e.g., claiming that a Fuchsian group can have a normalizer inside of a finite group, but what I propose should work.

Answer by stankewicz for Supersingular Elliptic Curves with rational isogeny?

2012-12-20T13:49:44Z

You can't prove it because it is untrue.

Let $E$ be an elliptic curve with CM by $\mathbf{Z}[\sqrt{-p}]$ defined over a number field $K$ which

Contains $\mathbf{Q}(\sqrt{-p})$ so that the action of $\mathbf{Z}[\sqrt{-p}]$ is $K$-rational and
Over which $E$ has good reduction (in fact, since $E$ has CM, there's a number field over which it has everywhere good reduction)

By Deuring's Criterion, $E$ has supersingular reduction above $p$, and since $K$ contains the CM field, $[\sqrt{-p}]$ is a $K$-rational $p$-isogeny.

In general, you need much finer information about how rare two quantities are in order to make the "both of these are rare, so they should be impossible together!" argument. The more likely occurrence is that if Condition A happens with probability $1/a$ and Condition B happens with probability $1/b$ then they both occur with probability $1/ab$ (and similarly in the probability zero cases).

Answer by stankewicz for How does a moduli interpretation give an analytic object an algebraic structure?

2012-11-09T15:20:31Z

Even on the level of sets, the idea that any compact Riemann surface gives rise to an algebraic curve over $\mathbf{Q}$ should feel resoundingly false. There are uncountably many compact Riemann surfaces and only countably many algebraic curves over $\mathbf{Q}$.

I think you may be confusing one or more of the following statements:

First, although a compact Riemann surface does not necessarily give rise to an algebraic curve over $\mathbf{Q}$, an algebraic curve over $\mathbf{Q}$ does give rise to an algebraic curve over $\mathbf{C}$, simply by extending the base of your curve to $\mathbf{C}$. Then we have an equivalence of categories between Riemann surfaces with analytic maps and algebraic curves over $\mathbf{C}$ with algebraic morphisms. You can prove this with the Riemann Existence theorem, but this is true in much more generality by Serre's GAGA. For the analytic theory I like Rick Miranda's book, but there are lots of potentially great references as it's an extremely classical subject.

Then, which Riemann surfaces give rise to algebraic curves over $\mathbf{Q}$? Well, that's a complicated question, but the start of the answer is Belyi's Theorem:

An algebraic curve $C$ over $\mathbf{C}$ is isomorphic to the base change of an algebraic curve over $\overline{\mathbf{Q}}$ if and only if there exists a finite map $C \to \mathbb{P}^1$ ramified only at 3 points.

You asked for references and at least with this one, Koeck's "Belyi's Theorem revisited" http://arxiv.org/abs/math/0108222 is pretty canonical.

Moving from $\overline{\mathbf{Q}}$ to $\mathbf{Q}$ is an exercise in Galois cohomology, and although you're talking about general algebraic curves, Chapter X and Appendix B of Silverman's Arithmetic of Elliptic Curves are as good as any.

If you want to bypass all of that and just be given an algebraic curve from on high, moduli spaces are great ways to do so! All you have to do is to cook up a functor taking schemes $S$ over $\mathbf{Q}$ to isomorphism classes of certain objects over $S$ and call it a "moduli problem." If the problem is rigid - there are no nonidentity automorphisms of the objects - then by certain general nonsense your moduli problem will be representable - i.e., will give rise to a scheme over $\mathbf{Q}$. If you pick the right problem, you get an algebraic curve over $\mathbf{Q}$.

Now it's worth noting that the moduli problems that Elkies references are not quite rigid. Still they are not so far from being rigid, so we can still get algebraic curves out of them. See Edidin's article for details on this process - http://arxiv.org/abs/math/9805101

Finally I'll make a note about S. Carnahan's comment: There is a little bit of a subtle issue which people are fond of "passing over in silence" - Moduli problems give algebraic curves over $\mathbf{Q}$, which give algebraic curves over $\mathbf{C}$, which give Riemann surfaces. Which Riemann surface? Well the Riemann surfaces that Elkies works with are chosen because over $\mathbf{C}$ they form certain analytic moduli spaces - so if we start off with "the analogue over $\mathbf{Q}$" we ought to get back to that very special quotient of upper half space, right? Well, we do, but there's something to prove here and that's been mentioned in comments on this site in the past - http://mathoverflow.net/questions/21755/is-there-a-schemetical-construction-for-modular-curves-over-the-rationals

Answer by stankewicz for Imaginary quadratic field contained in Hecke orbit field?

2012-10-05T19:26:36Z

"No" for both questions about CM elliptic curves and "I'm not even sure I know what the question would be" about general Shimura varieties.

