Timeline for Computing the inner automorphism group of a finite Lie algebra
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| May 24, 2018 at 17:59 | comment | added | Russ Woodroofe | Added below (posted as an answer, since it was getting long)! One such Lie element in sl(3,3) is v.1 + v.6, in GAP 4.9.1's notation. | |
| May 24, 2018 at 17:54 | answer | added | Russ Woodroofe | timeline score: 0 | |
| May 24, 2018 at 10:51 | comment | added | Paul Levy | Example please? | |
| May 24, 2018 at 10:09 | vote | accept | Russ Woodroofe | ||
| May 23, 2018 at 22:53 | comment | added | Russ Woodroofe | I can confirm by computation that exponentials of adjoints with nilpotency less than $p$ (in characteristic $p$) may not be isomorphisms. (And this also appears elsewhere in the literature.) | |
| Mar 5, 2018 at 12:25 | comment | added | Paul Levy | I haven't looked at the Mattarei paper in detail, but they are dealing with derivations of an arbitrary derivation of an arbitrary non-associative algebra. In your case, you have an inner derivation of a Lie algebra. So I am fairly sure that (e.g. by using the adjoint representation to reduce to the case of $\mathfrak{gl}_n$) if $({\rm ad}\, x)^p=0$ then ${\rm exp}({\rm ad}\, x)$ is (well-defined and is) an automorphism of ${\mathfrak g}$ (corresponding, in the right setting, to conjugating by $e^x$). This is completely algebraic and the proof goes through as in characteristic zero. | |
| Mar 3, 2018 at 23:06 | comment | added | Russ Woodroofe | Unless I'm missing the point, a main point of the Mattarei paper is that the exponential in characteristic $p$ of a nilpotent element need not be an automorphism, even if the $p$-th power is 0. I haven't yet asked GAP to look for a non-automorpism in the group, however -- it might be worthwhile to do so. | |
| Feb 23, 2018 at 23:48 | comment | added | Paul Levy | On your update 2: the exponential of a nilpotent element of $\mathfrak{g}=\mathfrak{sl}(3,3)$ is an automorphism of $\mathfrak{g}$, since all nilpotent elements have $p$-th power zero, so the ordinary exponential equals the exponential truncated at the $p$-th power, and in particular is well-defined. So I think your group of order 9285337152 will consist of automorphisms of $\mathfrak{g}$. | |
| Feb 21, 2018 at 19:16 | history | edited | Russ Woodroofe | CC BY-SA 3.0 |
update 2 contains more information on exponentials of adjoints over finite fields
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| Feb 17, 2018 at 18:59 | answer | added | Paul Levy | timeline score: 2 | |
| Feb 13, 2018 at 2:33 | answer | added | Jim Humphreys | timeline score: 2 | |
| Feb 12, 2018 at 14:52 | history | edited | Russ Woodroofe | CC BY-SA 3.0 |
added insight from further computations in clearly marked section at beginning; minor updates and clarifications elsewhere
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| Feb 11, 2018 at 20:30 | history | asked | Russ Woodroofe | CC BY-SA 3.0 |