Timeline for Simple lie algebras, (almost-)simple groups of Lie type
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 25, 2016 at 16:02 | comment | added | H A Helfgott | See my newest question above. | |
| Feb 20, 2016 at 7:53 | comment | added | H A Helfgott | A smooth subscheme - or in any event an affine (closed) subvariety that is nonsingular at the origin. Yes, I am sure Jim Humphreys can see a way out. | |
| Feb 20, 2016 at 5:31 | comment | added | Mikhail Borovoi | @HAHelfgott: In any case you should try to specify what you mean by a subvariety (smooth subscheme?). | |
| Feb 20, 2016 at 5:22 | comment | added | Mikhail Borovoi | @HAHelfgott: Concerning the current version of the question, after "What I really need is the following": I am not expert, hopefully Jim Humphreys can help. Otherwise you will have to read Steinberg's proof and Hogeweij's counterexamples. | |
| Feb 20, 2016 at 5:12 | comment | added | Mikhail Borovoi | @HAHelfgott: This is not an extra dimension. $\mathrm{dim}_K\,\mathfrak{sl}(3,K)=8$ in any characteristic, in particular in characteristic 3. | |
| Feb 19, 2016 at 17:20 | comment | added | H A Helfgott | But is there a relatively straightforward proof that works in all characteristics? I mean a proof of the statement in the current version of the question, after "What I really need is the following". | |
| Feb 19, 2016 at 16:13 | comment | added | Mikhail Borovoi | @HAHelfgott: According to this review Steinberg proved that in characteristic $>3$ such a Lie algebra is simple modulo center. | |
| Feb 19, 2016 at 15:06 | comment | added | H A Helfgott | But are all ideals of small dimension? That is, of dimension smaller than the tangent space to any variety $V$ of positive dimension in $G$? I've edited the question; see above. | |
| Feb 19, 2016 at 14:55 | comment | added | H A Helfgott | Oh, I see. You get an extra dimension! | |
| Feb 19, 2016 at 8:54 | comment | added | Mikhail Borovoi | @HAHelfgott: Yes, it still occurs! Consider the scalar matrix $\mathrm{diag}(1,1,1)$, in characteristic 3 it has trace 0, hence it is contained in the Lie algebra $\mathfrak{g}=\mathfrak{sl}(3,K)$, and it is clearly central. Therefore, $\mathfrak{g}$ is not simple. | |
| Feb 19, 2016 at 6:49 | comment | added | H A Helfgott | ... and, whatever the problem is: does it still occur for $p=3$, $n=3$, and $\text{SL}_n$ defined over a large enough field $K$ of characteristic $p$ (i.e., $K$ having more than $C$ elements, $C$ a constant)? | |
| Feb 19, 2016 at 6:42 | comment | added | H A Helfgott | Ah, I was thinking of $p|n-1$. Well, I still do not see the problem for $p=3$, $n=3$, and $\SL_n$, say... | |
| Feb 18, 2016 at 23:46 | comment | added | Jim Humphreys | For $p=5$, the Lie algebra of $\mathrm{SL}_6$ is simple, and similarly for $p=2, n=3$. I worked out more details in my 1966 thesis, mostly published as an AMS Memoir in 1967. But when $p$ divides $n$, there are several cases of groups and Lie algebras of type $A_{n-1}$. Here $n$ is the Coxeter number of the Weyl group $S_n$. | |
| Feb 18, 2016 at 21:33 | comment | added | H A Helfgott | Wait - what is it for $p=5$ and $n=6$, say? (Or for $p=2$ and $n=3$?) | |
| Feb 18, 2016 at 16:46 | comment | added | Jim Humphreys | @HA Helfgott: Even when $p>3$, there is still a problem with special linear groups $G=\mathrm{SL}_n$ in case $p|n$, since then the Lie algebra of $G$ has a one-dimensional center consisting of scalar matrices. | |
| Feb 18, 2016 at 9:12 | comment | added | H A Helfgott | Just to be clear: I am more than willing to discard fields of cardinality rather than a constant, but I would like to have a general result that does cover the case of fields of characteristics 2 or 3 (or anything else) and more than C elements, where C is a constant. | |
| Feb 18, 2016 at 8:20 | comment | added | H A Helfgott | Interesting. Well, is there a conceptual proof for characteristics $>3$? Taking a quick look at Hogeweij, it seems to me as if the ideal structure of the counterexamples for characteristics $2$, $3$ were very easy. Is there a statement slightly weaker than simplicity that can be proved directly? Corollary (2.7) in Hogeweij's first paper seems to be in this general direction, but its wording gives the impression (to me) that it is not as strong as possible (and the proof does involve some case-work). | |
| Feb 18, 2016 at 6:57 | comment | added | Mikhail Borovoi | @HAHelfgott: This is not true in characteristics 2 and 3, see the two papers by Hogeweij. | |
| Feb 17, 2016 at 23:52 | comment | added | H A Helfgott | Thank you very much for this - unfortunately I just realized that what I meant to ask is what you call the more difficult direction. Is it possible to give a conceptual proof of the statement in that direction (that is: the Lie algebra of a group that is simple in the sense of algebraic groups is simple)? It need not cover twisted groups and the like - only algebraic subgroups of $\text{SL}_n$ need be considered - but I would like to avoid case-work as much as possible. | |
| Feb 17, 2016 at 6:16 | comment | added | Mikhail Borovoi | This is so simple! | |
| Feb 16, 2016 at 23:31 | history | answered | Jim Humphreys | CC BY-SA 3.0 |