Basic idea: If $j$ is the $j$-invariant of a CM elliptic curve, then there is some imaginary quadratic discriminant $\Delta \equiv 0$ or $1 \bmod 4$ such that $\mathbf{Q}(j) \cong \mathbf{Q}[X]/H_\Delta(X)$ where $H_\Delta(X) \in \mathbf{Z}[X]$ is the Hilbert Class Polynomial of discriminant $\Delta$, whose roots are the $j$-invariants of elliptic curves over the complex numbers (actually $\overline{\mathbf{Q}}$ is enough) with CM by $\mathbf{Z}\left[\dfrac{ \Delta + \sqrt \Delta}{2}\right]$.

The point is that now we can see that there is an embedding $\mathbf{Q}(j) \hookrightarrow \mathbf{R}$, for all possible $j$. Therefore there is an embedding $\mathbf{Q}(S) \hookrightarrow \mathbf{R}$. To see this, it's enough to note that for any two CM $j$-invariants $j_1$ and $j_2$ that there exists an embedding $\mathbf{Q}(j_1,j_2)\hookrightarrow \mathbf{R}$. Let $J_1$ and $J_2$ be the canonical image of $j_1$ and $j_2$ in the real numbers. Then $\mathbf{Q}(j_1)$ embeds into the real numbers as $\mathbf{Q} + \mathbf{Q}J_1 + \dots + \mathbf{Q}J_1^{h_1 -1}$ and $\mathbf{Q}(j_2)$ embeds into the real numbers as $\mathbf{Q} + \mathbf{Q}J_2 + \dots + \mathbf{Q}J_2^{h_2 -1}$. Therefore $\mathbf{Q} + \mathbf{Q}J_1 + \mathbf{Q}J_2 + \dots + \mathbf{Q}J_1^{h_1 -1}J_2^{h_2 -1}$ is a copy of $\mathbf{Q}(j_1,j_2)$ inside of $\mathbf{R}$. Notice that I didn't use direct sums because $\mathbf{Q}(j_1)$ and $\mathbf{Q}(j_2)$ might not be linearly disjoint over $\mathbf{Q}$! This is also the reason I didn't use a tensor product argument. In any case, this inductive step allows us to work with direct limits and embed $\mathbf{Q}(S)$ into $\mathbf{R}$.

Therefore if we assume that there is an embedding $K\hookrightarrow \mathbf{Q}(S)$ then there must be an embedding $K\hookrightarrow\mathbf{Q}(S) \hookrightarrow \mathbf{R}$, which is absurd. Therefore, there is no embedding $K\hookrightarrow \mathbf{Q}(S)$.

To show that we have an embedding $\mathbf{Q}(j)\hookrightarrow\mathbf{R}$, consider that there is some $\tau \in \mathcal{H}$ of the form $\dfrac{1}{2}\sqrt\Delta$ or $\dfrac{ 1 + \sqrt \Delta}{2}$ such that $j(\tau)$ is a root of $H_\Delta(X)$. But then $j(\tau)$ is real, because the inverse image of the reals under the $j$-function contains the lines $\lbrace iy : y \ge 1\rbrace$ and $\lbrace 1/2 + iy : y \ge (1/2)\sqrt 3\rbrace$. Therefore we have our embedding $\mathbf{Q}(j) \hookrightarrow \mathbf{R}$.

Answer by stankewicz for The significance of modularity for all Galois representations

2012-10-04T22:46:27Z

Your question reminds me of a current strain of research whose starting point is Serre's conjecture, now the Khare-Wintenberger Theorem:

any continuous odd irreducible two-dimensional Galois representation over a finite field arises from a modular form

The question one might ask is then "Where are the even Galois representations?"

The answer given (mostly by F. Calegari) is that they just don't exist when you put certain additional restrictions on your Galois representation. Suppose then that you somehow have an even two-dimensional Galois representation in your hands. Well then this Galois representation is very special in some ways that might not be apparent! You can then ask: where did this representation come from? Is it modular in some non-obvious way?

So if I were to answer your question "What would we lose if we decided to focus only on those Galois representations that are attached to automorphic forms and ignore the possibility that some do not?" I would say that you lose a lot of knowledge about what makes modular Galois representations special. In your terminology, you might lose the scope of just how interesting certain representations can be!

Answer by stankewicz for Intersection of Hilbert class fields of imaginary quadratic fields

2012-09-20T12:20:12Z

The generalization of the phenomena you see is genus theory. If $K = \mathbf{Q}(\sqrt{-d})$ and $H = K(j_d)$ then $H$ contains the Genus field $G$.

If $d = \prod_{i=1}^n p_i$ is squarefree (and odd for convenience's sake) then $G = K(\sqrt{p_i^*})$ where $p_i^* = (-1)^{(p_i-1)/2}p_i$. In particular $p_i^* = p_i$ if and only if $p_i \equiv 1\bmod 4$. Therefore if $d \equiv 1\bmod 4$ then $d = \prod_i p_i^*$. Therefore if $d \equiv 1\bmod 4$ then $G$ and thus $H$ contains $K(\sqrt d) = K(i)$.

