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Previously we studied moduli of sections of X/C. Here we apply this to the study of rational curves on a smooth projective variety Y. In particular, if Y is rationally simply connected by chains of free lines, then the sequence Ze from Section 8gives a canonical sequence of irreducible components of the moduli spaceM0,0(Y,e).

Hypothesis10.1. — Let k be an uncountable algebraically closed field of characteristic 0. Let Y be a smooth projective variety over k. Let L be an ample invertible sheaf on Y. We assume that Y→Spec(k)satisfies Hypothesis7.8(where S is Spec(k)and f :X→S is YSpec(k)).

Assume that Hypothesis10.1holds. Let X=P1×Y and f =pr1:X=P1×Y−→P1.

The first thing to remark is that a morphism g:P1Y gives rise to a section s:P1→X, and conversely. Also, the section s is 0-free, i.e., unobstructed (see Definition4.7), if and only if the rational curve g:P1→Y is unobstructed (i.e., every invertible sheaf summand in the direct sum decomposition of gTYhas degree at least−1). Similarly, s is 1-free, i.e., free (see Definition4.7), if and only if the rational curve g:P1→Y is free. Arguing in this way we see that we can interpret many of the previous results for the family f :X→P1 in terms of the moduli spaces of lines on Y.

Instead of reproving everything from scratch in this setting we make a list of state-ments and we point out the corresponding lemmas, and propositions in the more general treatement above. We denoteM0,0(Y,e)the Kontsevich moduli space of degree e ratio-nal curves in Y. Also, let Ypax denote the set of peaceful points (w.r.t. Y→Spec(k)), see Definition7.4.

(1) There is a sequence of irreducible components ZeM0,0(Y,e), e≥1 uniquely characterized by the property that every comb in Y whose handle is contracted to a point of Ypaxand which has e teeth mapping to lines in Y is in Ze. This is the exact analogue of Lemma 8.4, starting with the fact that a constant map P1→Y is free!

(2) A general point of Ze corresponds to a mapP1→Y which is free and meets Ypax. This is the exact analogue of Lemma8.5.

(3) Consider a comb in Y whose handle has degree e with ee teeth which are lines. Such a comb is in Ze if the handle is free, is in Ze, and the attachment points map to Ypax. This is the exact analogue of Lemma8.6.

(4) Any stable map f :C→Y inM0,0(Y,e)which has exactly one component a free curve corresponding to a point in Ze and all other components free lines corresponds to a smooth nonstacky point of Ze. This is the exact analogue of Lemma8.7.

(5) Given any integer m0 there exists e 0 such that the general point of Ze

corresponds to a map f :P1→Y with H1(P1,fTY(m))=0. This is the exact analogue of Lemma8.8.

(6) Let ZM0,0(Y,e0) be an irreducible component whose general point cor-responds to a free rational curve. There exists an integer E0 such that for all e≥E we have the following property: Consider a comb in Y whose handle has degree e0with ee0teeth which are lines. Such a comb is in Ze if the han-dle is free, is in Z, and the attachment points map to Ypax. (And in particular

any smoothing of this comb defines a point of Ze.) This is the exact analogue of Corollary9.8, once you observe that such a comb defines an unobstructed nonstacky point ofM0,0(Y,e).

It is perhaps not necessary, but we point out that instead of defining Ze as in (1) above, we could just define Ze as the unique irreducible component of M0,0(Y,e)which contains all trees of free lines. This follows from (4) (with e=0).

Lemma10.2. — In Situation10.1 above. Let Ze,1M0,1(Y,e)be the unique irreducible component dominating ZeM0,0(Y,e). The general fibre of ev:Ze,1Y is irreducible.

Proof. — Let us temporarily denote by Ze,1⊂Ze,1 the dense open locus of stable 1-pointed maps

Ci →Y all of whose irreducible components Ci→Y are free and whose nodes map to points of Ypax. The morphism ev:Ze,1→Y is smooth. Consider the locus We,1⊂Ze,1 whose points correspond to those combs whose handles are con-tracted to a point of Ypaxand whose teeth are (automatically) free lines and whose marked point is on the handle. By definition of the irreducible components Ze the space We,1 is nonempty. It is smooth and We,1→Y is still smooth onto Ypax. Clearly, Hypothesis7.8 for Y→Spec(k)implies the general fibre of We,1→Y is irreducible. From this, and the irreducibility of Ze,1it follows that the general fibre of Ze,1→Y is irreducible, for example

by [Gro67, Proposition 4.5.13].

Lemma10.3. — In Situation10.1above. Let ZM0,0(Y,e0)and E be as in (6) above.

Suppose that g:C→Y is a genus 0 Kontsevich stable map such that C=A

i=1CiB j=1Lj is a decomposition into irreducible components with the following properties: (1) each CiY is free and corresponds to a point of Z, (2) each LiY is a free line and a “leaf of the tree”, i.e., it intersects the rest of the curve in a single node, and (3) B≥E−e0. Then g defines an unobstructed point of ZAe0+B.

Proof. — We will show there is a connected chain of curves contained in the un-obstructed locus ofM0,0(Y,Ae0+B)which connects g to a point of ZAe0+B. As the un-obstructed locus is smooth this will imply that the entire curve is contained in ZAe0+B

and hence g is in it. We may first move the map a little bit such that all the nodes of C are mapped into Ypax. Since the lines are leaves, we may 1 by 1 slide all the lines Li along the curves Ci and onto one of the curves Ci0 which is a leaf of the tree

We may assume (after possibly moving the map a little bit) that the attachment point p is mapped to a point y∈Y such that the fibre of Ze0+B,1→Y is irreducible over this point, see Lemma10.2. Hence we may connect the 1-pointed stable map(C,p)→Y inside the fibre of Ze0+B,1Y over y to a curve which is made out of a comb C whose handle is contracted and whose teeth are free lines. Since the marked point is throughout

mapped to the same y∈Y(k)this connects the original stable map C∪C→Y to a stable map C∪C→Y where the number of lines in Cis now e0+B. We may again slide these lines over to some irreducible component Ci of C and continue until all the Ciare gone, and so obtain a point of ZAe0+B. This proves the lemma.

The notation in the section conflicts with the notation introduced in Lemma8.4 in case Y is a fibre of a family as in Situation8.1 such that Hypothesis 7.8 holds over a nonempty open of C. In that situation we will use the notation Ze(Xt)to denote the irreducible component defined in this section for the irreducible fibre Xt.

Lemma10.4. — In Situation8.1, assume in addition Hypothesis7.8for the restriction of f to some nonempty open SC. Suppose that s:C→X is a free section, and let Z(X/C/k) be the unique irreducible component containing the moduli point[s]. Let Ze as in Lemma8.4. For any t1, . . . ,tδ∈S(k), for any free rational maps si:P1→Xti such that si(0)=s(ti), if sicorresponds to Zei(Xti), then the comb C

P1→X/C defines an unobstructed point of Zdeg(s)+ ei.

Proof. — This is very similar to the proof of Lemma10.3above. Namely, we first move the comb (as a comb) such that the points s(ti)are in the locus where the fibres of Zei,1→Xti are irreducible. then we connect the si in these fibres to chains of lines. After

this we can apply Lemma8.7for example.