User ricky - MathOverflow

Answer by Ricky for How to refer to a theorem that you have shown to be wrong

2013-04-05T14:40:29Z

You can have a look to the paper "A counterexample to a 1961 “theorem” in homological algebra" by Neeman and use his style. By the way, I think that the paper is very very good.

Answer by Ricky for Is there non-simple-connected projective variety(over C) with trivial etale fundamental group?

2013-04-04T12:15:42Z

I rewrite the answer taking into account anon's comment below.

The are two questions here, the one in the title and the one in the body of the question. The difference is that in the title the word variety is used, while in the body the word manifold is used (so to answer the question in the title we can use singular varieties).

Let me start with the question of the title. The answer is yes. Any finitely presented group is the topological fundamental group of a complex algebraic variety. This is Theorem 12.1 in the paper "local systems on proper algebraic $V$-manifolds" by Carlos Simpson. Now consider the so called Higman's group $G$ defined by $$ G := \langle a,b,c,d \;| \; aba^{−1} = b^2 , \; bcb^{−1} = c^2 , \; cdc^{−1} = d^2 , \; dad^{−1} = a^2 \rangle. $$

One can prove that $G$ is not trivial (it is infinite) but the only finite index subgroup of $G$ is $G$ itself (see for example Serre's "Trees", Proposition 6 in Section 1.4 of Chapter I). In particular, the profinite completion of $G$ is trivial. Then any complex manifold with $G$ as fundamental group gives an answer to your question, since the algebraic fundamental group is the profinite completion of the topological one.

The question for manifolds is of course harder, and I think it is open.

State of the art for integral models of PEL type Shimura varieties with deep level structure

2012-10-12T16:42:39Z

The theory of PEL type Shimura varieties is nowadays well developed, but it is not easy to be updated with the latest results. Here I am particularly interested in integrals models. Let me describe what I understand.

Let $B$ be a (semi)simple algebra over $\mathbb Q$, of finite dimension. We suppose that $B$ is endowed with a positive involution $\ast$. Let $\mathcal O_B$ be an order of $B$, preserved by $\ast$. Let $(V,\Psi)$ be a finitely generated symplectic left $(B, \ast)$-module. Let $h \colon \mathbb C \to End_{B_{\mathbb R}}(V_{\mathbb R})$ be an $\mathbb R$-algebras homomorphism that gives an Hode strucure of type $(1,0),(0,1)$ on $V_{\mathbb R}$. We can now define a Shimura datum $(G,X)$ in the usual way. We obtain in particular a family of complex varieties $S_K$ parametrized by compact open subgroup $K \subseteq G(\mathbb A_f)$. We will assume that $K$ is 'small enough', in particular these varieties are moduli spaces of abelian varieties with additional (PEL) structure. It turns out that there is a number field $E$, called the reflex field, such that $S_K$ admits a canonical model defined over $E$.

In arithmetic it is very interesting to consider integral model of $S_K$. We fix a rational prime $p$. We assume that there is a lattice $\Lambda \subseteq V_{\mathbb Q_p}$ that is self-dual for $\Psi$ and we fix a compact open, small enough, subgroup $K^p \subseteq G(\mathbb A_f^p)$. Assuming that $B$ splits over an unramified extension of $\mathbb Q_p$, we have that $G(\mathbb Q_p)$ admits an hyperspecial subgroup, that we denote $K_{0,p}$. We assume that $B$ is of type A or C (another question is what can be done in the case D). It is well known that $S_{K^pK_{0,p}}$ admits a canonical integral model over $\mathcal O_E$, that is smooth over $\mathcal O_E \otimes Z_p$ and solves a very reasonable moduli problem. This goes back to Kottwitz.

Question 1 What can be done without the unramifiedness assumption? Of course in this case we do not have an hyperspcial subroup of $G(\mathbb Q_p)$, so we need a level strucure also at $p$. Rapoport and Zink have defined some integral models that satisfy a moduli problem, but it seems that their models are not even flat over the base.

