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3 Double points of plane models in M 6,1

3.2 The divisor D 2 d

Remember that PicQ(Mg,1) is generated by the Hodge class λ, the cotangent class ψ corresponding to the marked point, and the boundary classes δ0, . . . δg−1 defined as follows. The classδ0 is the class of the closure of the locus of pointed irreducible nodal curves, and the class δi is the class of the closure of the locus of pointed curves [CiCg−i, p] whereCi andCg−i are smooth curves respectively of genusiand gimeeting transversally in one point, and p is a smooth point in Ci, for i = 1, . . . , g −1. In this section we prove the following theorem.

Theorem 3.2.1. Let g = 3s and d= 2s+ 2 for s≥1. The class of the divisor D2d in PicQ(Mg,1) is

h

D2di=+

g−1X

i=0

biδi

3.2 The divisor D2d

where

a= 48s4+ 80s3−16s2−64s+ 24 (3s−1)(3s−2)(s+ 3) Ng,2,d c= 2s(s−1)

3s−1 Ng,2,d

b0 = 24s4+ 23s3−18s2−11s+ 6 3(3s−1)(3s−2)(s+ 3) Ng,2,d b1 = 14s3+ 6s2−8s

(3s−2)(s+ 3)Ng,2,d bg−1 = 48s4+ 12s3−56s2+ 20s

(3s−1)(3s−2)(s+ 3) Ng,2,d. Moreover for g= 6 and for i= 2,3,4,we have that

bi = −7i2+ 43i−6.

3.2.1 The coefficient c

The coefficientccan be quickly found. LetCbe a general curve of genusg and consider the curve C = {[C, y] : yC} in Mg,1 obtained varying the point y on C. Then the only generator class having non-zero intersection withC isψ, andC·ψ= 2g−2. On the other hand,C·D2dis equal to the number of triples (x, y, l)∈C×C×G2d(C) such thatx and y are different points andh0(l(−x−y))≥2. The number of such linear series on a generalC is computed by the Castelnuovo number (remember thatρ = 0), and for each of them the number of couples (x, y) imposing only one condition is twice the number of double points, computed by the Plücker formula. Hence we get the equation

D2d·C= 2

(d−1)(d−2)

2 −g

Ng,2,d =c(2g−2) and so

c= 2s(s−1) 3s−1 Ng,2,d. 3.2.2 The coefficients a and b0

In order to compute a and b0, we use a Porteous-style argument. Let Gd2 be the family parametrizing triples (C, p, l), where [C, p]∈ Mirrg,1 andlis ag2donC; denote byη: Gd2 → Mirrg,1 the natural map. There existsπ:Yd2 → Gd2 a universal pointed quasi-stable curve, with σ: Gd2 → Yd2 the marked section. Let L → Yd2 be the universal line bundle of relative degree dtogether with the trivialization σ(L) ∼=OG2

d, and V ⊂π(L) be the sub-bundle which over each point (C, p, l = (L, V)) in Gd2 restricts to V. (See [Kho07,

§2] for more details.)

Furthermore let us denote byZd2 the family parametrizing ((C, p), x1, x2, l),

where [C, p]∈ Mirrg,1,x1, x2C andl is ag2d on C, and letµ, ν:Zd2 → Yd2 be defined as the maps that send ((C, p), x1, x2, l) respectively to ((C, p), x1, l) and ((C, p), x2, l).

Now given a linear series l= (L, V), the natural map ϕ:VH0(L|p+x) globalizes to

ϕ:e V →µL ⊗O/IΓ

σ+∆) =:M

as a map of vector bundles over Yd2, where ∆ and Γσ are the loci in Zd2 determined respectively byx1 =x2 and x2 =p. ThenD2d∩ Mirrg,1 is the push-forward of the locus in Yd2 where ϕe has rank≤1. Using Porteous formula, we have

h D2di

Mirrg,1 =ηπ V

M

2

(3.1)

=ηπ πc2(V) +πc1(Vc1(M) +c21(M)−c2(M).

Let us find the Chern classes ofM. Tensoring the exact sequence 0→I/I∆+Γ

σ →O/I∆+Γ

σ →O→0

byνL and applying µ, we deduce that ch(M) =ch(µ(OΓ

σ(−∆)⊗νL)) +ch(µ(OνL))

=ch(µ(OΓ

σ(−∆))) +ch(µ(OνL))

=e−σ+ch(L) hence

c1(M) =c1(L)−σ c2(M) =−σc1(L).

The following classes

α=πc1(L)2hYd2i γ =c1(V)∩hGd2i

3.2 The divisor D2d

have been studied in [Kho07, Thm. 2.11]. In particular 6(g−1)(g−2)

dNg,2,d η(α)|Mirr

g,1 = 6(gd−2g2+ 8d−8g+ 4)λ + (2g2gd+ 3g−4d−2)δ0

−6d(g−2)ψ, 2(g−1)(g−2)

Ng,2,d η(γ)|Mirr

g,1 = (−(g+ 3)ξ+ 40)λ +1

6((g+ 1)ξ−24)δ0

−3d(g−2)ψ, where

ξ = 3(g−1) +(g+ 3)(3g−2d−1) gd+ 5 . Plugging into (3.1) and using the projection formula, we find

h D2di

Mirrg,1 =η−γ·πc1(L) +γ·πσ+α+πσ2π(σc1(L))

= (1−d)η(γ) +η(α)−Ng,2,d·ψ.

