# Is the theta characteristic attached to an etale double cover of a plane quintic arising from a cubic threefold even in characteristic two?

We work over an algebraically closed field $$k$$ of characteristic $$2$$. Let $$X$$ be a cubic threefold realized as a conic bundle via $$f: X \to \mathbb{P}^2$$ (after blowing up some line). Let $$C \to \mathbb{P}^2$$ be a plane quintic curve describing the locus of degenerate conics of $$f$$, and let $$\pi: \widetilde{C} \to C$$ be the etale double cover parametrizing the lines on these degenerate conics. Let $$H$$ be the hyperplane section on $$C$$, and consider $$M = \pi^*H$$. I need to work out the dimension $$h^0(M)$$. If $$k$$ had characteristic not $$2$$ then we know $$h^0(M) = 4$$, since $$h^0(M) = h^0(\pi_*M) = h^0(H) + h^0(H + \eta)$$ where $$\eta$$ is some line bundle satisfying $$\eta^2 \cong \mathcal{O}_C$$, and we know $$h^0(H) = 3$$, and $$h^0(H + \eta) = 1$$ as in Beauville's paper on singularities of the Theta divisor.

However this approach does not work in characteristic $$2$$. Here we simply get a short exact sequence $$0 \to \mathcal{O}_C \to \pi_*\mathcal{O}_{\widetilde{C}} \to \mathcal{O}_C \to 0$$ because in characteristic $$2$$ we deal with Artin-Schreier extensions. After twisting by $$H$$ and taking long exact sequence, this bounds $$3 \leq h^0(M) \leq 6$$ for us. It would suffice in my case to know that this quantity is even, so essentially I want to ensure that even in characteristic $$2$$, the theta divisor is $$\textbf{even}$$.

I know that $$det(\pi_*M) = 2H$$ has $$h^0(2H) = 11$$ since $$C$$ is genus $$6$$ and $$2H$$ is the canonical divisor, but I am not sure if this is of any use. I also tried to manually compute the global sections from cocycle data but that seems like a nightmare.

• Even in characteristic 0, $h^0(M)$ depends on the choice of $\pi$ — it can be 3 or 4.
– abx
Commented Jul 26, 2022 at 18:37
• It seems likely that the possible values in characteristic $2$ are also $3$ or $4$, with $3$ generic. Commented Jul 27, 2022 at 11:47
• @abx I edited the question. I am interested in the case where $\pi$ comes from a conic bundle structure on a cubic threefold. I would probably have to use the geometry of the cubic threefold then. Commented Jul 27, 2022 at 12:11
• @WillSawin How would one produce an upper bound of $4$ in characteristic $2$? I think that would be enough to solve all my problems. In this case it doesn't hold that $\pi_*\pi^*H = H + H(\eta)$ Commented Aug 1, 2022 at 10:16

I'll show the rank is at most 4 in characteristic 2, modulo some claims that I think can be justified.

$$0 \to \mathcal{O}_C \to \pi_*\mathcal{O}_{\widetilde{C}} \to \mathcal{O}_C \to 0$$

induces a long exact sequence on cohomology

$$0 \to H^0(C, \mathcal{O}_C(1) ) \to H^0(C, \pi_*\mathcal{O}_{\widetilde{C}}(1) )\to H^0(C, \mathcal{O}_C(1) ) \to H^1(C ,\mathcal O_C(1))$$

and it suffices to show that the rank of $$H^0(C, \mathcal{O}_C(1) ) \to H^1(C ,\mathcal O_C(1))$$ is at least $$2$$.

It's not hard to check that this map is given by cup product with the extension class of the extension $$\pi_* \mathcal O_{\tilde{C}}$$ of $$\mathcal O_C$$ with $$\mathcal O_C$$ in $$\operatorname{Ext}^1(C,\mathcal O_C) = H^1 ( C, \mathcal O_C) = H^0 (C, K_C)^\vee$$.

Now by adjunction, $$K_C = \mathcal O_C(2)$$ and $$H^0(C, K_C) = H^0(C,\mathcal O_C(2) ) = H^0( \mathbb P^2, \mathcal O_{\mathbb P^2}(2))$$ is the space of quadratic polynomials in three variables. So the extension class is a linear form on quadratic polynomials in three variables, i.e. a three-by-three symmetric matrix.

The cup product $$H^0(C, \mathcal O_C(1)) \times H^0 (C, K_C)^\vee \to H^1(C,\mathcal O_C(1)) = H^0(C, K_C(-1))$$ is dual to the product map $$H^0(C,\mathcal O_C(1)) \times H^0(C, K_C(-1)) \to H^0(C,K_C)$$ so the rank of the cup product map is equal to the rank of this symmetric matrix. So it suffices to prove the symmetric matrix can't have rank $$\leq 1$$.

The symmetric matrix is certainly nonzero as the extension is nontrivial, so it suffices to prove it can't have rank exactly $$1$$. By $$GL_3$$-symmetry, we can rule out a specific $$3\times 3$$ matrix, i.e. $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$, that corresponds to the linear form on quadratic polynomials that extracts the coefficient of $$x^2$$.

What do we know about this extension class? By the Artin-Schreier exact sequence $$H^1(C, \mathbb F_2) \to H^1(C,\mathcal O_C) \to H^1(C, \mathcal O_C)$$, it must be stable under the Frobenius map $$H^1(C,\mathcal O_C) \to H^1(C,\mathcal O_C)$$, which is dual to the Verschiebung map $$H^0(C, K_C) \to H^0(C,K_C)$$.

How does Verschiebung act on quadratic polynomials in $$x,y,z$$? I claim it acts by multiplying by the degree $$5$$ polynomial $$f$$ defining your curve, then ignoring all monomials where the exponent of $$x,y$$, or $$z$$ is even, then dividing by $$xyz$$, so all the exponents are even, and then taking the square root. This sends a degree $$2$$ polynomial to a degree $$\frac{ 2+5 - 3}{2} =2$$ polynomial.

If the linear form that extracts the coefficient of $$x^2$$ is stable under Frobenius, then the coefficient of $$x^4 \cdot xyz$$ in $$fq$$ must match the coefficient of $$x^2$$ in $$q$$ for an arbitrary quadratic polynomial $$q$$. This can only happen if the coefficients of $$x^5, x^4y , x^4z$$ in $$f$$ all vanish, since otherwise we could multiply by $$yz, xz, xy$$ respectively and get a contradiction. But this forces the vanishing locus of $$f$$ to have a singularity at $$(1:0:0)$$, contradicting the assumption that it is smooth.

So indeed the rank of the matrix is $$\geq 2$$, making the rank of global sections $$\leq 4$$.

• This looks fantastic! However I am still not fully sure why Verschiebung acts as you say. Could you say a little more or provide a source for how one might figure this out? Thanks! Commented Aug 4, 2022 at 14:02
• @TCiur Equation (2.5) of people.maths.bris.ac.uk/~jb12407/ANTS-XV/papers/… (counting points on smooth plane quartics by Edgar Costa, David Harvey, and Andrew v. Sutherland), noting that the differential they call $\omega_{k\ell}$ corresponds to the monomial $x^{k-1} y^{\ell-1} z^{ \deg F - k - \ell -1 }$ as a global section of $H^0(\mathbb P^2, \deg F-2)$, and the Verschiebung is another name for the Cartier operator on differential forms. Commented Aug 4, 2022 at 14:43
• Thank you, this is perfect! Commented Aug 5, 2022 at 11:48