Let me go back to the unramified case. For different reasons, it is interesting to consider level structures at $p$ (for example of type $\Gamma_1(Np^m)$ or $\Gamma_1(N) \cap \Gamma_0(p^m)$ in the case of modular curves). Now there is no hope for a smooth model, but of course one wants a good integral models. I am in particular interested in Iwahoric (or, more generally, parahoric) level structure. In the Siegel case, for example, we have good integral models.

Question 2 Under which assumptions we have good (flat and with a moduli interpretation) model of Shimura varieties with Iwahoric level structure at $p$? We have the models of Rapoport and Zink, but I do not if they are flat in general. Some cases are studied by Görtz (http://arxiv.org/abs/math/9912064 and http://arxiv.org/abs/math/0011202), but it seems that the general case is open (here I am always assuming that $G$ is quasi-split).

In general, I am interested in various condition 'at $p$' one have to put in order to obtain good integral models of PEL type Shimura varieties.

Thank you!

Non emptyness of ordinary locus for PEL type Shimura varieties

2012-11-05T13:45:11Z

We let $B$ be a simple algebra over $\mathbb Q$, with the usual notations for PEL type Shimura varieties. In his paper "Ordinariness in good reductions of Shimura varieties of PEL-type" (available here http://archive.numdam.org/article/ASENS_1999_4_32_5_575_0.pdf), Wedhorn proved, under the assumption that $p$ is unramified in $B$, that the ordinary locus in (the reduction) a PEL type Shimura variety is non-empty if and only if it is dense if and only if $p$ splits completely in the reflex field.

So my question is the following:

Are there similar results, even partial, without the unramifiedness assumption?

Newton point and Newton polygon stratifications

2013-02-11T15:51:51Z

Let $k$ be a field of characteristic $p>0$, with absolute Galois group $\Gamma$. Let $Y$ be a Shimura variety of PEL type, defined over $k$, with associated reductive (connected) quasisplit group $G$. We fix a maximal torus $T$ of $G_{\overline k}$ and a Borel subgroup $B \supset T$. We get a root datum $(X^\ast,R^\ast,X_\ast,R_\ast, \Delta)$, and let $\Omega$ be the Weyl group.

We have the so called Newton stratification of $Y$, that is defined in terms of the Newton point of any $y \in Y$. This is defined taking the F-isocrystal associated to $y$ and considering its image in $$ (X_{\ast,\mathbb Q}/\Omega)^\Gamma $$ via the "Newton map". See "On the classification and specialization of F-isocrystals with additional structure", by Rapoport and Richartz.

On the other hand, we can consider the stratification given by the Newton polygon of the F-isocrystal (i.e. looking at the classical Newton polygon of the abelian variety given by $y$). This stratification can be obtained as above forgetting the $G$-structure via the natural morphism $G \to \operatorname{GL}(V)$ (where $V$ is part of the PEL datum). In particular, the Newton point stratification is finer than the Newton polygon stratification.

Question: are these two stratifications equal?

This is the case if $G=\operatorname{GL}$ or $G$ is the symplectic group (I think), so, using the standard terminology of PEL Shimura variety, in case (A)linear or in case (C). It remains the unitary case, where I think the answer is in genera "no". Can someone give an example?

Thank you very much!

p-rank stratification in unitary Shimura variety

2013-01-26T20:34:20Z

Let $K$ be a quadratic extension of $\mathbb Q$ and let $p \neq 2$ be a prime that is inert in $K$. Let $X$ be the Shimura variety associated to the unitary group $\operatorname{U}(2,1)$ over $K$ (after a choice of a suitable integral PEL datum). We have an integral model $\mathcal X$ of $X$ defined over $\mathcal O_E \otimes \mathbb Z_p$, where $E$ is the reflex field. Let $A$ be a abelian variety corresponding to a $k$-point of $\mathcal X$, where $k$ is a field $k$ of characteristic $p$. Its $p$-torsion $A[p]$ has rank $6$ and it is equipped with an action of $\mathbb F_{p^2}$ (this is true regardless the assumption on $k$ of course).

Question: What can be said about the $p$-rank of $A$? It must be $0$ or $2$ (since $A[p^\infty]$ is principally polarized), but I do not know whether both these cases really appear (I believe so) or not.

