User harry - MathOverflow

Do all curves have Néron models

2012-10-22T19:38:57Z

Let $X$ be a smooth projective geometrically connected curve over a number field $K$. Assume that $g\geq 2$.

Does there exist a Néron model $\mathcal X$ for $X$ over $O_K$?

By a Néron model, I mean a smooth model (not necessarily proper) with the "Néron universal property": for any smooth $O_K$-scheme $\mathcal Y$,

$$\mathrm{Hom}(\mathcal Y, \mathcal X) = X(\mathcal Y_K).$$

Is the smooth locus of the minimal regular model of $X$ over $O_K$ a Néron model? Does base change help? That is, does there exist a Néron model for $X_L$ after some suitable base change $L/K$?

Note that if $\mathcal{X}$ is the smooth locus of the minimal regular model of $X$ over $O_K$, we have the "Néron" property $\mathcal{X}(O_K) = X(K)$. (In fact, the image of a section in the minimal regular model lies in the smooth locus.)

This question was asked on stackexchange four months ago:

http://math.stackexchange.com/questions/153369/do-neron-models-of-hyperbolic-curves-exist

A question on Cebotarev's density theorem

2012-10-18T22:09:25Z

Let $K$ be a number field, $d$ a positive integer and $S$ a finite set of places of $K$.

By Cebotarev, there exists a finite set of finite places $T$ disjoint from $S$ such that the conjugacy classes of geometric Frobeni $F_v$ ($v\in T$) fill up $\mathrm{Gal}(K^\prime/K)$ for any $K^\prime/K$ Galois of degree at most $d$ (Edit) and unramified outside $S$.

For a finite field extension $L/K$, let $T_L$ be the set of places of $L$ lying over $T$. We use similar notation for $S_L$.

Question. Let $L/K$ be a finite extension, not necessarily Galois. Then $T_L$ is disjoint from $S_L$. Do the conjugacy classes of geometric Frobeni $F_w$ ($w\in T_v$) fill up $\mathrm{Gal}(L^\prime/L)$ for any $L^\prime/L$ Galois of degree at most $d$ (Edit) and unramified outside $S_L$?

Does a curve over a number field have a finite etale cover of given degree

2012-10-14T10:26:26Z

Let $X$ be a (smooth projective geometrically connected) curve over a number field $K$ of genus $g\geq 2$. Let $d\geq 2$ be an integer.

Does there exist a curve $Y$ over $K$ with a finite etale $K$-morphism $Y\to X$ of degree $d$?

I know how to do this over $\bar{K}$ (and thus over some finite extension of $K$). In fact, it suffices to find a finite degree topological cover $Y_{\mathbf{C}} \to X_{\mathbf{C}}$. This is easy to achieve by embedding $X_{\mathbf{C}}$ into its Jacobian and taking a degree $d$ topological cover of the Jacobian $J = \mathbf{C}^g/\Lambda$ of $X_{\mathbf{C}}$. The latter can be constructed easily by taking a sub-lattice of $\Lambda$ of index $d$.

Two problems arise.

The curve $X$ might not embed into its Jacobian, i.e., it might happen that $X(K)$ is empty. So if it helps, assume $X(K)$ is non-empty.

Also, the etale cover of $J$ constructed over $\mathbf{C}$ might not be defined over $K$ a priori. Can one show that it actually is defined over $K$?

What is the reduction of this hyperelliptic curve

2012-10-13T12:34:25Z

Let $K$ be a number field and $E/K$ an elliptic curve with equation $Y^2Z = X^3 +AXZ^2+BZ^3$ in $\mathbf{P}^2_K$, where $A,B\in K$.

Let $S$ be non-empty finite set of finite places of $K$ and suppose that $E$ has bad reduction over $S$ and good reduction outside $S$. Moreover, let $L/K$ be a finite field extension such that $E_L$ has good reduction over $O_L$. (In particular, $E/K$ has potential good reduction.)

How similar is the reduction of the hyperelliptic curve $H$ of genus $g\geq 2$ given by $$Y^2 Z^{2g-1} = X^{2g+1} + AX Z^{2g} + B Z^{2g+1}$$ to the reduction of $E$?

