User andrea mori - MathOverflow

Answer by Andrea Mori for Trichotomies in mathematics

2013-02-03T13:49:45Z

Did anybody mention yet the different arithmetic-wise behaviour of algebraic curves of genus $0$, $1$, or $\geq2$?

Explicit Casselman theory: reference needed

2013-01-28T10:40:54Z

Let $K$ be a nonarchimedean local field with ring of integeres $R_K$, maximal ideal $m_K$ and finite residue field $\bf k$. Let $\pi$ be an admissible irreducible complex representation of ${\rm GL}_2(K)$ with central character $\epsilon$. A fundamental result of Casselman says that there is a largest ideal $J\subseteq R_K$ such that the subspace $W_J$ of vectors in $\pi$ such that $$ \gamma\cdot v=\epsilon(a)v\qquad \forall\gamma=\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)\in{\rm GL}_2(R_K) \ \text{with}\ c\in J $$ is non-trivial and in fact $1$-dimensional. As every expert knows, this result is of paramount importance for the theory of modular forms.

Let $v_0$ be a generator of the $1$-dimensional space $W_J$. In some cases, it is rather easy to obtain $v_0$ explicitly. For instance if $\pi=\pi(\mu_1,\mu_2)$ is a class $1$ principal series representation with trivial central character (for which $J=R_K$) it is immediate to check that any generator of $W_{R_K}$ is of the form $$ v_0(g)=|a|^{s_1}|d|^{s_2}|a/d|^{1/2}v_0(1)\quad \text{where}\quad g=\left(\begin{array}{cc} a & *\\ & d \end{array}\right)r,\quad r\in{\rm GL}_2(R_K) $$ and $\mu_i=|\cdot|^{s_i}$, $i=1$, $2$.

My question is that if a table of generators $v_0$ has been tabulated explicitly anywhere, in particular for the supersingular representations and in other cases in which $J\subseteq m_K^2$.

Notation for a canonical quotient of an abelian variety in positive characteristic

2012-04-29T11:42:54Z

This is a light question about notation, but I received no answer in Stackexchange.

Let $k$ be an algebraically closed field of characteristic $p>0$ and let $A=A_{/k}$ be an ordinary abelian variety of dimension $g\geq1$ which we assume principally polarized for the sake of simplicity. One knows that the $p$-torsion of $A$ is a product: $$A[p]=\hat A[p]\times T_p(A)\otimes(\Bbb Z_p/p\Bbb Z_p).$$ Here $\hat A[p]$ is the maximal connected subgroup, $T_p(A)\otimes(\Bbb Z_p/p\Bbb Z_p)$ is etale, both factors are subgroups of rank $p^g$ and they're Cartier dual of each other.

The completely inseparable isogeny obtained by taking the quotient by $\hat A[p]$ is the relative Frobenius $F_k:A\rightarrow A^{(p)}$ where $A^{(p)}$ is the abelian variety obtained by twisting the $k$-structure of $A$ by the geometric Frobenius i.e. the automorphism $x\mapsto x^{1/p}$ of $k$.

The isogeny $A\rightarrow A^\prime$ obtained by taking the quotient by $T_p(A)\otimes(\Bbb Z_p/p\Bbb Z_p)$ is etale and after an identification $A\simeq(A^\prime)^{(p)}$ is the Verschiebung $V_k:(A^\prime)^{(p)}\rightarrow A^\prime$ associated with $A^\prime$.

Is there a standard notation for the abelian variety that I denoted $A^\prime$? I have checked a few textbooks and lecture notes about abelian varieties and/or group schemes, but I seemed not to be able to find any.

Divisors of solutions of elliptic problems

2010-05-07T08:43:15Z

I learned very recently that in the early-mid 90s Gromov and Shubin proved a generalization of the Riemann-Roch theorem. Let $X$ be a compact closed ${\cal C}^\infty$-variety of dimension $n\geq2$. Suppose that $\cal E$ and $\cal F$ are complex vector bundles on $X$ of rank $q$ and let $A:{\cal E}\rightarrow{\cal F}$ be an elliptic differential operator of order $d$. Then for a divisor $D$ on $X$ one can define the space $L(D,A)$ of sections $f$ of $\cal E$ such that $Af=0$ and the zeroes and poles of $f$ are subordinated to $D$ in the usual sense. Then their basic result is that $\dim(L(D,A))-\dim(L(-D,A^t))={\rm ind}(A)+q\deg(D)$ where ${\rm ind}(A)$ is the index of $A$ and $\deg(D)$ is the degree of $D$ (whose definition is slightly different then the usual "algebraic" one).