In general, if $d_1$ and $d_2$ have lots of prime factors in common, then the genus fields of their corresponding imaginary quadratic fields will also have large intersections. Outside of the genus field, I don't know of any studies into the intersections of Hilbert Class Fields.

The standard reference for this is Cox's wonderful book "Primes of the form $x^2 + ny^2$."

Answer by stankewicz for Isogeny classes of elliptic curves

2012-09-16T13:52:01Z

We say that an elliptic curve $E$ over a number field $K$ is an elliptic $\mathbf{Q}$-curve if it is is isogenous to its Galois conjugates $E^\sigma$. These were first studied by Benedict Gross, but were later studied by Elkies, Ellenberg, Ribet, and many others. It's possible to show that $\mathbf{Q}$ curves are modular, independently of the BCDTW modularity theorem, and that even though they are defined over $K$, their $p$-torsion Galois representations (at least away from the degrees of said isogenies) descend down to $\mathbf{Q}$. Elkies also famously showed if $E_{/K}$ is a $\mathbf{Q}$-curve without CM, there must be a geometrically isogenous curve $E'_{/L}$ such that $Gal(L/\mathbf{Q}) \cong (\mathbf{Z}/2\mathbf{Z})^r$ for some integer $r$. This comes down to showing that the moduli space of elliptic $\mathbf{Q}$-curves of degree $N$ form a certain very special quotient of $X_0(N)$. For details on most of this, see Jordan Ellenberg's "$\mathbf{Q}$-curves and Galois representations" ( http://www.math.wisc.edu/~ellenber/MCAV.pdf )

Answer by stankewicz for elliptic curves with and without CM

2012-08-30T18:12:33Z

1)If an elliptic curve has integral $j$-invariant it absolutely DOES NOT NEED to have CM. The class of curves with integral $j$-invariant (let's call that the class of IM Elliptic curves for Integral Modulus) is MUCH MUCH larger than the class of CM Elliptic curves. In fact, one can use Heilbronn's Theorem that class numbers of imaginary quadratic fields tend to infinity to show that over any given number field, there are only finitely many elliptic curves with CM. In particular over $\mathbf{Q}$, there are only 13 CM $j$-invariants. Even for all number fields of a certain degree, there are only a finite number $N(d)$ of elliptic curves with CM over any number field of degree $d$. This gives a way to enumerate all the CM $j$-invariants or "singular moduli," which is about as good as you can hope for in terms of describing the complex numbers which are $j$-invariants of CM elliptic curves. To do this explicitly (say for number fields of degree up to 100) see Mark Watkins' enumeration of imaginary quadratic fields of class number up to 100.

Meanwhile, for any regular integer $n$ (1,2,3, etc) there is at least one elliptic curve over $\mathbf{Q}$ whose $j$ invariant is $n$. Therefore over $\mathbf{Q}$ and therefore over any number field, there are infinitely many non-isomorphic IM elliptic curves.

If you want to say something general about elliptic curves with IM, consider the theorem of Deuring that an elliptic curve has IM if and only if it has potential good reduction. For this and much much more see Serre-Tate's "Good reduction of abelian varieties"

2) Easy proof that the answer is yes, at least as long as you mean "has a CM $j$-invariant" when you say CM: $y^2 = x^3 + 1$ is a CM elliptic curve with $j$-invariant zero defined over any number field. On the other hand, if we take an elliptic curve with $j$-invariant equal to 1/2, say this one: y^2 + x*y = x^3 + 72*x + 13822 , well it doesn't have integral $j$-invariant and therefore can't have CM!

Answer by stankewicz for Volumes of fundamental domains of maximal orders in definite quaternion algebras over Q

2012-08-27T19:22:12Z

See chapters II and III of the classic book of Vigneras. Beware however that in the later parts of chapter III she assumes widely that C.E. (the "Eichler condition") is verified, and C.E. is explicitly not verified for quaternion algebras $B$ over $\mathbf{Q}$ such that $B\otimes \mathbf{R} \cong \mathbb{H}$ (so called "definite" algebras). For lots of computation with definite algebras please see Chapter V of her book.

Answer by stankewicz for Uniform bounds for the order of a rational torsion point on CM elliptic curves

2012-08-24T18:58:45Z

Hey, I probably should have answered this one some time ago. It was proved in 1989 by J.L. Parish that the order of an $H$-rational torsion point is 1,2,3,4 or 6, and this also can be deduced from work of either Silverberg or Prasad-Yogananda. In any case the statement you want is at the beginning of section VI in the paper below and the proof is done in section V.

http://www.sciencedirect.com/science/article/pii/0022314X89900127

Is there a connected $k$-group scheme $G$ such that $G_{red}$ is not a subgroup?

