Timeline for (Non-)Rationality of a certain quotient of the symmetric square of the Fermat sextic (quartic) curve
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 3, 2023 at 15:36 | comment | added | Dimitri Koshelev | Perhaps, you are right. Thank you! | |
| Oct 3, 2023 at 12:51 | history | edited | Francesco Polizzi | CC BY-SA 4.0 |
added 546 characters in body
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| Oct 3, 2023 at 12:14 | comment | added | Francesco Polizzi | I mean, $$\tau = p^* \frac{dx_1}{y_1^3} \wedge q^*\frac{dx_2}{y_2^3},$$ where $(x_1, \, y_1)$ are the coordinates on the first factor and $(x_2, \, y_2)$ those in the second factor. | |
| Oct 3, 2023 at 12:05 | comment | added | Francesco Polizzi | Ok, now I will write down the correct involutions. However, it seems to me that, setting $\tau:=p^* \psi \wedge q^* \psi$, the tensor $\tau \otimes \tau$ is invariant for all your involutions (since $\tau$ itself is invariant for $i_x$ and $i_y$, whereas $\iota^* \tau = - \tau$), and so it provides a non-zero section of $K_S^{\otimes 2}$. Do you agree? | |
| Oct 3, 2023 at 11:59 | history | undeleted | Francesco Polizzi | ||
| Oct 3, 2023 at 11:42 | history | deleted | Francesco Polizzi | via Vote | |
| Oct 2, 2023 at 17:25 | comment | added | Dimitri Koshelev | With respect to my definition of $i$, the indicated $2$-form is not invariant, because $p^*\psi \wedge q^*\psi = -q^*\psi \wedge p^*\psi$. Hence, your argument of non-rationality of $S$ does not seem to work. | |
| Oct 2, 2023 at 17:09 | comment | added | Dimitri Koshelev | I am not sure that you consider the correct involutions. In my question there are the following ones: the diagonal actions $i_x \times i_x$, $i_y \times i_y$ and the permutation $i(P,Q) := (Q,P)$ for $P, Q \in F$. In turn, your definition of $i$ changes the coordinates on each copy of $F$ rather than the points. | |
| Oct 2, 2023 at 8:13 | history | answered | Francesco Polizzi | CC BY-SA 4.0 |