Does $H_L$ have good reduction over $O_L$?

Does $H$ have good reduction outside $S$?

Does $H$ have bad reduction over $S$?

Jacobians defined over smaller fields

2012-10-04T22:06:18Z

Let $L/K$ be an extension of number fields.

Let $X$ be a curve over $L$ which can not be defined over $K$. Let $J(X)$ be the Jacobian of $X$ over $L$.

In general, the Jacobian $J(X)$ probably doesn't admit a model over $K$. But one could imagine that this does happen sometimes.

Question. Does there exist an example where $J(X)$ admits a model over $K$?

In other words, can "the" field of definition of $J(X)$ be smaller than "the" field of definition of $X$?

Defining isogenies over smaller fields

2012-10-05T18:53:47Z

I'm having some issues with abelian varieties and fields of definition. This already became clear in my previous question on Jacobians. Here's another question. If somebody can explain some nice facts on fields on definition this would help me a lot (because these aren't the only questions I have concerning fields of definition).

Let $A$ be an abelian variety over a number field $K$. Let $L/K$ be a finite field extension. Suppose that there exists an isogeny $A_L\to B$ defined over $L$.

Is this isogeny defined over $K$ if $B$ can be defined over $K$?

I think the answer is negative, but I have to admit that my set of examples is very scarce and, therefore, I can't find an easy counterexample.

A question about the Tannakian etale fundamental group of a curve

2012-04-28T08:43:37Z

Let $X$ be a smooth connected quasi-projective curve over $\mathbf{Q}$. Let $U$ be the pro-unipotent etale fundamental group of $X$ over $\mathbf{Q}_p$.

$U^1 = U$ and let $U^n =[U,U^{n-1}]$.

Let $n\geq 1$.

Why is $U^n/U^{n+1} \cong \mathbf{Q}_p(n)^{r_n}$ for some positive integer $r_n$?

I "know" this is true for $X=\mathbf{P}^1-\{0,1,\infty\}$ because M. Kim uses it in his article on Siegel's theorem and the motivic fundamental group.

Why is this true in general? (Here we should probably ask our curve to be hyperbolic.)

Is this true if we replace $X$ by a higher-dimensional variety in general?

Does the above property also hold for the "other" unipotent fundamental groups such as the pro-unipotent de Rham fundamental group of $X$? (Again, the answer is yes if $X$ is the projective line minus three points.)

Does semi-stable reduction behave well with Weil restriction of scalars

2012-08-05T14:11:29Z

Let $A$ be an abelian variety over a number field $K$ with semi-stable reduction over $O_K$.

Does the Weil restriction $\textrm{Res}_{K/\mathbf{Q}}A$ of $A$ to $\mathbf{Q}$ have semi-stable reduction over $\mathbf{Z}$?

Note that $\textrm{Res}_{K/\mathbf{Q}}A$ is an abelian variety over $\mathbf{Q}$ of dimension $\dim A \cdot [K:\mathbf{Q}]$.

Does the self-product of a $g$-dimensional abelian variety contain an abelian variety of dimension smaller than $g$ at some point

2012-08-01T15:10:53Z

Let me be more precise than the title. (This will be my last attempt to do something with abelian varieties. Sorry for all the basic questions. The answers have been great!)

Let $A$ be a simple abelian variety over a field $k$. Let $g\geq 2$ be the dimension of $A$.

Does there exist an integer $n\geq 1$ such that $A^n = A\times_k A\ldots\times_k A$ contains an abelian variety of dimension less than $g$?

It suffices to prove that $A^n$ contains a curve of genus strictly smaller than $ g$ for some $n\geq 1$.

I'm afraid that this is not true. In fact, if $B\subset A^n$, then $B$ is isogenous to $A^m$ probably. Therefore, $\dim B =mg$. I'm just asking to be sure.

Is any simple abelian variety covered by a non-simple abelian variety

2012-08-01T13:39:10Z

Let $A/k$ be a simple abelian variety.

Does there exist a non-simple abelian variety $B/k$ and a finite homomorphism $f:B\to A$ over $k$?

