bio | website | math.u-psud.fr/~fouquet |
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location | ||
age | ||
visits | member for | 4 years, 10 months |
seen | 13 hours ago | |
stats | profile views | 4,477 |
Oct 21 |
comment |
Intuition behind $\zeta(2) = \frac{\pi^2}{6}$
Euler's first proof is very intuitive to me: find a "polynomial" whose roots are the non-zero integers (so your "polynomial" is sin/x) then use the relation between the sum of the inverse of the square of the roots and the first coefficients. Voilà. |
Oct 21 |
revised |
Definition of CM modular form
Corrected typo |
Oct 21 |
awarded | Nice Answer |
Sep 30 |
awarded | Explainer |
Sep 26 |
comment |
Special values of $\zeta$ outside the real line and the critical strip
At least, the motivation (pun intended) between the formulae expressing the values at integers seems to disappear, so if there exists interesting expressions of the values elsewhere, it will be for completely different reasons of which I am unaware. |
Sep 23 |
comment |
Adelic open image for modular forms?
There is an immediate obstruction to the naïve generalization: the determinant of $\rho_{\ell}$ has values in a smaller ring than the ring of coefficients of $\rho_{\ell}$ for an infinite number of $\ell$ in general so $\rho_{\ell}$ is not surjective for an infinite number of $\ell$ and so the products of the $\rho_{\ell}$ cannot have open image. But you are aware of this in your question, so you seem to be thinking about a more sophisticated generalization. |
Sep 18 |
comment |
Examples of intuition from fields other than Physics to solve math problems
That Aczel was exaggerating is a fixture of the reviews I have read. As for the thesis, I did look seriously into this question (albeit just once) and concluded that there was no influence of structuralism on mathematics. Of course, proving a negative is hard. |
Sep 17 |
comment |
Examples of intuition from fields other than Physics to solve math problems
The one time I seriously looked into this question, I concluded that structuralism in philosophy, anthropology and psychology had zero impact on Bourbaki, but that Bourbaki had a very slight impact on structuralism. Besides, reviews that I read of Aczel's book usually say 1) that this book is not reliable and 2) that the thesis of the book was that Bourbaki influenced structuralism, not the other way round. So I think this answer is incorrect. |
Aug 8 |
comment |
Atkin--Lehner operators in Hida theory
Yes, exactly. I have a feeling we are talking past each other, so let me try to be clearer: what you will find in Nekovar-Platter is a group-theoretic computation and a proof of the compatibility of this computation with inverse limits (of course in the sense that one projection is turned in the other) that is strictly parallel (though slightly harder because of the interchanging) to the one you need for $q\neq p$. Both cases (of course with the appropriate modification when $q=p$) are treated (maybe) in Ohta and my article. |
Aug 8 |
comment |
Atkin--Lehner operators in Hida theory
What I meant is that in order to prove these kind of results, you need a definition of the operators which is compatible with inverse limit on the level. This is possible at $p$ and the fact that it interchanges different ordinary spaces is a virtue, not a vice: it is absolutely crucial in the correct definition of twisted self-dual Hida families (but someday I'll tell you an amusing story about this my advisor told me once). |
Aug 7 |
answered | Atkin--Lehner operators in Hida theory |
Jul 8 |
comment |
About the restriction of a modular representation to a decomposition subgroup
Ah! So the Weil-Deligne representation is not enough. Interesting. |
Jul 8 |
comment |
About the restriction of a modular representation to a decomposition subgroup
TL;DR David Loeffler's and Ricky's comments are spot on. |
Jul 8 |
answered | About the restriction of a modular representation to a decomposition subgroup |
Jul 2 |
awarded | Curious |
Jun 26 |
awarded | Civic Duty |
Jun 26 |
comment |
Absolutely irreducible p-adic representation of the absolute Galois group of Q_p
If $\rho$ is continuous and has coefficients in $\mathbb F_{p}$, then its kernel is such an $H$, so that case is no problem. |
Jun 26 |
revised |
Absolutely irreducible p-adic representation of the absolute Galois group of Q_p
added 4 characters in body |
Jun 25 |
comment |
What is the probability that a random sequence of polynomials is regular?
I share your expectation, and this seems to follow from the principle of avoidance from sub-varieties, but at the moment, I am unable to write a full proof. |
Jun 12 |
comment |
Example of a non-smooth irreducible component of the generic fibre of a Hida family?
@Ramsey Not so easily I'm afraid, especially if you want real math but imagine what would happen if it were always possible to resolve the (putative) singularities of the irreducible components of an ordinary Hecke algebra by raising the level at varying auxiliary primes. |