Compositio Math. 143 (2007) 900–908 doi:10.1112/S0010437X07002849
Cycle relations on Jacobian varieties
Gerard van der Geer and Alexis Kouvidakis (with an appendix by Don Zagier)
Dedicated to Jozef Steenbrink on the occasion of his 60th birthday
Abstract By using the Grothendieck–Riemann–Roch theorem we derive cycle relations modulo algebraic equivalence in the Jacobian of a curve. The relations generalize the relations found by Colombo and van Geemen and are analogous to but simpler than the rela- tions recently found by Herbaut. In an appendix by Zagier, it is shown that these sets of relations are equivalent.
1. Introduction
Beauville showed in [Bea86] that the Chow ring with rational coefficients of an abelian variety i possesses a double grading CH Q(X)= CH (j)(X)wherei refers to the codimension and j refers i { ∈ i ∗ 2i−j } to the action of the integers: CH (j)(X)= x CH (X):k (x)=k x .Thequotient A(X)of i the Chow ring modulo algebraic equivalence inherits this double grading A(X)= A(j)(X)and carries two multiplication laws, the intersection product x · y and the Pontryagin product x ∗ y. If X =Jac(C) is the Jacobian of a curve C of genus g, then we can decompose the class [C]of g−1 ∈ g−1 the image of the Abel–Jacobi map of C as [C]= j=0 C(j) with C(j) A(j) (Jac(C)). Colombo and van Geemen proved (cf. [CV93]) that for a curve C with a map of degree d to P1, the component C(j) vanishes for j d − 1. In [Her07] Herbaut extended this result and found cycle relations for r curves having a gd, i.e. a linear system of degree d and projective dimension r,withr 2. It is the purpose of this note to show that one can use the Grothendieck–Riemann–Roch theorem to derive, in an easy way, the Colombo–van Geemen result as well as simple relations of higher degree. Let C be a smooth projective curve of genus g over an algebraically closed field K.
Theorem . r 1.1 If C has a base-point-free linear system gd,then ··· ∗···∗ (a1 +1)! (ar +1)!C(a1) C(ar) =0, (1) a1+···+ar=N for every N d − 2r +1.
For N = d − 2r + 1, this relation coincides with Herbaut’s relation (cf. [Her07,Theorems1 and 8]). For higher values of N they are in general different, but in an appendix we present a proof by Don Zagier that shows that Theorem 1.1 and Herbaut’s Theorem 1 are equivalent.
Received 12 September 2006, accepted in final form 27 November 2006. 2000 Mathematics Subject Classification 14C25, 14H40. Keywords: Jacobian variety, algebraic cycle. This journal is c Foundation Compositio Mathematica 2007. Cycle relations on Jacobian varieties
2. Preliminaries
Let C be a smooth projective curve of genus g over an algebraically closed field K. We suppose r that the curve C has a base-point-free linear system gd of degree d and projective dimension r.This defines a morphism γ : C → Pr.LetJ =Jac(C) be the Jacobian of C. We consider the incidence variety Y ⊂ C × Pˆ r defined by
Y = {(p, η) ∈ C × Pˆ r : γ(p) ∈ η}, where Pˆ r is the dual projective space of Pr. It has dimension r and possesses the two projections φ˜ andα ˜ onto C and Pˆ r.Notethat˜α is finite of degree d and φ˜ is a Pr−1-fibration. We shall write Pr for Pˆ r. We have the following diagram of morphisms
o v oα π / α˜ / Pr Pr × J Y × J Y Pr
p φ φ˜ o q π˜ / J C × J C where the morphisms v,p,q,π˜ and π are projections, and α =˜α × idJ and φ = φ˜ × idJ . ∗ Let P be the Poincar´e bundle on C ×J and set L := φ P , a line bundle on Y ×J.Put := c1(L) and Π := c1(P ). The Chow ring of Pr × J is generated over CH ∗(J)bytheclassξ = v∗h with h ahyperplane in Pr with ξr+1 =0.Foraclassβ ∈ CH ∗(Pr × J)wehavetherelation(cf.[Ful98, Theorem 3.3, p. 64]) r r−i i β = βiξ with βi = p∗(β · ξ ), (2) i=0 ∗ where, by abuse of notation, we write here and hereafter βi for p (βi). We let x = α∗(ξ) be the pull back of ξ.Weletρ = π∗φ˜∗(point) be the pull back class of a point on C. We work in the Chow ring up to algebraic equivalence. There we have the relations
xr = dπ∗(point),xr+1 =0,ρ2 =0,xr−1ρ = π∗(point).
