Relative differential forms and complex polynomials

Bull. Sci. math.124, 7 (2000) 557–571 2000 Éditions scientifiques et médicales Elsevier SAS. Tous droits réservés

RELATIVE DIFFERENTIAL FORMS ANDCOMPLEX POLYNOMIALS

BY

PHILIPPE BONNETa,1, ALEXANDRU DIMCA b,2

a Laboratoire de Topologie, Université de Bourgogne, 21000 Dijon, Franceb Laboratoire de Mathématiques pures, Université de Bordeaux-1,

351cours de la Liberation, 33405 Talence, France

Manuscript presented by J.-P. FRANCCOISE, received in July 1999

1. The main results

Let f :Cn→ C be a non-constant polynomial function and setX =Cn, S = C. Let B ⊂ S be a finite set such that if we setS∗ = S \ B andX∗ = f −1(S∗), thenf :X∗ → S∗ is a locally trivial fibration with fibertypeF . F is called the generic fiber of the polynomialf .

For anyk, 06 k 6 n− 1, we have a local systemV k =Rkf∗(CX∗) onS∗ and hence a corresponding analytic Gauss–Manin connection(Vk,∇)which can be constructed using the relative differential forms inΩ∗X∗/S∗ ,see Deligne [4].

Let (Ω∗, d) denote the de Rham complex of global polynomialdifferential forms onX and(Ω∗f , d) the corresponding truncated relative

de Rham complex. HereΩjf =Ωj/df ∧Ωj−1 for 06 j < n andΩj

f = 0otherwise, and the relative differentiald is induced by the differential ofthe de Rham complex(Ω∗, d). The cohomology groupsHk(Ω∗f , d) havea naturalC[t]-module structure, induced byt[ω] = [fω].

There is a natural morphism transforming relative forms into globalsections of the bundlesVk , namely

sk :Hk(Ω∗f , d

)→H 0(S∗,Vk)1 E-mail: [email protected] E-mail: [email protected].

558 P. BONNET, A. DIMCA / Bull. Sci. math. 124 (2000) 557–571

given bysk(ω)(t)= [j ∗t (ω)] ∈ Hk(Ft ) wheret ∈ S∗, Ft = f −1(t) is thecorresponding fiber andjt :Ft→X is the inclusion.

Recall that f has (at most cohomological) isolated singularitiesincluding at infinity if there exists a relative compactificationf :X→S of f such thatf is proper and dimΣs = 0 for any s ∈ S, whereΣs :=suppEs with Es = ϕf−sRj∗CX. Here ϕ is the vanishing cyclefunctor andj :X→X is the inclusion, see [6]. This condition is satisfiedin particular if f has isolated singularities including at infinity in thesense of Siersma and Tibar [18,20]. Moreover,f is cohomologicallytame ifΣs ⊂X for anys ∈ S, see Sabbah [16] and Némethi and Sabbah[13].

We say that a spaceY is cohomologicallyp-connected ifbk(Y ) = 0for 06 k 6 p, wherebk(Y ) denotes thekth reduced Betti number ofY .

When f has isolated singularities including at infinity then thegeneric fiberF is cohomologically(n − 2)-connected and the specialfibers are cohomologically(n− 3)-connected, see Dimca and Saito [5],Theorem (0.2).

On the other hand, iff is cohomologically tame or M-tame, acondition introduced by Némethi and Zaharia in [14] which is moregeneral than Broughton’s tameness condition in [3], then all the fibersof f are cohomologically(n− 2)-connected.

Our main results are the following.

THEOREM 1. –Assume thatf has isolated singularities onCn andthat the generic fiberF is cohomologically(n− 2)-connected. Then

(i) H 0(Ω∗f , d)=C[f ] andHk(Ω∗f , d)= 0 for 0< k < n− 1;(ii) if all the fibers off are cohomologically(n− 2)-connected, then

theC[t]-moduleHn−1(Ω∗f , d) is torsion free and the morphismsn−1 is injective;

(iii) the C[t]-moduleHn−1(Ω∗f , d) is free and of finite rank iff iscohomologically tame or M-tame.

