bio | website | people.brandeis.edu/~jbellaic |
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location | Stony Creek, États-Unis | |
age | ||
visits | member for | 4 years, 2 months |
seen | 1 hour ago | |
stats | profile views | 12,048 |
Professor of Mathematics at Brandeis University
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 |
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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. |
Nov 7 |
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Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$
Oh you're right for the absolute irreducibility question, which was not thoughtful of me to ask. But what really interests me is the irreducibility (not absolute). Can you explain you argument for $e$ odd in more details? |
Nov 7 |
revised |
Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$
deleted 1 character in body |
Nov 7 |
asked | Restriction of scalars for the adjoint representation of $SL_2(\mathbb F_q)$ |
Nov 7 |
awarded | Popular Question |
Nov 4 |
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What are the higher homotopy groups of Spec Z ?
@Tormer "the universal Coeff Theorem, the hurwitz Theorem and the profiniteness theorem for \'etale homotopy" are all to be found in Artin-Mazur's paper ? |
Nov 2 |
awarded | Taxonomist |
Oct 31 |
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A curious property of Ramanujan's function $\tau(n)$
"42 is the Answer to the Ultimate Question of Life, the Universe and Everything. This Answer was first calculated by the supercomputer Deep Thought after seven and a half million years of thought. This shocking answer resulted in the construction of an even larger supercomputer, named Earth, which was tasked with determining what the question was in the first place." |
Oct 28 |
awarded | Good Question |