Timeline for Automorphism group of real orthogonal Lie groups
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 28, 2020 at 14:39 | comment | added | annie marie cœur | also is this correct that: $$Out(𝑆𝑝𝑖𝑛(8;ℝ) )=S_3$$ $$Out(SO(8;ℝ) )=Z_2?$$ $$Out(SO(8;ℝ)/Z_2 )=S_3 $$ | |
| Sep 28, 2020 at 14:38 | comment | added | annie marie cœur | the answer is long, thanks --- is this true that $$Inn(𝑆𝑝𝑖𝑛(8;ℝ) )= Inn(SO(8;ℝ) )=Inn(SO(8;ℝ)/\mathbb{Z}_2 )=SO(8;ℝ)/\mathbb{Z}_2 ?$$ | |
| Apr 12, 2016 at 7:14 | vote | accept | Bilateral | ||
| Apr 12, 2016 at 1:17 | comment | added | nfdc23 | A linear algebraic group over a field $k$ (especially not algebraically closed; e.g., finite or $\mathbf{R}$) is a very different thing from its group of $k$-points. Please talk with a colleague or friend who knows about algebraic groups to understand the difference, which is technically very essential. For example, the algebraic groups ${\rm{SL}}_3$ and ${\rm{PGL}}_3$ over $\mathbf{R}$ are not isomorphic but ${\rm{SL}}_3(\mathbf{R}) = {\rm{PGL}}_3(\mathbf{R})$! Likewise, it is "wrong" to consider ${\mathbf{SO}}(p,q)$ as another viewpoint on ${\rm{SO}}(p,q)$; they're very different objects. | |
| Apr 12, 2016 at 0:05 | comment | added | nfdc23 | My notation carefully distinguishes these very different objects: ${\mathbf{SO}}(p,q)$ is a smooth $\mathbf{R}$-scheme with the Zariski topology, and ${\rm{SO}}(p,q)$ is its group of $\mathbf{R}$-points (a Lie group, with "analytic" topology coming from the topology of $\mathbf{R}$). Each has a Lie algebra, and those coincide, so we can use Lie-theoretic methods very well with ${\mathbf{SO}}(p,q)$ but less so with ${\rm{SO}}(p,q)$ since the former is Zariski-connected but the latter is disconnected. The group ${\rm{SO}}(p,q)^0$ is in no sense "algebraic". | |
| Apr 11, 2016 at 19:33 | comment | added | Sebastian Goette | I am a bit confused by the notion of a "connected algebraic group". From an algebraic point of view, $SO(p,q)$ is connected, isn't it? From a topological point of view, of course only if either $p=0$ or $q=0$. In other words, is $SO(p,q)^0$ (the real points) actually algebraic? | |
| Apr 11, 2016 at 12:46 | comment | added | nfdc23 | My assertions are consistent since non-triviality of the outer automorphism group is claimed only for ${\rm{O}}(p,q)$ with odd $n$ in the indefinite case (with $p, q > 0$), using that ${\rm{SO}}(p,q)$ is disconnected in such cases (whereas it is connected in the definite case, used crucially in my argument for triviality of the outer automorphism group of ${\rm{O}}(n)$ for odd $n$ -- I am following the standard notation that ${\rm{O}}(n)$ corresponds to the definite, equivalently compact, case). I haven't made any claims concerning ${\rm{O}}(n)$ for even $n$. Hopefully that clarifies things. | |
| Apr 11, 2016 at 8:36 | comment | added | Bilateral | Thanks again for your detailed answer. You say that for $n$ odd $O(p,q) = SO(p,q)\times <-1>$ does have a non-trivial outer automorphism given by $(g,z)\mapsto (g,f(z)g)$. However, then you say that for $n$ odd the outer automorphism group of $O(n)$ is trivial. In addition, @SebastianGoette says that $(g,z)\mapsto (g,f(z)g)$ is a non-trivial outer-automorphism in the $n$-even case. Am I missing something? | |
| Apr 11, 2016 at 4:04 | history | edited | nfdc23 | CC BY-SA 3.0 |
added 347 characters in body
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| Apr 11, 2016 at 3:47 | comment | added | nfdc23 | I have added a treatment of ${\rm{SO}}(p,q)$ for odd $p, q$ with $p+q \ge 3$ at the end of the answer (again, all automorphisms turn out to be algebraic, so the outer automorphism group is of order 2). Analyzing the case of ${\rm{SO}}(p,q)$ for even $p, q \ge 2$ requires a better idea (especially when $n \ge 6$). | |
| Apr 11, 2016 at 3:44 | history | edited | nfdc23 | CC BY-SA 3.0 |
treated p, q odd as well (for SO).
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| Apr 11, 2016 at 0:57 | comment | added | nfdc23 | Of course, in the definite case for odd $n$ the triviality of the outer automorphism group for ${\rm{O}}(n)$ reduces to that of ${\rm{SO}}(n)$ since ${\rm{O}}(n) = {\rm{SO}}(n) \times \langle -1 \rangle$ with ${\rm{SO}}(n)$ connected. So that is an alternative (albeit perhaps much heavier) approach to what is done in Goette's answer for odd $n$. The case of even $n$ requires more serious effort (as in the indefinite case). | |
| Apr 11, 2016 at 0:46 | comment | added | nfdc23 | Hochschild's book won't help with the algebraic aspects of my argument (e.g., Dieudonne's theorem, admittedly only needed in characteristic 0); I was freely using anything I needed from the theory of linear algebraic groups, so perhaps ask a friend who knows about algebraic groups if you need some assistance with that. In the indefinite case for odd $n$ the group ${\rm{O}}(p,q) = {\rm{SO}}(p,q) \times \langle -1 \rangle$ does have a non-inner (non-algebraic!) automorphism $(g,z) \mapsto (g, f(g)z)$ for the unique surjective homomorphism $f:{\rm{SO}}(p,q) \rightarrow \langle -1 \rangle$! | |
| Apr 11, 2016 at 0:09 | comment | added | Bilateral | Thank you very much for your answer. I have to go carefully through it and try to prove the $O(p,q)$ case along similar lines (in case this is possible). Just to know if I get the correct answer: do you know by heart if $O(p,q)$ has non-trivial outer automorphisms? The relevant reference that you recommend to understand the details of your proof is then Hochschild's book or there are others more appropriates? | |
| Apr 10, 2016 at 23:57 | history | answered | nfdc23 | CC BY-SA 3.0 |