When the conditions in (iii) are fulfilled, then the rank of theC[t]-moduleHn−1(Ω∗f , d) is equal toµ(f )= dim(Ωn/df ∧Ωn−1), the totalMilnor number of the polynomialf .

In some cases one can produce an explicit basis of thisC[t]-module.Here is such a situation. Letw = (w1, . . . ,wn) be a system of rationalpositive weights such thatf = fa0 + fa1 + · · · + f1 with 06 a0 < a1 <

· · ·< 1, fa being weighted homogeneous of degreea with respect to the

P. BONNET, A. DIMCA / Bull. Sci. math. 124 (2000) 557–571 559

weightsw andf1 having an isolated singularity at the origin ofX. Whensuch a system of weights exists, we call the polynomialf semi-weightedhomogeneous of typew. Then we have the following.

PROPOSITION 2. –Let f be a semi-weighted homogeneous polyno-mial. Thenf is M-tame andµ(f )= µ(f1)=∏i=1,n(w

−1i − 1).

Let xa be a monomial basis of the Milnor algebra for the polynomialf1 and ωa = xaΩ whereΩ = ∑i=1,n(−1)i−1wixi dx1 ∧ · · · ∧ ˆdxi ∧· · · ∧ dxn . Then theµ(f ) formsωa give rise to a basis of theC[t]-moduleHn−1(Ωf , d) such that any[ω] ∈Hn−1(Ωf , d) can be written as[ω] =∑a ca(t)[ωa] for some polynomialsca(t) ∈ C[t] with degca(t) 6degω− degωa.

Here the degree of a form is computed with respect to the weightswby setting degxi = degdxi =wi for all i = 1, . . . , n.

Our proof of Theorem 1 is based on recent results on the algebraicGauss–Manin system associated to a polynomial, see [6,13] and [16],while the proof of Proposition 2 is essentially elementary and depends ona classical result by Pham [15].

These relations with D-module theory are described in Section 2 wherewe prove Theorem 1 and Proposition 2. In the final two sections ofthis paper we describe in detail the kernel of the morphisms1 andthe torsion of theC[t]-moduleH 1(Ω∗f , d) in the casen = 2 when thepolynomialf has non-isolated singularities and/or non-connected fibers,see Proposition 7 and Proposition 11.

L. Gavrilov has obtained some of the above results in the casen = 2using completely different techniques, see [8]. On the other hand, ourTheorem 1(iii) gives a positive answer to a conjecture raised by Gavrilovin [9].

Note also that recently A. Douai has obtained closely related results toProposition 2 in the case of polynomials which are non-degenerated andconvenient with respect to their Newton polyhedron at infinity, see [7].

2. Relative differential forms and D-modules

The algebraic Gauss–Manin system off :X→ S is the direct imageKf = f+OX[−n] of theDX-moduleOX, see Borel [2]. We shift it by(−n) to get a complex in positive degrees, as it is more usual in algebraictopology.


At the level of global sections onS, the algebraic Gauss–Manin systemof f is represented by the complex ofA1-modulesK∗f = (Ω∗[∂], df )whereA1=C[t]< ∂ > is the Weyl algebra and theC -linear differentialdf is defined bydf (ω∂m)= dω ∂m − df ∧ω∂m+1, see [16,5] and [6] formore on this complex.

The cohomology sheavesGif = Hi(Kf ) are regular holonomicDS-modules and the Riemann–Hilbert correspondence [2] implies thatDRS(Gif )= pRif∗CX.

HerepRif∗CX = pHi(f∗CX) andpHi denotes the perverse cohomol-ogy functor. Basic properties of perverse sheaves gives the following iso-morphism

pRif∗CX =Ri−1f∗CX[1],(∗)see [16] or [6] for a proof. In this way theDS-modules Gif , orequivalently, theA1-modulesGi

f = Hi(K∗f ) are related to the topologyof the polynomialf .

In particular, Lemma 1 in [1] implies the following.

LEMMA 3. – If the generic fiberF of the polynomialf is cohomolog-ically (n− 2)-connected, thenGif =Gi

f = 0 for i = 0 and1< i < n andG1f =OSdf ,G1

f =C[t]df .

