bio | website | boolesrings.org/nickgill |
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location | San Jose, Costa Rica | |
age | 38 | |
visits | member for | 5 years, 10 months |
seen | 16 hours ago | |
stats | profile views | 2,086 |
I'm a visiting professor at the Universidad de Costa Rica.
Aug
19 |
comment |
How much of the ATLAS of finite groups is independently checked and/or computer verified?
Oh, and is there an implication that "old computer calculations" are worse than new ones?!? |
Aug
19 |
comment |
How much of the ATLAS of finite groups is independently checked and/or computer verified?
This is an interesting question, and worthy of discussion. But I don't really see why the ATLAS has been picked out here. Notionally it's because "the content has not been completely independently verified". But one could say similar things about results in many other areas of mathematics -- one is really asking how we know whether such-and-such a theorem is true, and this is (at least to some extent) a vexed and deep philosophical question. Actually the ATLAS seems to me to be a particularly bad target because it has stood the test of time so well. The level of accuracy is pretty astonishing. |
Aug
6 |
comment |
Generalization of a theorem of Steinberg
Ah! Thanks, that's helpful. |
Aug
6 |
comment |
What is the permutation group generated by those three given morphisms of the affine space $\mathbb{F}_q^3$?
@GeoffRobinson, Good point. The set you describe of size $3(q-1)$ breaks into orbits of size $6$ (provided $q>2$), and on these orbits these involutions generate a group of size $48$ I believe (this would need checking). So one needs to analyze the action on the remaining $q^3-3q+1$ points. |
Aug
6 |
comment |
What is the permutation group generated by those three given morphisms of the affine space $\mathbb{F}_q^3$?
BTW, I assume that @JoeSilverman's comment explains why the algebraic-geometry tag has been used? If you can give some more context, that would be nice. |
Aug
6 |
comment |
What is the permutation group generated by those three given morphisms of the affine space $\mathbb{F}_q^3$?
In fact the permutations generate a subgroup of $S_{q^3-1}$, since $(0,0,0)$ is fixed. Do you know if the group they generate is transitive / primitive on non-zero vectors? It would also help to know the answer for $q=2,3,4,5$ for instance... In general if you pick permutations at random, you'll either get $A_n$ or $S_n$, so it would be worth seeing if this is the case for small $q$... |
Aug
6 |
revised |
What is the permutation group generated by those three given morphisms of the affine space $\mathbb{F}_q^3$?
added a tag, and fixed a couple of typos. |
Aug
6 |
comment |
Generalization of a theorem of Steinberg
@DenisChaperondeLauzières, Thank you! I will look up the article you mention, and follow up the references there. |
Aug
6 |
revised |
Generalization of a theorem of Steinberg
added 727 characters in body |
Aug
6 |
comment |
Generalization of a theorem of Steinberg
Also, ludicrously, I don't have access to MathSciNet... So, can I check that the Douglass article you mention is this - hilbert.math.unt.edu/downloads/papers/douglass/levi.pdf - and the Lehrer paper is entitled Rational Tori, semisimple orbits and the topology of hyperplane complements. |
Aug
6 |
comment |
Generalization of a theorem of Steinberg
Thanks Jim. I'll edit the original question to address some of your remarks. I was being a little sloppy with dimension, since if $q\gg r$ I believe I can transfer notions of dimension pretty straightforwardly to the finite case... But, anyway, I don't need this, so will edit accordingly. |
Aug
4 |
asked | Generalization of a theorem of Steinberg |
Jul
31 |
comment |
SO$(4)$ (& SO$(n)$) characterization?
These notes assert that you can find a list of the finite subgroups of $SO(4)$ in Homographies, Quaternions and Rotations by P. Du Val. I don't have a copy of this reference, so can't confirm it. It would be nice to have an explicit list as an answer to the OP's question! |
Jul
29 |
comment |
Is every nonabelian finite simple group a quotient of a triangle group $(a,b,c)$ with $a,b,c$ coprime?
This result (about minimal FSG's) is a special case of Theorem B of A NEW SOLVABILITY CRITERION FOR FINITE GROUPS by Dolfi, Guralnick, Herzog and Praeger. They also state a conjecture in the final section which pertains directly to the OP's original question. It's slightly too long for me to write down in a comment, but you can find it here - arxiv.org/pdf/1105.0475.pdf |
Jul
28 |
comment |
Is every nonabelian finite simple group a quotient of a triangle group $(a,b,c)$ with $a,b,c$ coprime?
You should look at Claude Marion's lovely paper On triangle generation of finite groups of Lie type which studies triangle groups $(a,b,c)$ where $a,b$ and $c$ are prime. You may well be able to extract what you need from that paper. |
Jul
27 |
comment |
Isotropic subspaces in a symplectic vectorspace over $GF(q)$
You only need to prove the statement for $r=n$, as the rest will follow directly... And the statement is true when $2n=4$ and the field is ${\mathbb F}_3$ - see p.33 of this: math.lsu.edu/~hoffman/papers/spreads4.pdf (Whether that provides any evidence for the statement in general, I couldn't say. The group ${\rm Sp}_4(3)$ is a bit special.) |
Jun
17 |
comment |
Transitivity on $\mathbb{N}_0$ — a 42 problem
That statement looks crucial to me! If you can write down a proof of that, then this would be a most illuminating addition to the question. To me, the number "42", despite Douglas Adam's assertions to the contrary, seems a rather random element in the set-up that you have in this question. Your comment changes all that, though (for me). |
Jun
17 |
comment |
minimal polynomial of unipotents in orthogonal group
BTW, I've voted for this question to be reopened. In its current form it seems fine to me. |
Jun
17 |
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minimal polynomial of unipotents in orthogonal group
You'd be better of using the matrix with $1$'s on the anti-diagonal. Then you can take your unipotents to be upper-triangular, and this question should yield to direct calculation - my first thought is that the minimal polynomial of a unipotent $g$ will just depend on the dimension of the largest Levi subgroup that contains $g$ |
Jun
17 |
comment |
Transitivity on $\mathbb{N}_0$ — a 42 problem
Thanks for your replies Stefan. One last thing: can you prove your conjecture for the situation where no generator interchanges residue classes with different moduli? In light of your last comment, I guess you want to show that no such group can have an orbit containing {0,...,42}. This would already seem kind of surprising to me. (Although maybe that's just my ignorance...) |