Timeline for Which Lie groups are a central extension of an algebraic group?
Current License: CC BY-SA 4.0
22 events
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| Jul 27 at 14:30 | answer | added | Joshua Mundinger | timeline score: 1 | |
| Mar 9, 2022 at 22:58 | comment | added | Luis | @YCor. Thanks. Since you suggested before to fix the question, I added these examples too. | |
| Mar 9, 2022 at 22:57 | history | edited | Luis | CC BY-SA 4.0 |
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| Mar 9, 2022 at 22:21 | comment | added | YCor | @Luis no: consider $\mathrm{SL}_3(\mathbf{R})$. | |
| Mar 9, 2022 at 19:32 | history | edited | Luis | CC BY-SA 4.0 |
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| Mar 9, 2022 at 19:04 | comment | added | Luis | @YCor About the simple case: is it a dichotomy compact/non-compact? | |
| Mar 9, 2022 at 13:50 | comment | added | YCor | You should fix the question to take the counterexamples into account: to cover the semisimple groups, you need to relax the algebraicity definition (e.g., being the unit component in the Lie group of real points). Also for the nilpotent case, the point is that the quotient of every connected nilpotent Lie group (which may be non-linear) by its center is always simply connected. | |
| Mar 9, 2022 at 11:57 | comment | added | Luis | Thanks for all the examples | |
| Mar 9, 2022 at 11:56 | history | edited | Luis | CC BY-SA 4.0 |
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| Mar 9, 2022 at 11:26 | comment | added | Venkataramana | @YCor: of course you are well aware of all this. Mine was a "tongue in cheek" comment. | |
| Mar 9, 2022 at 11:24 | comment | added | YCor | @Venkataramana yes of course it's the connected component of $\mathrm{PGL}_2(\mathbf{R})$ (which is the group of real points of $\mathrm{PGL}_2$). In any case as you pointed out, the assertion "If $G$ is semisimple, this is always the case" of the OP is wrong. | |
| Mar 9, 2022 at 11:23 | comment | added | Venkataramana | @YCor: you are right. But is it not the case that $PSL_2(\mathbb R)$ the connected component of identity of $SO(1,2)$? | |
| Mar 9, 2022 at 11:19 | comment | added | YCor | The answer is negative for $\mathrm{SL}_2(\mathbf{R})$ itself: $\mathrm{PSL}_2(\mathbf{R})$ is not isomorphic to the group of $\mathbf{R}$-points of any algebraic $\mathbf{R}$-group. | |
| Mar 9, 2022 at 5:21 | comment | added | Venkataramana | Actually, even for semi-simple Lie groups this is false: the connected component of identity of SO(1,n) is not an algebraic group. | |
| Mar 9, 2022 at 3:48 | comment | added | David E Speyer | How about Allen Knutson's example mathoverflow.net/a/23594/297 : $\mathbb{R}^4 \rtimes \mathbb{R}$ with $\theta$ acting by $\left[ \begin{smallmatrix} \cos \theta & -\sin \theta && \\ \sin \theta & \cos \theta && \\ && \cos (a \theta) & - \sin(a \theta) \\ && \sin(a \theta) & \cos (a \theta) \\ \end{smallmatrix} \right]$ for irrational $a$? This is simply connected and, if I am not mistaken, centerless. | |
| Mar 9, 2022 at 3:35 | comment | added | Will Sawin | The image of $G/Z(G)$ in the automorphisms of the Lie algebra of $G$ is not always algebraic, so that proof strategy won't work, but the counterexample I have for that has $G / Z(G)$ itself algebraic: It's $ \mathbb R^3 \rtimes \mathbb R^2$, with the action of $\mathbb R^2$ on $\mathbb R^3$ by a rank two subtorus of the diagonal matrices defined by a non-algebraic equation. | |
| Mar 9, 2022 at 3:34 | comment | added | Will Sawin | What about the group of $3 \times 3$ upper-triangular matrices with diagonal entries positive? This is a connected component of an algebraic group but I don't think is an algebraic group itself, and the same thing is true mod center. | |
| Mar 9, 2022 at 3:27 | history | edited | Luis | CC BY-SA 4.0 |
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| Mar 9, 2022 at 2:29 | history | edited | Luis | CC BY-SA 4.0 |
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| Mar 9, 2022 at 2:11 | history | edited | Luis | CC BY-SA 4.0 |
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| S Mar 9, 2022 at 2:00 | review | First questions | |||
| Mar 9, 2022 at 2:13 | |||||
| S Mar 9, 2022 at 2:00 | history | asked | Luis | CC BY-SA 4.0 |