User bruno - MathOverflow

Is there an algebraic curve over Q which is not modular?

2013-05-11T17:35:06Z

Every elliptic curve $E/\mathbf Q$ is modular, in the sense that there exists a nonconstant morphism $X_0(N) \to E$ for some $N$.

It is tempting to extend this definition in a naïve way to an arbitrary projective curve over $\mathbf Q$; if $Y$ is such a curve, we might say that $Y$ is modular if there exists a nonconstant morphism $X_0(N) \to Y$ for some $N$. A necessary condition for $Y$ to be modular is that it should have at least one rational point, since $X_0(N)$ always has a rational point. For an elliptic curve, this condition is satisfied by definition.

There are certainly a great deal of curves which are modular in this sense. But is there an example of a curve (with a rational point) which is not modular?

Category with a "metric" for arrow composition

2011-07-21T07:08:39Z

Consider a category $\mathcal C$ with a "distance" function $d:\mathcal C^2 \to \mathbb{R}_{\geq 0}$ satisfying the "triangle inequality"

$$d(x \to z)\leq d(x \to y) + d(y \to z)$$

for every pair of composable arrows $(x\to z)=(x \to y \to z)$.

Let's call $(\mathcal C,d)$ a "metric" category.

The first example is to take any category $\mathcal{C}$, and define

$$d(f)=\begin{cases} 0 & \text{ if $f$ is an isomorphism} \\ 1 & \text{ otherwise.}\end{cases}$$

Then the triangle inequality simply translates the statement : "If $f=gh$, then $f$ is an isomorphism if $g$ and $h$ are isomorphisms."

Also, it's clear that every metric space can be made into a metric category in a canonical way.

We can define "open balls" in $\mathcal{C}$: for $c \in \mathcal{C}$, $r\geq 0$, let

$$B(c, r) = \{d \in \mathcal{C} | \text{ there exists }f: c \to d\text{ such that }d(f) < r \}.$$

In the category of number fields and monomorphisms, we can let $d(K \hookrightarrow L)=\log ([L:K])$. Then the triangle inequality is actually an equality. It's clear that $d$ is a good measure of "how far" $L$ is from consisting of just $K$. The open ball of radius $r$ around $K$ is the set of extensions of $K$ of degree $< e^r$.

Is it possible to endow a big category like $\text{Top}$ or $\text{Grp}$ with a meaningful distance?

Where do the product expansions of modular forms come from?

2013-05-03T13:33:19Z

It is well-known that many modular forms can be expressed as infinite products. For instance, the most famous one is probably the expansion

$$\Delta(q) = q \prod_{n=1}^\infty (1-q^n)^{24}$$

for the discriminant cusp form of weight $12$ and level $1$. Another example is the cusp form of weight $2$ and level $11$

$$f(q) = q\prod_{n=1}^\infty(1-q^n)^2(1-q^{11n})^2,$$

which is attached to the elliptic curve $X_0(11) : y^2-y = x^3-x^2$. Such product expansions can be derived from the product expansion of the Dedekind $\eta$ function, by taking suitable combinations.

But why should such product expansions exist? Is there a reason to expect that they should exist, say, from the point of view of Galois representations?

Elliptic curve over a scheme is a group scheme?

2013-04-08T17:00:31Z

In Katz's article p-adic properties of modular schemes and modular forms in the Antwerp proceedings, the following definition of an elliptic curve over a base scheme $S$ is given:

By an elliptic curve over a scheme $S$, we mean a proper smooth morphism $p: E \to S$, whose geometric fibres are connected curves of genus one, together with a section $e : S \to E$.

Now this is a quite reasonable definition, which coincides with the usual notion of an elliptic curve when $S$ is the spectrum of a field. However, it does not seem (to me) to follow directly from the definition that such an elliptic curve over $S$ should be a group scheme over $S$ (an obviously desirable property which Katz seems to take as an obvious fact). When $S= \text{Spec }k$, I understand that this essentially follows from Riemann-Roch...

So, what principle allows one to come to this conclusion in the general case?

Thank you!

A divergent series related to the number of divisors of of p-1

2013-04-25T05:16:55Z

Let $d(n)$ denote the number of divisors of $n$. Is it known that the series $$\sum_{p \text{ prime}} \frac{1}{d(p-1)}$$ diverges?

