Timeline for Are the automorphism groups of simple symmetric cones algebraic groups?
Current License: CC BY-SA 4.0
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| Oct 4, 2022 at 8:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 4, 2022 at 7:18 | answer | added | user473423 | timeline score: 0 | |
| Sep 3, 2022 at 11:58 | answer | added | Mingchen Xia | timeline score: 1 | |
| Sep 3, 2022 at 8:58 | history | edited | Mingchen Xia | CC BY-SA 4.0 |
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| Sep 3, 2022 at 5:34 | comment | added | Mingchen Xia | @Echo I checked the definition on Wallach's book 2.1.1. He required $G_{\mathbb C}$ to be Zariski closed. But in our question here, we only know the closeness in the Euclidean topology! | |
| Sep 2, 2022 at 9:30 | comment | added | user473423 | @Mingchen Xia: It depends on what you mean by 'real reductive group'. In the OP's setting we can adopt the definition of Wallach in his book "Real Reductive Groups I" which is the one the OP uses. Wallach shows that for such a group the connected component of the center is indeed a torus. | |
| Sep 1, 2022 at 8:05 | comment | added | Mingchen Xia | @Echo I believe that in order to make this argument work, we need to know a priori that the identity component of the centre of $G_{\mathbb C}$ is a torus, see math.stanford.edu/~conrad/papers/luminysga3.pdf Example D.3.3. This is not obvious to me at all. | |
| Aug 30, 2022 at 18:17 | comment | added | user473423 | @Mingchen Xia: You can complexify the Lie algebra and take the compact form of that Lie algebra, apply exp and get a compact form of your original Lie group. | |
| Aug 30, 2022 at 9:22 | comment | added | Mingchen Xia | @Echo I’m not sure if I understand your explanation. You talk about complexify G, and then as a complex reductive Lie group it gets an algebraic structure. But usually for the latter claim, a complex reductive Lie group refers to the complexification of a compact real Lie group. I’m not sure if we start from a general real reductive group G, $G_{\mathbb C}$ is reductive in this sense or not. If there is a canonical algebraic structure on $G_{\mathbb C}$, then we don’t really need the classification etc, it suffices to do a Galois descent. | |
| Aug 30, 2022 at 8:49 | comment | added | YCor | @Echo yes of course, I don't mean the $\mathbf{R}$-points of $\mathrm{PSL}_2$ (one should never refer to $\mathrm{PSL}_n$ as a group scheme because of these ambiguities and because it's the same as the group scheme $\mathrm{PGL}_n$ which doesn't prompt these issues). | |
| Aug 30, 2022 at 6:20 | comment | added | user473423 | @YCor You mean the real points of the group scheme $PSL_2$? No probably you mean $SL_2({\mathbb R})/\pm 1$, which is the connected component. Yes indeed, I interpreted the question as a question on the connected component. My thinking was that you complexify the group and then you are in the classification of complex algebraic groups which essentially coincides with the classification of complex Lie algebras. | |
| Aug 29, 2022 at 18:47 | comment | added | Mingchen Xia | @Echo I think both approaches only work if we know that G is algebraic a priori. For a real reductive group defined as a a matrix group closed under conjugate transpose, I do not really think there is a full classification (Maybe I'm wrong?). In the book jmilne.org/math/CourseNotes/LAG.pdf P160, Milne claimed what you claim. But the reference he referred to is totally irrelevant. So if what you claim is correct, could you please point out a reference? btw, I also noticed that there is a related discussion mathoverflow.net/questions/28849/… | |
| Aug 29, 2022 at 15:59 | comment | added | YCor | @Echo it depends on the exact meaning of "algebraic". For instance, there is no real algebraic group $G$ such that $G(\mathbf{R})$ is isomorphic as a Lie group to $\mathrm{PSL}_2(\mathbf{R})$. | |
| Aug 29, 2022 at 15:58 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag
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| Aug 29, 2022 at 15:52 | comment | added | user473423 | Any linear reductive Lie group is algebraic. One way so see this is to use the classification. The other is to use reductivity on the action of G on the polynomials to find that the Zariski closure of G has the same dimension as G. | |
| Aug 29, 2022 at 15:21 | history | asked | Mingchen Xia | CC BY-SA 4.0 |