Timeline for Isomorphism between two K3 surfaces in characteristic $11$ and the action of $\operatorname{PSL}(2, \, \mathbb{F}_{11}) $
Current License: CC BY-SA 4.0
10 events
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| Jun 21 at 16:10 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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| Jun 16 at 21:15 | comment | added | Francesco Polizzi | This would be a nice starting point. | |
| Jun 16 at 20:26 | comment | added | Noam D. Elkies | If you "just" want to find the slice of the Néron-Severi lattice of X_0 that's orthogonal to the hyperplane section H, that must be easier to do than constructing an explicit algebraic isomorphism. It's also surprisingly routine to check whether two positive-definite lattices of rank 21 are isomorphic, and if so then to find an isomorphism. Would it be enough to find a divisor D class on X_0 such that D.D=4 and (NS(X_0),D) is isomorphic with (NS(Y),H) ? | |
| Jun 16 at 20:11 | comment | added | Francesco Polizzi | @NoamD.Elkies: I have a certain $21$-dimensional lattice that (I think) is associated with $X_0$, and I would like to interpret it as a lattice associated with $Y$. Apart from this, I am also genuinely curious about such an isomorphism. | |
| Jun 16 at 19:46 | comment | added | Noam D. Elkies | That's a nice question. Do you ask out of curiosity or do you have some application for such a map and an action of an 11-cycle (or of the whole group PSL(2,F_11)) on Y? I think I know how to find one, starting from an elliptic fibration of Y followed by a series of "2-neighbor steps" to get to the fibration X_0; but it would take a while and would be rather complicated -- perhaps unavoidably so -- written out as homog.polynomials t0,t1,x,y in the x_i. One could then recover an 11-cycle acting on Y by composing this map with an easy 11-cycle on X_0 with the inverse map taking X_0 back to Y. | |
| Jun 15 at 18:34 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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| Jun 15 at 18:33 | comment | added | Francesco Polizzi | I corrected the typo, thanks for pointing it out. The action of $\operatorname{PSL}(2, \, \mathbb{F}_{11})$ on $X_0$ is defined over a quadratic extension of $\mathbb{F}_{11}$, indeed. | |
| Jun 15 at 18:28 | comment | added | Noam D. Elkies | The quartic Fermat surface is not $x_0^2+x_1^2+x_2^2+x_3^2 = 0$ (a quadric, which is a rational surface in odd characteristic) but $x_0^4+x_1^4+x_2^4+x_3^4 = 0$. I'm not sure that there must be an isomorphism defined over any $k$ of characteristic $11$, but it should be enough to work over a quadratic extension of ${\bf Z} / 11 {\bf Z}$. | |
| Jun 15 at 18:11 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
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| Jun 14 at 16:33 | history | asked | Francesco Polizzi | CC BY-SA 4.0 |