Answer by Ricky for New grand projects in contemporary math

2012-12-31T09:55:55Z

The Langlands program. It goes back to the sixties, but in the last years, with the proof of the fundamental lemma by Ngô Bảo Châu and with several results in the local case, it became one of the most active area in number theory, and I think there is no hope to finish the job in the next, say, 50 years.

Answer by Ricky for Math French Words

2012-12-30T16:05:51Z

Kai-Wen Lan has written a glossary for French and German. Quoting from his web page "These are prepared primarily for reading mathematical texts". You can find the French one here.

Answer by Ricky for Which local ringed spaces are schemes?

2012-12-10T22:14:42Z

I think the easiest condition is the fact that the natural morphism $$ (X, \mathcal O_X) \to \operatorname{Spec}(\mathcal O_X(X)) $$ is an isomorphism.

When is the degree of this number 3?

2012-08-04T18:08:26Z

I am helping a friend of mine, that works in history of mathematics. She is studying the story of the solution of the cubic equation by Cardano. Sometimes she asks me some mathematical questions, that are very hard to motivate from a modern point of view, but that were interesting to Cardano. So please do not ask for motivations. The question is the following. Let $a$, $b$ be rational numbers, with $b$ not a square. Consider the number $$ t=\sqrt[3]{a+\sqrt{b}}+\sqrt[3]{a-\sqrt{b}}-\sqrt{a^2-b} $$ Under what conditions on $a$ and $b$ is the degree (over $\mathbb{Q}$) of $t$ equal to $3$? A sufficient condition can be found as follows. Let $P(x) = x^3+\alpha_2 x^2 + \alpha_1 x + \alpha_0$ be a rational polynomial. The general expression of the roots of $P$ is $$ \sqrt[3]{- \frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} + \sqrt[3]{- \frac{q}{2} - \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} - \frac{\alpha_2}{3}, $$ where $$ q = \frac{2\alpha_2^3 - 9\alpha_2\alpha_1 + 27\alpha_0}{27} $$ and $$ p = \frac{3\alpha_1 - \alpha_2^2}{3}, $$ see here. So we can take $a = -\frac{q}{2}$ and $b = \frac{q^2}{4} + \frac{p^3}{27}$ and we need to force $\sqrt{a^2 -b} = \frac{\alpha_2}{3}$. This boils down to $\alpha_1 = \frac{\alpha_2^2 - 3 \sqrt[3]{3\alpha_2^2}}{3}$. We find that one of the solutions of $$ x^3 + \alpha_2 x^2+ \frac{\alpha_2^2 - 3 \sqrt[3]{3\alpha_2^2}}{3}x+\alpha_0=0 $$ has the required form (of course we need to assume that $\alpha_2$ is such that $\sqrt[3]{3 \alpha_2^2}$ is rational). In this case $a$ and $b$ are given by the above expressions.

I suspect that if $t$ has degree $3$, then its minimal polynomial must be of this form and that $a$ and $b$ are as above, but I am not able to prove it. Note that the condition that $t$ has degree $3$ implies that it can be written as the sum of two cubic root and a rational number (because of the formula), but it is not completely clear that this way of writing $t$ is unique.

Sum of the reciprocal of perfect numbers

2012-06-10T10:25:00Z

It is well known that the sum of the reciprocal of prime numbers is $+\infty$. This proves that there infinitely many prime numbers. On the other hand it is also known that the series of the reciprocal of twin prime numbers converges, so nothing can be said about the finiteness of the set of twin prime numbers. This is indeed an open problem. A similar open problem is the existence of infinitely many perfect numbers. So this is the question:

let $\mathcal{Pe}$ be the set of perfect numbers. What is known about the series $$ \sum_{n \in \mathcal{Pe}} \frac{1}{n} $$

This is just curiosity, and could be a trivial question. Note that if there are no odd perfect numbers (as it seems to be the case), the series converges. Indeed any $n$ perfect is of the form $2^{p-1}(2^p-1)$.

Answer by Ricky for Blackbox Theorems

2012-06-14T11:17:46Z

Faltings' almost purity theorem. The proof given, for the smooth case, in $p$-adic Hodge theory has some problems, and the proof of the general case in the Asterisque paper Almost Étale Extensions is completely unreadable (at least to me) and also contains some mistakes. We now finally have a very good proof (by Peter Scholze), but the almost purity theorem has been used as a black box for years.

