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We work over a field $k$ with $\operatorname{char}(k)=11$.

In the paper [1], Lemma 3.5, it is shown that the K3 surface $X_0$ defined as the weighted projective hypersurface of degree $12$ $$X_0=V(t_0^{12}+t_1^{12}+x^3+y^2) \subset \mathbb{P}(1, \, 1, \, 4, \, 6)$$ admits a symplectic action of a finite group of order 660, namely $L_2(11) \cong \operatorname{PSL}(2, \, \mathbb{F}_{11})$. Moreover, such an action also preserves the elliptic fibration $X_0 \to \mathbb{P}^1$ given by $[t_0:t_1:x:y] \mapsto [t_0:t_1]$.

In Remark 3.6 of the same paper it is claimed that, since $X_0$ is supersingular with Artin invariant $1$, then it must be isomorphic to the quartic Fermat hypersurface $$Y=V(x_0^4+x_1^4+x_2^4+x_3^4) \subset \mathbb{P}^3.$$

Questions.

  1. Is an explicit isomorphism between $X_0$ and $Y$ known? If so, what is its expression?
  2. What is an explicit symplectic action of $\operatorname{PSL}(2, \, \mathbb{F}_{11})$ on the Fermat hypersurface $Y$? For instance, how to write down explicitly a symplectic automorphism of $Y$ having order $11$?

Any answer or reference to the existing literature will be appreciated.

References

[1] Dolgachev, Igor V.; Keum, Jonghae, (K3) surfaces with a symplectic automorphism of order 11, J. Eur. Math. Soc. (JEMS) 11, No. 4, 799-818 (2009). ZBL1185.14035.

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    $\begingroup$ 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 ${\bf Z} / 11 {\bf Z}$. $\endgroup$
    – Noam D. Elkies
    Commented Jun 15 at 18:28
  • $\begingroup$ 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. $\endgroup$
    – Francesco Polizzi
    Commented Jun 15 at 18:33
  • $\begingroup$ 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. $\endgroup$
    – Noam D. Elkies
    Commented Jun 16 at 19:46
  • $\begingroup$ @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. $\endgroup$
    – Francesco Polizzi
    Commented Jun 16 at 20:11
  • $\begingroup$ 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) ? $\endgroup$
    – Noam D. Elkies
    Commented Jun 16 at 20:26

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