bio | website | math.u-psud.fr/~chambert |
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location | Université Paris-Sud (Orsay) | |
age | 42 | |
visits | member for | 4 years |
seen | 8 hours ago | |
stats | profile views | 2,582 |
Apr 15 |
awarded | Yearling |
Mar 30 |
comment |
When does a modular form satisfy a differential equation with rational coefficients?
@Dror: In general, of course no (think of a solution to $y'=y\sqrt{-1}$). But since you assume that the form has a rational $q$-expansion, yes. Form a differential equation $E$ of high degree with indeterminate coefficients. Check that $f$ is a solution of $E$ by looking at the expansion. That $f$ is an actual solution is a linear system with rational coefficients in the indeterminate coefficients of the equation. It has a solution in $\mathbf C$ by assumption. Hence it has a solution in $\mathbf Q$. (You can replace $\mathbf Q$ by the field generated by the coefficients of the $q$-expansion). |
Mar 28 |
comment |
A quotient stack question
Two questions on your question: 1) the normalization of a DM-stack is well-defined because normalization is an étale-local property, right ? is it well-defined for Artin stacks ? 2) What are the reasons for your proper/projective assumptions ? Do you have easy counterexamples without them ? |
Mar 26 |
comment |
When does a modular form satisfy a differential equation with rational coefficients?
@DrorSpeiser: If a form $F$ is solution of a differential equation with algebraic coefficients, then isn't it also solution of a differential equation with rational coefficients ? |
Mar 24 |
revised |
Modern algebraic geometry vs. classical algebraic geometry
typo - replace math mode by italics |
Mar 19 |
comment |
The Schottky group and the fundamental group of a compact Riemann surface
@user6818: Here (archive.org/details/introductiontoth00forduoft) is a preliminary version (1915) of Ford's book indicated by Alexandre. |
Mar 12 |
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Dirichlet density vs natural density
Cool & easy example which shows that I could have thought a little bit before making my comment above... |
Mar 12 |
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Dirichlet density vs natural density
I would guess that there exists an example of set of primes $A$ with a positive analytic density and such that the lower natural density ($\liminf_n (\#A\cap\{p\leq n\})/(\#P\cap\{p\leq n\})$) is zero. This would require the Dirichlet series to have precisely adjusted zeroes on the real line $\Re(s)=1$. |
Mar 12 |
comment |
Frobenius density theorem
@DavidSpeyer. Indeed. What one needs for the natural density is the absence of zero on the real line $\mathop{\rm Re}(s)=1$. The tauberian theorem of Ikehara shows that this condition is also sufficient. Newman's tauberian theorem is a bit weaker and requires some zero-free region; however, the standard (Hadamard) proof of non-vanishing implies the required estimate. For a more precise asymptotic expansion, larger zero-free regions are needed, with upper-bounds at infinity. The notion of analytic density is weaker and for that, one only needs to understand the pole at $s=1$. |
Mar 8 |
comment |
Deformation space form the point of view of intersection theory
Output of deformation theory (tangent space, obstruction space) is encoded in various cohomology groups. Vanishing theorems, when they hold, may imply that the dimensions of those groups are Euler characteristics. In principle, these Euler haracteristics can be computed via Riemann-Roch, within the Chow group (or cohomology) only. This program is likely to work for toric varieties but details have to be written. |
Mar 7 |
revised |
Do Hermitian metrics also split on the Riemann sphere?
Correcting the passage about automorphisms, following the comment of WillSavin. |
Mar 7 |
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Do Hermitian metrics also split on the Riemann sphere?
@WillSawin: You're perfectly right. I'll correct that. On the other hand, non-diagonal unipotent automorphisms are rarely isometric! |
Mar 7 |
answered | Do Hermitian metrics also split on the Riemann sphere? |
Mar 6 |
comment |
How many ways do we have to prove that a mapping is open?
There is classical nice sufficient condition (Brouwer): a continuous injective map from an open subset of $\mathbf R^n$ to $\mathbf R^n$ is open. However, it looks like it is superseded by the theorem of Titus-Young you mention. |
Mar 6 |
comment |
Frobenius density theorem
I say exactly that. I wrote the details in a mid-graduate course on number theory (4th year univ. in France). See the text on math.u-psud.fr/~chambert/enseignement/2007-08/h4/coursh4.pdf |
Mar 6 |
revised |
Understanding the definition of the quotient stack $[X/G]$
Completing reference |
Mar 6 |
comment |
Understanding the definition of the quotient stack $[X/G]$
@SimonRose: the use of the work "collection " is misleading, for one could interpret (wrongly) as the collection of $G$-torsors on the point, which is often (but not always) trivial. It has to be understood as the fibered category of $G$-torsors over schemes over the point, a fibered category which is never trivial (unless $G=1$). |
Mar 6 |
comment |
Understanding the definition of the quotient stack $[X/G]$
@BrianFitzpatrick: Negative answers to your questions are already given in the comments of Scott Carnahan. For a positive one: if $X$ is quasi-projective or, more generally, if every orbit is contained in an open affine subscheme, then the categorical quotient $X/G$ exists as a scheme. By a theorem of Mori and Keel, the categorical quotient $X/G$ exists in the larger category of algebraic spaces (an unavoidable category, if you're working with stacks, anyway). |
Mar 6 |
revised |
Understanding the definition of the quotient stack $[X/G]$
corrected spelling (principle->principal) |
Mar 6 |
comment |
Frobenius density theorem
@IgorRivin: To pass from Dirichlet density to natural density, you just need powerful enough tauberian theorems. I would bet that Newman's "easy tauberian theorem" (see Zagier's AMM paper) is enough for Frobenius. |