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?
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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 HolmesMay 21, 2020 at 8:33
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$\begingroup$ Yeah you are right, I should remove the word "smooth" In fact I just listed all the divisors for n=3,4,5 and 6 of $\overline{\mathbf{M}}_{0,n}.$ $\endgroup$– JosephMay 26, 2020 at 21:45
1 Answer
Here are some examples.
- Consider $\epsilon: \overline{M}_{0,4} \to \overline{M}_{0,3}$. Since $\overline{M}_{0,3}$ is a point, $F_3 = 0$. On the other hand, the boundary of $\overline{M}_{0,4}$ consists of three points $0,1,\infty$. These are exactly the images of the three sections.
- Consider $\epsilon : \overline{M}_{0,5} \to \overline{M}_{0,4}$ forgetting the $5$th marked point. The boundary of $\overline{M}_{0,5}$ consists of ten divisors $D_{ij,kmn}$. The divisors $D_{i5,jkl}$ are the images of the four sections of $\epsilon$. The remaining 6 divisors look like $D_{ij,kl5}$. Forgetting the marked point $5$, each of these divisors is a stable 4-pointed curve $D_{ij,kl}$, which is exactly a boundary point of $\overline{M}_{0,4}$. So the pullback of the boundary of $\overline{M}_{0,4}$ consists of these 6 divisors. This gives the formula in this case.
The general case is basically the same as the $\overline{M}_{0,5}$ case. All the divisors on $\overline{M}_{0,n}$ look like $D_{I,J}$ where $I \cup J$ is a partition of $\{1,\ldots,n\}$ with $\lvert I \lvert, \lvert J \lvert \geq 2$. The divisors where $I = \{i,n\}$ are the images of the $n-1$ sections of $\overline{M}_{0,n} \to \overline{M}_{0,n-1}$. The images of the remaining divisors on $\overline{M}_{0,n}$ give all the divisors on $\overline{M}_{0,n-1}$.
I would like to echo what David Holmes says in the comments. You could explicitly write out all the boundary divisors and check the statement by hand probably up to $\overline{M}_{0,6}$. (After that point there's probably too many divisors to do by hand, and regardless the result should be clear by then.) I think it's very helpful to do so in order to see what's going on.