Not being really an expert of elliptic operators, I wonder how much one can push analogies with the holomorphic (or algebraic) situation. For instance, suppose (to make things simple) that $X$ is a compact Riemann surface ($n=2$) and that $\cal E$ and $\cal F$ are line bundles. Let $f$ be a global ${\cal C}^\infty$-section of $\cal E$ such that $Af=0$. Is there any hope that the degree of the divisor of $f$ (possibly suitably modified) is 0?

"probabilistic" density of primes?

2010-09-16T10:22:01Z

A certain set $\cal P$ of primes is defined by two assumedly independent conditions:

The first condition on a prime $p$ can be characterized in terms of the type of splitting of $p$ in certain Galois extensions of $\Bbb Q$. The natural density of the bigger set ${\cal P}^\prime$ of primes satisfying this condition can be estimated under the usual independence hypotheses using Cebotarev's theorem.

Let $v$ be a non-zero vector in ${\Bbb Z}^2$. For each $p\in{\cal P}^\prime$ a certain $1$-dimensional subspace $V_p\subset({\Bbb Z}/p{\Bbb Z})^2$ is defined. The second condition is then that $p\in{\cal P}$ if and only if $v\bmod p\in V_p$.

Thus, a prime $p\in{\cal P}^\prime$ is actually in $\cal P$ with "probability" $c_p=\frac1{p+1}$.

I am tempted to consider the quantity $$ \delta({\cal P})=\lim_{n\to\infty}\frac {\sum_{p\in{\cal P}^\prime_n}\ c_p} {|\hbox{primes $\leq n$}|} $$ where ${\cal P}^\prime_n={\cal P}^\prime\cap[1,\ldots,n]$. What is unclear to me is if this may be taken as a good estimate (or "guess") of the natural density of $\cal P$ or not.

Zeroes of Maass forms

2010-08-04T13:48:20Z

By a Maass form I just mean--maybe a bit loosely--any real analytic $\Bbb C$-valued function $f$ on the upper halfplane $\cal{H}$ which is automorphic of weight $k\in\Bbb Z$ with respect to a discrete subgroup $\Gamma<{\rm SL}_2(\Bbb R)$ such that $\Gamma\backslash\cal H$ has finite volume, and an eigenfunction for the Laplacian operator corresponding to the Casimir element in the universal enveloping algebra of the complexified $\rm{sl}_2$.

Is it true that the zeroes of these forms are isolated?

The answer is obviously affirmative in the case of holomorphic modular forms.

$\textbf{Edit}$: Scott's comment and Matt's answer below show that the answer is generally negative when the weight is $0$. Then, one can construct real valued Maass forms which have nodal curves.

Thus, let's make the assumption that $k\neq0$ and in particular that the Maass form $f$ belongs to a discrete series representation space (generated by a holomorphic modular form).

To make things as explicit as possible assume also that $\Gamma$ is either a congruence subgroup of ${\rm SL}_2(\Bbb Z)$ or the group of norm 1 elements in an Eichler order of an indefinite quaternion algebra over $\Bbb Q$

Answer by Andrea Mori for Roulette probability

2010-08-04T16:41:06Z

How come that past outcomes have zero influence? Simply because a roulette is just wood, plastic and brass and got no memory.

p-split Hecke characters

2010-03-16T17:23:19Z

Let $K$ be a quadratic imaginary field, $\bf n$ an ideal in the ring of integers ${\cal O}_K$ and $\xi$ an algebraic Hecke character of type $(A_0)$ for the modulus $\bf n$. One knows (from Weil) that there exists a number field $E=E_\xi\supseteq K$ with the property that $\xi$ takes values in $E^\times$.

Let $p$ be a prime that splits in $K$. Consider the following condition: there exists an unramified place $v\mid p$ in $E$ with residue field $k_v={\Bbb F}_p$ such that $\xi$ takes values in the group of $v$-units in $E$.

The condition implies the existence of a $p$-adic avatar of $\xi$ with values in ${\Bbb Z}_p^\times$.

I would like to know:

1) to the best of your knowledge, has been this condition considered somewhere? does it have a "name"?

2) I'm tempted to say that $\xi$ is $p$-split if the condition is satisfied (and that $v$ splits $\xi$). Would this name conflict with other situations that I should be aware of?

Infinite sets of primes of density 0

2010-03-02T10:17:46Z

Sorry if the question is too vague or if the examples I look for are too boringly well-known: my knowledge of analytic number theory is rather primitive......

