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

Basic question, but I found no reference.

Is the $\psi$ class the only one which is not a boundary class in the PIcard group of the Deligne-Mumford compactification of $\mathcal{M}_{0,n}$? Or can it be expressed in terms of boundary divisors? If yes what is its expression?

share|cite|improve this question
up vote 4 down vote accepted

Unless you mean something else when you write $\psi$ class, it is expressible in terms of boundary divisors.

That is, if $\psi_i$ is the $i$-th cotangent bundle, then you can write it in terms of boundary divisors. One reference for this is the tome "Mirror Symmetry" by Hori, Katz, Klemm, et al. on p. 513, the comparison lemma.

The idea is that you can consider the forgetful maps $\pi$ from $M_{0,n}$ to $M_{0,n-1}$ and look at the divisor $$ \psi_i - \pi^*\psi_i $$ (where the $\psi$ classes are, by abuse of notation, living on different spaces). This is expressible in terms of boundary divisors, and so we can inductively write out the $\psi$ classes on any $M_{0,n}$.

share|cite|improve this answer
Thank you Simon, excellent and precise answer. – IMeasy Feb 7 '12 at 22:01

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.