2010-09-15T22:31:36Z

I've been trying a learn a little more about group schemes by working through a set of exercises on Brian Conrad's website. Exercise 8.3 of http://math.stanford.edu/~conrad/papers/gpschemehw1.pdf reads:

If $k$ is a perfect field and $G$ is a locally finite type $k$-group prove that $G_{red}$ is a closed $k$-subgroup of $G$. Can you find a counterexample if $k$ is not perfect? (There's a third part to the question, but it's irrelevant here)

I'll try to not give away too much for the benefit of others who want to use these exercises, but the first part of the exercise is Lemma 7.10 of http://www.math.columbia.edu/algebraic_geometry/stacks-git/groupoids.pdf in DeJong's stacks project and the hint given for the omitted proof is that $G_{red}\times_k G_{red}$ is reduced if $k$ is perfect.

For the counterexample part, I was able to find a disconnected example over any imperfect field, but it really seemed to depend crucially on being disconnected.

This naturally suggests the question:

If $G_{red}$ is not a subgroup scheme of $G$, must $G$ be disconnected?

edit: After almost a week out an answer in the zero-dimensional case was given and then retracted, but there has not been much action otherwise. Perhaps this is an open problem if $G$ is not finite.

For any finite group scheme $H$, $H_{red}$ is a subgroup of $H$ if and only if $H/H^0$ is a subgroup of $H$ (and in fact if $k$ is perfect, $H \cong H^0 \oplus H_{red}$, see Pink's notes Lecture 6- note this very explicitly depends on $H$ being finite)

For group schemes of locally finite type (as we assume here) there has been much done, for instance in SGA 3 where they prove among other things that if $G$ is connected, it's actually quasi-compact and thus of finite type (Proposition 2.4). Moreso, for a locally finite type group $H$ over a field, $(H^0)_{red}$ is a group in the category of reduced schemes. What's unclear is whether they believed it to be a group in the category of schemes but weren't able to prove that or if they knew of a counterexample but didn't include it.

Answer by stankewicz for Is the number of twists of a curve with a section in a given field finite

2012-07-26T03:55:10Z

Fact 1 (The Hurwitz Bound): If $X$ is a smooth projective connected curve of genus $g\ge 2$ over $\mathbf{C}$ then $$| Aut_{\mathbf C }(X)| \le 84(g-1)$$

Fact 2: $Aut_\mathbf{C}(X) = Aut_{\overline K}(X)$ (The sentence that says "If $\phi$ is an automorphism of $X$ then $\phi$ must be one of these possibilites" is first order and thus by the first order completeness of algebraically closed fields of characteristic zero...)

Fact 3 (See Silverman's Arithmetic of Elliptic Curves chapter 10 or Serre's Galois Cohomology or Berhuy's notes or...): The twists $W_{/K}$ of a variety $V_{/K}$ are given up to isomorphism by the pointed set $H^1(Gal(\overline K /K),Aut_{\overline K}(V))$. Assume now that $L/K$ is Galois. If not you can just replace $L$ by its Galois closure. The twists which resolve over $L$ are given up to isomorphism by $H^1(Gal(L/K),Aut_{\overline K}(V))$

Fact 4 (Exercise): $$ |H^1(Gal(L/K),Aut_{\overline K}(V))| \le 84(g-1) | Gal(L/K)| $$

I must say however that I'm not sure what sections have to do with anything.

Answer by stankewicz for Open problems with monetary rewards

2012-07-26T08:31:59Z

Recently Ian Morrison issued a 100 dollar prize for the construction of an effective divisor on $\overline M_g$ with slope less than 6 (See the recent preprint of Chen, Farkas and Morrison).

Answer by stankewicz for Stacks in modern number theory/arithmetic geometry

2012-05-15T03:51:16Z

Perhaps this doesn't count as "modern" but stacks are ubiquitous in the 1972 Antwerp paper of Deligne and Rapoport. Recall that the $\Gamma_0(N)$ moduli problem is not representable, and so they must frequently work directly with stacks before moving to the coarse moduli scheme we all know and love.

Answer by stankewicz for Average rank of elliptic curves over $\mathbb{Q}$

2012-05-08T03:40:08Z

Color me surprised if people no longer believe that the distribution of elliptic curves is half rank zero, half rank one, and a density zero subset of higher rank curves. To my knowledge this is still a conjecture that people believe.

I believe the state of the art results are still due to Bhargava and Shankar, and are best summed up in the slides listed in your post. In particular:

The average size of a 3-Selmer group is 4
The average size of a 4-Selmer group is 7
The average size of a 5-Selmer group is 6
These all hold true up to a finite number of congruence conditions
A positive proportion of elliptic curves have rank zero
Assuming the finiteness of Sha, a positive proportion of elliptic curves have rank 1
Unconditionally, the average rank of an elliptic curve over $\mathbb{Q}$ is strictly less than one

There are lots of good expositions of this work (or at least the 2-Selmer result), for instance this Seminar Bourbaki article of Poonen ( http://www-math.mit.edu/~poonen/papers/Exp1049.pdf ) and this short note of Gross ( http://www.math.harvard.edu/~gross/preprints/manjul.pdf ) .