So, here it goes: suppose that you want to prove that the set $\Sigma$ of primes satisfying a certain condition $C$ is infinite. Then you may attempt to compute the density $$ \delta(C)=\lim_{x\to\infty}\frac{|\text{$p\leq x$ such that $C(p)$ holds}|}{|p\leq x|}. $$ If $\delta$ turns out to be positive, you're done. But it could as well be that $\delta=0$ and yet $\Sigma$ be infinite.

My questions are: (1) what are the main known examples of this occurrence? (2) in these examples, if any, the proofs of the infiniteness of $\Sigma$ did use ad hoc case-by-case "tricks" or there are somewhat standard techniques than can be employed with the situation? (3) is there a standard reference?

Answer by Andrea Mori for When and why did the postdoctoral position originate?

2010-03-02T18:33:29Z

Well, in Italy Ph.D. programs started about in mid-80s (1980s, I mean), so strictly speaking we had no post-doctoral positions or fellowships before that. :-)

On periods of algebraic integers modulo rational primes

2010-01-30T13:33:53Z

I run, somewhat indirectly, into the following problem and I have no hints where to look in the literature in search for answers or clues.

Let $K$ be a number field, which we may assume Galois if it helps, $\cal O$ its ring of integers and for each prime number $p$ let $R_p={\cal O}/p{\cal O}$ (a finite product of finite fields of characteristic $p$ for almost all $p$). Fix $\lambda\in{\cal O}$, $\lambda\neq0$ or a root of $1$. Then $\bar\lambda\in R_p^\times$ for almost all $p$ and the period $\pi_p=\pi_p(\lambda)$ of $\lambda$ is defined as the smallest positive $d$ such that $\bar\lambda^d=1$.

It is obvious that if we fix an integer $n$ the number of $p$'s such that $\pi_p\leq n$ is finite, since $\pi_p=d$ implies that $p|(\lambda^d-1)$ and there are only finitely many of those.

On the other hand, if $n\geq2$ the number of $p$'s such that $\max{\text{Supp}(\pi_p)}\leq n$ (the support $\text{Supp}(N)$ of an integer $N$ is the set of prime divisors of $N$) is infinite. For instance, the set of elements $\lambda^{2^k}-1$ has an infinite set of rational prime divisors because $\lambda^{2^{k+1}}-1=(\lambda^{2^k}-1)(\lambda^{2^k}+1)$ and the only common prime divisors in $\cal O$ to the 2 factors are primes of residual characteristic 2. Thus, as k grows, a new set of primes adds up at each step, so to speak.

Now the question is: fix an arithmetic progression ${\cal P}:a,a+d,a+2d,\dots$ with $(a,d)=1$, is it true that there are infinitely many primes in $\cal P$ such that $\max{\text{Supp}(\pi_p)}\leq n$? Conditionally on $n$?

In particular: suppose $K$ quadratic, and let $\ell>2$ a prime. Are there infinitely many primes $p\equiv 1\bmod\ell$ such that $\max{\text{Supp}(\pi_p)}\leq\ell-1$?

Answer by Andrea Mori for Two questions on isomorphic elliptic curves

2010-02-03T16:17:48Z

As for your first question: if you think of your elliptic curves as plane cubics (Weierstrass'model) the isomorphism between them is a polynomial function. Polynomials include only finitely many coefficients and the isomorphism is defined over the field generated by them, which is finite over $\Bbb Q$.

Cohomological characterization of CM curves

2010-02-01T14:36:19Z

In his 1976 classical Annals paper on $p$-adic interpolation, N. Katz uses the fact that if $E_{/K}$ is an elliptic curve with complex multiplications in the quadratic field $F$, up to a suitable tensoring the decomposition of the algebraic $H_{\rm dR}^{1}(E,K)$ in eigenspaces for the natural $F^\times$-action coincides with the Hodge decomposition of $H_{\rm dR}^{1}(E,{\Bbb C})$ and (for ordinary good reduction at $p$) with the Dwork-Katz decomposition of $H_{\rm dR}^{1}(E)\otimes B$ for $p$-adic algebras $B$.

Then, he asks for a converse statement. Namely, is it true that if the Hodge decomposition of $H_{\rm dR}^{1}(E,{\Bbb C})$, where $E_{/K}$ is an elliptic curve, is induced by a splitting of the algebraic de Rham, then $E$ has complex multiplications?

The question is left unanswered in that paper. Does anyone know if the question has been answered since?

Answer by Andrea Mori for Which popular games are the most mathematical?

2010-02-01T10:11:25Z

Well, most card games have mathematical implications, of course.

I'm disappointed at your considering chess non-mathematical. Wonder what Noam Elkies would think. :-)

When I was a teenager I would play a lot a board game named here "Risiko!" (I believe that the English name is "Risk"). My impression then was that there were some mathematical aspects that could be considered while planning a strategy.

