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CAPÍTULO II: MARCO TEÓRICO

2.11.5 Diagrama de flujo y Flujograma

In this section, we recall the notion of Koszul free divisors and their basic properties. Furthermore, we describe the theory of the logarithmic Spencer complex in the case of Koszul free divisors.

Definition 2.7.1. ([8], Definition 4.1.1) Let D ⊂Cn be a divisor. We say that D

is a Koszul free divisor atx∈Cn if it is free at x and there exists a basisδ1, . . . , δn

of Derx(−logD) such that the sequence of symbols σ(δ1), . . . , σ(δn) is regular in

GrF•(D

Cn,x). IfD is a Koszul free divisor at every point, we simply say that it is a

Koszul free divisor.

Notice that for a free divisorD, to be Koszul is equivalent to being holonomic in the sense of Definition 3.8 from [48], i.e. the logarithmic stratification ofDis locally finite. See [25], Theorem 7.4.

Remark 2.7.2. If a basis of Derx(−logD) satisfies the condition of Definition

2.7.1, then every basis does.

Example 2.7.3. 1. The normal crossing divisor of Example 1.1.20 is Koszul free.

2. ([11], Example 2.8)Consider the free divisor D=V(28z3−27x2z2+ 24x4z+ 2432xy2z−22x3y2−33y4)⊂C3 with Saito matrix

[δ1, δ2, δ3] =    6y 4x2−48z 2x 8z−2x2 12xy 3y −xy 9y216xz 4z   .

Then the sequence of symbolsσ(δ1), σ(δ2), σ(δ3) is regular in GrF•(D

C3,x) for

anyx∈C3 and so D is Koszul free.

3. ([11], Example 4.2)Consider the free divisorD=V(xy(x+y)(y+xz))⊂C3

with Saito matrix

   x x2 0 y −y2 0 0 −z(x+y) xz+y   .

ThenD is Euler-homogeneous but is not Koszul free.

Remark 2.7.4. ([11], Remark 2.4) LetD⊂Cn be a free divisor. ThenDis Koszul

atx if and only if depth((σ(δ1), . . . , σ(δn)),GrF•(D

Cn,x)) =n, where δ1, . . . , δn is a

basis forDerx(−logD). Furthermore, by coherence, if a divisor is Koszul free at a

point, then it is a Koszul free divisor near that point.

Proposition 2.7.5. ([10], Example 1.11) Let D ⊂ C2 be a reduced divisor. Then

Proof. Suppose that f is a local reduced equation of D at x ∈C2. Der

x(−logD)

is a reflexive OC2,x-module and hence, it is free. So, we have only to check that

the symbols σ1, σ2 of a basis δ1, δ2 of Derx(−logD) form a GrF•(D

Cn,x)-regular

sequence. Let us suppose they are not. Then they have a common factorg∈ OC2,x,

because they are symbols of order 1. If g is a unit, we divide one of them by g

and eliminate the common factor. If g is not a unit, it would be in contradiction with Proposition 1.1.16, because the determinant of the Saito matrix of the basis

δ1, δ2 would have as factor g2, with g not invertible, while this determinant has to

be equal tof multiplied by a unit.

The Koszul free divisor behave well under products. In fact, we have the following:

Proposition 2.7.6. ([10], Proposition 1.10 1)) Let D ⊂Cn be a divisor such that

there exists a divisorD0 ⊂Cn−1 and D=D0×

C. ThenD is a Koszul free divisor

if and only if D0 is a Koszul free divisor.

Proposition 2.7.7. ([10], Proposition 1.10 2)) Let D ⊂ Cn and D0

Cr be two

divisors. Then

1. the divisor(D×Cr)S(Cn×D0)⊂Cn+r is free if D and D0 are both free;

2. the divisor (D×Cr)S

(Cn×D0) ⊂Cn+r is Koszul free if D and D0 are both

Koszul free.

Proposition 2.7.8. ([10], Corollary 4.2)LetD⊂Cnbe a free divisor and letΣ⊂D

be a discrete set of points. If D is Koszul free at all y ∈ D\Σ, then D is Koszul free.

