• 沒有找到結果。

Relation with the irregular Hodge filtration of E (v:u)f (∗H)

Part II. The case of a rescaled meromorphic function

9. Relation with the irregular Hodge filtration of E (v:u)f (∗H)

In this section, we set E := E(v:u)f(∗H). We will compare the filtration Fα+Ewith the irregular Hodge filtration Fα+irrE as defined in §5, namely, we consider the case where N = OX(∗D) (notation of §7.a) with its differential d twisted by the exponential of the rational function vf : X × P1v = X - - → P1. The module FfN considered in

§5.b is obtained here by gluing Eτ vf /z[∗H] (notation of Proposition 3.3) in the v-chart with Eτ f /uz[∗H] in the u-chart, and we will regard these modules algebraically with respect to τ , (v:u) and z. We will use the notation introduced in §3.

Theorem 9.1. For each α ∈ [0, 1), we have

Fα+Evf(∗H) = Fα+irrEvf(∗H),

Fα+G0Ef /u(∗H) = Fα+irrEf /u(∗H) ∩ G0Ef /u(∗H).

The proof of the theorem will be done in various steps. For the sake of simplicity, we will only treat the case where H = ∅.

Firstly, one identifies Eτ vf /z as a submodule of OX(∗Pred)[v, τ, z] · eτ vf /z and Eτ f /uz as a submodule of OX(∗Pred)[u, u−1, τ, z] · eτ f /uz. According to Proposition 3.4, we may have a strict inclusion only near points of Pred× {v = 0} and points of {f = 0} × {u = 0}. For the latter set, the computation is much simplified because we only consider the intersection with G0E. For the former set, we will need explicit computations of the V -filtration entering the very definition of Eτ vf /z in Proposi-tion 3.3.

Secondly, one computes the terms VατEτ vf /z (resp. VατEτ f /uz) of the V -filtration relative to τ = 0, in order to apply Proposition 5.5. We will work analytically with respect to the variables of X and algebraically with respect to τ , (u : v) and z.

9.a. Computation in the v-chart. We use the algebraic version (with respect to v, τ, z) RF(DX[v, τ ]h∂v, ∂τi) of RX×Cv×Cτ. Recall that Eτ vf /z is a coherent RF(DX[v, τ ]h∂v, ∂τi)-submodule of OX(∗Pred)[v, τ ] · eτ vf /z. We will set e = eτ vf /z. Computation away from Pred. Since τ vf is holomorphic, we have Eτ vf /z = OXrP

red[v, τ ] · eτ vf /z. Then, from the relation ðτe = vf e, we conclude that Eτ vf /z is already (RFDX[v, τ ]h∂vi)-coherent, and hence the Vτ-filtration is given by VkτEτ vf /z = τmax(−k,0)Eτ vf /z: by uniqueness of the Vτ-filtration, it is enough to check the strictness of the grVkτEτ vf /z, which is clear. Therefore, only α = 0

(3)We thank É. Mann, Th. Reichelt and Ch. Sevenheck for providing us with the necessary arguments.

is relevant. In particular, V0τEτ vf /z = Eτ vf /z. Hence, the quotient modulo (τ − z)Eτ vf /z is equal to Evf[z].

On the other hand, we have Fα+pEvf = Evf for any α ∈ [0, 1) and p > 0, and Fα−1Evf = 0, that is, RFEvf = Evf[z].

Computation in a neighbourhood of Pred. Near a point of Pred, let us set g = 1/f , which is holomorphic in a neighbourhood of this point. In local coordinates we have g = xe.

First step: computation of Eτ v/gz. By the very definition of Proposition 3.3(1) we have, on this neighbourhood, Eτ v/gz = (OX(∗Pred)[τ, v, z] · eτ v/gz)[∗Pred]. Let Then (δ ⊗ e)1+α satisfies the following equations:

(9.3) Similarly for (δ ⊗ e)<1+α the last line of (9.3) reads

ðxi(δ ⊗ e)<1+α= −ei

We have

We have a similar identification by using (δ ⊗ e)<1+α. Let us write the last line of (9.3) as the empty set is equal to one):

pa(s, z) =Q deduce that each element of U1+α can be written as

(9.6) X

a>0

ha(xI(a), v, τ, η, z)x−apa(η, z),

and the coefficient ehaof x−ain its decomposition (9.5) is ha(xI(a), v, τ, η, z)pa(η, z).