I don't need $f:B\to A$ to be etale.

Are abelian varieties degree two covers of some projective space

2012-08-01T12:48:40Z

Let $A$ be an abelian variety over a field $k$ of dimension $g\geq 2$.

There exists a finite morphism $A\to \mathbf{P}^g_k$. Here's the question.

Does there exist a finite morphism $A\to \mathbf{P}^g_k$ of degree two?

Can we say something about the minimal degree of a finite morphism $A\to \mathbf{P}^g_k$?

The cohomology of the relative dualizing sheaf of a relative curve

2012-05-27T23:19:13Z

Let $X\to S$ be a curve over $S=\mathrm{Spec} \ \mathbf{Z}$. Let $\omega$ be the relative dualizing sheaf of $X\to S$. Let $g$ be the genus of the generic fibre. Assume that $g\geq 2$.

I know that $\omega$ is ample. Even better, a theorem of Deligne and Mumford says that $\omega^{\otimes 3}$ is very ample.

So there is some integer $n $ depending on $X\to S$ such that $H^1(X,\omega^{\otimes m}) =0$ for all $m\geq n$.

Can we find such an $n$? Is it independent of $g$? Does $n=3$ work?

If not, can we find such an $n$ and bound it in terms of $g$?

Schemes associated to vector spaces

2012-05-12T10:29:22Z

Let $k$ be a field. Let $F$ be a covariant functor on the category of $k$-algebras to the category of sets. Assume that the opposite functor $F^{op}$ on the category of affine $k$-schemes is a sheaf. (This last assumption might not be necessary.)

Now, suppose that $F(R)$ is a finitely generated $R$-module for all $k$-algebras $R$ and that

$F(R) = F(k)\otimes_k R$.

It seems that in this case $F$ is "representable" by the vector space $$F(k).$$

I don't know what this means and I'm probably missing some very standard construction here. It should mean that $F^{op}$ is representable by a finite type $k$-scheme.

I know one can associate to a finite-dimensional vector space $E$ over $k$ the affine variety $\mathrm{Spec} \ \mathrm{Sym}(E).$ Is it clear that in this case the functor $F$ is representable by the affine variety $\mathrm{Spec} \ \mathrm{Sym}(E)$?

Answer by Harry for trying to understand the support of the sheaf of relative differentials

2012-04-29T07:56:29Z

Let $\pi:X\to Y$ be a finite morphism of curves.

Then, for any point $x$ in $X$ lying over $y$ in $Y$, the coefficient $v_x(\pi)$ of $\Omega_{\pi}$ is the valuation of the different of the extension of dvr's $\mathcal{O}_{y}\subset \mathcal{O}_x$. If you are working in characteristic zero, then $$v_y(\pi) = e_x-1,$$ where $e_x$ is the ramification index. So you see that $\Omega_{\pi}$ is supported on the ramification points.

Also, you have a short exact sequence (it's on page 2 of Chapter IV.2 in Hartshorne) which relates $\Omega_\pi$ with $\Omega_X$ and $\Omega_Y$. The above actually shows the important Riemann-Hurwitz formula: $$K_X = \pi^{\ast} K_Y + R.$$ Here $R$ is the ramification divisor. This equals $\Omega_{\pi}$ in this case.

Pro-affine varieties over a local field

2012-04-25T13:43:33Z

Let $K$ be a (perfect) local field, and let $S = \lim (\mathrm{Spec} A_i)_{i=0}^\infty$ be a pro-affine variety over $K$. This means that each $A_i$ is a finite type $K$-algebra and that the affine varieties $\mathrm{Spec} A_i$ form a projective system. Note that $S$ is an affine $K$-scheme which is not of finite type in general.

I have some elementary questions about such schemes.

Q1.Is $S$ regular if and only if $\mathrm{Spec} A_i$ is regular for all $i=0,\ldots$?

Q1*.Is $S$ normal (resp. irreducible or reduced) if and only if $\mathrm{Spec} A_i$ is normal (resp. irreducible or reduced) for all $i=0,\ldots$?

