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Relation with iterated residues

2.5 Special values of the Abel-Jacobi map

3.1.12 Relation with iterated residues

Note thatZ and XL have the same dimension.

Proposition 3.1.16. If the restriction of πL to Z is not surjective, then the set FlL(Z) is empty.

Proof. If not, then dimπL(Z)< d. LetL={k1, . . . , kd}. LetZ FlL(Z). Consider the corresponding flag of irreducible subvarieties ofXL:

πL(Z) = πL(Z0)⊃πL(Z1)⊃ · · · ⊃πL(Zd).

Because of the dimension reasoning there must be an indexiwithπL(Zi) =πL(Zi−1).

This implies

πki(Zi−1) = πki(Zi) Ski.

ThereforeZi =Zi−1, so the flag is not strict, which is a contradiction.

We may therefore suppose without loss of generality that πL : Z XL is surjective. Hence the set of points on Z where this map is not ´etale is a proper closed subvariety. The irreducible components of this subvariety have dimension d−1 or less and are special. ThereforeπL−1RL∩Z belongs to its complement. This means the following is true.

Proposition 3.1.17. The projection

πL:πL−1RL∩Z→RL is an unramified covering.

Let p = πLZd SL. Since for a flag Z FlL(Z) the set RL,Z is open and closed inπ−1L RL∩Z we get

Proposition 3.1.18. The projection

πL:RL,Z →RpL is an unramified covering.

Remark 3.1.5. By the proposition we see that sinceRpLis a product of annuli,RL,Z is a disjoint union of products of annuli.

We are going to determine the cycle ha(Z,U0)∩RL,Z Hd(RL,Z). The d-th homology group of a product of dannuli isZ. Hence there is a canonical generator hcofHd(RpL). In fact hc can be defined as the productc1× · · · ×cd, whereci is the circle in Rpkkii going around pki counterclockwise.

Proposition 3.1.19. The cycle ha(Z,U0)∩RL,Z is the pullback of(−1)d(d−1)2 hcvia the projection πL:RL,Z →RpL.

Proof.

Remark 3.1.6. It is clear that we can decrease numbersεk and increase numbersε0k. The sequence obtained in this way will be also good. Moreover the open subsets of the new hypercover are contained in the corresponding open subsets of the old one. Therefore the residues computed with respect to these hypercovers are equal.

Therefore one can assume that the projection πL : RL,Z RpL extends to an unramified covering for the closures ofRL,Z in Z,RpL inXL.

Consider the commutative diagram:

H2d(X) −−−−→ H2d(RL,Z, ∂RL,Z) ←−−−−πL H2d(RLp, ∂RpL)

ha



y ha



y ha

 y

Hd(Ua0) −−−−→ Hd(RL,Z) π

←−−−−L Hd(RpL)

We see that it is enough to prove that the image of the fundamental class of H2d(RpL, ∂RpL) by the map ha in Hd(RLp) is (−1)d(d−1)2 hc. There is a direct product decomposition RpL = ×k∈LRpkk. The hypercover on RpL is the product hypercover.

For k L let φk be the fundamental class of H2(RpL, ∂RpL). Let us lift it to a hyperchainφfk.

The hypercover of RLp is the ˇCech hypercover associated to the cover with two open sets:

V0 = n(pk, εk)\n(pk, εk) corresponding to the cella(ηk), V1 = n(pk, εk)\n(pk, εk) corresponding to the cella(pk), V0∩V1 = n(pk, εk)\n(pk, εk) corresponding to the cella(ηk, pk).

Letε0 < r < ε. Consider topological chains

ck0 = n(pk, εk)\n(pk, r)∈C2(V0), ck1 = n(pk, r)\n(pk, ε0k)∈C2(V1), ck01 = ∂n(pk, r)∈C1(V0∩V1).

They define a hyperchain ck. We have c0 +c1 = φk. The hyperchain is closed, thereforeh01k) =ck01, which is the canonical generator ofH1(Rpkk).