Proof. –Lemma 1 in [1] shows that the tubes around the fibers off arecohomologically(n− 2)-connected when the generic fiberF is so. Thisis equivalent toRif∗CX = 0 for 0< i < n− 1 andR0f∗CX = CS . Theresult follows by the properties of the de Rham functorDRS mentionedabove. 2

Note thatdf∧ : (Ω∗f , d)→ (K∗f , df ) [1] is a complex morphism up-tosign. This induces morphismsdf∧ :Hk(Ω∗f , d)→Hk+1(K∗f , df ) whichareC[t]-linear.

LEMMA 4. –The morphismdf∧ :Hk(Ωf , d)→Hk+1(Kf , df ) is anisomorphism fork < n − dim Sing(f ) − 1 and a monomorphism fork = n−dimSing(f )−1, whereSing(f ) denotes the singular locus off .

Proof. –Let (Ω∗, df∧) be the Koszul complex of the partial deriv-atives off . Then the cohomology groupsHk(Ω∗, df∧) are trivial fork < n−dim Sing(f ). Indeed, the cohomology groups of the Koszul com-plex are finitely generatedC[x1, . . . , xn]-modules and to prove that one


of them is trivial it is enough to show that all its localizations at maximalideals are trivial.

To check this local property we can use GAGA and replace algebraiclocalization by analytic localization. At this level the result follows fromGreuel’s generalized version of the de Rham Lemma, see [10], (1.7) orfrom a general result in Looijenga’s book [11], namely Corollary (8.16),p. 157 (takeX a smooth germ andk = 1 in that statement).

The rest of the proof is straightforward.2Proof of Theorem 1. –The first claim (i) follows directly from the

Lemmas 3 and 4 above.LetGf denote theA1-moduleGn

f andG(0)f theC[t]-submodule ofGf

given byΩn/df ∧ dΩn−2. Note that the image of the monomorphismdf ∧ from Lemma 4 is exactly theC[t]-submoduleG(−1)

f = df ∧Ωn−1/df ∧ dΩn−2 in G(0)

f .If all the fibers off are cohomologically(n−2)-connected, it follows

from [6], Theorem (1.4) that theC[t]-moduleG(0)f is torsion free. This

implies thatHn−1(Ω∗f , d) is torsion free, i.e., the first part of (ii).To prove the second part of (ii), we need to explain the relation

between theA1-moduleGf and the vector bundleVn−1. Assume thatthe restriction of the correspondingDS-moduleGf to S∗ is a connection,i.e., a locally freeOS∗-module of finite type. This can always be achievedby replacingS∗ by a smaller Zariski open subset ofS.

Then we have an isomorphism(Gf |S∗)⊗OS∗ OanS∗ = Vn−1 which givesat the level of global sections a monomorphism

ι :H 0(S∗,Gf |S∗)=Gf ⊗OS(S)OS∗(S∗)→H 0(S∗,Vn−1).Let ` :Gf → Gf ⊗OS(S) OS∗(S∗) denote the localisation morphism,whereOS(S)=C[t].

Then the compositionα = ι ` is given by the following formula

α

[∑ak∂

k

](t)=Res

(∑k!ak(f − t)−k−1

)∈Hn−1(Ft ),

whereak ∈Ωn and Res :Hn(Cn \Ft)→Hn−1(Ft ) is the Leray–Poincaréresidue, see for instance [17].

This formula implies thatsn−1= α (df∧).


Finally, (df∧) is injective by Lemma 4. whileα is injective since isinjective as there is noC[t]-torsion inG(0)

f as we have seen above. Thiscompletes the proof of claim (ii).

The last claim is a direct consequence of the fact that theC[t]-moduleG(0)f is of finite type whenf is cohomologically tame or M-tame, see

[13,16] and [6].In fact, since the quotientG(0)

f /G(−1)f is finite dimensional, it follows

that for a polynomialf with isolated singularities including at infinity,the C[t]-module Hn−1(Ω∗f , d) is of finite rank if and only if f iscohomologically tame, see [6].2

Proof of Proposition 2. –The fact that a semi-weighted homogeneouspolynomial is tame, and hence M-tame, is well known, see [3].

