Involution of the symmetric square of a smooth plane quartic

Question

Let $C$ be a smooth plane quartic defined over a field $K$. Denote by $J$ its Jacobian, and by $C^{(2)}$ its symmetric square. Since $C$ is a smooth plane quartic, it is non-hyperelliptic, and hence the morphism $C^{(2)}\longrightarrow J$ is an embedding.

The symmetric square of $C$ has an involution defined as follows: let $(P,Q)\in C^{(2)}(K)$, let $L_{P,Q}$ be the line connecting $P$ and $Q$ in $\mathbb{P}^2_K$, and the tangent line at $P$ if $P=Q$. The scheme theoretic intersection of $L_{P,Q}$ with $C$ has multiplicity $4$, and hence there is a degree $2$ effective divisor $(S,T)\in C^{(2)}(K)$ such that $(L) = (P) + (Q) + (S) + (T)$. We define the involution $(P,Q)\mapsto (S,T)\in C^{(2)}(K)$.

My questions are:

1. Is the involution $C^{(2)}\longrightarrow C^{(2)}$, defined above, algebraic?

2. The Jacobian $J$ has a canonical involution coming from multiplication by $(-1)$. Under what circumstances does the involution defined above coincide with multiplication by $(-1)$. I know for example this would be the case if the curve $C$ has a flex, but is it necessary?

Answer 1

Pic(C), the group of divisor classes on C, has two distinguished elements, 0 and K, hence two natural involutions: E—> - E, and E—>K-E, where K is the canonical divisor class. The first involution leaves invariant the subgroup Pic(0) = the Jacobian variety of classes of degree zero, while the second leaves invariant the coset Pic(g-1). The second involution also leaves invariant (by Riemann - Roch) the image W(g-1) of the symmetric product Sym^(g-1)(C) in Pic(g-1). Both these involutions are algebraic.

Since a smooth plane quartic is a non - hyperelliptic curve of genus 3, on which a canonical divisor is cut by any line L in the plane, C^(2) imbeds isomorphically onto W(2) in Pic(2), and your construction is an example of the second involution; in particular it is algebraic. I.e. your involution is the restriction of E—> K-E to W(2) ≈ C^(2).

As Will Sawin and Sasha pointed out, if D is any divisor class of degree g-1, effective or not, then translation by D takes the second involution on Pic(g-1) to some involution on Pic(0), and it takes it to the first involution, E—> - E, if and only if 2D = K, if and only if D is a fixed point of the second involution, E—>K-E.

Indeed Riemann associated to any canonical homology basis of C, a “theta function” on Pic(0), i.e. an even function whose zero locus, the “theta divisor”, is thus invariant under the first involution E—> - E. (Actually the analytic theta function itself is defined only on the universal cover of Pic(0), but its zero locus is periodic there, hence defines a divisor on Pic(0).) Riemann’s famous theorem says that such a homology basis determines also a specific divisor class D, with 2D = K, called a “theta characteristic”, which translates W(g-1) isomorphically onto the theta divisor, carrying the involution E —> K-E to the involution E—> -E.

Since Pic is a complete variety, and the map E—>2E on Pic(C) has finite fibers, there always exist divisor classes D with 2D = K, in fact exactly 2^(2g) of them, and (I believe) Riemann showed that 2^(g-1).(2^g -1) of them are "odd", i.e. have an odd number of sections, in particular these are effective. In your case this implies 28 effective divisors D exist with 2D = K = L, where D = the pair of points of contact of one of the 28 bitangents. The 2^(g-1).(2^g +1) even theta characteristics are usually not effective, i.e. usually have zero sections, as is the case for your non hyperelliptic curve of genus 3.

In fact it seems that even in the hyperelliptic case, in genus 3, the involution E-->K-E on W(2) lifts to an involution on C^(2). I.e. if I:C-->C is the hyperelliptic involution, then sending {p,q} to {I(p),I(q)} is an involution J of C^(2) such that again [J(p+q)] = [I(p)+I(q)] = K-[p+q].

This generalizes your construction, since here the canonical map f is 2:1 from C to a plane conic, and the involution on C^(2) takes p+q to the complement of p+q, in the inverse image on C of the line in the plane spanned by f(p) and f(q). Here the involution fixes not only the 28 pairs of points {p,q} which are branch points of the canonical map, but also the pencil of pairs {p,q} whose images are equal on the conic, i.e. such that h^0(p+q) = 2.

Answer 2

The morphism $C^{(2)}\to J$ is not well-defined. You have to pick a degree $2$ divisor $D$ class on $C$ to define the morphism. It then sends a pair $(P,Q)$ to $(P)+(Q)-D$.

Since $(S)+(T)-D$ is $(P)+ (Q)+(S)+(T)- D - (P)-(Q)= (L)-D-(P)-(Q)$, this involution, viewed as a map from the image of $C^{(2)}$ inside $J$ to itself, sends $(P)+(Q)-D$ to $(L)-D-(P)-(Q)$. Hence it is coincides with the map from $J$ to itself that sends a divisor class $E$ to $(L)-2D-E$. In particular, it coincides with the map $-1$ if and only if $2D=(L)$.

So it is not really relevant whether the curve has a flex. If we choose $D$ to be twice a particular point, then this happens if and only if that particular point is a flex. But the choice of $D$ is what matters, rather than any point.