Timeline for Non-isomorphic complex Lie groups with the same exceptional Lie algebra for $\mathfrak{g_2,f_4,e_6,e_7,e_8}$?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Aug 24, 2020 at 19:49 | comment | added | LSpice | @MikhailBorovoi, in @SamHopkins's defence (not that they need it from me), I'm not sure if the fact that there are lots of interesting things to say about this means that it was an appropriate MO question. Yours is obviously a good answer, and I appreciate your writing it; but it's the sort of thing that can be and should be found in the standard references, as you mention, and I wouldn't want to see people come to MO to ask questions when they can and should first check in the standard references (for whatever field their question is in). | |
| Aug 24, 2020 at 12:54 | comment | added | Mikhail Borovoi | @SamHopkins: Real forms were classified by Cartan in this paper of 1914. | |
| Aug 24, 2020 at 12:45 | comment | added | Sam Hopkins | @MikhailBorovoi: Haha, indeed. Well I knew that in principle this was all worked out by Cartan 100 years ago, but there are many moving parts to say the least! | |
| Aug 24, 2020 at 9:52 | comment | added | Mikhail Borovoi | @SamHopkins: And you wrote that this question would be more suitable for Math StackExchange.... | |
| Aug 24, 2020 at 1:26 | comment | added | Sam Hopkins | Looking at en.wikipedia.org/wiki/List_of_simple_Lie_groups#List again, I see that what I suggested in my last comment is not true: the fundamental groups of the centerless real groups will be different from in the complex case. | |
| Aug 24, 2020 at 0:08 | comment | added | Sam Hopkins | @MikhailBorovoi: I have one last question about this very nice answer. If I want to compute all the real Lie groups associated with a complex Lie algebra, is it true that I can, so to speak, "multiply" the choices from the two steps of choosing a real form and choosing a center? E.g., for $\mathfrak{e}_6$ I would get $5\times 2=10$ groups? In other words, will I have $C(\mathfrak{g}_{\mathbb{R}})=C(\mathfrak{g})$ for any real form of my Lie algebra? | |
| Aug 23, 2020 at 21:50 | history | edited | Mikhail Borovoi | CC BY-SA 4.0 |
added 11 characters in body
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| Aug 23, 2020 at 21:44 | comment | added | Mikhail Borovoi | @SamHopkins: Oops! Thank you for noticing. I have removed the erroneous assertion. | |
| Aug 23, 2020 at 21:41 | history | edited | Mikhail Borovoi | CC BY-SA 4.0 |
A wrong assertion about Out(g) removed.
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| Aug 23, 2020 at 21:20 | comment | added | Sam Hopkins | One thing is wrong with this answer: the exceptional Lie algebra $\mathfrak{e}_6$ has a non-trivial outer automorphism. | |
| Aug 23, 2020 at 16:25 | comment | added | Mikhail Borovoi | I understood what you meant, but this is an answer on two pages! If somebody asks a separate question, I will answer it. | |
| Aug 23, 2020 at 16:22 | comment | added | Mikhail Borovoi | In particular, if a semisimple algebraic $\Bbb R$- group $\bf G$ is a connected (over $\Bbb C$), and if it either compact or simply connected (simply connected over $\Bbb C$), then the group of $\Bbb R$-points $\bf G(\Bbb R)$ is connected. | |
| Aug 23, 2020 at 16:20 | comment | added | LSpice | Certainly I didn't mean to suggest that you didn't know that, and I apologise if I gave that impression. I just meant that it might be worth it to make it part of the answer for the benefit of people who might not know it. | |
| Aug 23, 2020 at 16:18 | comment | added | Mikhail Borovoi | @LSpice: All what you say in your first comment - I do know that. If you ask a relevant question, I will answer it when I have time. | |
| Aug 23, 2020 at 16:11 | comment | added | LSpice | I see now that you edited the question to refer only to complex Lie groups, as the wording obliquely suggested and OP may or may not have intended. In this case I agree with you: we get very lucky—I think of it as luck, but perhaps design is a more appropriate way to describe it—and the confusions I worried about do not arise (which, of course, is how those alg.-gp. terms arose!). | |
| Aug 23, 2020 at 16:08 | comment | added | LSpice | I think it may be important to note what is paid for by using the language of algebraic groups (which I also prefer). For example, "there is only one algebraic group with Lie algebra $\mathfrak g$" (where $\mathfrak g$ is $\mathfrak e_8$, $\mathfrak f_4$, or $\mathfrak g_2$) is true only to the extent that we speak of groups over alg. closed field; it's not true over $\mathbb R$ (as you say later). Also, the alg. groups that we are calling connected and simply connected need not be either when regarded as real Lie groups (by taking real points). Finally, we miss non-linear groups this way! | |
| Aug 23, 2020 at 15:56 | history | answered | Mikhail Borovoi | CC BY-SA 4.0 |