Consider the degree of differential forms inΩ∗ with respect to theweightsw as introduced in Section 1. This degree induces increasingpositive rational filtrationsF onG(0)

f ,G(−1)f andHn−1(Ω∗f , d) by setting

for instance

FsG(0)f =

[ω] ∈G(0)f ;deg(ω)6 s

.

In this way these three modules become filteredC[t]-modules, where thefiltration on the polynomial ringC[t] is the obvious one. 2

Note also that the corresponding modulesG(0)f1

, G(−1)f1

andHn−1(Ω∗f1,

d) associated to the leading termf1 are in a similar way gradedC[t]-modules. These graded objects are related to the graded objects inducedby the filtrationF on G(0)

f , G(−1)f andHn−1(Ω∗f , d) as shown in the

following.

LEMMA 5. –We have the following isomorphisms of gradedC[t]-modules

GrF(G(0)f

)=G(0)f1, GrF

(G(−1)f

)=G(−1)f1

and

GrF(Hn−1(Ω∗f , d)

)=Hn−1(Ω∗f1, d).

The proof of this result is straightforward and it is therefore left to thereader.

Let ηa = xadx1∧· · ·∧dxn. Pham shows in [15, p. 165], that the classes[ηa] form a basis for the gradedC[t]-moduleG(0)

f1. Note that the exterior


differential d induces aC-linear isomorphismd :Hn−1(Ω∗f1, d)→ G

(0)f1

such thatd([ωa])= |a|[ηa] where|a| = (a1+ 1)w1+ · · · + (an + 1)wn.This formula (and a similar result for any weighted homogeneous

form) implies thatd is (up to some multiplicative constants) compatiblewith the gradedC[t]-module structure and hence the formsωa producea basis for the gradedC[t]-moduleHn−1(Ω∗f1

, d). Proposition 2 nowfollows from basic graded/filtered module theory: a basis of the gradedobject lifts to give a basis of the initial filtered object.

This proof gives also the following.

COROLLARY 6. –The formsdf ∧ωa induce aC[t]-basis for the mod-uleG(−1)

f with similar properties relative to degrees as in Proposition2.

3. Non-isolated singularities and connected fibers in the casen= 2

In this section we assume thatn= 2 and that the generic fiberF of thepolynomialf is connected. Our main result here is the following.

PROPOSITION 7. –Letf be a polynomial as above.(i) If f has non-isolated singularities, then the morphism

df∧ :H 1(Ω∗f , d)→G(−1)f

has a non-trivial kernel, sayK(f ). ThenK(f ) = H 1(Ω∗, df∧)and the localizationK(f )b of the moduleK(f ) at b ∈ C is non-zero if and only if the corresponding fiberFb has non-isolatedsingularities. One also has(t − b)K(f )b = 0 for anyb ∈ C.

(ii) If all the fibers off are connected, then the restriction of thelocalisation morphism |G(−1)

f is injective.In particular, when both(i) and(ii) holds, thenKer(s1)=K(f ).Before giving the proof of this result, we describe the cohomology

group H 1(Ω∗, df∧) of the Koszul complex forfx and fy . Here tosimplify notation we setx1 = x, x2 = y and letfx, fy denote the partialderivatives off .

Let h = g.c.d.(fx, fy) and note thath divides∏b(f − b), where

the product is taken over allb such that the fiberFb has non-isolatedsingularities. In fact, for anys ∈ S we have a factorization into a productof distinct irreducible factorsf − s = f k1

1 · · · · · f kpp and it is easy


to see (please refer to the proof of Lemma 8 below for details) thath =∏s f

k1−11 · · · · · f kp−1

p (the factors from differents’s being distinct).Whenf has non-isolated singularities, thenh 6= 1.

Defineω(f )= h−1df . With this notation we have the following.

LEMMA 8. –H 1(Ω∗, df∧) a free C[x, y]/(h)-module spanned by[ω(f )]. In particular

∏b(t − b) · H 1(Ω∗, df∧) = 0 where the product∏

b(t − b) has the same meaning as above.