Answer by stankewicz for 'Reference' request: Program to work with cyclic quotient singularities.

2012-02-26T22:46:45Z

On page 19 of http://www.cems.uvm.edu/~voight/notes/cfrac.pdf is a maple program for computing the so-called Hirzebruch-Jung continued fraction expansion. So that's (1).

For (2), at least when $n$ is prime, see page 93 (page 100 of the actual pdf) of Giancarlo A. Urzua's Thesis here: http://deepblue.lib.umich.edu/bitstream/2027.42/60657/1/gian_1.pdf

See also http://math.stanford.edu/~conrad/papers/j1p.pdf to see how far out the Hirzebruch-Jung resolution can take you. I think Corollary 2.4.3 should be the natural generalization of the above.

Answer by stankewicz for Applications for knowing the singularities parametrized by the boundary of a moduli space

2011-09-04T14:29:17Z

Here's an example: Suppose you'd like to know about the divisors on $X = \overline{M_{g,n}}$. Say for instance that you have a divisor $D$ and you'd like to know whether $D$ is ample or nef, that is, if for all curves $C$ on $X$, we have $D\cdot C > 0$ or $\ge 0$.

There's a conjecture out there called the $F$ conjecture which says that if we want to show $D$ is nef, it suffices to check $D\cdot C \ge 0$ for a smaller set of curves in the boundary strata, called the $F$-curves. Because that's a definition for which pictures help, I refer you over to http://www-irm.mathematik.hu-berlin.de/~larsen/talkM2Goettingen.pdf

Of course that's a conjecture to be proven, but it's related to a big circle of ideas surrounding the minimal model program, including the question of Hu and Keel on whether or not $\overline{M_{0,n}}$ is a Mori Dream Space ( http://arxiv.org/PS_cache/math/pdf/0004/0004017v1.pdf )

Sequences of Squares with all square differences

2011-08-21T18:31:50Z

Background

The following question was first asked by Alex Rice, who was thinking about small subsets $A\subset [1,\ldots , N]$ with lots of square differences. Certainly for any set $A$ the maximum number of square differences is going to be $\binom{|A|}{2}$. From the point of view of someone working in additive combinatorics, an infinite set of positive integers can't get much less substantial than the squares, and so it's natural to wonder if there are arbitrarily large sets $A$ inside the squares, all of whose differences are squares [edit: I apparently misunderstood the original motivation, see Alex's answer/comment below]. This question was asked of a few others, including Adrian Brunyate, Jacob Hicks and Nathan Walters before it was asked of me by Adrian in this form:

Definition: We say that a sequence $(a_1, \ldots, a_n) \in \mathbf{Z}^n_{\ge 1}$ is a Super-$n$ if for all $1 \le i \le n$, $a_i$ is an integer square and for all $1 \le i < j \le n$, $a_j - a_i > 0$ is also an integer square.

Clearly a Super-2 defines a Pythagorean triple.

Perhaps less clearly, a Super-3 defines an Euler Brick, and is strongly related the the question of whether there is a perfect rational cuboid.

Question 1: For which positive integers $n$ does there exist a Super-$n$ ?

If the answer is yes to the above question, we may also ask the following:

Question 2: For which positive integers $n$ do there exist infinitely many Super-$n$'s?

One may note that the following problems are related to some problems already asked on MO about rational polytopes and sequences of squares

http://mathoverflow.net/questions/72040/how-many-sequences-of-rational-squares-are-there-all-of-whose-differences-are-al

http://mathoverflow.net/questions/71949/totally-rational-polytopes

What seems to be known already

It has been known for millenia that there are infinitely many Pythagorean Triples.

Euler discovered in 1772 that there are infinitely many Super-$3$'s, and in fact he gave a parametrized family of them.

None of us have been able to find a Super 4 (although I haven't been searching myself).

The connection to algebraic geometry

Definition: The Super-$n$-variety is the intersection of the following $\binom{n}{2}$ quadratic polynomials in projective space over $\mathbf{Q}$. $$d_1^2 = c_2^2 - c_1^2$$ $$\vdots$$ $$d_{\binom{n}{2}}^2 = c_{n}^2 - c_{n-1}^2$$

Clearly the Super-2 variety is a copy of $\mathbb{P}^1_{\mathbf{Q}}$.

In Section 8 of the link given above for Euler's family of "Euler Bricks" we see that the Super-3 variety is birational to a singular K3 surface of Mordell-Weil rank 2. In this setting, one could say that Euler found a rational curve on this variety. It is also noted in the article that Narumiya and Shiga found a different rational curve on this variety.

