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

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

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

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.

• Thank you for your answer. I will check the details. Commented Mar 21, 2022 at 17:17
• 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? Commented Mar 21, 2022 at 17:46
• @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. Commented Mar 21, 2022 at 17:53
• 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. Commented Mar 21, 2022 at 18:13
• @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. Commented Mar 21, 2022 at 18:16