Since the product of φk is the fundamental class of H2d(RLp, ∂RpL), the prod-uct of the hyperchains c = ×k∈Lck lifts the fundamental class. The term of c at

×k∈Lk, pk) is, by the definition of the product for hyperchains, (−1)d(d−1)2 ×k∈Lck01. This is exactly (−1)d(d−1)2 hc.

Let Z be a strict L, p-special flag, L = {k1, . . . , kd}, p = (pk1, . . . , pkd). Let Z0 =Z1,k=k1,L0={k2, . . . , kd}. LetZ0 be the flagZ1 ⊃ · · · ⊃Zd,Z0 FlL0(Z0).

Letti be a local parameter onXki at the pointpki =πkiZd. Let p0 =πL0p. For Z0 we have the projection

πL0 :RL0,Z0 →RpL00,

which is also an unramified covering. Let

Ze = Z0×XL0 Z = (Xk1×Z0)×XLZ.

We obtain the canonical diagrams

Ze −−−−→ρ Z Ze −−−−→ρ Z

τ

 y



y eτ

 y

 y Z0 −−−−→ XL0 (Xk1×Z0) −−−−→ XL

The diagonal embedding Z0 →Z0×XL0 Z0 induces a morphism ∆ :Z0 →Z, whiche is a section to the natural projectionτ :Ze→Z0. Let

RZ = Z π−1L (Ukpk×RpL00) n(Z, ~δ).

Consider the fiber products overUkpk×RpL00:

Re −−−−→ρ RZ

e τ



y πL

 y Ukpk ×RL0,Z0 −−−−→ Ukpk×RpL00

Again, we have a section ∆ : RL0,Z0 R. Lete Re0 be the union of connected components of Re which intersect the image of ∆. Let ρ0 be the restriction of ρ to Re0.

Remark 3.1.7. The setRZ is open and closed inZ∩πL−1(Ukpk×RpL00) because Propo-sition 3.1.11 and the first sentence of Corollary 3.1.12 work if we replace RL by Ukpk ×RLp00.

Remark 3.1.8. All special subsets of Z which intersect RZ are contained in Z0. Therefore the mapπLon the diagram is an unramified covering outside {pk} ×RpL00

and πk−1pk∩RZ =RL0,Z0.

Proposition 3.1.20. The mapρ0 is an analytic isomorphism.

Proof. Since ρ0 is a base change of the unramified covering πL0 :RL0,Z0 RpL00, it is an unramified covering itself. Therefore it is enough to construct a continuous section s0 : RZ Re0 to ρ0 which extends the diagonal map. This is equivalent to constructing a retractions:RZ →RL0,Z0 which respects the projection πL0.

Take a compact connected setA⊂RpL00 such thatπ−1L0 (A)∩RL0,Z0 =A1∪· · ·∪Am is a disjoint union of spaces isomorphic to A.

One can choose disjoint open subsetsV1, . . . , Vm inπL−10 (A)∩RZ such thatAi Vi. The setC =πL−10 (A)∩RZ\(V1∪ · · · ∪Vm) is closed. ThereforeπL(C) is closed.

Take α >0 such that n(pk, α)×A does not intersect πL(C). This means that the open setπL−1(n(pk, α)×A)∩RZ is contained in the union of the setsVi.

LetVi0=πL−1(n(pk, α)×A)∩Vi. Let us prove that Vi0 is connected for eachi. If not, then Vi0 =B1tB2 withBj open, closed and nonempty. Since Ai is connected,

for somej Bj does not intersectAi. ThereforeπLBj is open, closed and nonempty.

Hence it must be the wholeUkpk×A. This contradicts the assumption thatBj does not intersectAi.

One can construct a deformation retract retractingπL−10 (A)∩RZinsideπL−1(n(pk, α)×

A)∩RZ. Therefore there are exactlymconnected components ofπL−10 (A)∩RZ, each containing exactly one Ai. Therefore there is a unique map sA : πL−10 (A)∩RZ π−1L0(A)∩RL0,Z0 which is identity on π−1L0(A)∩RL0,Z0 and makes the diagram below commutative.

πL−10 (A)∩RZ −−−−→ Ukpk×A

sA

 y

 y πL−10 (A)∩RL0,Z0 −−−−→ A

Patching these sA together gives s as required proving the first statement of the proposition.

