Timeline for Non-split extensions of $GL_n(F_q)$ by $F_q^n$ ?
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23 events
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| Jun 25, 2010 at 11:39 | vote | accept | BS. | ||
| Jun 25, 2010 at 11:39 | comment | added | BS. | Many thanks to Andy and Jack for your very interesting answer and comments. This is definitely an answer ! | |
| Jun 25, 2010 at 5:31 | comment | added | Jack Schmidt | G a group, X a generating set, f:G→X* a set-theoretic section of the evaluation X*→G such that f(G) is closed under subwords, V a G-module. Only consider transversals of V defined by the preimages of elements of X. Then a cocycle is completely determined by its values on X×G, and in fact it is almost constant. The only values needed are on (x,g) where xf(g) is not in the image of f. Pairs (xf(g),f(xg)) are called rewrite rules. This allows very efficient compression of cocycles, and evaluation on (g,h) is polynomial in the length of f(g) as long as one has an oracle to compute f. | |
| Jun 25, 2010 at 5:17 | comment | added | Jack Schmidt | I think most of it is folklore. H^2 is harder than H^1. H^1 depends on the number of generators, H^2 more on the number of relations (rewrite rules). In the soluble world, the number of rewrite rules is the square of the number of generators (composition length). In the finite world, if the field size of composition factors is bounded, then the number of rewrite rules stays polynomial in the max rank and composition length. Squier, Groves, and Anick described using rewriting systems for cohomology, but I guess people didn't know it actually worked until recently. | |
| Jun 25, 2010 at 2:52 | comment | added | Andy Putman | As a side note, this might be the longest on-topic comment thread I've seen in a long time! Though this comment probably breaks that... | |
| Jun 25, 2010 at 2:51 | comment | added | Andy Putman | @Jack : Interesting! Can you recommend some good references for those types of calculations? The only source I know is Holt-Eick-O'Brien, and I remember them saying that computing H^2 was pretty intense (though I haven't looked through that book in a while). | |
| Jun 25, 2010 at 1:03 | comment | added | Jack Schmidt | You could use the "cohomolo" package for n ≤ 6. H^2(GL(n,2),2^n) has dimension 1 when n=3,4,5, and 0 for n=1,2,6. Perhaps those calculations suffice. In case you are curious, H^1(GL(n,2),2^n) has dimension 1 when n=3, and 0 for n=1,2,4,5,6,7,8. You definitely don't need to keep track of every pair for a cocycle. For GL(n,2), you only need polynomially many values in n, even if you want to efficiently to compute any other value (and not just know it is determined). I think C*n^4 should suffice, and the calculations to get the cohomology are somewhere in the n^8 to n^12 range. | |
| Jun 25, 2010 at 0:24 | comment | added | Andy Putman | Computing H^2 of even modestly sized finite groups with twisted coefficients is pretty computationally intensive. The real problem is that a 2-cocycle is not determined by its values on a generating set for the group, so you have to actually keep track of EVERY pair of elements of the group. An amusing size effect of this is a theorem of Gordon that says that there is no algorithm for computing even untwisted H^2 of a group given by generators and relations (of course, it is easy to compute twisted H^1 of such a group...). | |
| Jun 25, 2010 at 0:13 | comment | added | Homology | I've been fighting with gap and the hap package to compute the H^2 when n is small, but it seems it is not possible with this package (it is more or less restricted to trivial actions...). | |
| Jun 24, 2010 at 23:52 | history | edited | Andy Putman | CC BY-SA 2.5 |
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| Jun 24, 2010 at 23:38 | comment | added | Ben Wieland | Andy's argument is not about char 2, but about nontrivial units; it applies to every field but $\mathbb F_2$. I think Dwyer proves $n=6$ suffices (leaving $n=4,5$ up in the air), but that's probably pessimistic. Incidentally, my notes say that Quillen proved that $H_k(GL_n(F);\mathbb Z)$ stabilizes about when $n=k$ for all fields except $\mathbb F_2$. For $\mathbb F_2$, you need $n=2k+1$. | |
| Jun 24, 2010 at 22:35 | comment | added | Andy Putman | @BS : I haven't had a chance to read Betley's paper carefully, but if Ben Wieland is correct, then the answer is that it can only happen in low dimensions. I'm not sure how small "low" means here. | |
| Jun 24, 2010 at 22:30 | comment | added | BS. | So, I understand that it is a char 2 phenomenon. Does it happens in higher dimensions (and char 2)? | |
| Jun 24, 2010 at 22:19 | comment | added | Andy Putman | @Homology : You can use the following duality (see Prop VI.7.1 in Brown's book). If $M$ is an abelian group, define $M'=Hom(M,\mathbb{Q}/\mathbb{Z})$. Then if $M$ is a finite rank $G$-module, we have $H^k(G;M')=(H_k(G;M))'$. | |
| Jun 24, 2010 at 22:03 | comment | added | Homology | The paper is about homology (not cohomology), how do you make it work? | |
| Jun 24, 2010 at 21:55 | comment | added | Ben Wieland | and for $q=2$, $n$ large, it doesn't happen, either, according to Betley. ams.org/mathscinet-getitem?mr=MR1004602 | |
| Jun 24, 2010 at 21:27 | history | edited | Andy Putman | CC BY-SA 2.5 |
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| Jun 24, 2010 at 21:09 | comment | added | Andy Putman | Good point Homology! I'll revise it to prove the more general result. I'm so used to working with GL_n(Z) that I sometimes forget that GL_n(R) can have central elements other than -1... | |
| Jun 24, 2010 at 20:50 | comment | added | Homology | I don't know how to edit, so it's $\ker D / \mathrm{im} N$ and $\ker N / \mathrm{im} D$, and a reference for this is Serre's Local fields, ch. VIII, § 4 | |
| Jun 24, 2010 at 20:45 | comment | added | Homology | I think your argument works with $F^{\times}$ (which is also a cyclic subgroup of $GL_n(F)$) instead of $\mathbb{Z}/2\mathbb{Z}$, if $F \neq \mathbb{F}_2$. Let $N = \sum_{t \in F^{\times}} t$ and $D=s-1$ where $s$ is a generator of $F^{\times}$. If $F \neq \mathbb{F}_2$, it is easy to check that $N=0$ and $D$ is a nonzero scalar. Then $H^0(F^{\times},V)=0$ and for $i \geq 1$, $H^i(F^{\times},V)= \ker D / \im N$ if $i$ is even, $\ker N / \im D$ if $i$ is odd (noncanonically), and both are zero. So the only possible case is $\mathbb{F}$_2 | |
| Jun 24, 2010 at 20:30 | comment | added | Homology | In this case $N=0$ on $V$ | |
| Jun 24, 2010 at 19:59 | history | edited | Andy Putman | CC BY-SA 2.5 |
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| Jun 24, 2010 at 19:52 | history | answered | Andy Putman | CC BY-SA 2.5 |