(added later)

Also Hex should be added to the list of mathematically interesting games.

Answer by Andrea Mori for Cool problems to impress students with group theory

2010-01-29T09:36:28Z

Another nice arithmetic application of (cyclic) group theory is the fact that the multiplicative group of a finite field is cyclic, or (in down-to-earth terms) that one can obtain every non-zero residue class modulo a prime just taking consecutive powers of a single well-chosen one.

I always give this example in my undergraduate Algebra course because it also gives me the occasion to show that in maths there are still questions which are unanswered but can be easily understood by a beginner (obviously, I'm referring to Artin's conjecture in this context).

Comment by Andrea Mori

2013-05-06T10:47:31Z

In the case in question one can just write "The canonical map $f:X\rightarrow Y$ is injective".

Comment by Andrea Mori

2013-02-03T13:45:18Z

Holy Trinity! That's a striking observation.

Comment by Andrea Mori

2013-01-28T18:48:38Z

@Giuseppe: Grazie! I don't understand what I was doing wrong...

Comment by Andrea Mori

2013-01-28T14:49:42Z

Class 1 principal series are induced from unramified characters. Sorry about the down-vote: I feel that down-voters should always give a justification.

Comment by Andrea Mori

2013-01-28T13:48:59Z

@Marc_Palm : I'm certainly primarily interested in complex representations. I left that somewhat implicit, also the remark about modular forms should be revealing! Please repost your answer.

Comment by Andrea Mori

2012-12-11T22:57:32Z

Who does actually conjectures that White has a winning strategy? One hundred and forty years of Grandmaster practice seems to suggest that the game is about even.

Comment by Andrea Mori

2012-04-29T15:24:49Z

@Kevin Buzzard: I actually like the idea of calling it $A^{(1/p)}$! I'm sort of surprised, though, that there is no standard notation for it, like if this object had not been given much consideration.

Comment by Andrea Mori

2011-10-09T17:20:35Z

Actually Italians have it both ways: "corpo" is usually a skew-field (like in "il corpo dei quaternioni") while "campo" (i.e. "field") is a "corpo commutativo".

Comment by Andrea Mori

2011-06-19T14:41:45Z

Ben, I've been away from MO for a while and I read your answer just now. The result you're pointing at seems very relevant for a possible application I have in mind. Thanks a lot!

Comment by Andrea Mori

2011-01-25T21:10:03Z

Sometimes the confusion has more tragic consequences: independent.co.uk/news/…

Comment by Andrea Mori

2011-01-25T14:16:16Z

I witnessed something like that happening at a conference some 20 years ago. About 15 minutes into a talk a Fields medalist in the audience commented loudly (and laughing....!) that a "theorem" just written on the board couldn't possibly be true for this and that reason. The speaker blushed, stared silently at the board for a couple of minutes and then had to admit that he was wrong. We had a longer-than-usual coffee break.......

Comment by Andrea Mori

2010-12-31T14:01:26Z

They have (had?) a Galton box in the math section of the Museum of Science in Boston with an added feature which I found intriguing (and clever): while most of the balls in the box are white, only a handful were black. After operating the machine the balls would overall arrange themeselves in a bell curve, BUT the few black balls would be scattered here and there in a unstructured random way. This shows that the expected distribution is reached only after a large amount of trials (=balls) while the theory is ineffective for a small amount. Unfortunately no panel on the exhibit explained this!

Comment by Andrea Mori

2010-09-22T08:16:03Z

On second thought, I'm not so sure. For instance because the sphere has two poles (and for the Riemann sphere interpretation a zero should be called a "southern pole"). So I thought that maybe the two names are not directly related and they both derive from some other older meaning of the term. So, a relevant side question may be "why the Earth's poles are called such and who was the first to name them so?". Is a question for cartographers too off topic for MO? :-)

Comment by Andrea Mori

2010-09-22T08:12:14Z

Upon reading the question, I figured something very close to Faisal's comment above. Poles are where $f(z)$ is the (North) Pole in the Riemann sphere. This, by the way, would make sense also in Italian (where "polo" is used both for Earth's poles and for, well, poles of meromorphic functions, whereas the "tall pole" in anon's answer is a "palo").

Comment by Andrea Mori

2010-08-24T13:18:40Z

(continued) It goes without saying that this makes no "proof" but it's just my personal conviction out of actual game practice. On the other hand would be very interesting to come up with positions that would be drawn in the regular 8x8 chessboard, but can be won in an infinite (or very large) board thanks to the possibility of manouvering in a wider space.