Timeline for Simple lie algebras, (almost-)simple groups of Lie type
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 29, 2016 at 20:29 | comment | added | H A Helfgott | Thanks for this. This is all for a treatment in front of a group of students I haven't met yet; it's unclear to me what I will be able to assume. I'll study the source, but I suspect there has to be a simpler way - one that does not use the simplicity of the Lie algebra, but rather some slightly weaker fact that can be proven more easily and without casework. | |
| Feb 29, 2016 at 12:33 | comment | added | nfdc23 | OK, it has been hard to tell what the main aim is since the focus of the question has kept changing (and there is no mention of $gVg^{-1}W$ in the question). The link I just gave provides a definitive answer to the simplicity question, showing that the strategy of going through the Lie algebra will work if $p>3$ and one avoids type A, but whether or not one considers the proof of that Lie algebra simplicity result to be clear, quick, and conceptual depends upon one's familiarity with the structure theory of semisimple groups (mainly roots and weights) in positive characteristic. | |
| Feb 28, 2016 at 2:49 | comment | added | H A Helfgott | Hi - aha. I am not trying to make progress there. My original motivation (as I thought had originally been made clear) was to give a clear, quick conceptual proof of several facts in simple groups of Lie type -- above all the fact that, if $G$ is irreducible and almost simple, and $V$, $W$ are a subvarieties of positive codimension of $G$, then, for all $g$ outside a subvariety of $G$ of positive codimension, $g V g^{-1} W$ has dimension larger than $V$. This is known, but proofs are a bit long and can be hard to motivate. If $\mathfrak{g}$ is simple, this is immediate. | |
| Feb 28, 2016 at 2:39 | comment | added | nfdc23 | I don't see how such a relationship helps us to make progress on ruling out the breakdown of simplicity of Lie algebras, so I hadn't considered things like that to be of interest. To make progress on such questions of simplicity one must use structure theory via roots and weights (as has been thoroughly worked out in the literature on modular Lie algebras). That explains conceptually why one should only consider $p \ge 5$ and $p \nmid (n+1)$ for type A$_n$, and happily those are the only obstructions: see Theorem 18.3 of homepages.warwick.ac.uk/~masdf/modular/ln_11mar.pdf | |
| Feb 27, 2016 at 22:47 | comment | added | H A Helfgott | Well, there is a reasonable relationship, in that, if $T_e(V)$ is not an ideal of $\mathfrak{g}$, then there is certainly a $g\in G$ such that $g V g^{-1} V$ (or $g V g^{-1} V^{-1}$) has dimension larger than $V$ (to be precise: its Zariski closure has at least one component of dimension larger than that of $V$). | |
| Feb 27, 2016 at 3:51 | comment | added | nfdc23 | Since $V$ is not a subgroup, $n_a V n_a V^{-1}$ has no reasonable relationship to ${\rm{T}}_e(V) \subset \mathfrak{g}$. The breakdown of simplicity for $\mathfrak{g}$ in characteristic $>0$ rests on the characteristic-free structure theory of semisimple groups. What is going on above is that nontrivial root group commutation laws degenerate in characteristics 2 and 3, leading to the existence of weird smooth subgroups with no analogue in other characteristics. In general positive characteristic, some infinitesimal subgroup schemes create weird subalgebras of the Lie algebra. | |
| Feb 26, 2016 at 13:48 | comment | added | H A Helfgott | Well, what is going on then, exactly? So $n_a \mathfrak{v} n_a^{-1} = \mathfrak{v}$ , yet the dimension of $n_a V n_a^{-1} V^{-1}$ is greater than the dimension of $V$? | |
| Feb 25, 2016 at 16:36 | comment | added | nfdc23 | @HAHelfgott: Yes, $G$ is Spin$_{2n+1}$ or Sp$_{2n}$ (or F$_4$ or G$_2$). Personally, I find descriptions in terms of root groups a much more illuminating to "see" what is going on than to calculate with matrices (which is more concrete but not necessarily instructive). For $n=2$ this is an exceptional endomorphism of Sp$_5$, so you can write down $V$ in that case using the standard root groups. (Also, in my preceding comment I should have chosen $\Phi^+$ relative to which $a$ is a simple root, by the way.) | |
| Feb 25, 2016 at 15:57 | comment | added | H A Helfgott | Interesting. Is $G$ still $\Spin_{2n+1}$ here? Just for concreteness - can you come up with a "small" enough example that you can write this in terms of matrices? I'd really like to see what is going on here. Also, see the new question at the end of the original post. Thanks! | |
| Feb 23, 2016 at 1:27 | comment | added | nfdc23 | @HAHelfgott: In the examples above $V$ is not normalized by $G$. Pick short $a\in\Phi^+$, so $V=U^{-}\times T\times U^{+}$ where $U^{-}$ and $U^{+}$ are direct products in any order of short roots in $\Phi^{\pm}$ (consider unipotent radicals of Borels of $G_{<}$ as in Prop. 7.1.7(1) in "Pseudo-reductive groups"). The reflection $r_a$ preserves $\Phi^{+} - \{a\}$, so conjugation on $V$ by a representative $n_a\in N_G(T)(k)$ of $r_a$ only swaps where $U_{\pm a}$ appear in the direct product. But $U_{-a}TU_a \ne U_aTU_{-a}$ by ${\rm{SL}}_2$-considerations, so $n_aVn_a^{-1}\ne V$. | |
| Feb 22, 2016 at 10:34 | comment | added | H A Helfgott | Aha, interesting - thanks! Now I'm at a loss, however. Say you have a proper, irreducible subvariety $V$ of $G$ ($\dim V>0$) such that its tangent space $\mathfrak{v}$ at the origin is invariant under $Ad_G$. Does it follow that $g V g^{-1} = V$ for all $g$ in $G$? Is that even possible, given that $G$ is simple? It certainly isn't when the stabilizer $Stab(V) = \{h\in G: h V = V\}$ is non-trivial, since then the stabilizer would be a non-trivial, normal, proper subgroup of $G$. | |
| S Feb 21, 2016 at 4:11 | history | answered | nfdc23 | CC BY-SA 3.0 | |
| S Feb 21, 2016 at 4:11 | history | made wiki | Post Made Community Wiki by nfdc23 |