On the Torsion of Brieskorn Modules of Homogeneous Polynomials.
KHURRAMSHABBIR(*)
ABSTRACT- Letf 2C[X1; :::;Xn] be a homogeneous polynomial andB(f) be t he corresponding Brieskorn module. We describe the torsion of the Brieskorn moduleB(f) forn2and show that any torsion element has order 1. Forn>2, we find some examples in which the torsion order is strictly greater than 1.
1. The Milnor algebra and the Brieskorn module.
Letf 2RC[x1; :::;xn] be a homogeneous polynomial of degreed>1.
Then the Koszul complex of the partial derivativesfj @f
@xj;j1; :::;ninR can be identified to the complex
0 !V0 df!^ V1 df!^ ::: df!^ Vn 1 df^!Vn !0;
(1:1)
whereVjdenotes the regular differential forms of degreejonCn. Let Jf be the Jacobian ideal spanned by the partial derivatives fj, j1; :::;n, inRandM(f)R=Jf be the Milnor algebra off. One has the following obvious isomorphism of graded vector spaces
M(f)( n) Vn df^Vn 1: (1:2)
Here, for any gradedC[t]-moduleM, the shifted moduleM(m) is defined by settingM(m)sMmsfor alls2Z:
We define the (algebraic) Brieskorn module as the quotient B(f) Vn
df ^d(Vn 2) (1:3)
(*) Indirizzo dell'A.: 68-B, New Muslim Town, School of Mathematical Sciences, GCU Lahore, Pakistan.
E-mail: khurramsms@gmail.com
in analogy with the (analytic) local situation considered in [3], see also [11], as well as the submodule
C(f) df^Vn 1 df^d(Vn 2): (1:4)
These modules are modules over the ringC[t] and the multiplication bytis given by multiplying by the polynomialf. SometimesB(f) is denoted byG(0)f andC(f) byG( 1)f , see [2], [6].
One has the following basic relation between the Milnor algebra and the Brieskorn module, see [7], Prop. 1.6.
PROPOSITION1.1.
df^Vn 1df^d(Vn 2)fVn: In particular
B(f)=f:B(f)'M(f)( n):
Let B(f)tors be the submodule of C[t]-torsion elements in B(f) and define the reduced Brieskorn moduleB(f)B(f)=B(f)tors.
REMARK 1.2. The reduced Brieskorn module B(f) is known to be a free C[t]-module of rank bn 1(F), where F fx2Cnjf(x)1g is the affine Milnor fiber off. Indeed, it follows from the section (1.8) in [7] that one has a canonical isomorphism j0;d 1B(f)qdj Hn 1(F;C) for any qn. If we assume theC[t]-basis ofB(f) to be formed by homogeneous elements (which is always possible), each basis element will contribute by 1 to the dimension ofj0;d 1B(f)qdj forqlarge enough.
Moreover, it exists an integerN>0 such thattNB(f)tors0, see [7], Remark 1.7. The leastNsatisfying this condition is called thetorsion order off and is denoted byN(f).
REMARK1.3. As noted in [7], Remark 1.7, there is a slight difference between the Brieskorn module defined above and the Brieskorn module considered by Barlet and Saito in [1]. In fact, the latter one is defined to be then-th cohomology groupHn(Af), whereAjf ker (df^:Vj !Vj1) and the differential df :Afj ! Afj1 is induced by the exterior differential d:Vj!Vj1. Sincedf ^d(Vn 2)df(Anf 1), one has an epimorphism
B(f)!Hn(Af);
(note that the direction of this arrow is misstated in [7], Remark 1.7).
Moreover, whenn2, it follows from Proposition 3.7 in [1] thatH2(Af) is torsion-free. Our results in section 2 show that the torsion module B(f)torscan even be not of finite type in this setting, and hence the above epimorphism is far away from an isomorphism in general.
DEFINITION1.4. Forb2B(f), we say thatbist-torsion of orderkb1 if tkbb0 andtkb 1b60:
It is clear that such a torsion orderkbdivides the torsion orderN(f) of f, sincetN(f)b0 and thatN(f) is the G.C.D. of all thet-torsion orderskb forb2B(f) a torsion element.
