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Sep 5, 2020 at 11:17 vote accept CommunityBot
Sep 5, 2020 at 11:15 history edited user164740 CC BY-SA 4.0
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Sep 5, 2020 at 9:18 history edited user164740
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Sep 5, 2020 at 3:32 history edited C.F.G
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Sep 5, 2020 at 0:21 answer added Robert Bryant timeline score: 15
Sep 3, 2020 at 22:13 comment added Qfwfq (I edited a bit incorporating the remark of abx in the OP's question)
Sep 3, 2020 at 22:12 history edited Qfwfq CC BY-SA 4.0
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Sep 3, 2020 at 21:53 comment added Jason DeVito - on hiatus @Qfwfq: If rotations by $\pi/2$ at every point are isometries, then so are rotations by $\pi$. Thus, the metric on $S^2$ exhibits $S^2$ as a symmetric space. This implies the metric on $S^2$ is isometric (up to scaling) to the usual round metric. So, no, there is no such metric distinct from the $FS$ metric.
Sep 3, 2020 at 21:46 history edited user164740
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Sep 3, 2020 at 21:18 history edited user164740 CC BY-SA 4.0
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Sep 3, 2020 at 21:07 comment added Qfwfq @alglearner: the metric on the sphere would have to be invariant under rotations of $\pi/2$ at every point (see comments to my wrong answer). Is there such a metric on the $2$-sphere, distinct from FS?
Sep 3, 2020 at 21:03 comment added aglearner Take $\mathbb CP^1$ with Fubini-Studi metric of area $1$, and take any other Kaher metric on $\mathbb CP^1$ which has area $1$ but has non-constant curvature. Then $\psi$ doesn't exist, because the manfiolds are not isometric and your condition implies that $\psi$ is an isomery. But they are both symplectomorphic and obviously biholomorphic.
Sep 3, 2020 at 21:00 history edited user164740 CC BY-SA 4.0
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Sep 3, 2020 at 20:04 history edited user164740 CC BY-SA 4.0
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Sep 3, 2020 at 19:56 comment added abx On a Hermitian manifold, the symplectic and the complex structure determine the metric.
Sep 3, 2020 at 19:28 comment added LeechLattice @Qfwfq It probably means an isometry that preserves the complex and the symplectic structure at the same time.
Sep 3, 2020 at 19:27 comment added Qfwfq What do you mean by "isomorphic" Kaehler manifolds? Is isometry is implied in your definition of isomorphism?
Sep 3, 2020 at 19:05 history asked user164740 CC BY-SA 4.0