Timeline for Are Generalized Involution Varieties hyperlane sections of Generalized Severi Brauer Varieties?
Current License: CC BY-SA 3.0
9 events
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Nov 29, 2017 at 20:52 | comment | added | Jason Starr | No, I did not use any computer. | |
Nov 29, 2017 at 19:40 | comment | added | nxir | I was under the assumption that you used a computer, instead of doing anything by hand. | |
Nov 29, 2017 at 0:22 | comment | added | Jason Starr | I do not know understand your last comment. | |
Nov 28, 2017 at 23:40 | comment | added | nxir | I didnt have the time yet to do any calculations using computers. What did you use to figure that out? | |
Nov 28, 2017 at 10:38 | comment | added | Jason Starr | I tried to find other solutions of the two Diophantine equations from my comment above, and it looks to me like there are very few solutions (maybe none, beyond the case $(k,2m)=(1,6)$ and $(\ell,n)=(2,4)$). So that might settle your question. | |
Nov 27, 2017 at 2:27 | comment | added | Jason Starr | . . . For the dimensions to work out, you need that $k(4m-3k-1)/2$ equals $\ell(n-\ell)$, resp. equals $\ell(n-\ell) - 1$. Then you can also compute the dimension of the Lie algebra of infinitesimal automorphisms of both sides. For the orthogonal Grassmannian, it is the dimension of the special orthogonal group, $m(2m-1)$. For the Grassmannian, it is the dimension of the special linear group $n^2-1$. For a hyperplane section that happens to be homogeneous (so that $h^1$ of the tangent bundle equals $0$), the dimension is $n^2-\binom{n}{\ell}$. | |
Nov 27, 2017 at 1:42 | comment | added | Jason Starr | If you had an isomorphism of $\text{Inv}(B,\sigma)_{k,2m}$ and $\text{SB}(A)_{\ell,n}$, resp. a hyperplane section thereof, then after an etale extension of your original base, you would have an isomorphism between the orthogonal Grassmannian of isotropic $k$-dimensional subspaces in a vector space of dimension $2m$ and the Grassmannian of $\ell$-dimensional subspaces of an $n$-dimensional vector space, resp. a hyperplane section thereof. As Krashen points out, this is actually the case for $(k,2m)=(1,6)$ and $(\ell,n)=(2,4)$, i.e., $\text{Grass}(2,4)$ is a quadric fourfold . . . | |
Nov 26, 2017 at 23:45 | history | edited | nxir | CC BY-SA 3.0 |
deleted 9 characters in body
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Nov 26, 2017 at 22:32 | history | asked | nxir | CC BY-SA 3.0 |