By uniqueness, we conclude that an element written as (9.5) belongs to U1+α if and only if pa(η, z) divides eha(xI(a), v, τ, η, z). In particular, the decomposition (9.6) is unique.

We wish to identify U1+α· (δ ⊗ e)1+αwith V1+αt (ig,+OX(∗Pred)[v, τ, z]) · eτ v/tzand U<1+α· (δ ⊗ e)<1+α with V<1+αt (ig,+OX(∗Pred)[v, τ, z]) · eτ v/tz. It is enough to check

(tðt+ (1 + α)z)mU1+α· (δ ⊗ e)1+α⊂ U<1+α· (δ ⊗ e)<1+α for m big enough and

U1+α· (δ ⊗ e)1+α/U<1+α· (δ ⊗ e)<1+α has no z-torsion (see [Sab05, Lem. 3.3.4 & §3.4.a]). For the first point, we set

Iα= {i | αei∈ Z} and, for a > 0, Iα(a) = Iα∩ I(a) and Iα(a)c= Iα∩ I(a)c. Then (δ ⊗ e)1+α= x−1(δ ⊗ e)<1+α, and we have the relation

Q

i∈Iα

xi− eixe−1iðvðτ) · (δ ⊗ e)<1+α= (−e)1tt+ αz)#Iα(δ ⊗ e)1+α

= (−e)1 tðt+ (1 + α)z#Iα

(δ ⊗ e)1+α. For the torsion-free assertion, let us consider a section (9.6) of U1+α and let us de-compose (in a unique way) ha(xI(a), v, τ, η, z) as

ha(xI(a), v, τ, η, z) = X

ε∈{0,1}Iα(a)

ha,α,ε(xI(a+1−ε), v, τ, η, z)xε,

where ha,α,ε is holomorphic in its x-variables and polynomial in v, τ, η, z. Then the decomposition (9.6) reads

X

a>0

X

ε∈{0,1}Iα(a)

ha,α,ε(xI(a+1−ε), v, τ, η, z)x−(a−ε)pa(η, z).

We now note that, for ε ∈ {0, 1}Iα(a), setting b = a + 1Iα− ε, we have pb,<α(η, z) = (η + αz)#Iα(b)c· pa(η, z).

The unique decomposition (9.6) can thus also be written uniquely as

(9.7) X

b>0

hb(xI(b), v, τ, η, z)x−b pb,<α(η, z)

(η + αz)#Iα(b)c · x1, with hb= ha,α,ε, where (a, ε) is defined by the following conditions:

ai= bi if i /∈ Iα,

ai= bi− 1 and εi= 0 if i ∈ Iαand bi>1, ai= 0 and εi= 1 if i ∈ Iα and bi= 0.

The condition that a section (9.7) · (δ ⊗ e)1+α= (9.7) · x−1(δ ⊗ e)<1+α belongs to U<1+α· (δ ⊗ e)<1+α now reads

∀ b > 0, (η + αz)#Iα(b)c divides hb(xI(b), v, τ, η, z).

It is therefore clear that a section of U1+α· (δ ⊗ e)1+αbelongs, when multiplied by z, to U<1+α· (δ ⊗ e)<1+α if and only if it already belongs to U<1+α· (δ ⊗ e)<1+α. In other words, U1+α· (δ ⊗ e)1+α

U<1+α· (δ ⊗ e)<1+α has no z-torsion.

We conclude that

V0tRF(DX[t, v, τ ]h∂t, ∂v, ∂τi) · (δ ⊗ e)1= U1· (δ ⊗ e)1

= V1t(ig,+OX(∗Pred)[v, τ, z]) · (δ ⊗ e)1, hence ig,+Eτ v/gzis generated by (δ ⊗ e)1. It follows that Eτ v/gzis generated by x−1e.