Question 2': Could there be rational curves on the Super-$n$ variety for all $n$?

But perhaps (probably) this is way too much to ask. More generally, I'd like to know:

Question 3: Is there any interesting geometry to the Super-$n$ variety for $n\ge 4$?

In general this seems like an interesting problem, and one that people may have studied before, but perhaps in some guise that I'm not familiar with, so any input is appreciated.

Answer by stankewicz for Curves which are not covers of P^1 with four branch points

2011-08-05T19:34:13Z

Without pretending to understand Ben Wieland's argument, let me take the Shimura curve suggestion and run with it. In particular, modulo a few details I think the following should be true:

The Shimura curve $X^{218}$ defines a complete curve in $M_2$

My first reaction upon seeing Jordan's Shimura curve suggestion was "No way". The reason is that a priori Shimura curves $X^D$ don't parametrize curves of genus 2, but abelian surfaces which admit the action of a maximal order in the unique $\mathbf{Q}$-quaternion algebra of discriminant $D$ (and the fixed choice of a polarization compatible with the action of the choice of maximal order). Indeed, infinitely many of these abelian surfaces are products of elliptic curves with complex multiplication, and these of course do not need to necessarily be Jacobians.

Thankfully we have at our disposal the work of Hayashida and Nishi who asked the question: For what pairs of isogenous elliptic curves $E,E'$ is there a genus 2 curve inside $E\times E'$?

In particular they show the following:

If $E$ does not have CM, $E\times E$ is not a Jacobian

If $E,E'$ have CM by a maximal order in $\mathbf{Q}(\sqrt{-m})$ then $E\times E'$ contains a genus 2 curve if and only if $m \ne 1,3,7,15$

by reducing the above question to one about a certain real valued 4-variable quadratic form over the integers. A priori this doesn't answer the question, because we have to concern ourselves with nonmaximal orders in $\mathbf{Q}(\sqrt{-m})$. However, it seems (at least in the cursory read of their paper I've been able to give this afternoon) that their methods can be generalized to nonmaximal orders with a little bit of care. If the apparent qualms people have about Ben Wieland's argument can be resolved, I'll do exactly that in a few days.

But assuming that all works out, life is great! In particular we can come up with a Shimura curve $X^D$ which avoids those particular products of CM elliptic curves if and only if there is a prime $p|D$ such that $\left(\dfrac{-m}{p}\right) = 1$ for $m= 1,3,7,15$. We can compute that the minimal such prime is $109$, which is $1 \bmod 12$, $4 \bmod 5$ and $4\bmod 7$.

We recall momentarily that $X^D$ is a complete curve if and only if the unique rational quaternion algebra of discriminant $D$ is a division algebra if and only if $D$ is the squarefree product of an even number of primes. For convenience we take $D = 2p$.

Therefore (again assuming the bit about nonmaximal orders) we have found $X^{218}$ to be a curve lying in the image of the Torelli map in $\mathcal{A}_2$, and thus in $M_2$.

Comments are very welcome!

Answer by stankewicz for Atkin-Lehner involution and class number

2011-06-29T21:23:46Z

Given that I don't know exactly which relation you're talking about, I'll give you something old and something new:

A priori, asking for a formula for the number of fixed points of Atkin-Lehner is asking for the trace of the matrix representing the Atkin-Lehner involution. Hence you're asking for a trace formula, in particular the Eichler-Selberg trace formula. The original reference for that, featuring many relations between class numbers is

Eichler, M. Modular correspondences and their representations. J. Indian Math. Soc. (N.S.) {\bf 20} (1956), 163-206.

On the other hand a more modern view of fixed points of an Atkin-Lehner involution $w_m$ is that they're in bijection with conjugacy classes of embeddings $\mathbf{Z}[\sqrt{-m}] \hookrightarrow \mathcal{O}_0(N)$, the order used to define the Shimura curve $X^D_0(N)$. You said you wanted me to sweep the CM theory under the rug, so I won't elaborate on Shimura curves.

Anyways, this can by done by counting conjugacy classes of optimal embeddings of either $R = \mathbf{Z}[\sqrt{-m}]$ or $\mathbf{Z}[\dfrac{1 + \sqrt{-m}}{2}]$ into your quaternion order.

For counting these things, probably the book of Vigneras is best, but I like the exposition of Santiago Molina here http://www.crm.es/Publications/10/Pr928.pdf or here http://arxiv.org/PS_cache/arxiv/pdf/0912/0912.5217v4.pdf (same paper in section 2)

In particular let's simplify things and say both that $N$ is squarefree and $\mathbf{Z}[\sqrt{-m}]$ is a maximal order so every embedding is optimal. In this case the number of fixed points of $w_m$ is

$$h(-4m)\prod_{p|D}\left(1 -\left(\dfrac{-4m}{p}\right)\right)\prod_{q|N}\left(1 +\left(\dfrac{-4m}{q}\right)\right)$$

Answer by stankewicz for What is the high-concept explanation on why real numbers are useful in number theory?