Q2. How can I determine the dimension of $S$ from the dimension of $A_i$. (Assume the schemes to be integral for this question.)

Unfortunately, I have little feeling for such pro-affine varieties at the moment.

A last (and bit vague) question:

Q3. Is there a moduli space of smooth connected pro-affine varieties of given dimension?

Answer by Harry for Second Chern Class of Blow-up

2012-04-09T10:49:53Z

Let me slightly expand on Dmitri's comment.

Let $X$ be a finite type separated $\mathbf{C}$-scheme. Let $e_c(X)$ be the compactly supported Euler characteristic. (We consider the singular cohomology of $X$ with $\mathbf{Q}$-coefficients.)

Then, if $X$ is irreducible and $n$-dimensional, by Gauss-Bonnet, we have $\deg c_n(X) = e_c(X)$.

Let $p:Y\to X$ be a proper birational surjective morphism. Let $s$ be the number of exceptional components of $p$. It is easy to see that $e_c(Y) = e_c(X) +s$.

This generalizes what you need.

Answer by Harry for Excellent uses of induction and recursion

2012-04-01T13:29:30Z

In pro-algebraic geometry you get to see some nice arguments by induction. For example, M. Kim proves that the continuous cohomology $$H^1(G_{\mathbf{Q}_p},\pi_{1,et}^{uni}(X))$$ is representable by induction on the terms in the lower central series of the $\mathbf{Q}_p$ pro-unipotent algebraic group associated to the etale fundamental group of a curve $X$. Not very surprising, but still crucial for the argument.

For a reference, see page 639 in http://www.ucl.ac.uk/~ucahmki/siegelinv.pdf

Q-factorial and rational singularities on surfaces

2012-03-26T15:18:12Z

Let $X$ be a normal surface. Is any rational singularity $\mathbf{Q}$-factorial? I've seen this somewhere for surfaces over fields, but what about the general case of an integral 2-dimensional excellent normal scheme?

In this generality it might not hold so what if we assume that $X$ is fibered ( = flat projective) over a Dedekind scheme? When can we hope for such a result to hold. Probably there are some problems depending on the characteristic.

What about the converse?

I know that every surface fibered over $\mathrm{Spec} \mathbf{Z}$ is $\mathbf{Q}$-factorial. Are all its singularities rational?

I know that one has to be careful with the base scheme. Probably if the base scheme is a smooth projective curve over a field things might not work so well, but maybe if the base is $\mathrm{Spec} \mathbf{Z}$ things might become better.

The use of embedding a curve into its Jacobian

2012-03-25T09:08:42Z

I'm looking for as many examples/applications as possible of the use of embedding a smooth projective geometrically connected curve $X$ over a number field $k$ with $X(k)\neq \emptyset$ into its Jacobian (via a rational point). Please do not take the meaning of "use" to seriously.

I know of

Chabauty's method of proving a special case of the Mordell conjecture.

Faltings' use of the Torelli map in his proof of the Shafarevich conjecture for curves.

Raynaud's theorem (previously Manin-Mumford conjecture).

The Bogomologov conjecture (proven by Ullmo and Zhang).

The Mordell-Lang theorem.

In some of these examples the embedding of X into its Jacobian is simply part of the statement. I also consider this as "useful".

Are there any other nice examples? They don't have to be as difficult as the ones mentioned above.

Can we define homotopy groups using Tannakian categories

2012-03-24T13:22:44Z

This is another vague question. Hope you guys don't mind.

Let $T$ be a Tannakian category. For any fibre functor $F$ on $T$ we define the fundamental group of $T$ at $F$, denoted by $\pi_1(T,F)$, to be the tensor-compatible automorphisms of $F$. This fundamental group is representable by an affine group scheme.

Can one give a meaningful definition of homotopy groups $\pi_n(T,F)$ using the Tannakian formalism?

Are torsors over unipotent groups trivial

2012-03-20T09:07:41Z

I might have misunderstood something I heard somewhere.

Are torsors over unipotent groups trivial?

I couldn't find this in some standard references.

