In this answer it is said that $\bar{M}_{2,1}\cong \bar{M}_{0,7}/S_6$. However, I cannot see this. Given a curve of genus $2$ and a marked point, quotienting by the involution surely gives a rational curve with a divisor of degree $6$ together with a distinct marked point. However, going the other way, from a rational curve with a divisor of degree $6$ with a distinct rational point we get a genus $2$ curve, but we (generally) have two choices to lift the distinguished marking. In other words, that map looks more like a (ramified) double cover that an isomorphism.

Where is my reasoning wrong? Is there a way to fix this (either my wrong reasoning, or Jim's wrong reasoning)? The objective is still to find a presentation of $\bar{M}_{2,1}$, as similar as possible to the one proposed.

Note: I agree with his proof in the case $\bar{M}_{1,2}$.


1 Answer 1


Yes: there are generically two choices when lifting a point on the rational curve to the genus two curve. But the two choices are interchanged by the hyperelliptic involution, and Jim only claims an isomorphism of coarse moduli spaces.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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