We will prove the analogue of (9.2 ∗) after ig,+, from which one deduces similarly (9.2 ∗). We first notice that the equality

V1t(ig,+Eτ v/gz) ∩ (τ − z)(ig,+OX(∗Pred)[v, τ, z]) · eτ v/tz= (τ − z)V1t(ig,+Eτ v/gz) immediately follows from the unique decomposition (9.6) of a local section of V1t(ig,+Eτ v/gz). To end the proof, it therefore suffices to produce a similar unique decomposition of local sections of V1+kt (ig,+Eτ v/gz) := Pk

j=0ðjtV1t(ig,+Eτ v/gz) for any k > 1. This is obtained by writing

ðkt(δ ⊗ e)1= x−keðktt′k(δ ⊗ e)1= x−ke

k−1Q

j=0

tt+ jz)(δ ⊗ e)1,

giving rise to a formula similar to (9.6) for sections of V1+kt (ig,+Eτ v/gz), which makes use of polynomials pa,k (k > 1), derived from pa,0like in [Sab97, Lem. 4.7].

Second step: computation of the Vτ-filtration of Eτ v/gz. For α ∈ [0, 1), let us set eα= eτ v/gz/x[αe]+1.

Lemma 9.8. The Vτ-filtration of Eτ v/gz satisfies

VατEτ v/gz= V0τRF(DX[v, τ ]h∂v, ∂τi) · eα ∀ α ∈ [0, 1).

Proof. Since we are only interested in giving the formula for VατEτ v/gz, we can as well work with the localized module Eτ v/gz−1] (see [Sab05, Lem. 3.4.1]). In such a way, we can write

eα= x−[αe](x−1e) = x⌈(1−α)e⌉τ−1ðv(x−1e),

showing that eα is a section of Eτ v/gz−1]. For α ∈ [0, 1) let us also set e= Q

i∈Iα

xi



eα=: x1eα and, for p ∈ Z,

Uα+pτ (Eτ v/gz−1]) = τ−pV0τRF(DX[v, τ ]h∂v, ∂τi) · eα

U<α+pτ (Eτ v/gz−1]) = τ−pV0τRF(DX[v, τ ]h∂v, ∂τi) · e, so that, clearly,

(9.9) U<α+pτ (Eτ v/gz−1]) ⊂ Uα+pτ (Eτ v/gz−1]).

For p 6 0, we will set Uα+pτ Eτ v/gz = Uα+pτ (Eτ v/gz−1]) and U<α+pτ Eτ v/gz = U<α+pτ (Eτ v/gz−1]). We will prove that Uτ(Eτ v/gz−1]) is the good Vτ-filtration of Eτ v/gz−1]. It is enough to prove that UατEτ v/gz = VατEτ v/gz for α ∈ [0, 1). The proof will be very similar to that of Lemma 9.2, although with the variable τ instead of the variable t.

By using (9.9), one first easily checks that Uα−1τ Eτ v/gz⊂ Uτ Eτ v/gz and (τ ðτ+ αz)#IαUατEτ v/gz ⊂ Uτ Eτ v/gz.

Indeed, the first point follows from the relation τ eα = xeðveα= xe−1ðve, and the second one follows from the relation

(τ ðτ+ αz)#Iαeα= (−1)#Iα

Due to the uniqueness of the Vτ-filtration, the assertion of the lemma would follow from the property that grUατEτ v/gz has no z-torsion. We will argue in a way similar to that of Lemma 9.2 by finding a suitable expression for the sections of UατEτ v/gz.

Let us decompose OX[x−1, z][v, τ ] as OX[x−1, z][v, vτ ] ⊕ τ OX[x−1, z][vτ, τ ]. Due We thus obtain an isomorphism of free OX[x−1, z]-modules:

OX[x−1, z][v, τ ] · e−→ O X[x−1, z, v][τ ðτ] ⊕ τ OX[x−1, z, τ ]hτ ðτi

· e.

We can replace e with eαor ein the above isomorphism. We will express UατEτ v/gz and Uτ Eτ v/gz as sub-OX[z]-modules of the right-hand side, and by using the gen-erator e in both cases, to make the computation of the quotient module easier.

We note first that UατEτ v/gz = OX[v, z]hðx, ðv, τ ðτi · eα, i.e., we can forget the for some nonzero constants ⋆ and with pa(s, z) defined by (9.4), we conclude that, through the isomorphism (9.10), holo-morphic in its x-variables and polynomial in τ ðτ, z.

