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bio website people.brandeis.edu/~jbellaic
location Stony Creek, États-Unis
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visits member for 4 years, 8 months
seen 2 hours ago

Professor of Mathematics at Brandeis University


1d
comment On sentences true in all finite groups
@Christian: what about $w=x^2$? or $w=x^n$ for any $n \in \mathbb Z$? or $w=y x y^{-1}$? All those words are $1$ for $x=1$ whatever the value of $y$.
May
14
accepted Upper bound on Betti numbers of an intersection of hypersurfaces (or quadrics)
May
14
revised Upper bound on Betti numbers of an intersection of hypersurfaces (or quadrics)
added 608 characters in body
May
14
comment Upper bound on Betti numbers of an intersection of hypersurfaces (or quadrics)
Thanks a lot. That doesn't leave much hope for a positive answer in the case of an intersection of quadrics, though I would love to see a counter-example. Very good blog post, by the way...
May
14
revised Upper bound on Betti numbers of an intersection of hypersurfaces (or quadrics)
added 1 character in body
May
14
revised Upper bound on Betti numbers of an intersection of hypersurfaces (or quadrics)
added 453 characters in body; edited tags
May
14
asked Upper bound on Betti numbers of an intersection of hypersurfaces (or quadrics)
Apr
24
awarded  Nice Question
Mar
24
comment Did Grothendieck write about modular forms?
Good quote! This is the way Grothendieck could have become interested in the Langlands program : not by the modular forms, too special, too peculiar, to computational for his taste, but by the great vision of Langlands using Motives, Tannakian Categories. How wonderful would have it been if Grothendieck had learnt this theory and worked on it.
Mar
22
awarded  Good Answer
Feb
25
awarded  Good Question
Jan
6
answered Examples of component crossing between families of modular forms
Jan
5
comment In “splendid isolation”
Very surprising and interesting story! It means that great physicists like Bohr and Heisenberg were not completely familiar with multiplication of matrices. Or do I understand wrongly this passage ? In general relativity for example, multiplication of matrices (and tensors) is everywhere...
Jan
1
awarded  Nice Question
Dec
8
comment The resolution of which conjecture/problem would advance Mathematics the most?
I am not an expert but there are counter-examples to the conjecture "with coefficients" since about 20 years (due to V. Lafforgue and others). Is the conjecture without coefficients almost as useful than the conjecture in general ?
Nov
25
comment short character sums averaged on the character
Thanks for this answer, Lucia.
Nov
13
comment Etale fundamental group of a curve in characteristic $p$
@jacob. The finite quotient of $\mathbb Z_p^I$ for $I$ infinite are all the abelian $p$-group. Now take two infinite sets $I$, one enumerable say and another say larger than the continuum, and you have two profinite groups non-isomorphic (because they don't have the same cardinality) having the same set of finite quotients. So Niels is right that there is more in understanding a $\pi^1$ than its understanding it's finie quotient. Now in algebraic geometry one could have the point of view that it is the finite quotients, corresponding to finite cover, that are what we want to understand.
Nov
13
comment Etale fundamental group of a curve in characteristic $p$
@Niels, Ah okay, interesting. But "completely wrong" is perhaps not the right term (or possibly I am more wrong that you explain in your comment): one at least knows all the finite quotients of this profinite group (the $pi^1$ of a non-complete curve), which is already something important, and sufficient for many applications. Then it depends of what we mean by "know", and apparently the OP was not completely sure about that. But, do we know the complete list of finite quotient of the $\pi^1$ for a curve of genus $\geq 2$ without any point removed ?
Nov
13
comment Etale fundamental group of a curve in characteristic $p$
Of course (you certainly know this), if you remove $t$ points to your curve, then the problem is solved, provided than $t>0$. It is Abyankhar's conjecture, now a theorem of Raynaud and Harbater.
Nov
12
comment Two rings…are they isomorphic?
@Qfwfq: Yes, they can. Or are you talking of something different? One can always recall the standard proof. If $q$ is your non-degenerate quadratic form over a finite dimensional $\mathbb C$-vector space $V$, you can find a $v \in V$ such that $q(v) \neq 0$ (otherwise $q=0$ and is already diagonal), and even such that $q(v)=1$. The orthogonal space $W$ of $v$ for $q$ has dimension $dim V-1$ since $q$ is non-degenerate and does not contain $v$. Hence to diagonalize $q$ on $V$ it is enough to do so by $W$, which is already done by induction.