Looking for a curve with a special, free $\mathbb{Z}/2$-action

Question

I am looking for a smooth curve $C$ of genus $g=2k+1 \geq 5$ over the complex numbers, endowed with a free $\mathbb{Z}/2$-action such that the following condition is satisfied: denoting by $$H^0(C, \, \omega_C) = V^+ \oplus V^{-},$$ the decomposition of $H^0(C, \, \omega_C)$ into invariant and anti-invariant subspaces, there are bases $\{ \omega_1, \ldots, \omega_{k+1} \}$ of $V^+$ and $\{\omega_{k+2}, \ldots, \omega_g \}$ of $V^{-}$ such that the divisors $\mathrm{div}({\omega_i})$, $\mathrm{div}({\omega_j})$ have disjoint supports for all $1 \leq i < j \leq g$.

Question. Does such a curve exist?

What I have tried. Let me explain one example that does not work. I considered a hyperelliptic curve $C$ of affine equation $$y^2=(x^2-a_1^2)\ldots (x^2-a_{2g+2}^2),$$ where the scalars $a_i$ are general. Since $g$ is assumed to be odd, such a curve admits a free $\mathbb{Z}/2$-action given by $(x, \, y) \mapsto (-x, \, -y)$. A basis for $H^0(C, \, \omega_C)$ is given by $$\left \{ \omega_i \; \; | \; \; 0 \leq i \leq g-1 \right \},$$ where $\omega_i=x^i \frac{dx}{y}$. Therefore we have $$V^+=\{\omega_i \, | \, i\textrm{ even} \}, \quad V^-=\{\omega_i \, | \, i\textrm{ odd} \}.$$ This shows that all the $1$-forms in $V^-$ vanish at the points of the curve with $x=0$, hence the condition above cannot be satisfied when $g \geq 5$ (that means $\dim V^- \geq 2$).

Answer 1

Your examples are the only curves with fixed-point-free involution that don't have such a basis.

Let $C_0$ be a curve of genus $k+1$ and let $L$ be a nontrivial line bundle on $C_0$ with $L^2\cong \mathcal O_{C_0}$. Then $\mathcal O_{C_0}+L$ has a natural algebra structure defined using that isomorphism. The relative Spec is a double cover $C$ of genus $2k+1$. Since it's a double cover, it has a natural involution. All curves with involution arise this way.

This involution acts on $H^0(C, \omega_C)$. The invariant subspace consists of 1-forms that are equal on each of the two fibers of $C \to C_0$, hence are pullbacks from $C_0$, and thus is isomorphic to $ H^0(C_0 , \omega_{C_0})$, while the anti-invariant subspace consits of 1-forms with opposite values on the two fibers, hence locally pullbacks of 1-forms from $C_0$ multiplied by section of $L$, and thus is isomorphic to $H^0(C_0, \omega_{C_0}\otimes L)$.

For your criterion, it suffices that $\omega_{C_0}$ and $\omega_{C_0}\otimes L$ are both base-point free. Indeed, for a base-point free line bundle, a generic section avoids any given finite set of points, so it is easy to choose a basis where each section in the basis successively avoids the vanishing locus of the previous ones.

By Serre duality, $\omega_{C_0}$ is always base-point-free and $\omega_{C_0} \otimes L$ is base-point free if and only if the divisor class of $L$ can be expressed as the difference of two points of $C_0$.

If $P-Q$ is a two-torsion divisor, then $(2P, 2Q)$ are two divisors of degree $2$ equivalent to each other, thus defining a $g^1_2$, so $C_0$ is hyperelliptic. Even in the hyperelliptic case, this can only happen if $P$ and $Q$ are Weierestrass, so there are $\binom{ 2k+4}{2}$ two-torsion line bundles where such a basis does not exist on the cover, out of $2^{2k+2}-1$ nontrivial two-torsion line bundles total, so there are plenty of line bundles that do work for any $k \geq 2$. For the ones that don't, the cover can be obtained from the double cover of $\mathbb P^1$ branched at the $x$ coordinates of $P$ and $Q$, meaning that, after a change of variables, it's given by your construction.