Timeline for Number of fixed points of an involution
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Dec 20, 2019 at 13:57 | vote | accept | abx | ||
| Dec 16, 2019 at 6:54 | answer | added | abx | timeline score: 9 | |
| Dec 14, 2019 at 18:34 | comment | added | abx | @EBz: $X=(\mathbb{P}^1)^n$, $\sigma =(\sigma _1,\ldots ,\sigma _n)$, where $\sigma _i$ is a nontrivial involution of $\mathbb{P}^1$. | |
| Dec 14, 2019 at 10:09 | answer | added | user108998 | timeline score: 1 | |
| Dec 14, 2019 at 8:51 | comment | added | user108998 | Can anyone find examples realizing this bound of 2^dim? 2^2dim is easy to construct (abelian varieties) | |
| Dec 13, 2019 at 18:31 | history | edited | Olivier Benoist | CC BY-SA 4.0 |
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| Dec 12, 2019 at 20:01 | comment | added | abx | Yes, that's right, great! I should have thought of that... If you want to write it as an answer I will accept it. | |
| Dec 12, 2019 at 19:33 | comment | added | SashaP | Doesn't the holomorphic Lefschetz fixed point formula imply that the number of fixed points has to be divisible by $2^{\dim X}$? On one hand, $\sum (-1)^i Tr(\sigma:H^i(O))$ is an integer because eigenvalues of an involution are $\pm 1$. On the other hand, for an isolated fixed point all the eigenvalues of the involution on the tangent space have to be equal to $-1$, so the local contribution of each fixed point is $1/\det(1-d\sigma)=1/2^{\dim X}$. | |
| Dec 12, 2019 at 17:46 | history | became hot network question | |||
| Dec 12, 2019 at 17:25 | comment | added | abx | @Nick L: Right, thanks. In fact, in the case of a holomorphic involution (or automorphism of finite order) isolated fixed points are always non-degenerate. | |
| Dec 12, 2019 at 16:42 | comment | added | Nick L | A very partial remark. A Kahler manifold with complex dimension $2n+1$ has even topological Euler characteristic by Hodge theory. So any diffeomorphism with non-degenerate fixed points has an even number of fixed points by the Lefschetz-Hopf formula (for an involution the Lefschetz number is equal to the Euler characteristic mod $2$). | |
| Dec 12, 2019 at 11:00 | comment | added | abx | Yes, thanks! Of course this is what I had in mind, but I forgot to mention it. Edited. | |
| Dec 12, 2019 at 11:00 | history | edited | abx | CC BY-SA 4.0 |
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| Dec 12, 2019 at 10:29 | answer | added | Francesco Polizzi | timeline score: 7 | |
| Dec 12, 2019 at 10:28 | comment | added | Francesco Polizzi | By the way, you must assume $X$ irreducible, otherwise one can consider an involution over three points fixing just one of them. | |
| Dec 12, 2019 at 10:07 | comment | added | Francesco Polizzi | $\mathbb{A}^1$ is not projective | |
| Dec 12, 2019 at 10:05 | comment | added | Jef | This seems false: the involution $x \mapsto -x$ on $\mathbb{A}^1$ has exactly one fixed point. | |
| Dec 12, 2019 at 9:42 | history | asked | abx | CC BY-SA 4.0 |