1
$\begingroup$

Let $\overline{\mathbf{M}}_{0,n}$ be the moduli space of stable $n-$pointed smooth rational curve of genus zero and $\overline{\mathbf{U}}_{0,n}$ the universal family described by $\pi_n:\overline{\mathbf{U}}_{0,n}\longrightarrow\overline{\mathbf{M}}_{0,n}$ with the n disjoints sections $\sigma_i: \overline{\mathbf{M}}_{0,n}\longrightarrow \overline{\mathbf{U}}_{0,n}.$ Joachim Kock in "An invitatation to quantum cohomology" Example 1.5.11 gave the relationship between a boundary $\mathbf{F}_n$ of $\overline{\mathbf{M}}_{0,n}$ with the boundary $\mathbf{F}_{n+1}$ of $\overline{\mathbf{M}}_{0,n+1}$ by the following formula $$ \mathbf{F}_{n+1} = \varepsilon^*\mathbf{F}_n + \sum_{i}\sigma_i$$ where $\varepsilon: \overline{\mathbf{M}}_{0,n+1}\longrightarrow \overline{\mathbf{M}}_{0,n}$ is the forgetful maps. Can someone explain me more how to get the formula?

| cite | improve this question | | | | |
New contributor
Joseph is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
  • 1
    $\begingroup$ You write `stable n-pointed smooth'; I think you want to omit the word smooth. As far as answering the question, have you worked out some examples with small n? I think if you work out the cases $n=3$ and $n=4$ by hand you will see what is going on. I think this is more useful than me writing out an argument, but feel free to disagree :-). $\endgroup$ – David Holmes May 21 at 8:33

Your Answer

Joseph is a new contributor. Be nice, and check out our Code of Conduct.

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.