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

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    $\begingroup$ Could we please call that group Z/2 or Z/2Z or $\{ \pm 1\}$ instead of Z_2 ? $\endgroup$ Commented Mar 21, 2022 at 15:59
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    $\begingroup$ @NoamD.Elkies: I changed the notation (which is however very common in Algebra textbooks) $\endgroup$ Commented Mar 21, 2022 at 16:11

1 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.

  • $\begingroup$ Thank you for your answer. I will check the details. $\endgroup$ Commented Mar 21, 2022 at 17:17
  • $\begingroup$ It seems to me that your argument uses that the involution is base-point free only when you say that $L^2= \mathcal{O}_C$. In the same vein, we could take a divisor $B$ on $C_0$ such that $L^2=\mathcal{O}_{C_0}(B)$, obtaining a double cover $C \to C_0$ branched at $B$. Then, if both $\omega_{C_0}$ and $\omega_{C_0} \otimes L$ are base-point free, we get a smooth curve $C$ with a non-free $\mathbb{Z}/2$-action and satisfying my condition. In fact, in this case, things are easier, since the degree of $\omega_{C_0} \otimes L$ is greater than the degree of $\omega_{C_0}$. Is it correct? $\endgroup$ Commented Mar 21, 2022 at 17:46
  • $\begingroup$ @FrancescoPolizzi Not quite. In this case $H^0(C_0,\omega_{C_0})$ is base-point-free but viewed as a linear system inside $H^0(C,\omega_C)$ has a base point, since when you pull back any 1-form it will vanish at the base points. In the case without branched points, this can't happen. $\endgroup$
    – Will Sawin
    Commented Mar 21, 2022 at 17:53
  • $\begingroup$ So, it seems to me that you are saying that in the branched case the condition cannot be satisfied, unless $g(C_0) \leq 1$ (namely, unless $\dim V^+ \leq 1$). In fact, the linear system corresponding to $V^+$ has always base points at the fixed locus of the involution $f \colon C \to C_0$, since $$\f^*\omega_p(v)=\omega_{f(p)}(df_p(v))$$ and $df_p$ is zero at the ramification points. $\endgroup$ Commented Mar 21, 2022 at 18:13
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    $\begingroup$ @FrancescoPolizzi Yes, exactly. Hyperelliptic curves clearly have $V^-$ base-point free, and branched hyperelliptic curves with two or more branch points do, but branched hyperelliptic curves with two branch points don't, since then $\omega_C \otimes L = L$ has degree $1$ and thus always (on an elliptic curve) has a base point. $\endgroup$
    – Will Sawin
    Commented Mar 21, 2022 at 18:16

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