Answer by Ricky for A profinite group which is not its own profinite completion?

2011-11-29T14:14:32Z

Yes, this is possible. Take as $G$ the product of countable many copies of $\mathbb Z_p$. It has a countable basis of open subgroups, hence only countably many open subgroups. But it has many more subgroups of finite index!

Why chain complexes?

2011-07-08T13:34:22Z

Possible Duplicate:
Motivating the category of chain complexes

Chain complexes (of, say, abelian groups) are fundamental in homological algebra and algebraic geometry. For example, using the derived category of an abelian category is very natural nowadays. From another point of view, a lot of people are working on generalizing this ideas to the non abelian world (using simplicial stuff), see for example this answer.

So my question is: why chain complexes? I already know that they are extremely useful, but I'm looking for some theoretical reason for study them.

Ricky

trace(xy)=trace(yx) in full generality

2011-07-06T21:31:52Z

It is well known that, for square matrix $x$ and $y$, we have $\operatorname{tr}(xy)=\operatorname{tr}(yx)$. Here of course the trace of a matrix is just the sum of the elements of the diagonal.

The notion of trace has a lot of generalization. As I know, the most general definition is the following: let $(\mathcal C, \otimes, 1, ^\vee)$ be a rigid symmetric monoidal category, $X$ an object of $\mathcal C$ and $f$ an endomorphism of $X$. Then $\operatorname{tr}(f) \in \operatorname{End}(1)$ is defined by the following composition $$ 1 \longrightarrow X^\vee \otimes X \stackrel{\operatorname{id}_{X^\vee} \otimes f}{\longrightarrow} X^\vee \otimes X \longrightarrow X \otimes X^\vee \longrightarrow 1 $$ So my questions is: it is true, in this generality, that $\operatorname{tr}(f\circ g)=\operatorname{tr}(g \circ f)$, for $f$ and $g$ in $\operatorname{End}(X)$?

Ricky

Answer by Ricky for Which 'well-known' algebraic geometric results do not hold in characteristic 2?

2011-07-06T13:13:57Z

In characteristic 2 (and 3) a lot of familiar result about elliptic curves are false (usually not essentially false, but quite different). For example the Weierstrass equation is more complicated.

In Silverman's book "The arithmetic of elliptic curves" there is a whole appendix dedicated to this subject.

Tensor product of regular ring (with some conditions)

2011-05-23T10:25:57Z

Basically, my question is whether this answer is correct. Here is the point. Let $R$ be a ring, and let $A$ and $B$ be $R$-algebras. Suppose that $A$ is regular and $B \otimes_R B$ is regular too. Does it follow that $A \otimes_R B$ is regular?

What if we suppose that $R$ is regular and $B$ a smooth $R$-algebra?

Answer by Ricky for Which book would you like to see "texified"?

2011-05-13T19:20:15Z

All the SGA's. Note that SGA 1 and 2 already exists in TeX, and there is something for SGA 3 and 4.

Flatness of normalization

2011-05-12T13:00:22Z

Let $X$ be a noetherian integral scheme and let $f \colon X' \to X$ be the normalization morphism. It is known that, if non trivial, $f$ is never flat (see Liu, example 4.3.5).

What happens if we suppose $X$ normal, and we take the normalization in a finite (separable) extension of the function field of $X$? Note that in the easiest case, namely $X=\rm{Spec}(R)$, with $R$ a Dedekind domain, we have that $f$ is flat.

When are these rings regular?

2011-05-11T16:21:30Z

Let $R$ be a noetherian regular domain. Suppose that $a, b \in R$, with $b \neq 0$, and consider the ring $S:=R[\frac{a}{b}]=R[X]/(bX-a)$. Is $S$ regular? If this is not the case are there some conditions on $a$ and $b$ (or on $R$) that imply regularity? For example $a=1$ is enough.

Answer by Ricky for Theorems that are 'obvious' but hard to prove

2011-01-09T12:43:12Z

The Jordan curve theorem! Of course in this case the real problem is the meaning of "closed curve".