Take a meromorphic form ω on Z which is holomorphic outside the special subvarieties ofZ. The formω can be written as

ω = f dt1∧dt2∧ · · · ∧dtn,

where f is a rational function on Z. Let K be the field of fractions of Z0. Then SpecK is a curve and ∆(SpecK) is a point. Therefore the algebraic residue res∆(SpecK)ρf dt1 is defined. We have

Proposition 3.1.21. Put

ω0 = (res∆(SpecK)ρf dt1)dt2∧ · · · ∧dtn. Then

resintL,Zω = (−1)d−12πi resintL0,Z0 ω0. Proof. By the definition

resU,L,Zω = Z

ha∩RL,Z•

ω.

Sinceρ0 is an isomorphism, Z

ha∩RL,Z•

ω = Z

ρ0∗(ha∩RL,Z•)

ρω.

We have

ρ0∗(ha∩RL,Z) = (−1)d(d−1)2 Re0 ρπLhc

= (−1)d(d−1)2 Re0 τe(hck1×(RL0,Z0 ∩πL0h0c)),

wherehck1 is the circle inRpkk11 and h0c is the product of circles inRpL00. By Fubini’s theorem

Z

Re0fτ(hck1×(RL0,Z0

∩πL0h0c))

ρω = Z

RL0,Z0

∩πL0h0c

g(z0)dt2∧ · · · ∧dtd,

where forz0 ∈RL0,Z0

g(z0) = Z

Re0∩τ−1z0∩π−1k1hck1

ρf dt1. The last integral is nothing else than

2πi res∆(z0)ρf dt1. Therefore

g(z0)dt2∧ · · · ∧dtd = 2πiω0 Taking into account that

RL0,Z0 ∩πL0h0c = (−1)(d−1)(d−2)2 RL0,Z0 ∩ha0,Z0,

wherea0 is the cell obtained fromaby replacing the componenta(ηk1, pk1) with the component a(pk1), we obtain the statement.

Let us denote by resL,Zω the iterated algebraic residue ofω with respect to the flag Z. This is defined by induction on the dimension of Z by the formula

resL,Zf dt1∧ · · · ∧dtd = resL0,Z0(res∆(SpecK)ρf dt1)dt2∧ · · · ∧dtd. Thus we obtain a formula for our residue.

Corollary 3.1.22.

resintL,Zω = (−1)d(d−1)2 (2πi)dresL,Zω.

For any subvariety Z X and any hyperform using the decomposition of the residue according to flags we also can state a formula for the trace:

Corollary 3.1.23. For a meromorphic hyperformω of degree 2donZ⊂X1× · · · × Xn which is holomorphic outside the special subvarieties of Z, dimZ =d,

TrintZ ω= (−1)d(d−1)2 (2πi)d X

L⊂{1,...,n},|L|=dimZ

X

Z∈FlL(Z)

resL,ZωaL(Z).

Dropping the coefficient (−1)d(d−1)2 (2πi)dwe may also define the algebraic version of the trace:

TrZω := X

L⊂{1,...,n},|L|=dimZ

X

Z∈FlL(Z)

resL,ZωaL(Z). Then the latter corollary can be reformulated as follows:

Corollary 3.1.24. For a meromorphic hyperformω of degree 2donZ⊂X1× · · · × Xn which is holomorphic outside the special subvarieties of Z, dimZ =d,

TrintZ ω = (−1)d(d−1)2 (2πi)dTrZω.

Now we make a summary of the computation of TrZω. The necessary notation is explained in Section 3.1.10.

(i) List all the subsetsL⊂ {1, . . . , n}of size dimZ.

(ii) For each Llist all the flags Z FlL(Z).

(iii) For each flag find the cell aL(Z) of ∆m1 × · · · ×mn, which is a product of several vertices and edges (vertices for k∈Lc and edges fork∈L).

(iv) Compute the iterated residue corresponding to the flag Z of the form which is the component of ω on the open setUaL(Z) =×kUk, where Uk is an open set from the cover of Xk fork∈Lc, or an intersection of two open sets of the cover ofXk fork∈L.

(v) Add these residues.