1.5. The case of an isolated singularity.
Assume that f 2C[x1; :::;xn] is a homogeneous polynomial having an isolated singularity at the origin ofCn. Then the dimension of the Milnor algebraM(f) (as aC-vector space) is the Milnor number off atthe origin, denoted bym(f). One has in this casem(f)bn 1(F), see for instance [5]. In this case, the structure of the moduleB(f) is as simple as possible.
PROPOSITION1.6. TheC[t]-moduleB(f)is free of rankm(f):
PROOF. Indeed, a homogeneous polynomial f having an isolated sin- gularity at the origin induces a tame mapping f :Cn!C, to which the results in [6], [12] and [13] apply. Indeed, the Gauss-Manin systemGf off, which is anA1C[t]h@ti-module, contains aC[t]-submoduleG(0)f , which is known to be free of finite type for a tame polynomial, see for instance Re- mark 3.3 in [6]. As we have already mentionned above,G(0)f B(f). p
COROLLARY1.7. There is an isomorphism B(f)(n)M(f)CC[t]
of graded modules over the graded ringC[t]. In particular, one has, at the level of the associated Poincare series, the following equality.
PB(f)(t)PM(f)(t) tn
1 ttn(1 td 1)n (1 t)n1 :
EXAMPLE1.8. Forn3;takef(x;y;z)x3y3z3:Then aC-basis ofM(f) is given by 1;x;y;z;xy;yz;xz;xyz:The same 8 monomials form a
basis ofB(f) as a freeC[t]-module. In this case PB(f)(t)t3(1 t2)3
(1 t)4 t34t47t58t68t78t8 : COROLLARY1.9. TheC[t]-module B(f)is torsion free if and only if0 is an isolated singularity of the homogeneous polynomial f .
PROOF. If 0 is not an isolated singularity, then the Milnor algebra M(f) is an infinite dimensional C-vector space. The isomorphism in Proposition 1.1 implies that in this case B(f) is notfinitely generated overC[t]. By Remark 1.2, it follows that the canonical projection
B(f)!B(f)
is notan isomorphism, henceB(f)tors60: p Proposition 1.1 implies that f B(f)C(f), which in turn yields the following.
COROLLARY1.10. Assume that0is not an isolated singularity of the homogeneous polynomial f . Then N(f)1if and only if theC[t]-module C(f)is torsion free.
PROOF. Let b2B(f)tors be a non-zero element, which exists by our assumption. By Remark 1.2, it follows thatbist-torsion, say of orderk:If k>1;then
0tkbtk 1(tb):
By Proposition 1.1, we know thattb2C(f):IfC(f) is torsion free, we get thattb0 inC(f), i.e.tb0 inB(f), a contradiction. Hencek1 for anyb2B(f)tors, in other wordsN(f)1. p
2. The casen2.
In this section we suppose thatf 2C[x;y] is a homogeneous polynomial of degree d>1, which is notthe power gr of some other polynomial g2C[x;y] for somer>1:This condition is equivalent to asking the Mil- nor fiberFoffto be connected and such polynomials are sometimes called primitive. For more on this, see [8], final Remark, part (I).
PROPOSITION2.1. The submoduleC(f)is a freeC[t]-module of rank b1(F).
PROOF. The facttheC(f) is torsion free follows from Proposition 7. (ii) in [2]. Indeed, it is shown there that the localization morphism
`:C(f)!C(f)C[t]C[t;t 1]
is injective, which is clearly equivalent to the fact that there are not-torsion elements.
For the second claim, note that the composition of the inclusion C(f)!B(f) and the canonical projection B(f)!B(f) gives rise to an embedding of C(f) into B(f) whose image is exactly tB(f) (use again Proposition 1.1 and the fact that C(f) is torsion free). Since C[t] is a principal ideal domain, the claim follows from the structure theorem of submodules of free modules of finite rank over such rings. p
Corollary 1.10 implies the following:
COROLLARY 2.2. If f 2C[x;y] is a homogeneous polynomial with a non-isolated singularity at the origin, then N(f)1.