Constructing rational functions with ramification locus the divisor of some $n$-form

2012-03-19T17:36:54Z

I'm still busy learning the theory of linear systems for compact Riemann surfaces. If the answer to the following question is negative, then there might not be any point in continuing.

Let $X$ be a compact connected Riemann surface and let $\omega$ be an $n$-form on $X$. That is, $\omega$ is a global section of the canonical sheaf $\omega_X^{\otimes n}$.

Now, let $D$ be the divisor of $\omega$ on $X$.

Can we construct a morphism $X\to \mathbf{P}^1$ such that the support of the ramification locus equals the support of $D$ for some choice of $n$? If yes, the degree of such a morphism equals the degree of $\omega_X^{\otimes n}$, right?

Slightly weaker: can we construct a morphism $X\to \mathbf{P}^1$ such that the support of the ramification locus is contained in the support of $D$?

As Francesco points out, this is not possible if $g=2$ and $n=1$

Probably, if $g$ is small compared to $n$, the answer will be negative.

How do subgroups of fundamental groups relate to torsors

2012-03-18T17:15:09Z

Fix a nice complex algebraic variety $X$ with base point $x$. (We will work in the analytic topology.)

Let $G$ be a finite group. Then the set of normal subgroups of $\pi_1(X,x)$ with quotient $G$ corresponds to the set of $X$-torsors for $G$. (We write $G$ for the constant sheaf on $X$ associated to $G$.)

This is just saying that Galois covers of $X$ correspond to torsors for $G$.

Now, not all covers of $X$ are Galois. In general, finite covers of $X$ correspond (in some sense) to finite index subgroups of $\pi_1(X,x)$ and the degree is given by the index.

I'd like to know if we can replace the set $H^1(X,G)$ of torsors for $G$ by something "torsorish" and still have a correspondence with the set of finite index subgroups of $\pi(X,x)$ of some fixed degree.

Does Nori's fundamental group scheme appear in Kim's work

2012-03-18T17:41:05Z

This is a very vague question. I was just reading the introduction to M. Kim's article on motivic fundamental groups and the theorem of Siegel and noticed that there are essentially three fundamental groups appearing in his work: the De Rham fundamental group, the cristalline fundamental group and the etale fundamental group.

Now, does Nori's fundamental group scheme also appear somewhere in his work? If no, why not?

Construction of Kummer map for abelian variety

2012-03-18T14:00:37Z

Let $A$ be an abelian variety over the rational numbers $\mathbf{Q}$. Let $V=T_p A \otimes \mathbf{Q}_p$ be the $\mathbf{Q}_p$-Tate module of $A$. Let $G$ be the absolute Galois group of $\mathbf{Q}$. (added in edit)

I keep seeing a natural map $A\to H^1(G,V)$. How is this map constructed? What does it have to do with "Kummer theory"?

What is the image of this map? That is, how can one describe it? Does it have to do with Selmer groups?

Sorry for the vagueness.

Why do twists of an algebraic group over k correspond to k-torsors over G

2012-03-17T19:28:55Z

Let $G$ be an algebraic group over a field $k$. Let $k^s$ be the separable closure of $k$.

I can't seem to figure out why isomorphism classes of twists of $G$ correspond to $k$-torsors over $G$.

It's easy to see that a $k$-torsor over $G$ gives a twist of $G$, but the other way around isn't clear to me. It is bound to involve something from descent theory that I don't know.

In fact, my problem is that I can't seem to define an action of $G$ on $X$. I can only seem to get an action of $G_{k^s}$ on $X_{k^s}$ via the isomorphism of $X_{k^s}$ with $G_{k^s}$. Why does this action descend to $k$?

Let me precise that a twist of $G$ is a variety $X$ over $k$ such that $X_{k^s}$ is isomorphic to $G_{k^s}$ as a variety.

Comment by Harry

2012-10-23T07:24:53Z

My apologies. In the course of editing the original question posted on stackexchange I reversed some of the statements. What I meant is that it's easy to see that the weak Neron property holds for $\mathcal{X}$ when $X$ has good reduction over $O_K$. This statement is only slightly harder to prove when $X$ doesn't have good reduction. I'll edit the question accordingly.