This would follow immediately from the Sophie Germain Conjecture. Indeed, if there are infinitely many primes of the form $2p+1$ ($p$ a prime), then infinitely many terms of the series are equal to $1/4$, so the series doesn't even satisfy the most basic requirement for convergence! So, surely there must be a direct proof?

Answer by Bruno for A divergent series related to the number of divisors of of p-1

2013-04-27T22:33:26Z

Answering my own question, because I totally overlooked the following ridiculous idea:

Obviously $d(n)\leq n$ for every $n$. Thus $d(p-1)\leq p-1 < p$, so $1/d(p-1) > 1/p$ and the divergence follows from the divergence of $\sum 1/p$ (if one is willing to assume that).

Is the primitive element theorem a cohomological statement?

2013-03-21T02:52:02Z

If $K$ is a field, then as is well known every finite separable extension $L$ of $K$ is of the form $L=K(\alpha)$ for some $\alpha \in L$.

A similar statement can be made about an extension of discrete valuation rings with separable residue field extension.

These statements very much resemble the statement "every projective module over a principal ideal domain $A$ is free". This last statement can be interpreted as the vanishing of a certain cohomology group. Now my question is: can the primitive element theorem be interpreted as the vanishing of a cohomology group?

Thank you!

Which formulae of Euler is Fröhlich referring to?

2012-05-07T04:30:30Z

In A. Fröhlich's article Local Fields in Algebraic Number Theory, the following claim is made: if $R$ is a Dedekind domain with field of fractions $K$, $L$ is a finite separable extension of $K$ and $S$ is the integral closure of $R$ in $L$, and $x$ is an element of $S$ with minimal polynomial $g$, then, "by Euler's formulae",

$$\text{tr}_{L/K}(x^i/g'(x)) \in R$$ for each $0 \leq i \leq n-1$, where $n=\text{deg } g$.

Which formulae of Euler are being referred to? The claim can be proven by the Lagrange interpolation formula; in fact the given quantity is $1$ if $i=n-1$, and $0$ for $0 \leq i < n-1$. However, I have no idea what proof Fröhlich has in mind. I also cannot resist pointing out the humor in appealing to Euler's "formulae" without further precision. Perhaps the formulae in question are well-known, and I am the only one who has not been invited to the party? In any case, more details would be greatly appreciated!

Thank you.

Patching together homeomorphisms: how badly can it fail?

2011-11-03T04:19:14Z

Suppose we have a set $X$ with $X=U \cup V$. If we pick a permutation $f$ of $U$ and a permutation $g$ of $V$ which agree on the intersection $U \cap V$, we can coalesce them into one big endo-map $F$ of $X$. In general, of course, $F$ is no longer a permutation - we can only say that it is onto. A counter-example is fun to think of - for example, if $X$ is finite, then $F$ is always a permutation.

Now let $X, U$ and $V$ be open subsets of $\Bbb{R}^n$, and suppose $f$ and $g$ are smooth homeomorphisms. I have been unable to come up with a single example for which the above patching fails "badly", say for which the set of points $x$ having more than one pre-image had positive measure.

I'd love to know if anyone knows anything related to this problem, or could provide a counter-example. Thank you!

Multiplying functions on the unit square as generalized matrices

2011-09-03T23:25:46Z

Consider the $\mathbb{R}$-vector space of sufficiently nice real-valued functions on the unit square $I^2$, where "sufficiently nice" could be taken to mean any one of a number of things - say continuous for now.

In analogy with matrix multiplication, we can define the product of two such functions $F$ and $G$ as

$$(F\times G)(i,j) = \int_0^1F(i,t)G(t,j)dt.$$

We can check immediately that this operation is associative (the proof is exactly the same):

$$((F\times G)\times H)(i,j) = \int_0^1(F\times G)(i,t)H(t,j)dt$$ $$=\int_0^1\left(\int_0^1F(i,s)G(s,t)ds \right)H(t,j)dt$$ $$=\int_0^1F(i,s)\left(\int_0^1 G(s,t)H(t,j)dt \right)ds$$ $$=\int_0^1F(i,s)(G\times H)(s,j)ds = (F\times (G\times H))(i,j)$$

Also, $\times$ is obviously bilinear with respect to usual addition of real-valued functions, and hence defines a ring structure on $C(I^2)$ which is considerably different from the usual ring structure (but addition is the same).