The above Corollary can be restated by saying that 0!B(f)tors!B(f)!t C(f)!0
is an exactsequence of gradedC[t]-modules. We getthus an isomorphism of gradedC[t]-modules
B(f)( d)'C(f):
EXAMPLE 2.3. Let f xpyq with (p;q)1. In order to compute the torsion of the Brieskorn moduleB(f);we start by findingC-vector space monomial bases forB(f)
C(f)'M(f) (up-to a shift in gradings) andC(f). Note that the Jacobian ideal is given by
Jf hxp 1yq;xpyq 1i xp 1yq 1hx;yi:
Itfollows thatB(f) C(f)' S
Jf is an infinite dimensionalC-vector space with a monomial basis given by xayb with ap 2 or bq 2 or (a;b)
(p 1;q 1):
We have to compute df ^dV0 fdf^dgg; where g is a polynomial function onC2. Since we are working with homogeneous polynomials, it is enough to work with one monomial, say (xayb);ata time. We have
df^d(xayb)(pb qa)xap 1ybq 1dx^dy:
Since
C(f) Jf V2
df^dV0xp 1yq 1hx;yiV2 df ^dV0 ;
a system of generators of theC-vector spaceC(f) is given by the classes of the elementsw2xp 1yq 1hx;yiV2;which do notbelong todf^dV0:To find them, it is enough to look at monomial differential forms i.e.xaybdx^dy withap 1,bq 1 andabpq 1:
So, ifap 1a;bq 1b andpb qa60;then df^d xayb
pb aq
xaybdx^dy:
Hence, the only elements xaybdx^dy which are notin df^dV0 are xap 1ybq 1dx^dy;wherepbqa, i.e.akp;bkqfor somek0:It follows thatC(f), as aC-vector space, has a basis given byx(k1)p 1y(k1)q 1 wherek0, which can be written as
C(f)C[t]x2p 1y2q 1dx^dy i.e.C(f) is a freeC[t]-module of rank 1.
Moreover, the corresponding monomial basis as aC-vector space for the Brieskorn module B(f) is given by xaybdx^dy with ap 2 or bq 2 ora(k1)p 1;b(k1)q 1 for somek0:
With respect to this basis,B(f)torsis the linear span ofxaybdx^dywith ap 2 orbq 2. Indeed our computation above shows that
df^d(xa1yb1)(p(b1) q(a1))fw i.e.tw0 if the coefficientp(b1) q(a1) is non-zero.
Itfollows that
B(f)C[t][xp 1yq 1dx^dy];
a freeC[t]-module of rank 1.
Itremains to explain whyb1(F)1, whereF f(x;y)2C2;xpyq1g:
Consider the covering
Zq,!F!f C wherefmaps (x;y) t ox:
Now a covering yields an exactsequence (see for instance [10], page 376), 0,!p1(F;(x0;y0))!f# p1(C;x0)!p0(Zq;(x0;y0))Zq!0:
Hencep1(F)'Z, which shows that the first Betti numberb1(F) is 1:
3. Eigenvalues of the monodromy and torsion of Brieskorn modules.
Any homogeneous polynomialf 2C[x1; :::;xn] induces a locally trivial fibration f :Cnnf 1(0)!C, with fiber F and semisimple monodromy operators
Tfk:Hk(F;C)!Hk(F;C)
fork0; :::;n 1:The eigenvalues ofTf ared-roots of unity, wheredis the degree of f, and for each such eigenvaluel we denote byH(F;C)l the corresponding eigenspace.
According to [7], see the discussion just before Remark 1.9, one has for q<nan inclusion
tn q:B(f)qd j!B(f)nd j and, forqn, an identification
B(f)qd jHn 1(F;C)l;
wherelexp 2pj
p 1 d
!
;withj0;1; :::;d 1:Letvndx1^:::^dxn
and note that [vn]2B(f)qd jfor someqif and only ifn jis divisible byd.
This yields the following, see [7], Corollary 1.10.
COROLLARY3.1. Assume thatlexp 2pn
p 1 d
!
is not an eigenva- lue of the monodromy operator Tf acting on Hn 1(F;C). Then[vn]is a non-zero torsion element in B(f):
Here are some examples of torsion elements inB(f) obtained using this approach, and having torsion orderN(f)>1.