Comment by Harry

2012-10-22T19:41:48Z

@David Speyer. Thank you for the example.

Comment by Harry

2012-10-20T12:32:05Z

This shows that there exists some $T_L$ with the right property. It doesn't show that I can take $T_L$ to be the places lying over $T$, though. And as David Speyer's example shows, this is not possible in general.

Comment by Harry

2012-10-19T06:34:46Z

@GH. You're right. I forgot to mention why I fix $S$. I'll edit the question. My apologies.

Comment by Harry

2012-10-14T11:32:32Z

Fair enough. That solves the problem for $d=2$. Does your remark also generalize to $d>2$ in some way? I have the feeling that the existence of a $d$-cover of $X$ implies the existence of an element of order $d$ in its Jacobian. This means that it should be possible that $X$ has no finite etale covers at all. Is that true? (I think it's true for an elliptic curve with $X(K)$ trivial.)

Comment by Harry

2012-10-14T11:15:37Z

Right. I forget to mention that $X$ is hyperbolic. But probably, the answer will be "not in general" again. I just can't think of a good reason.

Comment by Harry

2012-08-01T14:56:31Z

I think I see what you mean. Thanks for correcting my English.

Comment by Harry

2012-07-29T08:35:14Z

Theorem 1.1. in "UNIFORMITY OF RATIONAL POINTS" by CAPORASO HARRIS and MAZUR states that, assuming the weak Lang conjecture, there exists a real number $c(K,g)$ such that for all curves $X$ over $K$ of genus $g\geq 2$, the number of $K$-rational points of $X$ is bounded by $c(K,g)$. So conjecturally, there really is a uniform bound on the number of rational points.

Comment by Harry

2012-07-15T22:14:31Z

By the adjunction formula, a rational curve $E$ on $\mathcal X$ will have self-intersection $-E^2 = -2 - K\cdot E$, where $K$ is a canonical divisor of $\mathcal X /O_K$. So a necessary and sufficient condition to have no rational curves is that there aren't any vertical divisors $E$ such that $E^2 = -2 - (K,E)$.

Comment by Harry

2012-06-12T06:47:52Z

The following MO thread might be relevant: mathoverflow.net/questions/77644/…

Comment by Harry

2012-05-28T09:31:32Z

@myself. I think I got it. The push-forward f_\ast coincides with taking global sections and then exact functor "tilde" which associates to a $\mathbf{Z}$-module $M$ the coherent sheaf $\widetilde{M}$ on $\mathrm{Spec} \mathbf{Z}$.

Comment by Harry

2012-05-28T09:23:59Z

@ulrich. Let $n\geq 2$ and let $f:X\to S$ be a semi-stable curve. It's clear that the higher cohomology of $\omega^{\otimes n}$ vanishes on the generic fibre by duality. By loc. cit., this implies that $R^1 f_\ast \omega^{\otimes n} =0$. But why does this imply that $H^1(X,\omega^{\otimes n}) =0 $? I don't see how $f_\ast $ and the global sections functor are related in this case, because the base is $\mathrm{Spec} \mathbf{Z}$.

Comment by Harry

2012-05-28T08:07:45Z

Here's the link : math.ucdavis.edu/~osserman/math/… There was just a small typo in your hyperlink.

Comment by Harry

2012-05-28T08:05:09Z

Thanks for your answer. The link doesn't work, unfortunately. What do you mean when you answer by yes? $n=3$ works?

Comment by Harry

2012-05-27T23:28:18Z

"If you take a ring that contains $\mathbf{Q}[\alpha]$, with $\alpha$ a unit not in $\mathbf{Q}$, and tensor with $\mathbf{C}$ you will get $\mathbf{C}[\alpha]$." What if I tensor $\mathbf{Q}[\sqrt{2}]$ with $\mathbf{C}$? Then I get $\mathbf{C}\otimes \mathbf{C}$. This is not $\mathbf{C}[\sqrt{2}] = \mathbf{C}$. Moreover, why "with $\alpha$ not a unit in $\mathbf{C}$"? What am I missing here? I shouldn't be able to take $\alpha = \sqrt{2}$?