Hence

a= 48s4+ 80s3−16s2−64s+ 24 (3s−1)(3s−2)(s+ 3) Ng,2,d b0 = 24s4+ 23s3−18s2−11s+ 6

3(3s−1)(3s−2)(s+ 3) Ng,2,d and we recover the previously computed coefficient c.

3.2.3 The coefficient b1

Let C be a general curve of genusg−1 and (E, p, q) a two-pointed elliptic curve, with pq not a torsion point in Pic0(E). Let C1 := {(C ∪y∼q E, p)}y∈C be the family of curves obtained identifying the pointqE with a moving pointyC. Computing the intersection of the divisorD2dwithC1 is equivalent to answering the following question:

how many triples (x, y, l) are there, withyC,xCy∼qE\ {p} and l ={lC, lE} a limitg2d onCy∼qE, such that (p, x, l) arises as limit of (pt, xt, lt) on a family of curves {Ct}t with smooth general element, where pt and xt impose only one condition on lt a g2d?

Let alE(q) = (a0, a1, a2) be the vanishing sequence of lEG2d(E) at q. Since C is general, there are nog2d−1onC, hencelC is base-point free anda2 =d. Moreover we know a1d−2. Let us supposexE\{q}. We distinguish two cases. Ifρ(E, q) =ρ(C, y) = 0, then wlE(q) = ρ(1,2, d) = 3d−8. Thus alE(q) = (d−3, d−2, d). Removing the base

point we have that lE(−(d−3)q) is a g23 and lE(−(d−3)q−px) produces a g11 on E, hence a contradiction. The other case is ρ(E, q) = 1 andρ(C, y)≤ −1. These force alE(q) = (d−4, d−2, d) and alC(y)≥(0,2,4). OnEwe have thatlE(−(d−4)q−px) is ag12.

The question splits in two: firstly, how many linear serieslEG24(E) and points xE\{q}are there such thatalE(q) = (0,2,4) andlE(−p−x)∈G12(E)? The first condition restricts our attention to the linear series lE = (O(4q), V) where V is a tridimensional vector space andH0(O(4q−2q))⊂V, while the second condition tells usH0(O(4q−p− problem discussed in [Far09b, Proof of Thm. 4.6]. The answer is

(g−1)15Ng−1,2,d,(0,2,2)+ 3Ng−1,2,d,(1,1,2)+ 3Ng−1,2,d,(0,1,3)

= 24(2s2+ 3s−4) s+ 3 Ng,2,d. Now let us suppose xC\ {y}. The condition on x and p can be reformulated in the following manner. We consider the curveCyE as the special fiberX0 of a family of curves π:XB with sections x(t) and p(t) such that x(0) = x, p(0) = p, and with smooth general fiber having l = (L, V) a g2d such that l(−xp) is a g1d−2. Let VV be the two dimensional linear subspace formed by those sectionsσV such that div(σ) ≥x+p. Then V specializes onX0 to VCVC and VEVE two-dimensional

3.2 The divisor D2d that the map

ϕ:H0(lC)→H0(lC|2y+x)

has rank ≤1. By Thm. 3.0.2 there is only a finite number of such triples, and clearly the case alC(y)>(0,2,3) cannot occur. Moreover, note that xand y will be necessarily distinct.

Let µ=π1,2,4:C×C×C×Wd2(C) →C×C×Wd2(C) andν =π3,4:C×C×C× Wd2(C) → C×Wd2(C) be the natural projections respectively on the first, second and fourth components, and on the third and fourth components. Let π:C×C×Wd2(C)→ Wd2(C) be the natural projection on the third component. Now ϕ globalizes to

e

Let us find the Chern classes ofM. First we develop some notation (see also [ACGH85,

§VIII.2]). Letπi:C×C×C×Wd2(C)→C fori= 1,2,3 andπ4:C×C×C×Wd2(C)→

Moreover

γijγjk = ηjγik, for 1≤i < j < k≤4. With this notation, we have

ch(νL ⊗O/ID) = (1 +3+γ34η3θ)1−e−(η1133+2η2+2γ23+2η3), hence by Grothendieck-Riemann-Roch

ch(M) =µ((1 + (2−g)η3)ch(νL ⊗O/ID))

= 3 + (d−2)η1+ (2g+ 2d−6)η2−2γ12+γ14+ 2γ24

η1θ−2η2θ+ (8−2d−4g)η1η2−2η1γ24−2η2γ14+ 2η1η2θ.

Using Newton’s identities, we recover the Chern classes ofM:

c1(M) = (d−2)η1+ (2g+ 2d−6)η2−2γ12+γ14+ 2γ24, c2(M) = (2d2−8d+ 2gd+ 8−4g)η1η2+ (2g+ 2d−8)η2γ14

+ (2d−4)η1γ24+ 2γ14γ24−2η2θ, c3(M) = (4−2d)η1η2θ−2η2γ14θ.