Let us check that the decomposition (9.11) is unique. The coefficient h(n)of τn(n > 0) is uniquely determined by the section. If n = 0, the function h(0)∈ OX,0[x−1, v, η, z]

decomposes uniquely asP

a>0h(0)a (xI(a), v, η, z)x−a. Thus h(0)a must be divisible by pa(η, z) and this determines uniquely ha(xI(a), v, η, z). We argue similarly for n > 0 and h(n)∈ OX,0[x−1, η, z].

There is a similar decomposition for sections of Uτ Eτ v/gz, by replacing pa(τ ðτ, z) · eα with pa,<α(τ ðτ, z) · e. In order to check whether a section (9.11) belongs to Uτ Eτ v/gz, we replace eα with x−1e. Let us decompose (in a unique way) ha(xI(a), v, τ ðτ, z) as

ha(xI(a), v, τ ðτ, z) = X

ε∈{0,1}Iα(a)

ha,α,ε(xI(a+1−ε), v, τ ðτ, z)xε,

where ha,α,ε is holomorphic in its x-variables and polynomial in v, τ ðτ, z. We have a similar decomposition for ga,α,n(xI(a+ne), τ ðτ, z). Then the decomposition (9.11) reads

X

a>0

X

ε∈{0,1}Iα(a)

ha,α,ε(xI(a+1−ε), v, τ ðτ, z)x−(a+1−ε)pa(τ ðτ, z) · e

+X

a>0

X

n>0

X

ε∈{0,1}Iα(a+ne)

τnga,α,n(xI(a+ne+1−ε), τ ðτ, z) ·

· x−(a+ne+1−ε)pa(τ ðτ+ nz, z) · e. We now note that, for ε ∈ {0, 1}Iα(a), setting b = a + 1Iα− ε, we have

pb,<α(s, z) = (s + αz)#Iα(b)c· pa(s, z).

The unique decomposition (9.11) can thus also be written uniquely as

(9.12) X

b>0

hb(xI(b), v, τ ðτ, z)x−b pb,<α(τ ðτ, z)

(τ ðτ+ αz)#Iα(b)c · e

+X

b>0

X

n>0

τngb,α,n(xI(b+ne), v, τ ðτ, z)x−(b+ne) pb,<α(τ ðτ+ nz, z)

(τ ðτ+ (n + α)z)#Iα(b)c · e

with hb= ha,α,ε, where (a, ε) is defined by the following conditions:

ai= bi if i /∈ Iα,

ai= bi− 1 and εi= 0 if i ∈ Iαand bi>1, ai= 0 and εi= 1 if i ∈ Iα and bi= 0,

and similarly for gb,α,n. The condition that a section (9.12) belongs to Uτ Eτ v/gz now reads

∀ b > 0, (τ ðτ+ αz)#Iα(b)c divides hb(xI(b), v, τ ðτ, z),

∀ b > 0, ∀ n > 0, (τ ðτ+ (n + α)z)#Iα(b)c divides gb,α,n(xI(b+ne), τ ðτ, z).

and

It is therefore clear that a section (9.12) of UατEτ v/gzbelongs, when multiplied by z, to Uτ Eτ v/gzif and only if it already belongs to Uτ Eτ v/gz. In other words, grUατEτ v/gz has no z-torsion.

Third step: End the proof of Theorem 9.1 in the v-chart. We will now use the expression of Lemma 9.8 to regard VατEτ v/gz as an OX[v, τ, z]-submodule of OX[x−1, v, τ, z] · eα. We can write (locally on X near P ):

V0τ(RFDX[v, τ ]h∂v, ∂τi) = OX[v, τ, z]hðx, ðy, ðv, τ ðτi

and we notice that the action of τ ðτ on eα is equal to that of vðv, so we can for-get τ ðτ. We will also forget (y, ðy) which plays no significant role. Recall that the variables x are indexed as x1, . . . , x. Working now within OX[x−1, v, τ, z] · eα, we have by induction on |a|,

(xðx)aðcveα≡ z|a|+cx−ce

X|a|

j=0

qa,c,j(x)x−jevj



eα mod (τ − z)OX[x−1, v, τ, z], where qa,c,j(x) is some polynomial and qa,c,|a|(x) is a nonzero constant. From this one concludes that there exist polynomials ra,c,j(x, z) with ra,c,|a|(x, z) constant, such that

ðaxðcveα≡ z|a|+cx−(|a|+c)e

X|a|

j=0

ra,c,j(x, z)x(|a|−j)e−avj



x−[αe]−1e

mod (τ − z)OX[x−1, v, τ, z], and since for 0 6 j 6 |a| we have |a − (|a| − j)e|+ 6j (see the end of the proof of Lemma 7.15), the coefficient of vjbelongs to FjOX(∗Pred)([(α+p)P ]) with p = |a|+c.