2011-04-14T15:17:33Z

I would like to say that at least to some extent we need the real numbers and not just rational approximations.

Suppose we were being completely formal and we asked the question, ``Is there some algebraic number $\beta>0$ such that for all $n$ there exist $p_n , q_n \in \mathbf{Z}_{> 0}$ with $\beta \ne \frac{p_n}{q_n}$ and there exists a fixed $\delta >1$ such that $|\beta - \frac{p_n}{q_n}| < \frac{1}{q_n^{1+\delta}}$'' ?

The answer of course is no by the Thue-Siegel-Roth Theorem. But moreover, we know that we can only discover a measure zero subset of the transcendental numbers in this way. So while we can approximate real numbers by rationals all day long, the amount of information we get out of a rational approximation can vary wildly.

Answer by stankewicz for "Bad" reduction of Shimura curves via dual graphs

2011-01-13T14:16:52Z

Inkspot is indeed correct that the component graphs are indeed not generally trees.

As you seem to have deduced for yourself, Cerednik-Drinfeld uniformization is a highly nontrivial concept, and it really helps to have some examples to set it in your mind. Most helpful in this direction is Ogg's "Mauvaise réduction des courbes de Shimura" where he draws out a few of these dual graphs.

In general the dual graphs have $2h$ vertices $x$ where $h$ is the class number of $\mathcal{O}_x$, a level $H$ Eichler order in $D$ (your totally definite quaternion algebra, so note there's a choice of which $x$ to make here, but as long as the level is squarefree all orders are hereditary and it doesn't make a difference).

An orbit-stabilizer theorem computation then shows that for instance when $B$ is a quaternion algebra over $\mathbf{Q}$, $p+1$ (the size of the set of edges $y$ stemming from a particular vertex $x$ in the Bruhat-Tits tree) is equal to $\sum_{e(y) = x} \frac{\mathcal{O}_x^\times}{\mathcal{O}_y^\times}$. So it's not just that there are edges, but we know exactly how many there are! (For a readable account of details of this, see Kurihara's paper on Equations defining Shimura Curves)

Also, if I may take issue with 1.(ii) and 1.(iii), you've given a good description of the special fiber over $\overline \kappa_v$ (which is what I'm taking the dual graph to represent the data for), not necessarily $\kappa_v$. What you've claimed is that the special fiber is a Mumford Curve, that is, the transverse union of a number of copies of $\mathbb{P}^1$'s. The truth is that the special fiber is a quadratic twist of a Mumford curve, where the Galois action is not simply given by the $| \kappa_v|$-Frobenius, but where the action of Frobenius is identified with the action of the Atkin-Lehner operator $w_p$, which interchanges some of the components (if you want to think about the graph, its vertex set can be partitioned into $ {x_1, \dots , x_h, x_1', \dots, x_h'}$ where $w_p(x_i) = x_i'$).

All that said, some of the best advice I've heard for trying to understand this stuff is to first completely understand what happens when $v$ divides the LEVEL because in that case the moduli problem is much easier (if an abelian variety here is isogenous to a product of supersingular elliptic curves, it's isomorphic to a product of supersingular elliptic curves)

Here are a few additional references:

http://www.math.mcgill.ca/cfranc/documents/bctranslation.pdf (a translation of Boutot-Carayol)

http://math.berkeley.edu/~ribet/Articles/bimodules.pdf (this includes a somewhat more intuitive description of the components of the Mumford curve)

http://math.uga.edu/~pete/thesis.pdf (a comprehensive introduction to Shimura Curves and the action of the Atkin-Lehner group)

http://www.springerlink.com/content/gj8365486214l141/ (this is Oort's "which abelian surfaces are products of supersingular elliptic curves", see also his book on moduli of supersingular abelian varieties, as when $v$ ramifies in $B$, you're asking a question about moduli of supersingular abelian varieties, see the appendix on Honda-Tate theory to the thesis above)

Answer by stankewicz for Shimura datum of family of fake elliptic curves

2010-11-12T14:07:53Z

Question 1(what is the group for the Shimura datum):

Well, remember that $H^\times$ is just a bare group. A Shimura datum requires an algebraic group over $\mathbf{Q}$: that is, a functor from $\mathbf{Q}$-schemes to groups. Assuming you mean the group whose $\mathbf{Q}$ points, yes and you can see this in example 5.24 of http://www.jmilne.org/math/xnotes/svi.pdf although you can also use the algebraic group whose $\mathbf{Q}$ points are the norm one units of $H$ if you were interested in the connected Shimura datum (which is another example in milne's notes).

