Timeline for Fixed locus of a Kahler $S^1$-action
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 12, 2022 at 13:48 | comment | added | Filip | Hmm, maybe it still works: given a non-trivial element $t \in S^1,$ the spectrum of $t_y$ consists of a discrete set $\{t^{w_i} \mid w_i \text{ are weights on }y\}$ so one cannot "jump" from one set of weights to an another, while on a path from $y$ to $y'.$ Thus I would still think that the weight decomposition is uniform on a path component $F_\alpha,$ even in the case of smooth actions. And possibly even compactness of $F_\alpha$ is not important. Then the eigenspaces $E_k=ker(t^k \cdot I - t_y)$ are smooth in $y$ hence constitute bundles, for each $k$ that appears as a weight. | |
| Jul 12, 2022 at 13:14 | comment | added | abx | Yes, the argument definitely uses the holomorphicity, and also the compactness. I would guess that the result does not hold in the smooth category, though I don't see an obvious counter-example. | |
| Jul 11, 2022 at 20:42 | comment | added | Filip | Does this decomposition of the tangent bundle generalises to a compact Lie group action on a smooth manifold? The decomposition of a tangent space according to characters is just representation theory so generalises easily, but the continuation argument from the answer above uses the holomorphicity... | |
| Jun 27, 2020 at 9:57 | history | edited | abx | CC BY-SA 4.0 |
added 21 characters in body
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| Jun 27, 2020 at 9:21 | vote | accept | Filip | ||
| Jun 27, 2020 at 6:54 | history | answered | abx | CC BY-SA 4.0 |