(Non-)Rationality of a certain quotient of the symmetric square of the Fermat sextic (quartic) curve

Question

Consider the Fermat sextic curve $F: x^6 + y^6 + 1 = 0$ over an algebraically closed field of characteristic $0$. It has the two order $3$ automorphisms $\omega_x(x,y) := (\omega x, y)$ and $\omega_y(x,y) := (x, \omega y)$, where $\omega$ is a primitive cubic root of unity. Obviously, the quotient $F/\langle \omega_x, \omega_y \rangle$ is the diagonal conic $x^2 + y^2 + 1$, i.e., this quotient is a rational curve. Further, consider the symmetric square $F^{(2)}$ of the curve $F$. The automorphisms $\omega_x$, $\omega_y$ also act on $F^{(2)}$ in a natural way. Finally, we come to the quotient surface $S := F^{(2)}/\langle \omega_x, \omega_y \rangle$.

Is there a (simple) way to determine if $S$ is a rational surface or not? It is reasonale to apply Castelnuolo's rationality criterion, but I don't know how to compute the irregularity $q$ and second plurigenus $P_2$ of $S$. Is it necessary to construct explicit equation(s) of $S$ to answer my question?

I am also interested in a similar simpler question when $F\!: x^4 + y^4 + 1 = 0$ is the Fermat quartic. Instead of $\omega_x$, $\omega_y$ we have the two involutions $i_x(x,y) := (-x,y)$ and $i_y(x,y) := (x,-y)$. As above, the quotient $F/\langle i_x, i_y \rangle$ is the diagonal conic. What about (non-)rationality of the quotient surface $S := F^{(2)}/\langle i_x, i_y \rangle$?

Thank you in advance for your help!

Answer 1

Let me consider the case of the Fermat quartic. Since you are talking about Castelnuovo's criterion, I guess you are interested in the projective situation.

The $1$-form $$\psi \colon =\frac{dx}{y^3}=-\frac{dy}{x^3}$$ is holomorphic on the projective curve $F$, so the $2$-form $$\tau:=p^* \psi \wedge q^* \psi = p^* \frac{dx_1}{y_1^3} \wedge q^* \frac{dx_2}{y_2^3}$$ (where $(x_1, \, y_1)$ are the coordinates on the first factor, $(x_2, \, y_2)$ those on the second factor and $p, \, q$ are the natural projections) is holomorphic on $F \times F$.

One easily checks that $\tau$ is invariant by the two involutions $i_1, \, i_2$ of $F \times F$ defined by $$i_1:=i_x \times i_x \colon ((x_1,\, y_1), \, (x_2, \, y_2)) \mapsto ((-x_1,\, y_1), \, (-x_2, \, y_2)),$$ $$i_2:=i_y \times i_y \colon ((x_1,\, y_1), \, (x_2, \, y_2)) \mapsto ((x_1,\, -y_1), \, (x_2, \, -y_2)).$$ Moreover, denoting by $\iota$ the involution switching the two factors, namely $$\iota \colon ((x_1,\, y_1), \, (x_2, \, y_2)) \mapsto ((x_2,\, y_2), \, (x_1, \, y_1)),$$ we have $\iota^* \tau = - \tau$.

Thus, the tensor $\tau \otimes \tau$ is invariant by the action of the subgroup $G \subset \operatorname{Aut}(F \times F)$ given by $G = \langle i_1, \, i_2, \, \iota\rangle$, hence it descends to a non-zero, global holomorphic tensor on $S=(F \times F)/G$.

This shows that $h^0(S, \, K_S^{\otimes 2}) \geq 1$, in particular $S$ is not rational.