Proof of Lemma 8. –As in the proof of Lemma 4, we can localize inthe analytic sense and check the property at the origin ofC2. Assume thatf (0)= 0 and letf = f k1

1 · · · · · f kpp be the factorization into a product ofdistinct irreducible factors in the local ringOC2,0.

Note that the number of local branches at the origin can be larger thatthe number of irreducible components of the affine curveF0. In spiteof this, the germ of the formω(f ) at the origin is given byω(f )0 =∑i=1,p kif1 · · · · · fi · · · · · fpdfi (up to units inOC2,0).Since the nearby fibersFt for t 6= 0 are smooth, this form can be zero in

a neighborhood of the origin only at pointsa ∈ F0. There are two cases:(1) There are two indicesi 6= j such thatfi(a) = fj(a) = 0. Then

a = 0 since the branches corresponding tofi and fj meet only at theorigin.

(2) There is only one index, sayi = 1, such thatf1(a) = 0. Thenω(f )0(a)= 0 impliesdf1(a)= 0. Again we geta = 0 since the branchcorresponding tof1 is smooth outside the origin.

This shows that the two coefficients ofdx and dy in ω(f )0 vanishsimultaneously only at the origin, i.e., they form a system of parametersin the local ringOC2,0 and hence a regular sequence. At the origin,df ∧ ω = 0 is equivalent toω ∧ ω(f )0 = 0 and the result follows bySerre’s Theorem, see for instance [12, Section 18].2

Proof of Proposition 7. –(i) Let [ω] ∈K(f ). Thendf ∧ω= df ∧ dg for some polynomialg. It

follows thatdf ∧ (ω− dg)= 0, i.e., the class[ω] ∈K(f ) is the same asthe class[α] ∈K(f ) if [ω− dg] = [α] in H 1(Ω∗, df∧). It follows fromLemma 8 that any such formα is a multiple of the formω(f ).

To prove the isomorphismK(f )=H 1(Ω∗, df∧) we argue as follows.Let us denote byH the GCD off and its partial derivatives, and put:ω(f ) = df/H . Then, any 1-formω = dR + Qω(f ) is easily seen to


have torsion order6 1 with respect tof . Let us determine when such a1-form gives rise to the trivial class inK(f ).

If ω = dR + Qω(f ) = dR1 + Q1df , then: d(R − R1) ∧ df = 0;since the generic fiber is connected,(R − R1) belongs toC[f ], and(Q/H −Q1)df = P(f )df , whereP lies inC[t]. Therefore,H dividesQ and soω= dR+Qω(f ) is trivial if and only ifH dividesQ.

(ii) Let ω ∈ Ω2 be such that ([ω]) = 0. This means that there isa rational functionR ∈ C(x, y) such thatR = A/((f − b1)

m1 · · · (f −bq)

mq ) whereA ∈ C[x, y], the bifurcation setB is given byb1, . . . , bqandR satisfyingω = df ∧ dR.

We may assume thatb1 = 0, m1 > 0 and that∑i mi > 0 is minimal

with this property. Whenω = df ∧ α, then we may writeω =Hg dx ∧dy, whereg ∈ C[x, y], f = f k1

1 · · ·f kpp is the factorization off into

distinct irreducible factors, andH = f k1−11 · · ·f kp−1

p .To get a contradiction, it is enough to prove the following.

LEMMA 9. – If the fiberF0 and the general fiberF are connected anddf ∧ dA= fHg dx ∧ dy for someg ∈ C[x, y], then there is a constantc ∈C such thatA− c is divisible byf .

Proof. –We havedf ∧dA= 0 onF0, in particularA has to be constantalong the smooth part of each irreducible component ofF0 ( usefi = 0as a local equation for the smooth part ofF0). By continuity and sinceF0 is connected,A has to be constant onF0, sayA = c. ReplacingAby A − c we may assume thatA = 0 on F0. We have a factorizationA = Bf n1

1 · · · · · f npp , whereB ∈ C[x, y] is not divisible by anyfi .Moreoverki > 0 andni > 0 for all i = 1, . . . , p.

Assume that there is ani such thatni < ki and letu/v =minni/ki:i = 1, . . . , p< 1, with u, v relatively prime positive integers,v > 1.

