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?

  • 4
    $\begingroup$ 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. $\endgroup$
    – abx
    Sep 20 at 17:27
  • $\begingroup$ What if I am particularly interested in projective plane quartics then? Is it still the case? $\endgroup$
    – kindasorta
    Sep 20 at 17:32

1 Answer 1


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.

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    $\begingroup$ 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). $\endgroup$ Sep 20 at 18:54
  • 1
    $\begingroup$ A similar result for curves of arbitrary genus was proved by Martens in the 60's. See "An extended Torelli theorem", Amer. J. Math, 87 (1965), pp. 257–261. There is also a more recent result of Ciliberto and Sernesi for symmetric products of degree greater than the genus.) $\endgroup$
    – naf
    Sep 21 at 2:03

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