Using that

(τ − z)VατEτ v/gz = (τ − z)Eτ v/gz∩ VατEτ v/gz (see [ESY15, Proof of Prop. 3.1.2])

= (τ − z)OX[x−1, v, τ, z] ∩ VατEτ v/gz (Lemma 9.2), we conclude that the coefficient of zp in the gr VατEτ v/gz/(τ − z)VατEτ v/gz

(graded with respect to the z-adic filtration) is contained in Fα+pEvf, so Fα+pirr Evf ⊂ Fα+pEvf, according to Remark 5.6.

In order to obtain the reverse inclusion, we remark that, for |a| + c = p fixed, x−av|a|evf/x[(α+p)e]+1∈ F|a|OX(∗Pred)([(α + p)P ])v|a|evf is equal, up to a nonzero constant and modulo P

j<|a|FjOX(∗Pred)([(α + p)P ])vjevf, to the class of ðaxðcveα. We conclude by induction on |a|, the case where |a| = 0 being clear.

9.b. Computation in the u-chart

Computation away from Pred. We have Fα+pG0Ef /u = G0Ef /u = OXrP[u]ef /u for p > 0. We set similarly e = eτ f /uz.

Lemma 9.13. With respect to the inclusion Eτ f /uz⊂ OXrPred[u, u−1, τ, z] · e, we have e∈ Eτ f /uz.

Proof. The question is local near a point of {f = 0}, since otherwise we have equality in the previous inclusion, according to Proposition 3.4, and it amounts to proving that e ∈ V1u(OXrPred[u, u−1, τ, z] · e), so we are reduced to computing the order of e with respect to the Vu-filtration.

Let us first assume that the divisor {f = 0} has normal crossings. Let us choose local coordinates x1, . . . , xn such that f (x) = xm with m ∈ Nn (a local setting not

to be confused with that of §7.a). From the relation uðue= −(τ f /u)e we obtain ðxi(f e) = miz

xi

xme+mi

xi

τ f

u xme= −mi(uðu− z)xm−1ie, and iterating the process we find

ðmx (f e) = (−m)m Qn

i=1 mQi

j=1

(uðu− jz/mi) · e.

Since f e = uðτe, this gives a Bernstein relation for e showing that e ∈ V<0u (OXrPred[u, u−1, τ, z] · e).

When {f = 0} is arbitrary, the proof proceeds exactly like in [Kas76]. We work locally near a point of {f = 0} and we choose a projective birational morphism π : X → X which is an isomorphism away from {f = 0} and such that f := f ◦ π defines a normal crossing divisor. Using the global section eτ f/uzof V<0u Eτ f/uz(first part of the proof), one constructs a global section e of H0π+V<0u Eτ f/uz which coincides with e away from {f = 0}. This is done by using the global section 1X←X of RX←X[u, τ ]huðui. Because Eτ f/uz underlies a mixed twistor module, H0π+Eτ f/uz is strictly specializable along u = 0 and we have V<0u H0π+Eτ f/uz= H0π+V<0u Eτ f/uz. Therefore, e is a section of V<0u H0π+Eτ f/uz, and thus satisfies a non-trivial Bernstein equation of the form

Q

β<0

(uðu+ βz)νβ· e = uP (x, u, ðx, ðτ, uðu) · e.

We conclude that e satisfies the same equation away from {f = 0}, hence everywhere, since OX[u, u−1, τ, z] has no OX-torsion.

Due to the relation f τ ðue= −τ2ð2τewe conclude that

V0τEτ f /uz⊃ V0τRF(D(XrP )[u, τ ]h∂u, ∂τi) · e ⊃ OXrP[u, τ, z] · e.

Then, computing modulo (τ − z)Eτ f /uz, Fα+pirr Ef /u contains OXrP[u, z] · ef /u = G0Ef /u, hence Fα+pG0Ef /u= Fα+pirr G0Ef /u away from Pred.

