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Jul 13, 2022 at 11:24 comment added Jason Starr "Is there a way to tell which hypersurfaces in $\mathbb{P}^n$ have such an involution?" For all $n\geq 4$, there is no such hypersurface. For $n\geq 4$, the restriction map on Picard groups from $\mathbb{P}^n$ to the hypersurface is an isomorphism. Thus, every automorphism extends to $\mathbb{P}^n$. The fixed locus of the involution of $\mathbb{P}^n$ equals two linear subspaces, at least one of which has dimension $>1$. The intersection of this linear subspace with the hypersurface is nonempty.
Jul 12, 2022 at 19:56 answer added abx timeline score: 7
Jul 12, 2022 at 19:26 history edited Mohammad Farajzadeh-Tehrani CC BY-SA 4.0
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Jul 12, 2022 at 14:16 comment added Mohammad Farajzadeh-Tehrani Thanks, Jason. That's helpful. Is there a way to tell which hypersurfaces in $\mathbb{P}^n$ have such an involution? (or better said, for what $d\geq 0$, there is a degree $d$ hypersurface in $\mathbb{P}^n$ admitting such an involution.
Jul 12, 2022 at 2:16 history became hot network question
Jul 11, 2022 at 20:45 answer added Francesco Polizzi timeline score: 4
Jul 11, 2022 at 20:44 comment added Sasha In the opposite direction --- if $Y$ is a variety such that $\pi_1(Y)$ has a surjection onto $\mathbb{Z}/2$, the corresponding etale double covering $X$ of $Y$ has such an involution.
Jul 11, 2022 at 18:59 comment added Moishe Kohan Many compact ball quotients admit fixed-point free holomorphic involutions.
Jul 11, 2022 at 18:40 comment added Jason Starr I looked up that comment. There is no fixed-point-free involution acting on a hyper-Kaehler manifold whose (complex) dimension is divisible by $4$, e.g., there are such actions on K3 surfaces (with quotient an Enriques surface), but not on $4$-dimensional hyper-Kaehler manifolds.
Jul 11, 2022 at 18:33 comment added Jason Starr Given one such variety $X$, for every variety $Y$, the product $X\times Y$ is another such variety. Starting from $X$ a curve of genus $g>0$, you can produce examples in every dimension $n$, e.g., $X\times \mathbb{P}^{n-1}$. I wrote a bit more about this once in a comment to an MO post by user abx . . .
Jul 11, 2022 at 18:13 history asked Mohammad Farajzadeh-Tehrani CC BY-SA 4.0