Consider the rational functionC = Av/f u and note that in factCis a polynomial, by the definition ofu and v. Moreover, using therelationdf ∧ dC = vAv−1f −u df ∧ dA, we see that for anyi one hasdfi ∧ dC = pifi dx ∧ dy with pi a polynomial.

This implies thatC is constant on the curveF0. If say u/v = n1/k1,then the functionC is non-zero on the component ofF0 given byf1= 0.As a result,C has to be non-zero on the other components as well, andthis is possible only ifu/v = ni/ki for all i = 1, . . . , p. But then we haveki = vri for some positive integersri , and hencef = (f r11 · · · · · f rpp )v


with v > 1, in contradiction with the fact that the generic fiberF isconnected. 2

Remark10. – It is not true that the restriction of the localisation mor-phism `|G(0)

f is injective. As an easy example consider the polynomial

f = x2y. Thenω = dx ∧ dy has a non-zero class inG(0)f but `([ω])= 0

sinceω= df ∧d(xy/f ). In fact the same argument as in Lemma 9 showsthat the elements in the kernelK of `|G(0)

f have torsion order at most 1in the sense of the following section.

4. Non-connected fibers in the casen= 2

We are going to see that there are torsion elements of any order inthis case, and moreover we describe exactly how this torsion behaves.We start with some notation. For anyC[t]-moduleM we setT6n0 (M)=Ker(tn :M→M) andT n0 (M)= T 6n0 (M)/T

6n−10 (M).

Note that multiplication byt induces a monomorphismT n0 (M)→T n−1

0 (M).Whenm ∈ T6n0 (M) we say thatm has torsion order6 n. If m has

torsion order6 n but not6 n−1, then we say thatm has torsion ordern.In all the above notations we may of course replace 0 by anya ∈C and

t by t − a to obtained the corresponding torsion at the pointa.Our main result in this section is the following.

PROPOSITION 11. –Letf :C2→C be a polynomial with a connectedgeneric fiberF . Letr > 1 be the number of connected components of thefiberF0.

(i) If the fiber F0 is reduced, then any integern > 1, the spaceT n0 (H

1(Ω∗f , d)) of torsion elements of ordern in H 1(Ω∗f , d) hasdimensionr − 1.

(ii) If the fiber F0 is not reduced, then the first torsion quotientT 1

0 = T 10 (H

1(Ω∗f , d)) is infinite dimensional and the rest of the torsionquotients T n0 = T n0 (H 1(Ω∗f , d)) for n > 1 are finite dimensional ofdimension at leastr − 1.

More precisely, in the case (i) above we can describe an explicitbasis forT n0 = T n0 (H 1(Ω∗f , d)) as follows. In this casef factors asf = g1 . . . gr , where each polynomialgi is reduced and the curves


g−1i (0) correspond exactly to the connected components ofF0. For

any integern and any indexi = 1, . . . , r , there exist by Hilbert’sNullstellensatz some polynomialsΠn,i

1 ,Πn,i2 such that:Πn,i

1 (gi)n+1 +

Πn,i2 gn+1

1 . . . gn+1i−1 g

n+1i+1 . . . g

n+1r = 1. Let us put:

ωn,i = d(1−Πn,i1 gn+1

i

)/f n.

Then the classes of the formsωn,i, i = 1, . . . , r−1, yield a basis forT n0 inthe case (i) and linear independent elements in the case (ii). On the otherhand Example 15 below shows that the inequalities in Proposition 11(ii)can be either equalities or strict inequalities.

To prove these claims we need a couple of lemmas.

LEMMA 12. –Let ω a polynomial 1-form such thatf nω = dR +Qdf , withR,Q polynomials, i.e.,ω ∈ T6n0 (H 1(Ω∗f , d)). If ω has torsionorder< n then there exists a constantc such thatR− c is divisible byf .The converse implication holds when the fiberF0 is reduced.

Proof. –Let ω be such that:f nω = dR +Qdf, R,Q ∈ C[x, y], andassume that:f n−1ω = dR1+Q1df, R1,Q1 ∈C[x, y]. Then, we get thatd(R− fR1)∧ df = 0.