Remark 9.14. The explicit computation of Fα+pirr Ef /u in the neighbourhood of f = 0 would be more complicated, and restricting to G0Ef /uallows us to avoid this compu-tation. Let us however note that, in the neighbourhood of the smooth locus of f−1(0), an explicit formula for Fα+pirr Ef /u can be obtained from Lemma 9.8 by setting there g = u and v = f . Since the order of the pole at u = 0 is one, the only interesting α is zero, and the result is:

FpirrEf /u= 1 up+1

Xp

j=0

OX[u]fk uk

· ef /u (f smooth).

This formula extends in a natural way to Fα+pirr Ef /u(∗H), provided that moreover f−1(0) has no common component with H and that f−1(0) ∪ H has normal crossings.

Computation near Pred. The computation is similar to, and even simpler than, the computation done in the v-chart. Indeed, due to Proposition 3.4, we have Eτ /uxez= OX(∗Pred)[u, u−1, τ, z] · eτ /uxez, and there is no need for an analogue of Lemma 9.2.

We will consider the variable u as part of the x-variables, and the divisor u = 0 of X (see §1.b). For α ∈ [0, 1), we set eα = eτ /uxez/ux[αe]+1. The following lemma is similar to Lemma 9.8.

Lemma 9.15. The Vτ-filtration of Eτ /uxez satisfies

VατEτ /uxez= V0τRF(DX[u, τ ]h∂u, ∂τi) · eα ∀ α ∈ [0, 1).

We also obtain

ðaxðbueα∈ z|a|+bF|a|+b OX(∗Pred)[u, u−1]

[(α + |a| + b)P]

· eτ /uxez

mod (τ − z)Eτ /uxez, where F OX(∗Pred)[u, u−1]

is the filtration by the order of the pole along Pred. Moreover, the coefficient of (uxe)−(|a|+b)· (ux[αe]+1)−1· eτ /uxezis a nonzero constant.

It follows that

Fα+pirr E1/uxe = Fp OX(∗Pred)[u, u−1]

[(α + p)P]

· e1/uxe. Intersecting with G0E1/uxe= OX(∗Pred)[u]e1/uxe gives

Fα+pirr E1/uxe∩ G0E1/uxe= FpOX(∗Pred) [(α + p)P ]

[u] · e1/uxe = Fα+pG0E1/uxe. This ends the proof of Theorem 9.1.

9.c. Another approach of Theorem 9.1 at the de Rham level. Let us assume that the zero divisor f−1(0) of f : X → P1 is smooth, that it has no component in common with D, and that f−1(0) ∪ D still has normal crossings. We have a filtration Fα+E (by using the formula given in Remark 9.14 in the u-chart). Then the proof of Theorem 9.1 gives in fact the equality Fα+E = Fα+irrE.

Let π : eX → X be a projective birational morphism such that vf extend as a morphism fvf : eX → P1 and eD := π−1(D) is a normal crossing divisor in eX. The pole divisor of vf is Pred and that of fvf , that we denote by ePred, is contained in π−1(Pred). We denote by eH the remaining components of eD. The construction of [ESY15] produces a filtration Fα+DelEfvf(∗ eH). Note that

π+Efvf(∗ eH) = H0π+Evff(∗ eH) = E(v:u)f(∗H) =: E.

By Theorem 1.3, the push-forward π+ Efvf(∗ eH), Fα+DelEfvf(∗ eH)

is strict, since Fα+DelEvff(∗ eH) = Fα+irrEfvf(∗ eH), and produces the filtration Fα+irrE, which is nothing but Fα+Eby Theorem 9.1 in the present setting. The strictness of the push-forward implies a quasi-isomorphism at the de Rham level:

(9.16) FαpDR E ≃ RπFDel,αp DR Efvf(∗ eH), where, as usual, we set for a filtered D-module (M, FM),

FpDR M = {F−pM−→ Ω1⊗ F−p+1M−→ · · · }.

We will show how to recover the quasi-isomorphism (9.16) for a suitable modification π : eX→ X by a direct computation. This will give, in the present setting, a proof of the degeneration at E1 of the spectral sequence attached to the hypercohomology of Hk(X, Fα+DR E) which only relies on [ESY15] for fvf : eX → P1, and not on the finer results of Theorem 1.3. However, the identification at the level of filtered D-modules, and not only at the level of filtered de Rham complexes, is needed for the application to Kontsevich bundles given in Theorem 1.11.

