Timeline for Maximum symmetry metric on $ \mathbb{C}P^n $
Current License: CC BY-SA 4.0
18 events
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| Nov 28, 2022 at 0:13 | comment | added | Ian Gershon Teixeira | Is there a trivial way to extend this argument to quaternionic projective space $ \mathbb{HP}^n $? Or is that best asked as a separate question? Also wondering if there is any easy way to make a counter example of the form $ G/H $ for $ H $ semisimple, since the counterexample $ SU_3/T^2 $ is not of that form. | |
| Nov 25, 2022 at 17:25 | comment | added | Ian Gershon Teixeira | Oh I see its $ Sp_n/Sp_{n-1} \cong S^{4n-1} $ but then you mod out by $ U_1 $ to get $ Sp_n/(Sp_{n-1}\times U_1) \cong \mathbb{CP}^{2n-1} $. | |
| Nov 25, 2022 at 17:15 | comment | added | mme | It acts on quaternionic space H^n, transitively on its unit sphere. It commutes with the left action of H (it is the group of H-linear isometries) and this descends from a transitive action on S^{4n-1} to a transitive action on HP^{n-1}. In between, it factors through a transitive action on CP^{2n-1}. | |
| Nov 25, 2022 at 16:30 | comment | added | Ian Gershon Teixeira | My bad I just didn't read carefully. So in general $ PSp_n $ acts transitively on $ \mathbb{CP}^{2n+1} $? I known $ PSp_1=SO_3 $ is transitive on $ \mathbb{CP}^1 $ and I guess you just said that $ PSp_2=SO_5 $ is transitive on $ \mathbb{CP}^3 $. That's a very cool fact! Is there some intuitive way to understand why $ Sp_n $ has this action? What is the stabilizer of $ Sp_2 $ on $ \mathbb{CP}^3 $? Is it $ U_2 $? | |
| Nov 25, 2022 at 12:01 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Put in a little more detail to make the structure of the argument more clear.
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| Nov 24, 2022 at 21:05 | comment | added | Robert Bryant | @IanGershonTeixeira: There's no need to assume 'transitive', so I didn't assume that. If a connected, compact group $G$ acts effectively on $\mathbb{CP}^n$ then the dimension of $G$ is at most $n(n{+}2)$. If equality holds, then the action is transitive and $G$ is the quotient of $\mathrm{SU}(n{+}1)$ by its center. As to the final question in your comment, the answer is 'no'. The 10-dimensional group $\mathrm{Sp}(2)$ acts transitively on $\mathbb{CP}^3$, and the action is almost effective, the center (isomorphic to $\mathbb{Z}_2$) acts trivially. Similar results for any $\mathbb{CP}^{2n+1}$. | |
| Nov 24, 2022 at 19:06 | comment | added | Ian Gershon Teixeira | Wow loving the new update to this answer! In the spirit of Thanks giving I want to thank Ramiro Lafuente for a thought provoking comment and big thanks to Robert Bryant for putting in this beautiful new-and-improved argument! Also small typo I think you omitted "transitively" from the sentence "Suppose that a connected, compact group 𝐺 acts effectively on ℂℙ𝑛." Also does this prove that any compact connected group acting transitively effectively on $ \mathbb{CP}^n $ must be (isogeneous to?) $ SU_{n+1} $? | |
| Nov 24, 2022 at 15:06 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Removed the old torturous argument and misguided remark and put in a proof valid for all $n$.
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| Nov 24, 2022 at 14:24 | comment | added | Robert Bryant | @RamiroLafuente: Of course you are right. That remark was wrong-headed. I'll remove it. Anyway, I have realized that there is a much simpler argument for the maximum symmetry of $\mathbb{CP}^n$ that works for all $n$, so I'll replace that entire segment. | |
| Nov 23, 2022 at 22:58 | comment | added | Ramiro Lafuente | A comment on that last remark about the general case: this could be more subtle as it is not true for the sphere $S^{2n}$. There is an effective action of $SO(2n-1)$ on $S^{2n}$ by simply rotating the last 2n-1 coordinates in $\mathbb{R}^{2n+1}$. For $n\geq 5$ the dimension of $SO(2n-1)$ is greater than $n(n+2)$. | |
| Nov 4, 2022 at 20:29 | vote | accept | Ian Gershon Teixeira | ||
| Nov 4, 2022 at 20:29 | comment | added | Ian Gershon Teixeira | Ok this answer is super interesting I think I'll just accept it and spin of my guess about irreducible compact symmetric spaces having unique up to equivalence maximum symmetry metric as it's own question. | |
| Nov 4, 2022 at 11:39 | comment | added | Robert Bryant | @JasonDeVito: It's not hard to show that the identity component the isometry group of each of these $\mathrm{SU}(3)$-invariant metrics is $\mathrm{SU}(3)/\mathbb{Z}_3$ (where $\mathbb{Z}_3$ is the center of $\mathrm{SU}(3)$). There might be other components, especially in some special cases, but I'm not sure about that. What I suspect is that no compact group of dimension greater than $8$ can act smoothly and effectively on $\mathrm{SU}(3)/\mathbb{T}^2$, and I can imagine how one could try to prove it, but I haven't tried myself. If it's true, it's probably in the literature somewhere. | |
| Nov 3, 2022 at 23:59 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added the argument for n=2
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| Nov 3, 2022 at 23:49 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added the argument for n=2
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| Nov 3, 2022 at 23:40 | comment | added | Jason DeVito - on hiatus | Is it easy to see thay none of the $SU(3)$-invariant metrics in your answer "accidentally" has a larger isometry group? E.g., if one considers the same problem on $SU(3)/SU(2)$, there is a 2-parameter family of invariant metrics, but for a 1-parameter sub-family, the isometry group is $O(6)$. | |
| Nov 3, 2022 at 23:06 | comment | added | Ian Gershon Teixeira | Ok so my guess was a little ambitious. What about the title question (reworked as suggested by you in the comments)? If $ g $ is a metric on $ \mathbb{CP}^n $ with isometry group of dimension $ n(n+2) $ then must $ g $ be isometric to a constant scalar multiple of the Fubini-Study metric? And if it's true for $ \mathbb{CP}^n $ and spheres then maybe it's true for all compact symmetric spaces? | |
| Nov 3, 2022 at 22:52 | history | answered | Robert Bryant | CC BY-SA 4.0 |