Since the generic fiberF is irreducible,R − fR1 belongs toC[f ]by Bertini’s first Theorem, see for instance Stein [19]. Therefore,R isdivisible byf up to a constant. Conversely, assume there exists a constantc such thatR − c is divisible byf . Then we writeR − c = fR1 and getthatf nω= dR +Qdf = f dR1+ (R1+Q)df .

Therefore, asf is reduced by assumption, it must divide(R1+Q). Put(R1+Q)= fQ1; we have thatf n−1ω = dR1+Q1df and the lemma isproved. 2

LEMMA 13. –ωn,i is a polynomial1-form of torsion ordern. More-over, if the linear combinationα1ω

n,1 + · · · + αrωn,r has torsion order< n thenα1 = · · · = αr . The converse implication holds when the fiberF0 is reduced.

Proof. –First, the existence ofΠn,i1 ,Π

n,i2 is obvious, asgi corresponds

exactly to a connected component ofF0, so thatgi and gj cannotvanish simultaneously; on one hand,d(1 − Π

n,i1 gn+1

i ) is divisibleby gn1 . . . g

ni−1g

ni+1 . . . g

nr and gni by differentiation, and soωn,i is a


polynomial 1-form. On the other hand, by definition, this form has torsion6 n with respect tof .

Second, denote for convenienceHn,i = 1− Πn,i1 gn+1

i , and note thatby constructionHn,i equals 1 ong−1

i (0) and vanishes on the othercomponents ofF0. If we consider a linear combinationα1ω

n,1 + · · · +αrω

n,r , then we can write that:

f n(α1ω

n,1+ · · · + αrωn,r)= d(α1Hn,1+ · · · + αrHn,r

).

Therefore, this 1-form has torsion order< n only if α1Hn,1+· · ·+αrHn,r

is divisible byf up to a constant. But, this implies thatα1Hn,1+ · · · +

αrHn,r is constant on the fiberF0, which leads toα1= · · · = αr . Whenf

is reduced, the converse implication is similar.2Proof of Proposition 11. –(i) First, we see by the previous lemma that the classes ofωn,ii=1,..,r−1

are linearly free inT n0 (H1(Ω∗f , d)). It suffices to check that they generate

this space.To that purpose, letω be any 1-form of torsion6 n, and write

f nω = dR + Qdf . Then, it is clear thatR is singular onF0, so it islocally constant on it. Therefore, by construction, there exists a linearcombinationα1H

n,1+ · · · + αr−1Hn,r−1 such thatR − (α1H

n,1+ · · · +αr−1H

n,r−1) is constant onF0; that implies by Lemmas 11 and 12 thatω − (α1ω

n,1 + · · · + αr−1ωn,r−1) has torsion order< n, which is the

required result.(ii) It is clear by the proof of Proposition 7(i) that the following map:

L :C[x, y]/(H)→ T 10 , [Q] 7→ [Qω(f )],

is injective, so that the quotientT 10 is infinite dimensional.

Since all the torsion in the kernelK(f ) is of order one, we have fork > 1 the following inclusion

T k0 = T k0(H 1(Ω∗f , d)

)= T k0 (G(−1))⊂ T k0 (Gf ).

SinceGf is a holonomicA1 module, it follows thatT k0 (Gf ) is a finitedimensional vector space, see [2, pp. 191–193].


Remark14. – When the fiberF0 has only isolated singularities, itfollows from Proposition 11(i) that

r − 1= dimT n0(H 1(Ω∗f , d)

)= dimT n0(G(−1)f

)for anyn. On the other hand we know from [6] that dimT 1

0 (G(0)f )= r−1

and dimT 10 (G

(1)f ) = n(F0) − 1, wheren(F0) denotes the number of

irreducible components ofF0 andG(m)f = ∂mG(0)

f for any m ∈ Z. Wealso know that dimT n0 (Gf )= n(F0)− 1 for anyn> 1, see [6]. It wouldbe interesting to have information on the dimension of all torsion spacesT n0 (G

(m)f ).

Here are two examples whose behaviour in terms of torsion is quitedifferent although the reduced curve corresponding to the fiberF0 is thesame in both cases. The second example shows that we cannot expectT k0 to have always dimensionr − 1 for k > 1, while this may happen asdepicted in the first one.

