Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I would like to explicitly compute the limit of a family of stable maps in $\overline{M}_{0,n}(\mathbb{P}^r,d)$. I know in principle how this works for families of curves without maps as I found a lot of literature on stable reduction for this case, like Harris and Morrison's "Moduli of Curves". Is there any literature which explains how to do this in case of stable maps?

I don't know if it helps but I am mostly interested in families of the form $\mathbb{C}^*\times\mathbb{P}^1$ and interested in the special fibre over $0$. For example a family like $(t,(z_0:z_1))\mapsto(tz_1^3:(z_1-z_0)(z_1+z_0)^2)$ in $\overline{M}_{0,3}(\mathbb{P}^1,3)$ where the marked points are $(1:0),(1:1)$ and $(1:-1)$ in each fibre. What I tried is blowing up at the points where the map is ill-defined for $t=0$, namely $(0,(1:1))$ and $(0,(1:-1))$ but that didn't work out and I just don't see what could have gone wrong.

I would be very grateful if anybody had suggestions what I did wrong, how to do it right or hints to the literature.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

I don't know any literature for how to do this, but I think I can do this example by hand in an ad hoc way.

As you say you have to blow up $\mathbb C \times \mathbb P^1$ in the two points you gave, so the special fiber becomes a chain of three $\mathbb P^1$s, with a marked point on each component. On the middle one of these components you get the constant map to $(0:1)$ when $t = 0$. On the other two you will have degrees 1 and 2.

The evaluation maps will send $(1:0) \mapsto (0:1)$ for all $t$, [which is a good sanity check since this will be the marked point on the middle component which maps constantly to $(0:1)$], and will map $(1:1) \mapsto (1:0)$ and $(1:-1) \mapsto (1:0)$ for $t \neq 0$. But then this must be true in the limit too, so on both of the two new components we will have two special points, one which meets the middle component and must go to $(0:1)$ and one marking which must go to $(1:0)$.

Now we consider the branch points of the stable map. There are four branch points counted with multiplicity by Riemann-Hurwitz, and one computes that $(0:1)$ is a double branch point, $(1:0)$ is simple, and the third one has the form $(t\alpha:1)$ for some nonzero $\alpha$. So in the limit we will have two branch points, one of multiplicity three and one of multiplicity one.

For a stable curve, the image of a node should be always counted as a branch point with multiplicity one, which means that the point $(0:1)$ needs one more ramification point over it.

Now by looking at the equation one is led to think that the map should have degree 1 on the component given by blowing up at $(1:1)$ and degree 2 on the component given by blowing up at $(1:-1)$. I think one way of seeing this is that the point $(1:-1)$ is a ramification point of the stable map for $t \neq 0$, hence this should be true in the limit, which would be impossible if it were on a degree 1 component. So the map on the $(1:1)$-component is an isomorphism, and the map on the $(1:-1)$-component is the unique branched double cover of $\mathbb P^1$ branched over $(1:0)$ and $(0:1)$.

share|improve this answer
    
Thank you, this was very helpful to at least understand the example :) Do you also know a way that is more like a recipe that also works for cases where the map is not just a cover of the projective line? –  Dennis Ochse Oct 5 '11 at 9:25
    
No, I don't know that. In this case you could figure out the stable limit (almost) only by looking at the evaluation maps and the Lyashko-Looijenga map; I don't know if this is always true. –  Dan Petersen Oct 5 '11 at 9:37

Your Answer

 
discard

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.