Extending the matrix analogy, we see that each $F$ also defines a linear operator $C(I) \to C(I)$ in the usual way, as

$$F(f)(i)=\int_0^1 F(i,t)f(t) dt$$

for each $f : I \to \mathbb{R}$.

Also, all of this actually generalizes usual matrix multiplication if we subdivide the square $I^2$ into a bunch of small rectangles and let $F$ be constant on each subrectangle, being more or less careful on boundaries.

The only candidate for a unit element for $\times$ is the distribution which has a weight $1$ dirac delta on the diagonal, and is $0$ everywhere else (in other words, the product of the dirac delta with the Kronecker delta! :))

Now my question is: what is this? Is it of any interest, or a mere curiosity? For example, could a notion of "determinant" be assigned to these objects?

Answer by Bruno for Bounding the modular discriminant of an elliptic curve in the j-invariant

2011-07-25T15:48:24Z

As it stands, I think this question is still too vague to be answerable in generality. What kind of expression are you permitting for the bound? Certainly one can construct an artificial bound which doesn't even involve $j$ at all, which would be completely silly and certainly not what you have in mind.

As a first step, it's easy to see that no rational function of $j$ bounds $\Delta$. Indeed, if $R$ is a rational function such that $|\Delta| \leq |R(j)|$ everywhere on $X=\mathbb{H}/PSL(2, \mathbb{Z})$, then the bounded meromorphic function $\Delta/R(j)$ must be a constant. But this implies that $\Delta$ has weight $0$, which is false.

Combinatorial interpretations of integral transforms

2011-06-17T19:15:24Z

It is well known that the ordinary and exponential generating functions of a sequence of numbers are related by an integral transform (the Borel transformation).

Does there exist a combinatorial theory of integral transforms? The example above indicates that something might be going on "behind the scenes". Has anyone been able to formulate a precise combinatorial explanation of this phenomenon? If not this one, might other types of integral transforms have combinatorial interpretations?

Thanks for your interest!

Answer by Bruno for Why does the parameterization (F:F':1) happen?

2011-06-14T23:21:41Z

I would say it's a consequence of the Riemann-Roch theorem. Let's take a look at what happens in the case of an elliptic curve. The $\wp$ function has a double pole at a prescribed point $\mathcal{O}$. The function $\wp'$ thus has a triple pole there. By the Riemann-Roch theorem, the vector space $H^0(6\mathcal{O})$ has dimension $6+1-g=6$. Thus, the functions ${1, \wp, \wp^2, \wp^3, \wp\wp', \wp', \wp'^2}$ must be linearly dependent. A nontrivial relation of linear dependence for this set translates into a parametrization of the curve using $\wp$ and $\wp'$ - this is pretty much a tautology.

Note that the integers $n$ and $n+1$ are always relatively prime. This implies that every sufficiently large integer can be written as a linear combination of $n$ and $n+1$ with non-negative integer coefficients (this is the so-called postage stamp theorem). Take any meromorphic differential $\omega$ on a curve $X$, with a pole at $\mathcal{O}$. The derivative $d\omega$ has a pole of order $n+1$ at $\mathcal{O}$. The exact same construction can be carried out!

What is a "best" transcendence basis for R/Q ?

2011-06-13T21:02:28Z

It is easy to show, using the axiom of Zorn, that there exists a transcendence basis for $\mathbb{R}/\mathbb{Q}$, i.e. a set $S$, algebraically independent over $\mathbb{Q}$, such that $\mathbb{R}/\mathbb{Q}(S)$ is an algebraic extension.

What can we say about $T=\mathbb{R} - \mathbb{Q}(S)$? It is easy to show that $\mathbb{R}/\mathbb{Q}$ is not purely transcendental, so that $T \neq \emptyset$. (Indeed, no element of $\mathbb{R}-\mathbb{Q}$ algebraic over $\mathbb{Q}$ may be contained in $\mathbb{Q}(S)$ - an algebraic relation for such an element over $\mathbb{Q}$ would immediately translate into a relation of algebraic dependence in $S$ over $\mathbb{Q}$.)

Thus, all real, non-rational algebraic numbers are contained in $T$. Is it possible, for example, to choose $S$ so that this inclusion is an equality? If not, how "small" can we make $T$?

I'm sorry if this turns out to be trivial, or if the answer to my question can be easily found in the literature. I tried!

Thank you very much and have a pleasing day.

