bio | website | people.brandeis.edu/~jbellaic |
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location | Stony Creek, États-Unis | |
age | ||
visits | member for | 4 years, 6 months |
seen | 9 hours ago | |
stats | profile views | 13,412 |
Professor of Mathematics at Brandeis University
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 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 |
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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 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 |
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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 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 |
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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 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. |
Nov 9 |
accepted | Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$ |
Nov 9 |
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Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$
Thanks for all the three good answers to my question. I accept Jim's because it is the most general, but the others are very good too. |
Nov 7 |
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Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$
Thanks a lot! That's a very nice criterion, and simple to remember since one direction is easy (at least in the case where $p \not \mid |G|$). |
Nov 7 |
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Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$
Thanks a lot, that's nice ! |
Nov 7 |
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Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$
Dear Venkaramana, I understand your strategy. But I am not sure how to make the "small check" you mention. So basically you are saying that in the space $M_3(\mathbb F_q)$ the elements of $Ad(SL_2(\mathbb F_q)$ generates everything as an $\mathbb F_p$-vector space, i.e. as an abelian group. Those elements are matrices $((-a^2,-2ab,-b^2),(-ac,ad+bc,bd),(-c^2,2cd,d^2))$ for $((a,b),(c,d)) \in SL_2(\mathbb F_q)$, and it isn't obvious to me what's the $\mathbb F_p$ linear span of those matrices. It would be enough to prove that the scalar matr. are in that span, but even this I find not clear. |
Nov 7 |
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Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$
Thanks, but I am a little lost. It looks like for you $V$ is the natural 2-dimensional representation while I defined it as the adjoint, 3-dimensional, representation. |
Nov 7 |
revised |
Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$
added 64 characters in body |
Nov 7 |
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Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$
Meanwhile, let me edit my question to remove the part about absolute irreducibility. |