MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For a hyperelliptic curve H, the mod 2 monodromy is smaller than $GSp_{2g}(F_2)$ -- since the two torsion of the Jacobian H is generated by differences of Weierstrass point, the monodromy of a generic curve is the symmetric group.

The degree 3 map $C \to \mathbb{P}^1$ associated to a trigonal curve may only have ramification of type (2,1), and thus may not give any "easy to see" 3 torsion points on the Jacobian. So it is reasonable to ask if a generic such curve has maximal mod l monodromy for every l.

share|cite|improve this question

Set $\tilde{C}:=\overline{C\times_{\mathbb{P}^1}C\smallsetminus\mathrm{diagonal}}$, then $J\tilde{C}$ is a quotient of $JC\times JC$, and $JC$ maps into $J\tilde{C}$, where the kernel is a subgroup of the $2$-torsion points. Moreover, $J\tilde{C}/\mathbb{P}^1$ is Galois (with Galois group $S_3$). Now let $C'$ be the unique intermediate cover of $\tilde{C}/\mathbb{P}^1$ which is degree $2$ over the $\mathbb{P}^1$, then $\tilde{C}/C'$ is a cyclic degree $3$ cover, so you get some $3$ torsion data.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.