Timeline for Maximum symmetry metric on $ \mathbb{C}P^n $
Current License: CC BY-SA 4.0
10 events
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| Jan 26, 2023 at 21:59 | comment | added | Jason DeVito - on hiatus | @IanGershonTeixeira: I find your guess quite reasonable and would generally expect it to be true. But I have no idea how to prove it at this level of generality. I'm not even sure I could prove the corresponding result for $\mathbb{H}P^n = Sp(n+1)/Sp(n)\times Sp(1)$. Also, I wouldn't be surprised if there were a few examples where your method does not give the most symmetric metric, say, because that example is accidentally diffeomorphic to something else which is know to have a large symmetry group. | |
| Jan 26, 2023 at 21:11 | comment | added | Ian Gershon Teixeira | What do you think about the conjecture that if $ M=G/K $ is isotropy irreducible and $ g $ is the pushforward onto $ M $ of the biinvariant metric then $ g $ is a maximum symmetry metric on $ M $ in the sense described above and moreover it is the unique such metric (up to equivalence)? Is there anything about that conjecture that strikes you as immediately or obviously wrong? | |
| Jan 26, 2023 at 20:48 | comment | added | Jason DeVito - on hiatus | It is not isotropy irreducible. The group $SU(n)$ is contained in a copy of $U(n)$ lying in $SU(n+1)$. The isotropy action splits into two irreducible summands: a trivial rep (corresponding to the vectors in tangent to $U(n)$), and the usual $n$-dim complex rep of $SU(n)$. | |
| Jan 26, 2023 at 20:11 | comment | added | Ian Gershon Teixeira | Is $ S^{2n+1}=SU(n+1)/SU(n) $ not isotropy irreducible? | |
| Nov 28, 2022 at 0:55 | comment | added | Jason DeVito - on hiatus | @Ian: Well, I have a way of generating guesses, but proving that any of them works will require some effort. Pick any $H\subseteq G$ for which the isotropy action of $H$ on $\mathfrak{g}/\mathfrak{h}$ splits into has precisely two irreducible summands. (This situation has been classified: link.springer.com/article/10.1007/s10455-008-9109-9). Each of these admits $G$-invariant metrics which are not isometric, even up to scaling. I would bet that generically, the isometry group of each of these metrics is $G$ (perhaps up to components and up to cover). | |
| Nov 28, 2022 at 0:45 | comment | added | Ian Gershon Teixeira | What I said was unclear, let me clarify. In my question I mention that "My guess is that there is always such a unique [maximum symmetry] metric for manifolds of the form $ G/H $ for $ G $ a compact connected simple Lie group and $ H $ a closed subgroup." $ SU_3/T^2 $ is a counterexample to this. $ SU_{n+1}/SU_n \cong S^{2n+1} $ is not a counterexample because spheres do admit a unique maximum symmetry metric (although this is a counterexample to the other guess I made that for simple $ G $ the push forward of the biinvariant metric onto $ G/H $ always induces a maximum symmetry metric.) | |
| Nov 28, 2022 at 0:35 | comment | added | Jason DeVito - on hiatus | @IanGershonTeixeira: A counterexample to what? My examples in point 1. above already have $G$ simple and $H$ semisimple (simple, in fact). | |
| Nov 28, 2022 at 0:15 | comment | added | Ian Gershon Teixeira | Do you know a counterexample of the form $ G/H $ where $ G $ simple and $ H $ is semisimple? Just wondering since $ SU_3/\mathbb{T}^2 $ is not of that form. | |
| Nov 25, 2022 at 17:23 | comment | added | Robert Bryant | Thanks! I could imagine how to prove the special case of $\mathrm{SU}(3)/\mathbb{T}^2$, but the details were messy. It's good to know that it works for all $G/T$, which I wouldn't even have thought of attempting to prove. | |
| Nov 25, 2022 at 17:10 | history | answered | Jason DeVito - on hiatus | CC BY-SA 4.0 |