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 $\dim V^- \geq 2$).