Let us set, for each p and α ∈ [0, 1), FN,αp DR E =n

pX(log D)([αP]) d + d(vf )

−−−−−−−−−→ Ωp+1X (log D)([(α + 1)P]) −→ · · ·o [−p].

Such a filtration already appeared in [Yu14] in the study of the toric case, where the notation FNPλ (∇) was used (NP for Newton polygon).

Lemma 9.17. The natural morphism FN,αp DR E → FαpDR E is a quasi-isomorphism.

Proof. Let us prove the lemma in the v-chart for instance (the proof in the u-chart is similar), and let us assume that H = ∅ for the sake of simplicity, so that D = Pred(the general case is obtained by a Kunneth formula). Recall that n = dim X. Everything below is thus only valid on X×Cv. Consider the following complexes, with differentials induced by d + d(vf ):

Φ1(p) :n

pX(log Pred) [αP]

−→ Ωp+1X (log Pred) [(α + 1)P]

−→ · · ·

−→ Ωn+1X (log Pred) [(α + n − p + 1)P]o [−p]

and, for k such that 1 < k 6 n − p + 2, Φk(p) :n

Φ<n−k+21 −→ F0OX(∗Pred)Ωn−k+2X [(α + n − p − k + 2)P]

−→ · · ·

−→k−1X

j=0

FjOX(∗Pred)vj

n+1X [(α + n − p + 1)P]o .

Then Φ1(p) = FNP,αp DR E and Φn−p+2(p) = FαpDR E. On each successive quotient grΦk = Φkk−1, the induced differential becomes −(v/xe)Peidxi/xi. Except at the first non-zero term, the complex grΦk decomposes into many parts of the Koszul complex associated with −(v/xe){e1dx1/x1, . . . , edx/x}. By a direct computation, the first non-zero chain map of grΦk is injective. In particular, grΦk is quasi-isomorphic to zero.

Let us now end the direct proof of (9.16). In the discussion of the toric case in [Yu14, §4], a specific resolution π : eX→ X of vf is constructed inductively by taking blowups along irreducible components of the intersection of the pole set Pred of vf with its zero set (f−1(0) × P1v) ∪ (X × {v = 0}). Then it is shown in loc. cit. that (9.16) holds when we replace its left-hand side with FN,αp DR E. Lemma 9.17 allows us to conclude.

Appendix. Brieskorn lattices and Hodge filtration

A.a. Brieskorn lattices in dimension one. Let (M, FM ) be a holonomic C[t]h∂ti-module equipped with a good filtration. We denote by G the holonomic C[t]h∂ti-module C[∂t, ∂t−1] ⊗C[∂t] M . If we identify C[t]h∂ti with C[v]h∂vi by the Laplace correspondence t 7→ ∂v, ∂t 7→ −v, we also regard G as a holonomic C[v]h∂vi-module on which the multiplication by v is bijective. It is therefore also a C[v, v−1]-module. We will denote by cloc the natural morphism M → G.

The Brieskorn lattice G(F0 ) of the filtration FM is defined as the saturation of the filtration by the operator ∂t−1, that is,

(∗) G(F0 ):=X

j

t−jloc(Fc jM ) ⊂ G.

It is naturally a C[∂t−1]-module (equivalently, a C[v−1]-module). We will also set Gp(F

)= v−pG(F0 )for any p ∈ Z. Let us make the link with the definition in [Sab08,

§1.d]. Let pobe an index of generation, so that Fpo+ℓM = FpoM + · · · + ∂tFpoM for any ℓ > 0. Then the definition in loc. cit. is

(∗∗) G(F0 )= ∂t−poX

j>0

t−jloc(Fc poM ).

Let us check that both definitions give the same result. Let us write (∗) as G(F0 )= ∂t−poX

j

t−jloc(Fc po+jM ).

Firstly, for j 6 0, we have

t−jloc(Fc po+jM ) = cloc(∂t−jFpo+jM ) ⊂ cloc(FpoM ), so we can also write

G(F0 )= ∂−pt oX

j>0

t−jloc(Fc po+jM )

= ∂−pt oX

j>0

t−j cloc(FpoM ) + · · · + ∂tjloc(Fc poM )

= ∂−pt oX

j>0

t−jloc(Fc poM ) = (∗∗).

