bio  website  people.brandeis.edu/~jbellaic 

location  Stony Creek, ÉtatsUnis  
age  
visits  member for  4 years, 8 months 
seen  2 hours ago  
stats  profile views  14,019 
Professor of Mathematics at Brandeis University
1d

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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 counterexample. 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 
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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 
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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 
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The resolution of which conjecture/problem would advance Mathematics the most?
I am not an expert but there are counterexamples 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 
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short character sums averaged on the character
Thanks for this answer, Lucia. 
Nov 13 
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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 nonisomorphic (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 
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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 noncomplete 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 
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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 
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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 nondegenerate 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 $dim V1$ since $q$ is nondegenerate 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. 