$\psi$ class in $\overline{M}_{0,n}$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T06:27:26Z http://mathoverflow.net/feeds/question/87669 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87669/psi-class-in-overlinem-0-n $\psi$ class in $\overline{M}_{0,n}$ IMeasy 2012-02-06T14:24:19Z 2012-02-06T17:50:59Z <p>Basic question, but I found no reference.</p> <p>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?</p> http://mathoverflow.net/questions/87669/psi-class-in-overlinem-0-n/87696#87696 Answer by Simon Rose for $\psi$ class in $\overline{M}_{0,n}$ Simon Rose 2012-02-06T17:50:59Z 2012-02-06T17:50:59Z <p>Unless you mean something else when you write $\psi$ class, it is expressible in terms of boundary divisors.</p> <p>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.</p> <p>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 $M_{0,n}$.</p>