# Isomorphic symmetric squares of non-isomorphic curves

Is it possible for the symmetric squares of a pair of non-isomorphic curves $$C_1, C_2$$ defined over a field $$K$$ to be isomorphic?

EDIT: as the user @abx has mentioned, there exist such examples in the case of genus 2 curves. I would like to add further restrictions which correspond to the special case I am interested in: what if $$C_1$$ and $$C_2$$ are both smooth plane quartic curves?

• Yes. There are many examples of pairs of non-isomorphic curves of genus 2 whose Jacobians are isomorphic. Therefore their symmetric squares are isomorphic, because they are obtained by blowing up one point in their Jacobian.
– abx
Sep 20 at 17:27
• What if I am particularly interested in projective plane quartics then? Is it still the case? Sep 20 at 17:32

If $$C_1$$ and $$C_2$$ are smooth plane quartics and $$f \colon S^2C_1 \to S^2C_2$$ is an isomorphism then it induces an isomorphism $$J(C_1) \cong \mathrm{Alb}(S^2C_1) \stackrel{\mathrm{Alb}(f)}\to \mathrm{Alb}(S^2C_2) \cong J(S^2C_2)$$ is an isomorphism that takes the theta-divisor of the first (equal to the image of $$S^2C_1$$ under the Albanese morphism) to the theta-divisor of the second (equal to the image of $$S^2C_2$$ under the Albanese morphism). Thus, we have an isomorphism of Jacobians compatible with principal polarizations, hence, by Torelli theorem, an isomorphism of the curves.
• I think that essentially the same argument works for hyperelliptic curves of genus 3 as well (using that $S^2C$ contains a unique rational curve in this case). Sep 20 at 18:54