Timeline for A characterisation of $\mathbb{P}^n$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| May 22, 2018 at 14:32 | comment | added | Jason Starr | I believe Tom Graber told me once that it is an open problem whether every finite group action on affine space (even over $\mathbb{C}$) is conjugate to a linear action via a conjugation that is a regular automorphism of affine space (not necessarily affine linear). | |
| May 22, 2018 at 14:28 | comment | added | Jason Starr | I do not know whether or not that is true. (I realize that my comment above suggested that I did -- sorry!) | |
| May 22, 2018 at 14:14 | comment | added | Will Sawin | @JasonStarr Do you know if the other approach (If $Y_{\overline{k}} = \mathbb A^n_{\overline{k}}$ then $Y_k = \mathbb A^n_k$) is true? I don't know how to compute Galois cohomology of an automorphism group that's so large. Edit: looks like an open problem. mathoverflow.net/questions | |
| May 22, 2018 at 14:08 | history | edited | Will Sawin | CC BY-SA 4.0 |
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| May 22, 2018 at 14:05 | comment | added | Jason Starr | Yes, the point-counting should work using the Grothendieck-Lefschetz fixed point formula and the fact that the compactly supported etale cohomology of affine space is concentrated in a single degree. | |
| May 22, 2018 at 14:02 | comment | added | Will Sawin | @JasonStarr Thanks! You are right. I somehow convinced myself that one only needs to pass to an etale cover of the base, which is not correct. Probably the cleanest way to deal with it is not to show isomorphism with $\mathbb A^n$ but to show the number of points is still $q^n$. | |
| May 22, 2018 at 13:56 | comment | added | Jason Starr | The issue in my comment also recurs in the second paragraph of your answer: how do we choose an isomorphism with $\mathbb{A}^n$ over the stratum? | |
| May 22, 2018 at 13:46 | comment | added | Jason Starr | I think that you are using cohomology somewhere (even if you are not directly computing the cohomology of $X$ or an associated incidence variety). You need to know that each hyperplane complement in $X$ that is defined over $k$ is $k$-isomorphic to affine space $\mathbb{A}^n_k$ (so that you can count its $k$-points). We are assuming that the hyperplane complement is geometrically isomorphic to affine space. I suspect the existence of a $k$-isomorphism uses cohomology. Anyway, this is a fantastic argument. | |
| May 22, 2018 at 13:42 | history | edited | Will Sawin | CC BY-SA 4.0 |
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| May 22, 2018 at 13:23 | history | answered | Will Sawin | CC BY-SA 4.0 |