Timeline for Real automorphisms of the "quaternionic" real group ${\rm SO}^*(4m)$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 19, 2018 at 15:37 | vote | accept | Mikhail Borovoi | ||
| Jun 19, 2018 at 15:30 | answer | added | Skip | timeline score: 2 | |
| May 11, 2018 at 19:10 | comment | added | YCor | If I understand correctly Gungodan says that the Out of the Lie algebra is cyclic of order 2. So exactly one of the following holds for $H=G(\mathbf{R})$. (a) $H$ has two real (Lie) components; then $H$ has trivial Out. (b) $H$ is connected as a Lie group. Then $H$ has Out cyclic of order 2. (What I don't clearly see in Gundogan's list is what the homomorphism from the real Out to the complex Out is. Both surjectivity and injectivity can fail.) | |
| May 11, 2018 at 18:55 | comment | added | YCor | Just to illustrate my previous comment, consider the Lie algebra $\mathfrak{sl}_2(\mathbf{R})$. Then conjugation by a matrix with determinant $-1$ is not an inner automorphism (but it's an inner automorphism in the adjoint group $\mathrm{PGL}_2(\mathbf{R})$, which is connected in the Zariski sense but has 2 real (Lie) components. | |
| May 11, 2018 at 18:53 | history | edited | Mikhail Borovoi | CC BY-SA 4.0 |
added 293 characters in body
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| May 11, 2018 at 15:01 | comment | added | YCor | One has to be careful about the meaning of outer automorphism. Let $G$ be the corresponding adjoint (connected) group. One can consider the group $G(\mathbf{R})$ and say that its elements are inner automorphisms. But one can say that inner automorphisms are only those in the unit component of $G(\mathbf{R})$ in the real topology. Unlike in the complex case these can be distinct. From the question it seems that one uses the first definition to define the group of outer automorphisms. But I'm clarifying, especially if one talks of the outer automorphism group of the Lie algebra... | |
| May 11, 2018 at 11:23 | comment | added | Mikhail Borovoi | @Vincent: Abusing notation, I meant the real algebraic group $G$ over $\Bbb R$ with the group of real points $G(\Bbb R)={\rm SO}^*(4m)$. However, if you answer for the Lie algebra, I will be quite happy. | |
| May 11, 2018 at 11:07 | comment | added | Vincent | Just to clarify the notation, $G$ is a real Lie algebra here, right? Otherwise I do not know how to make sense of a Lie-group homomorphism being defined 'over a field'. But maybe there is a definition I don't know about? | |
| May 11, 2018 at 10:21 | comment | added | YCor | This has been systematically looked in "Gundogan, The component group of the automorphism group of a simple Lie algebra and the splitting of the corresponding short exact sequence", J. Lie Theory 20 (2010), no. 4, 709-737. But I'm not sure what the conclusion is. Apparently, the automorphism group of the real Lie algebra, as Lie group, has 2 components. But I don't see if the resulting homomorphism of the component group to the complex Out is an isomorphism or is the trivial map; this might come from a more careful reading. | |
| May 11, 2018 at 10:05 | history | asked | Mikhail Borovoi | CC BY-SA 4.0 |