Question 2(what is the map from this group to the symplectic group):

I don't know. Is it even clear that a forgetful map of coarse moduli spaces which happen to be Shimura varieties induces a morphism of Shimura data? Either way your question sounds strongly related to the work of Victor Rotger's thesis which asks about the irreducibility of the quaternionic locus in $\mathcal{A}_2$.

Answer by stankewicz for is there any way to bound the number of CM points by height functions?

2010-11-02T02:36:17Z

For your titular question, Beats me. Personally I'm not aware of anyone who's studied the distribution of CM points with respect to height in the way you describe.

What I have seen papers that study the distribution of CM points with respect to things other than height and papers that look at the height of CM points (as well as papers that say quite a lot about the Faltings height of a CM abelian variety).

Here are a few different papers of note on those topics:

Equidistribution of CM points:

http://www.math.ucla.edu/~wdduke/preprints/modud.pdf - Gives an amazing asymptotic result on the trace of a CM $j$-invariant (i.e. a result on $X(1)$).

http://www.math.columbia.edu/~szhang/papers/ZhangIMRN.pdf - Gives an analogue on more general modular curves and quaternionic Shimura Curves as well as a connection to the Andre-Oort conjecture

Also good for an overview is the "Equidistribution in Number Theory" volume edited by Granville and Rudnick http://www.springer.com/mathematics/numbers/book/978-1-4020-5403-7

Heights of CM points/cycles:

http://www.math.columbia.edu/~szhang/papers/HCMI.pdf - Heights of CM Points I--The Gross-Zagier formula

http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.32.4204 - This attempts to identify affine elliptic modular curves by the presence of CM points of large enough height(logarithmic height on $\overline{\mathbf{Q}}$).

I'd like to emphasize that this does not pretend to be a complete reference list and I don't pretend to know much about the theory of heights. Rather this is a smattering of exciting ideas in the area.

Comment by stankewicz

2013-05-22T02:36:30Z

Well perhaps I'm being stupid here, but it seems to me that Pastorini is asking an existence question and that the theory of moduli as given in Katz-Mazur is sufficient to say that yes, there exist schemes and a map defined over some number field which base extend to the canonical map. I know quite well that you could twist any part of this data and screw something up, but for the question as asked I don't see the objection.

Comment by stankewicz

2013-05-21T18:56:35Z

For information about exactly which number field you can use, please see Chapter 9 of the same book.

Comment by stankewicz

2013-04-15T20:52:34Z

I suppose then that the $s$ is at the end of "Douglas" ?

Comment by stankewicz

2013-04-05T01:29:17Z

They are often special in arithmetic situations because all the square roots of unity happen to be rational, which is certainly not the case for higher roots of unity.

Comment by stankewicz

2013-04-03T15:25:03Z

Desargues' book is quite short and actually available online: bibnum.education.fr/math%C3%A9matiques/…

Comment by stankewicz

2013-03-31T19:54:42Z

Ah, it really is a bad habit to think of the field over which a morphism is really defined. carry on then.

Comment by stankewicz

2013-03-31T18:45:53Z

@wccanard: Do you have any examples where $R$ and $S$ are finite-dimensional $k$-algebras and $R \to S$ is a homomorphism of rings but not of $k$-algebras?

Comment by stankewicz

2013-03-20T08:33:37Z

As a reference for this ramification stuff, see the book of Vigneras, especially section 2.1

Comment by stankewicz

2013-03-11T06:52:49Z

It's also worth noting that additional work by Kolyvagin proves that under these conditions, the Tate-Shafarevich group of $E$ over $\mathbf{Q}$ is also finite.

Comment by stankewicz

2013-02-25T22:38:28Z

D'oh! Sorry Misha, completely didn't see your answer. I will gladly make this answer CW if requested.

Comment by stankewicz

2013-02-18T14:51:16Z

This question is not really well-defined. See for instance the FAQ ( If you have a very broad question (like "Please explain topic X"), try searching Google, Wikipedia, nLab, or looking for survey articles on the arXiv ). What is it that you would like to know about orthogonal modular forms?

Comment by stankewicz

2013-01-15T19:15:27Z

You are of course correct. I'm being far too glib.

Comment by stankewicz

2012-12-20T16:26:16Z

Slick proof, kreck!

Comment by stankewicz

2012-12-20T15:13:30Z

My previous comment was trivially wrong. Nonetheless, I still doubt that it's true over $\mathbf{Q}$. The basic reason is that the modular curves $X_0(p)$ for $p=2,3,5$ are all isomorphic to $\mathbb{P}^1$ and so have tons of rational points. It seems like there should be a counterexample somewhere in there.

Comment by stankewicz

2012-12-13T20:58:43Z

Could you give some motivation? For you, what would qualify as not resorting to algebraic number theory?