We now express the Brieskorn lattice of the filtration as obtained by a push-forward operation. We consider the holonomic C[t, v]h∂t, ∂vi-module M [v, v−1]evt. The (t, v)-Brieskorn lattice is the C[t, v−1]-module defined by the following formula (see [Sab99, §1]):

G0(M, FM ) =L

j

FjM · v−jevt⊂ M [v, v−1]evt, Gp(M, FM ) = v−pG0(M, FM ).

We have ∂tGp(M, FM ) ⊂ Gp−1(M, FM ) since ∂tFjM ⊂ Fj+1M and ∂tevt= vevt. The relative de Rham complex

DR(M [v, v−1]evt) :=n

M [v, v−1]evtt

−−−→ M [v, v−1]evto

has cohomology in degree one only, and we have a natural identification as C[v]h∂v i-modules

cokerh

t: M [v, v−1]evt−→ M [v, v−1]evti

≃ G by sending P

jmjv−jevt to P

j(−∂t)−jloc(mc j). This relative de Rham complex is filtered by the subcomplexes

(A.1) GpDR(M [v, v−1]evt) :=n

Gp(M, FM )−−−→ G∂t p−1(M, FM )o .

Lemma A.2 (Push-forward). The relative de Rham complex is strictly filtered by the G-filtration and, through the previous identification, the filtration on its H1≃ G is equal to G(F−1

).

Proof. With respect to the previous identification, Gp(M, FM ) is sent onto Gp(F

)

according to the definition (∗). It remains to show that (im ∂t) ∩ Gp(M, FM ) =

tGp+1(M, FM ) and it is enough to check this for p = 0.

We have ∂t P

jmjv−jevt

=P

j(∂tmj+ mj+1)v−jevt and by induction on j we deduce that (∂tmj+ mj+1) ∈ FjM for all j implies mj∈ Fj−1M for all j.

Remark A.3 (Rees modules). It will also be useful to have the following interpretation in terms of Rees modules (see Proof of Theorem 4.1), for which we use the variable u instead of z here. We can twist the Rees module RFM by et/u by changing the action of u∂tto that of u∂t+ 1. We denote the corresponding RFC[t]h∂ti-module by RFM · et/u. This is nothing but G0(M, FM ) by the change of variable u = v−1.

The push-forward of an RFC[t]h∂ti-module M by the constant map q : Ct = Spec C[t] → Spec C is nothing but the de Rham complex ∂t: M → u−1M, where the latter term is in degree zero. The push-forward q+(RFM · et/u) is thus equal to the complex (with the in degree zero):

RFM · et/u−−−→ u∂t −1RFM · et/u

. Setting u = v−1 we thus have an identification

G0(M, FM ) ∂t

// G−1(M, FM )

RFM · et/ut

//u−1RFM · et/u so we can interpret G−1(F

)as H0q+(RFM · et/u), while H−1q+(RFM · et/u) = 0.

A.b. Brieskorn lattices in arbitrary dimension. We fix k and we will apply the previous result to (M, FM ) = Hk−dim Xf+(OX(∗D), (FOX(∗H))(∗Pred)). Here, we identify filtered C[t]h∂ti-modules and DP1(∗∞)-modules filtered by OP1(∗∞)-modules.

We know that the latter underlies a mixed Hodge module (up to a shift of the filtra-tion), according to [Sai90]. Working with Rees modules, the strictness property for the push-forward f+ of mixed Hodge modules can also be stated by saying that the push-forward f+

(RFOX(∗H))(∗Pred)

is strict, and thus Hk−dim Xf+ (RFOX(∗H))(∗Pred)

= RFM.

On the other hand, one checks that Hk−dim Xf+ (RFOX(∗H))(∗Pred) · ef /u

≃ Hk−dim Xf+(RFOX(∗H)(∗Pred)

· et/u, and since Hjq+(RFM · et/u) = 0 for j 6= 0, we conclude

Hk−dim X(q ◦ f )+ (RFOX(∗H))(∗Pred) · ef /u

≃ H0q+(RFM · et/u).

The left-hand term is by definition equal to

Hk X, (ΩXOX(uRFOX(∗H))(∗Pred), d + u−1df ) ,

that is, to G0Hku as defined by (8.12), while the right-hand term is equal to G−1 as defined above.

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