Timeline for Maximum symmetry metric on $ \mathbb{C}P^n $
Current License: CC BY-SA 4.0
13 events
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| Nov 25, 2022 at 17:10 | answer | added | Jason DeVito - on hiatus | timeline score: 5 | |
| Nov 4, 2022 at 20:46 | history | edited | Ian Gershon Teixeira | CC BY-SA 4.0 |
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| Nov 4, 2022 at 20:29 | vote | accept | Ian Gershon Teixeira | ||
| Nov 4, 2022 at 20:27 | history | edited | Ian Gershon Teixeira | CC BY-SA 4.0 |
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| Nov 4, 2022 at 20:20 | history | edited | Ian Gershon Teixeira | CC BY-SA 4.0 |
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| Nov 4, 2022 at 12:42 | comment | added | Robert Bryant | Also, for your 'new guess', you need to restrict to irreducible compact symmetric spaces in order to avoid the obvious counterexamples: $M=\mathbb{R}^n/\Lambda$ where $\Lambda\subset\mathbb{R}^n$ is a lattice. These compact symmetric spaces have $N(M)=n$ realized by the translations in $\mathbb{R}^n$, but there is an $(n(n+1)/2-1)$-parameter family of inequivalent ones. Also, if $M$ is a non-trivial product of irreducible symmetric spaces, you'll have a positive dimensional family of inequivalent $G$-invariant metrics. | |
| Nov 4, 2022 at 12:30 | comment | added | Robert Bryant | Unfortunately, 'unique up to isometry' is not what you want, since, on a compact manifold at least, a metric $g$ is not isometric to $cg$ for any constant $c\not=1$, while the non-zero constant multiples of $g$ all have the same isometry group. I would recommend to say something like, 'Two metrics are considered to be equivalent if they are isometric up to a constant multiple.', and then, after that, say 'unique up to equivalence'. | |
| Nov 3, 2022 at 23:00 | history | edited | Ian Gershon Teixeira | CC BY-SA 4.0 |
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| Nov 3, 2022 at 22:52 | answer | added | Robert Bryant | timeline score: 13 | |
| Nov 3, 2022 at 22:47 | comment | added | Robert Bryant | A plausible guess might be that any metric on $\mathbb{CP}^n$ whose isometry group has dimension $n(n{+}2)$ must be isometric to a constant scalar multiple of the Fubini-Study metric. While it's possible (when $n=1$) for a non-constant scalar multiple of the Fubini-Study metric to be isometric to a constant scalar multiple of the Fubini-Study metric, most of the time, when you have a non-constant scalar multiple of the Fubini-Study metric, the result is not even Kähler. | |
| Nov 3, 2022 at 22:32 | comment | added | Ian Gershon Teixeira | @RobertBryant Is there a good way I could rephrase my question to fix that problem? For example just requiring that they are isometric to Fubini-Study not necessarily a constant scalar multiple? | |
| Nov 3, 2022 at 22:28 | comment | added | Robert Bryant | You are leaving out the diffeomorphism group. There are many (even Kähler) metrics on $\mathbb{CP}^n$ that are isometric to 'the' Fubini-Study metric but are not equal to a (constant) scalar multiple of it. | |
| Nov 3, 2022 at 22:21 | history | asked | Ian Gershon Teixeira | CC BY-SA 4.0 |