Answer by Ricky for What are Galois Categories used for?

2011-01-04T19:34:01Z

At the end, a Galois category is equivalent to the category $\pi$-sets, of finite sets with a continuous action of a profinite group $\pi$, so once you know this is easy to study Galois categories. The interesting part is that sometimes you a have a category, you can prove that it is Galois, so you have your $\pi$, but this is the only way you have to define the group. This is the method used by Grothendieck to define the fundamental group of a scheme (w.r.t. to a geometric point. used to define the fibre functor). See here link text for the details.

Answer by Ricky for Generalizing a problem to make it easier

2010-11-10T14:12:36Z

The free cocompletion. A lot of adjoint couples of functors are just particular case of this construction, and often it is easier to use the general theorem than working out a particular case by hand (for example for $i_{!}$ and $i^{!}$).

Answer by Ricky for Examples of common false beliefs in mathematics.

2010-10-19T20:57:20Z

Related to this answer: $$ \pi=\left(\frac{1}{10^5}\sum_{-\infty}^{+\infty}e^{-n^2/10^{10}}\right)^2. $$ Proof: With a computer one can verify that the first 42 billions digits of the two numbers are the same, see J. Borwein and P. Borwein, Strange series and high precision fraud, in The American Mathematical Monthly, 1992, pages 622-640.

What is the completion at a family of ideals?

2010-08-05T13:46:59Z

Let $A$ be a (commutative with unit) noetherian ring. If $I$ is an ideal of $A$, the $I$-adic completion of $A$ is by definition $$ \widehat{A} := \underset{\leftarrow}\lim A/I^n. $$ This operation is well known and has a lot of good properties, for example it is $I$-adically complete.

In an article I'm reading I found the sentence "completion at all prime ideals such that...". My question is: what is the completion with respect to a family of ideals?

My guess is the following: suppose we have finetely many ideals, say $I_1,\ldots,I_n$. Then the final result is obtained taking the completion wrt $I_1$, then the completion wrt (the ideal generated by) $I_2$ and so on. If we have infinitely many ideals we have to take a direct limit.

Is this construction discussed somewhere? Has it good properties? For example it's not totally clear to me that the order of the ideals doesn't affect the final result.

Answer by Ricky for Facts from algebraic geometry that are useful to non-algebraic geometers

2010-08-23T17:55:33Z

Any compact Riemann surface is projective and algebraic.

Riemann surfaces are studied in analysis an differential geometry, and of course is easier to work with polynomial equations. This statement is useful also for studying non compact Riemann surfaces.

Answer by Ricky for A morphism from proper to affine is constant?

2010-08-07T13:09:42Z

The following example does not work, it hasn't geometrically connected fibers. Sorry.

In general the answer is no. Take $k \subset K$, a finite extension of field (so the morphism $\operatorname{Spec}(K)\to\operatorname{Spec}(k)$ is proper). Let $X$ be an affine variety over $k$. Let now $x$ be a $K$-point of $X$ that is not defined over $k$. The corresponding morphism $\operatorname{Spec}(K)\to X$ does the job.

Local characterization of semistability

2010-07-29T13:44:04Z

It is known that a morphism of schemes $f\colon X \to S$ is smooth at a point $x \in X$ if and only if there is an open neighborhood $U$ of $x$ and an étale map $g \colon U \to \mathbb A^n_S$ such that $g \circ p=f_{|U}$, where $p \colon \mathbb A^n_S \to S$ is the natural projection.

I'm looking for a similar characterization for semistable curves $f \colon X \to S$. I'm interested in the case $S=Spec(k)$, with $k$ a field, and in the case $S=Spec(V)$, with $V$ a discrete valuation ring, where now $X$ is generically smooth.

In particular my question is: in the second case it is true that we can find $\lbrace Spec(R_i)\rbrace _{i \in I}$, an affine open covering of $X$, such that for each $i$, there is an étale map $V[x,y]/(xy-\pi) \to R_i$, where $\pi$ is a uniformizer of $V$?

Thanks.