Recall the Fourier transform F : A(X) → A(X) for a principally polarized abelian variety (X, θ) of dimension g (cf. [Bea86, Bea04]). It has the properties (i) F ◦ F =(−1)g(−1)∗, (ii) F (x ∗ y)= · · − g ∗ i g−i+j F (x) F (y)andF (x y)=( 1) F (x) F (y), (iii) F (A(j)(X)) = A(j) (X). Π We have the relation q∗(e )=F [C](cf.[Bea04, § 3]). Comparing terms gives that F [C(j)]= j+2 (1/(j +2)!)q∗(Π )forj =0,...,g− 1. Note, also, that q∗1=q∗Π = 0. More generally, extending scalars to Q we have the relation
kΠ 2g −1 ∗ q∗(e )=k F [(k ) C]fork ∈ Z1. g−1 −1 ∗ j+2−2g In fact, writing [C]= j=0 C(j) we have (k ) [C]= j k C(j), hence 2g −1 ∗ j+2 j+2 j+2 kΠ k F [(k ) C]=F k C(j) = k q∗(Π /(j +2)!)=q∗(e ). j j
901 G. van der Geer and A. Kouvidakis
3. The proof
We shall prove that if C has a base-point-free linear system gr,then d ··· ··· (a1 +1)! (ar +1)!F [C(a1)] F [C(ar )]=0, (3) a1+···+ar=N for every N d − 2r +1. We are going to apply the Grothendieck–Riemann–Roch theorem to the morphism α and the ⊗k line bundle L.Fork 1 we put Vk := α∗(L ). Since α is a finite morphism of degree d this is a vector bundle of rank d and we get ⊗k k ch(Vk)=ch(α!L )=α∗(e tdα), with tdα the Todd class of the morphism α. The Todd class td α is algebraically equivalent to a class of the form A(x)+B(x)ρ.HereA = r−1 j r−1 j j=0 ajx and B = j=0 bjx are polynomials in x and a0 = 1. In fact, tdα is the pull back under r−1 π of tdα˜,anelementofA(Y ). The ring A(Y ) is generated as an A(C)-module by 1,x1,...,x1 ∗ with x1 =˜α (h)andA(C) is generated by 1 and the class of a point.
r Proposition 3.1. For k ∈ Z1 we have in A(P × J) the relation 2g −1 ∗ ch(Vk)=dA(ξ)+ξB(ξ)+k F [(k ) C]ξA(ξ).
In particular, all chj(Vk) are divisible by ξ for j 1. Before we give the proof of Proposition 3.1 we state a corollary and a lemma.
Proposition 3.1 gives an expression of the Chern characters of the bundles Vk. We can express the Chern classes of the bundles Vk of rank d in terms of the Chern characters by using the well-known formula (cf. [Mac95, ch. I (2.10)]) d j−1 j 1+c1(Vk)t + ···+ cd(Vk)t =exp (−1) (j − 1)! chj(Vk)t . (4) j1 Formula (4) combined with Proposition 3.1 will give us the vanishing relations we are asking for. For example, applying these formulas for r =1andk = 1 immediately gives us the theorem of Colombo–van Geemen [CV93] as we now show. Corollary . 1 − 3.2 If C has a gd,thenC(j) =0for j d 1.
Proof. Put V = V1.Weseech(V )=d + nξ + F [C]ξ for some n (actually n =1− d − g). Since 2 chj(V ) is divisible by ξ for j 1andξ =0,formula(4) becomes in this case d 2 j−1 j 1+c1(V )t + ···+ cd(V )t =1+ch1(V )t − ch2(V )t + ···+(−1) (j − 1)! chj(V )t + ··· .
Therefore, chj(V ) vanishes for j>d. Hence, F [C] has no terms of codimension d or more. Since F [C(j)]isofcodimensionj + 1, it follows that C(j) =0forallj d − 1. Lemma 3.3. In A(Pr × J) the following relations hold for ν 0: µ ν+1 µ ν q∗(Π )ξ µ>0, µ ν 0 µ>0, α∗( · x )= and α∗( · x · ρ)= dξν µ =0 ξν+1 µ =0.
Proof. For the first relation. By (2) the coefficient of ξr−j is given by µ ν j µ ν ∗ j ∗ µ ν+j p∗(α∗( x )ξ )=p∗(α∗( x α (ξ) )) = p∗(α∗(φ (Π) x )) ∗ µ ν+j µ ν+j = q∗(φ∗(φ (Π) x )) = q∗(Π φ∗(x )).
902 Cycle relations on Jacobian varieties
If ν + j = r,thenxr is algebraically equivalent to d times point × J and since Π is algebraically equivalent to 0 on point ×J, we get that any term with µ>0andν +j = r contributes 0. The term µ r−1 µ r−1 with ν + j = r − 1 contributes q∗(Π φ∗(x )) = q∗(Π )sinceφ∗(x )) = 1C×J .Ifν + j
We now give the proof of Proposition 3.1.