Proposition 2.7.9. ([10], Theorem 4.3)Every locally quasi-homogeneous free divi- sor is Koszul free.

Corollary 2.7.10. Every free divisor that is locally quasi-homogeneous at the com- plement of a discrete set is Koszul free.

Proposition 2.7.11. ([8], Proposition 4.1.2) Let D⊂ Cn be a Koszul free divisor

atx∈Cn and considerδ

1, . . . , δn a basis of Derx(−logD). Then we have that

σ(DCn,x1, . . . , δn)) = GrF•(D

Cn,x)(σ(δ1), . . . , σ(δn)).

The following Theorem is a generalisation of Proposition 4.1.3 from [8]. The statement, without a proof, is already present in Section 1.2 of [12].

Theorem 2.7.12. Let D ⊂Cn be a Koszul free divisor and let M be a V

0(DCn)-

module that is locally free of finite rank over OCn. Then the complex D

Cn⊗V0(DCn)

Sp•(logD)(M) is concentrated in degree 0.

Proof. We can work locally. Fix a pointx∈D and a reduced equationf forD at

x. To prove that the complex DCn,xV0(D

Cn,x)Sp

(logD)(M)

x is concentrated in

degree zero, we define a filtration G• such that the graded complex has the same property. Consider Gk(DCn,xO Cn,x p ^ Derx(−logD)⊗OCn,xMx) :=Fk−p(DCn,x)⊗O Cn,x p ^ Derx(−logD)⊗OCn,xMx.

Clearly, this filtration is compatible with the differentials of the complex. Consider now the complexDCn,xV0(D

Cn,x)Sp

(logD)

x with the filtration

Gk(DCn,xO Cn,x p ^ Derx(−logD)) :=Fk−p(DCn,x)⊗OCn,x p ^ Derx(−logD),

also in this case, this filtration is compatible with the differentials of the complex. Hence, we have that

GrG•(D

Cn,x⊗V0(DCn,x)Sp•(logD)(M)x)

= GrG•(D

Cn,x⊗V0(DCn,x)Sp•(logD)x)⊗OCn,x Mx.

Because M is free and hence flat over OCn,x, to conclude is enough to show that

GrG•(D

Cn,x⊗V0(DCn,x)Sp•(logD)x) is concentrated in degree zero. Because

GrG•(D Cn,x⊗OCn,x p ^ Derx(−logD)) = GrF•(D Cn,x)[−p]⊗OCn,x p ^ Derx(−logD), the complex GrG•(D

Cn,x⊗V0(DCn,x)Sp•(logD)x) looks like

0 //GrF•(D Cn,x)[−n]⊗OCn,x Vn Derx(−logD) ψ−n // · · · ψ−2 // GrF•(D Cn,x)[−1]⊗OCn,x V1 Derx(−logD) ψ−1 // GrF•(D Cn,x) //0,

where the local expression of the differential is defined by ψ−p(G⊗(δj1∧ · · · ∧δjp)) := p X i=1 (−1)i−1Gσ(δji)⊗(δj1∧ · · · ∧δcji∧ · ∧δjp),

forp= 2, . . . , n and forp= 1

ψ−1(g⊗δi) :=Gσ(δi),

whereδi, . . . , δn is a basis of Derx(−logD). This complex is the Koszul complex of

the ring GrF•(D

Cn,x) with respect to the sequence σ(δ1), . . . , σ(δn). By hypoth-

esis, D is Koszul free and hence, this sequence is regular and so, the complex GrG•(D

Cn,x⊗V0(DCn,x)Sp•(logD)x) is concentrated in degree zero.

We now present Proposition 4.1.2 from [8] as a corollary of the previous Theorem.

Corollary 2.7.13. Let D ⊂Cn be a Koszul free divisor and consider δ

1, . . . , δn a

basis ofDerx(−logD). Then the complexDCn,x⊗V0(DCn,x)Sp•(logD)xis a resolution

of the quotient module DCn,x DCn,x(δ1,...,δn).

Proof. It is enough to apply Theorem 2.7.12 to M=OCn.

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