Ricky

Answer by Ricky for Are morphisms of schemes generically affine

2010-07-24T17:51:06Z

I think the answer is no. Take any field $k$, and consider the natural morphism $\mathbb P^n_k \to spec(k)$, where $\mathbb P^n_k$ is the $n$-dimensional projective space. Since $spec(k)$ is, as topological space, a single point and $\mathbb P^n_k$ is not affine, your open subset cannot exist.

Answer by Ricky for Examples of common false beliefs in mathematics.

2010-07-23T17:58:36Z

Before reading about it, I really thought that if $f \colon [0,1] \times [0,1] \to [0,1]$ is a function with the following properties:

for any $x \in [0,1]$ the function $f_x\colon [0,1] \to [0,1]$ defined by $f_x(y)=f(x,y)$ is Lebesgue measurable, and also the function $f^y \colon [0,1]\to[0,1]$ defined by $f^y(x)=f(x,y)$ is Lebesgue measurable, for all $y \in [0,1]$;
both $\varphi(x)=\int_0^1 f_x d\mu$ and $\psi(y)=\int_0^1 f_y d\mu$ are Lebesgue measurable.

Then the two iterated integrals $$ \int_0^1\varphi(x)dx \mbox{ and } \int_0^1\psi(y)dy $$ should be equal. This is false (see Rudin's "Real and Complex Analysis", pag. 167), at least if you assume the continuum hypothesis.

Comment by Ricky

2013-05-08T17:02:24Z

This is probably not appropriate for this site. In any case the degree of the multiplication by $n$ on any elliptic curve is always $n^2$, for all $n$, regardless of the characteristic of the ground field. You can find this in any book/notes about elliptic curves.

Comment by Ricky

2013-04-18T15:15:34Z

@Joël: can you precise which Buzzard's paper are you talking about? Looking at the titles of his papers the word "Artin" appears just in one paper...

Comment by Ricky

2013-04-10T19:49:26Z

Doesn't this work only for projective abelian schemes?

Comment by Ricky

2013-04-04T13:04:58Z

Yes, you're right. I've added a reference.

Comment by Ricky

2013-04-04T12:37:19Z

Yes, of course, the manifold must be algebraic!

Comment by Ricky

2013-04-04T12:25:26Z

I don't understand your comment. The (topological) fundamental group depends only on the structure of topological space. A complex manifold is of course a topological space (with extra structure), so it has a topological fundamental group.

Comment by Ricky

2013-02-20T13:17:38Z

Thank you very much for pointing out Hartwig's paper!

Comment by Ricky

2013-01-27T09:22:57Z

Thank you very much Joël!

Comment by Ricky

2013-01-25T17:36:45Z

@pz Can you explain where spectral spaces are used in the theory of Adic spaces? I mean, I know that Huber proved that adic spaces are spectral (maybe with some conditions), but I always thought that he did this just to give an idea of how to visualize the topology on the adic spaces (in contrast with Berkovich spaces that have a "real" topology) and not to prove anything about adic spaces.

Comment by Ricky

2013-01-15T15:23:28Z

Really? Even for statements like Fermat last theorem?

Comment by Ricky

2013-01-10T12:10:27Z

Is there a particular reason to consider $\pi$ and not other transcendental numbers? For example, is the problem known for $e$ or $\sum_i 10^{-i!}$? Just for curiosity, can you give an example of a number $x$ such that $a^x=b$ but $x$ is not defined as $\log_a(b)$ (I know this not a precise question)?

Comment by Ricky

2013-01-04T12:40:04Z

Dear Joël, thank you for the comment. I would like to stress once again that my sentence was just a personal opinion (based on my, very limited, knowledge on the subject) and was not intended to be taken very seriously. In any case I am very happy to know about the results you cited!

Comment by Ricky

2013-01-02T17:44:03Z

The answer to my previous comment is: less than 5 minutes (but the question was voted to close not by quid). Amazing!

Comment by Ricky

2013-01-02T17:41:07Z

The real question: how many time will he need to close this question? Seriously: this question does not make sense and should be closed.

Comment by Ricky

2013-01-02T11:36:43Z

I really meant no. This is of course just a personal opinion (and am not an expert!). But I think there very important recent results in the local case, for example the local-global compatibility proved by Emerton.