µ ν Proof of Proposition 3.1. We have α∗( x ρ) = 0 for all µ 1, ν 0. So in the contributions k k α∗(e tdα) we get contributions of the form α∗(e A(x)) and α∗(ρB(x)) only. We have µ µ k k µ k µ α∗(e A(x)) = α∗(A(x)) + α∗( A(x)) = dA(ξ)+ q∗Π ξA(ξ) µ! µ! µ1 µ1 kΠ 2g −1 ∗ = dA(ξ)+q∗(e )ξA(ξ)=dA(ξ)+k F [(k ) C]ξA(ξ).
On the other hand, α∗(ρB(x)) = ξB(ξ).
By using the relation µ+2 2g −1 ∗ kΠ k µ+2 µ+2 k F [(k ) C]=q∗(e )= q∗Π = k F [C( )], µ +2! µ µ0 µ0 we get the following corollary of Proposition 3.1. r−1 j Corollary 3.4. We put a = b =0for every j r.WithA = a x where a0 =1and j j j=0 j r−1 j B = j=0 bjx we have for j 1,k 1 that j−1 j j−m+1 m chj(Vk)=(daj + bj−1)ξ + am−1k F [C(j−m−1)] ξ . m=1
With j 1 we put 1 r chj(Vk)=A1(j)ξ + ···+ Ar(j)ξ , where Am(j)isofcodimensionj − m. Please note that chj(Vk) is divisible by ξ for j 1by Proposition 3.1.
Remark 3.5. The coefficient Am(j) depends on k, but for simplicity of notation we do not involve the index k in the notation.
Then, for every j 1and1 m r,wehave daj + bj−1 m = j, j−m+1 Am(j)= a −1k F [C( − −1)] m
M+1 Since rank(Vk)=d, the coefficient of t of the right-hand side of (4) must be zero for every M d.Letuswrite j−1 j F (t)= (−1) (j − 1)! chj(Vk)t . j1 903 G. van der Geer and A. Kouvidakis
i Note that F (t) =0foreveryi r + 1 because chj(Vk) is divisible by ξ for j 1. Therefore, the r i M+1 i right-hand side of (4)isequalto i=0(1/i!)F (t) . The coefficient of t in the polynomial F (t) , for i 1, is equal to − M+1−i − ··· − ··· ( 1) (α1 1)! (αi 1)! chα1 (Vk) chαi (Vk). α1+···+αi=M+1
We denote the expression (α1 − 1)! ···(αi − 1)! by α{1,i}. Then we have that r (−1)i α{1,i} ch (V ) ···ch (V )=0 i! α1 k αi k i=1 α1+···+αi=M+1 for M d,andso r − i r r ( 1) m1 mi α{1,i} A (α1)ξ ··· A (α )ξ =0. i! m1 mi i i=1 α1+···+αi=M+1 m1=1 mi=1 Now the left-hand side of this is easily seen to be equal to r m − i ( 1) m α{1,i} [A (α1) ···A (α )]ξ . i! m1 mi i m=1 i=1 α1+···+αi=M+1 m1 +···+mi=m We therefore have, for every m =1,...,r and M d,that m (−1)i α{1,i}A (α1) ···A (α )=0. i! m1 mi i i=1 m1+···+mi=m α1+···+αi=M+1 With M d,thecasem = r gives the relation r (−1)i α{1,i}A (α1) ···A (α )=0. i! m1 mi i i=1 m1+···+mi=r α1+···+αi=M+1
If we write { } ··· BM (i)= α 1,i Am1 (α1) Ami (αi), +···+ = +···+ = +1 m1 mi r α1 αi M r − i then this relation becomes i=1(( 1) /i!)BM (i) = 0 for every M d. We analyze the dependence on k. Proposition . r − i s 3.6 We write i=1(( 1) /i!)BM (i)= s Γsk as a polynomial in k.WithM d we have that Γs =0for s>M+1and (−1)r Γ +1 = (a1 +1)!···(a +1)!F [C ] ···F [C ]. M r! r a1 ar a1+···+ar=M−2r+1 ··· Proof. If Am1 (α1) Ami (αi) contains a factor with mj >αj, then it vanishes. Otherwise, since αj −mj +1 Amj (αj)=amj −1k F [Cαj −mj −1], except when mj = αj,inwhichcaseAαj (αj)=daαj + ··· ··· − ··· bαj −1,thepowerofk contained in Am1 (α1) Ami (αi)isequalto(α1+ +αi) (m1+ +mi)+ν = M +1− r + ν,whereν is given by ν =#{mj = αj, j =1,...,i} i.Nowifi