Mathematics and autodidactism

2010-06-16T22:11:14Z

Mathematics is not typically considered (by mathematicians) to be a solo sport; on the contrary, some amount of mathematical interaction with others is often deemed crucial. Courses are the student's main source of mathematical interaction. Even a slow course, or a course which covers material which one already knows to some level, can be highly stimulating. However, there are usually a few months in the year when mathematics slows down socially; in the summer, one might not be taking any courses, for example. In this case, one might find themselves reduced to learning alone, with books.

It is generally acknowledged that learning from people is much easier than learning from books. It has been said that Grothendieck never really read a math book, and that instead he just soaked it up from others (though this is certainly an exaggeration). But when the opportunity does not arise to do/learn math with/from others, what can be done to maximize one's efficiency? Which process of learning does social interaction facilitate?

Please share your personal self-teaching techniques!

Picard-Fuchs equations for modular functions

2010-06-15T05:56:40Z

Hello, MathOverflow community!

Suppose we have a modular curve of genus $0$, whose rational function field is generated by the modular function $f$. We can view $f$ as the parameter for some pencil of elliptic curves over $\mathbb{C}$. Under certain conditions, it is possible to express $f$ as the inverse function of the ratio of two linearly independent solutions of a second-order linear differential equation. The prototypical example is the case of the period integrals of the Legendre elliptic curve $y^2=x(x-1)(x-\lambda)$, which satisfy the Fuchsian equation $$\lambda(1-\lambda)D^2y + (1-2\lambda)Dy - y/4=0,$$ where $D=d/d\lambda$. We can interpret this differential equation as measuring the variation of the periods of an elliptic curve, as the parameter $\lambda$ changes.

Now my question is : have other Picard-Fuchs equations been calculated for modular functions? In principle, there should be many such equations; the Picard-Fuchs equation for Klein's $j$ function, without the calculation, is given in (Harnad, McKay). I have seen the calculation for the $\lambda$ case carried out in a few books. But I have not seen such equations for the Hauptmodul associated to the other genus $0$ modular curves.

Any thoughts, comments, questions or references are much appreciated.

Please be kind, as I am only an undergraduate. (There seems to be much "tough love" here!)

Answer by Bruno for Sieve of Eratosthenes - eventual independence from initial values

2010-06-14T20:52:38Z

This is not exacly what you are asking for, but it's relevant enough to mention : lucky numbers.

Comment by Bruno

2013-05-12T17:34:55Z

Ah, thank you David

Comment by Bruno

2013-05-11T21:26:10Z

Dear David: thank you, that is very interesting. I will try to wrap my head around it. Why does the representation afforded by $H^1$ land inside the sympectic group? Is this a consequence of some kind of $\mathcal l$-adic Riemann relations? (Forgive me, I know next to nothing about $\mathcal l$-adic cohomology.)

Comment by Bruno

2013-05-03T15:18:18Z

Dear @Steve: that's true, but I don't think the above products are directly related to the Euler products (they're products over $n$, rather than over $p$). Different animals! I may be wrong, though. Regards,

Comment by Bruno

2013-04-27T22:48:09Z

Thank you, Nilotpal.

Comment by Bruno

2013-04-27T22:36:28Z

Interesting, @Greg - thanks for sharing.

Comment by Bruno

2013-04-25T05:57:21Z

Cool! Thanks zeb!

Comment by Bruno

2013-04-09T00:25:40Z

Dear anon, would you mind expanding a little bit? Thank you!

Comment by Bruno

2013-04-08T22:07:59Z

Thank you very much Joël!

Comment by Bruno

2013-04-08T22:07:40Z

Thank you Timo!

Comment by Bruno

2013-03-22T16:50:09Z

@Martin, I also have not completely given up on the idea. If you think of anything, please share! Regards,

Comment by Bruno

2013-03-21T03:07:12Z

That's a good point!

Comment by Bruno

2013-03-21T02:58:39Z

This is the sort of thing that I have in mind: For a Dedekind domain $A$, every projective $A$-module is free if and only if $\text{Pic}(A) = H^1(X, \mathcal O_X^*) = 0$ (where $X=\text{Spec }A$).

Comment by Bruno

2012-05-07T18:09:08Z

Wonderful! Thanks for writing that out, Faisal.

Comment by Bruno

2011-11-10T03:53:08Z

It is great, thank you again :-)

Comment by Bruno

2011-11-08T22:55:03Z

Thank you for this beautiful example, Ramiro!