Timeline for Reductive group with simply connected derived group has all root groups $\mathrm{SL}_2$
Current License: CC BY-SA 4.0
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| May 2, 2021 at 5:15 | history | edited | spin | CC BY-SA 4.0 |
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| May 2, 2021 at 3:49 | comment | added | spin | @LSpice: Yes, it is the same result, but mine is more elementary in the sense that it just uses the basic definitions of Chevalley groups from Steinberg's notes. Paul Levy gives a more general approach that works for all SL2's (labeled Dynkin diagrams, Jacobson-Morozov theorem etc). | |
| May 2, 2021 at 3:26 | comment | added | LSpice | @spin, this case-by-case analysis is the same as in @PaulLevy's and @nfdc23's answers to mathoverflow.net/questions/270205/…, right? | |
| May 2, 2021 at 3:20 | history | edited | spin | CC BY-SA 4.0 |
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| May 2, 2021 at 1:13 | history | edited | spin | CC BY-SA 4.0 |
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| May 1, 2021 at 16:44 | comment | added | LSpice | @MikhailBorovoi, since we can't have $\operatorname{PGL}_2$ as a direct factor of a simply connected group, what is the relevant result in the simply connected setting? | |
| May 1, 2021 at 16:43 | comment | added | Mikhail Borovoi | @LSpice: I mistakenly remembered my old work where I considered simply connected groups. | |
| May 1, 2021 at 16:34 | comment | added | LSpice | @MikhailBorovoi, not to pile on about it, but surely you had an interesting motivation to make that claim. What intuition suggested it? | |
| May 1, 2021 at 16:29 | comment | added | Mikhail Borovoi | @spin: you are right, and my comment about a direct factor is erroneous. | |
| May 1, 2021 at 16:11 | comment | added | LSpice | @spin, re, see mathoverflow.net/questions/270205/…. | |
| May 1, 2021 at 15:13 | comment | added | spin | @MikhailBorovoi: What if $G$ is of adjoint type $C_2$, and $\alpha$ is a short root? Anyway doing a case-by-case for all root systems to check which images are $\operatorname{PGL}_2$ should be routine. | |
| May 1, 2021 at 15:04 | comment | added | spin | The notation you used for the map $\operatorname{SL}_2 \rightarrow G$ is a bit unfortunate, since it clashes with the notation $h_{\alpha}(t)$ used in Steinberg. Lemma 19 is just where the definition of $h_{\alpha}(t)$ is. Lemma 28 (c) is what you want to use, since $\operatorname{SL}_2 \rightarrow G$ maps $\begin{pmatrix} t & 0 \\ 0 & t^{-1} \end{pmatrix}$ to $h_{\alpha}(t)$. | |
| May 1, 2021 at 14:49 | comment | added | Mikhail Borovoi | Let $G$ be a semisimple $\Bbb C$-group. I think that if the image of the homomorphism $h_\alpha\colon {\rm SL}_2\to G$ is isomorphic to ${\rm PGL}_2$, then this image is a direct factor of $G$. | |
| May 1, 2021 at 14:45 | comment | added | Mikhail Borovoi | Here the reductive group has nothing to do with the question, because the image of the homomorphism $h_\alpha\colon {\rm SL}_2\to G$ is contained in the derived group $G^{\rm der}=[G,G]$, which by assumption is simply connected. | |
| May 1, 2021 at 14:05 | history | answered | spin | CC BY-SA 4.0 |