EXAMPLE3.2. Let f x3y2z; n3; d3; j0 (this defines a cuspidal cubic curve C in P2). In this case l1 is notan eigenvalue of the monodromy acting on H2(F;C). Indeed, one has H2(F;C)1
H2(U;C)0. Here UP2n C and the last vanishing comes from
the following obvious equalities: b0(U)1, b1(U)0 and x(U)
x(P2) x(C)3 21 as C is homeomorphic to a sphere S2. Forwdx^dy^dz;to find the value ofksuch thattk[w]0 inB(f) is equivalent to saying thatfkw2df^dV1. We have to check for which value ofk;we have solutions of the equation:
fkw 3x2 @R
@y
@Q
@z
2yz @P
@z
@R
@x
y2 @Q
@x
@P
@y
dx^dy^dz;
whereP;Q;R2C[x;y;z] are homogeneous polynomials of degree 3k 1.
Fork1;we get a system of non-homogeneous linear equations in which we have 15 unknowns and 9 equations. Then using the software Mat Lab we compute the rank of the matrix corresponding to the homogeneous system (containing 9 rows and 15 columns) and get 8:On other hand, the rank of the matrix associated to the non-homogeneous system (containing 9 rows and 16 columns) is 9; which show that this system has no solution.
Fork2;we get another system of non-homogeneous linear equations containing 60 unknowns and 26 equations. Then using Mat Lab we com- pute the rank of the matrix corresponding to the homogeneous system (containing 26 rows and 60 columns) and get 26: This time the corre- sponding non-homogeneous system matrix (containing 26 rows and 61 columns) has also rank 26;which shows that this system of equations has a solution. An explicit solution for k2; is P32
3 (x3yz); Qxy2z2 and R1
3(x4y):Hence [w] hast-torsion order 2 inB(f):
The nextexample shows somehow whathappens whenHn 1(F;C)l60.
EXAMPLE 3.3. Let f x2y2xz3yz3, where n3; d4 and j3. The corresponding curveC:f 0 in P2 has two cusps as singula- rities. It follows from the study of the plane quartic curves, see [5], p. 130, thatp1(U)Z4and henceH1(F;C)0.
Nextx(U)x(P2) x(C)3 (2 2322)3. The zeta-function of the monodromy operatorTf looks like
det(tId T0f)det(tId Tf2)(t4 1)3
see for instance [5], p. 108. Hence we can not apply Corollary 3.1 to this polynomial to infer that [w][v3] is a torsion element inB(f).
However, we can try to find values ofk;for which we have solutions of the equation:
fkw
"
(2xy2z3) @R
@y
@Q
@z
(2x2yz3) @P
@z
@R
@x
(3xz23yz2) @Q
@x
@P
@y
#
w
where P;Q;R2C[x;y;z] are homogeneous polynomials of degree 4k 2.
Fork1 andk2 a similar computation of matrix ranks as above shows that the corresponding systems have no solution. Hence [w] is either a non-torsion element, or it has t-torsion order greater or equal t o 3 in B(f):
REMARK3.4. Note that in both examples above the elementt[w] is a non-zero torsion element inC(f). Hence Proposition 2.1 fails forn>2. It also shows that in these cases the moduleC(f) is nottorsion free, compare to Corollary 1.10.
The last example shows that even for rather complicated examples (here the zero set off is a surfaceS with non-isolated singularities) one may still have 1 as thet-torsion order of [vn].
REMARK 3.5. Let f x2zy3xyt; n4; d3; j1 be t he equation of a cubic surfaceSinP3. Itfollows from [4], Example 4.3, that H3(F;C)0. Hence we can apply Corollary 3.1 to this polynomial to infer that [w][v4] is a torsion element inB(f).
We have to check for which values of k; we have solutions of the equation:
fkw
"
(2xzyt) @S
@t
@T
@z@U
@y
(3y2xt) @Q
@t @R
@z
@U
@x
x2 @P
@t
@R
@y@T
@x
xy @P
@z
@Q
@y
@S
@x
#
dx^dy^dz^dt;
where P;Q;R;S;T;U2C[x;y;z;t] are homogeneous polynomials of de- gree 3k 1.
For k1; we get a system of non-homogeneous linear equations in which we have 42 unknowns and 15 equations. Using MatLab we compute
the rank of the corresponding homogeneous system (containing 15 rows and 42 columns) and get14: And the rank of non-homogeneous system (containing 15 rows and 43 columns) has the same rank 14;which show that this system has a solution.
Hencet[w]0 inB(f):
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Manoscritto pervenuto in redazione il 5 maggio 2007.