Example15. –(i) Let us consider the polynomialf = x(1+ xy)2. It is clear that its

generic fiberF is irreducible and thatF0 has two connected components.Moreover,f is singular over1+ xy = 0. We are going to check thatT n0 is 1-dimensional as soon asn > 1: To that purpose, let us see first thata polynomial solutionR of the equation

dR ∧ df = f (1+ xy)γ dx ∧ dy,that vanishes over the fibreF0 has to be divisible byf .

Let R be such a solution and write it asR = Sx(1 + xy). Then,following the proof of Lemma 9, we can consider the functionR2/f =S2x; it turns out that it is locally constant overF0. HenceS is divisibleby 1+ xy, i.e.,R has to be divisible byf .

Now, let us consider a torsion 1-formω, whose torsion order isn,wheren > 1, and writef nω= dR+Qdf . If we substract toω a multipleof ωn,1, we may assume thatR is divisible byx(1+ xy). SincedR has tobe divisible by(1+xy) and(1+xy) dividesdf , it is easy to see thatR isdivisible byf = x(1+ xy)2. If R = fR1, thendR1∧ df = f n−1ω∧ df .If we substract to it a multiple ofωn−1,1, thenR1 may be assumed to beconstant onF0, so that it is divisible byf andR is divisible byf 2. But


ωn−1,1 has torsion ordern− 1, thereforeω ≡ α1ωn,1+Ω in T n0 , where

f nΩ = d(f 2R1)+Q1 df .Such a 1-formΩ must have torsion order< n. Indeed,Q1 is divisible

by x(1+ xy) anddQ1 ∧ df = f (1+ xy)γ dx ∧ dy, soQ1 is divisibleby f ; putQ1= fQ2. Then, it is easy to check thatf n−1Ω = d(fR2)+(Q2 + R2) df , andΩ has torsion order< n. Soω ≡ α1ω

n,1 in T n0 andhenceT n0 is 1-dimensional for anyn > 1.

(ii) Let us consider the polynomialf = x2(1+ xy). It is clear that itsgeneric fiberF is irreducible and thatF0 has two connected components.Moreover,f is singular overx = 0. We are going to check thatT 2

0 is2-dimensional:

Using the same arguments as above, we can see that, for any 1-formω

of torsion6 2 we haveω≡ α1ω2,1+Ω in T 2

0 , whereΩ is a 1-form suchthatf 2Ω = d(f x(1+ xy)R1)+Q1 df .

Define

Ω1=−9y2 dx + (2− 3xy) dy.

An easy computation shows thatf 2Ω1 = d(4f x(1 + xy)) + Q2 df .Moreover, we get thatd(x(1 + xy)R1) ∧ df = fΩ ∧ df and that(R1)

2(1+ xy) is constant onx = 0. So, eitherR1 is zero onx = 0and we are led tof 2Ω = d(f 2R2) + Q1 df , either R1 is non-zeroconstant onx = 0, and there exists a constantµ1 such thatf 2Ω =µ1f

2Ω1 + d(f 2R2) +Q3df . As before, the last term of this equalitycorresponds to a 1-form of torsion order< 2, and we find the followingequality inT 2

0 :

ω≡ α1ω2,1+µ1Ω1.

So,T 20 is at most 2-dimensional, and there remains to check thatω2,1 and

Ω1 are linearly independent moduloT610 .

Assume that:ω = α1ω21 + µ1Ω1 is in T 62

0 ; thenf 2ω = dR +Qdf .As in the previous section, a necessary condition for this 1-form to havetorsion order< 2 is thatR be constant overF0; butR = α1(1−Π2,1x3)+µ14f x(1 + xy), so α1 = 0 and ω must be a multiple ofΩ1. SincefΩ1 = d(4x(1+ xy)) +Q2ω(f ), whereω(f ) = df/x andQ2 is notdivisible by x, we can see thatfΩ has torsion order 1 in virtue of theproof of Proposition 7(i). Therefore,Ω has torsion order 2,µ1 = 0, andthe result follows.


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Relative differential forms and complex polynomials