We finally find

[Y] =η1η2(c21(2d2−8d+ 2dg+ 4−4(g−1))

+c1θ(−12d−4g+ 40) +c2(−4d+ 16−8g) + 12θ2)

= (28s+ 48)(s−2)(s−1)

(s+ 3) Ng,2,d·η1η2θg−1

where we have used the following identities proved in [Far09b, Lemma 2.6]

c21=

1 + 2s+ 2 s+ 3

c2 c1θ= (s+ 1)c2

θ2= (s+ 1)(s+ 2)

3 c2

c2=Ng,2,d·θg−1.

We are going to show that we have already considered all non zero contributions.

Indeed let us suppose x = y. Blowing up the point x, we obtain Cy P1qE with x∈ P1\ {y, q} and pE\ {q}. We reformulate the condition on x and p viewing our curve as the special fiber of a family of curves π:XB as before. Let{lC, lP1, lE} be a limitg2d. Now V specializes toVC,VP1 and VE. There are three possibilities: either ρ(C, y) =ρ(P1, x, y, q) =ρ(E, p, q) = 0, orρ(C, y) =−1,ρ(P1, x, y, q) = 0,ρ(E, p, q) = 1, orρ(C, y) =−1,ρ(P1, x, y, q) = 1,ρ(E, p, q) = 0. In all these casesalC(y) = (0,2, al2C(y)) (remember that lC is base-point free) and alE(q) = (al0E(q), d−2, d). Hence alP1(y) =

3.2 The divisor D2d (al0P1(y), d−2, d) andalP1(q) = (0,2, al2P1(q)). Let us restrict now to the sections in VC, VP1 and VE. For all sections σP1VP1 since div(σP1)≥x, we have that ordyP1) < d and hence ordyP1) ≤ d−2. On the other side, since for all σEVE, div(σE) ≥ p, we have that ordqE) < d and hence ordqP1) ≥ 2. Let us take one section τVP1

such that ordy(τ) = d−2. Since div(τ) ≥ (d−2)y +x, we get ordq(τ) ≤ 1, hence a contradiction.

Thus we have that

D2d·C1= 24(2s2+ 3s−4)

s+ 3 Ng,2,d+ (28s+ 48)(s−2)(s−1) (s+ 3) Ng,2,d.

while considering the intersection of the test curve C1 with the generating classes we have

D2d·C1 =b1(2g−4), whence

b1= 14s3+ 6s2−8s (3s−2)(s+ 3)Ng,2,d.

Remark 3.2.2. The previous class [Y] being nonzero, it implies together with Thm.

3.0.2 that the scheme Gd2((0,2,3)) over Mg−1,1 is not irreducible.

3.2.4 The coefficient bg−1

We analyze now the following test curve E. Let (C, p) be a general pointed curve of genusg−1 and (E, q) be a pointed elliptic curve. Let us identify the pointspandq and let y be a movable point inE. We have

0 =D2d·E=c+b1bg−1, whence

bg−1 = 48s4+ 12s3−56s2+ 20s (3s−1)(3s−2)(s+ 3) Ng,2,d. 3.2.5 A test

Furthermore, as a test we consider the family of curves R. Let (C, p, q) be a general two-pointed curve of genusg−1 and let us identify the pointq with the base point of a general pencil of plane cubic curves. We have

0 =D2d·R=a−12b0+bg−1. 3.2.6 The remaining coefficients in case g = 6

Denote by Pg the moduli space of stable g-pointed rational curves. Let (E, p, q) be a general two-pointed elliptic curve and letj:Pg → Mg,1 be the map obtained identifying

the first marked point on a rational curve with the point qE and attaching a fixed elliptic tail at the other marked points. We claim thatj(D26) = 0.

Indeed consider a flag curve of genus 6 in the image of j. Clearly the only possibility for the adjusted Brill-Noether numbers is to be zero on each aspect. In particular the collection of the aspects on all components but E smooths to a g26 on a general one-pointed curve of genus 5. As discussed in section 3.2.3, the point x can not be in E.

Supposexis in the rest of the curve. Then smoothing we getlag26 on a general pointed curve of genus 5 such thatl(−2qx)) is ag13, a contradiction.

Now let us study the pull-back of the generating classes. As in [EH87, §3] we have thatj(λ) =j0) = 0. Furthermorej(ψ) = 0.

For i = 1, . . . , g −3 denote by ε(1)i the class of the divisor which is the closure in Pg of the locus of two-component curves having exactly the first marked point and other i marked points on one of the two components. Then clearly ji) = ε(1)i−1 for i= 2, . . . , g−2. Moreover adapting the argument in [EH89, pg. 49], we have that

jg−1) =−

g−3X

i=1

i(gi−1) g−2 ε(1)i while

j1) =−

g−3X

i=1

(g−i−1)(g−i−2) (g−1)(g−2) ε(1)i . Finally sincej(D26) = 0, checking the coefficient of ε(1)i we obtain

bi+1 = (g−i−1)(g−i−2)

(g−1)(g−2) b1+i(gi−1) g−2 bg−1 fori= 1,2,3.

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