aa r X i v : . [ m a t h . AG ] J a n Two finiteness theorem for (a,b)-modules.
Daniel Barlet ∗ .16/01/08 Summary.
We prove the following two results1. For a proper holomorphic function f : X → D of a complex manifold X on a disc such that { df = 0 } ⊂ f − (0), we construct, in a functorial way,for each integer p , a geometric (a,b)-module E p associated to the (filtered)Gauss-Manin connexion of f .This first theorem is an existence/finiteness result which shows that geometric(a,b)-modules may be used in global situations.2. For any regular (a,b)-module E we give an integer N ( E ), explicitely givenfrom simple invariants of E , such that the isomorphism class of E (cid:14) b N ( E ) .E determines the isomorphism class of E .This second result allows to cut asymptotic expansions (in powers of b ) ofelements of E without loosing any information.AMS Classification : 32-S-05, 32S 20, 32-S-25, 32-S-40.Key words : (a,b)-module or Brieskorn modules, Gauss-Manin connexion, vanishingcycles. ∗ Barlet Daniel, Institut Elie Cartan UMR 7502Nancy-Universit´e, CNRS, INRIA et Institut Universitaire de France,BP 239 - F - 54506 Vandoeuvre-l`es-Nancy Cedex.France.e-mail : [email protected] ontents A− structures. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 The existence theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Introduction.
The following situation is frequently met : we consider a vector space E of multi-valued holomorphic functions (possibly with values in a complex finite dimensionalvector space V ) with finite determination on a punctured disc around 0 in C ,stable by multiplication by the variable z and by ”primitive”. These functions aredetermined by there formal asymptotic expansions at 0 of the type X ( α,j ) ∈ A × [0 ,n ] c α,j ( z ) .z α . ( Logz ) j j !where n is a fixed integer, where A is a finite set of complex numbers whose realparts are, for instance, in the interval ] − , c α,j are in C [[ z ]] ⊗ V .Define e ( α, j ) = z α . ( Logz ) j /j ! and E ( α, n ) := ⊕ nj =0 ( C [[ z ]] ⊗ V ) .e ( α, j ). Then foreach α, E ( α, n ) is a free C [[ z ]] − module of rank ( n + 1) . dim V which is stableby b := R z . To be precise, b is defined by induction on j ≥ b [ e ( α, e ( α + 1 , α + 1 and for j ≥ b [ e ( α, j )] = e ( α + 1 , j ) α + 1 − α + 1 .b [ e ( α, j − . So we have an inclusion E ⊂ ⊕ α ∈ A E ( α ) which is compatible with a := × z andby b . Note that each E ( α ) is also a free finite rank C [[ b ]] − module which is stableby a .Let us assume now that E is also a C [[ b ]] − module which is stable by a . Then E has to be free and of finite rank over C [[ b ]]. Our aim in this situation is to under-stand and to describe the relations between the coefficients c α,j of the asymptoticexpansions of elements in E . This leads to construct ”invariants” associated to thegiven E .As in the ”geometric” situations we consider the complex numbers α ∈ A arerationnal numbers, a change of variable of type t := z /N , with N ∈ N ∗ , allowsto reduce the situation to the case where all α are 0. Then one can use the”nilpotent” operator ddt and the nilpotent part of the monodromy e α,j → e α,j − to construct some filtrations in order, in the good cases, to build a Mixte Hodgestructure on a finite dimensional vector space associated to E . For instance, this isthe case in A.N. Varchenkho’s description[V. 80] of the Mixte Hodge structure builtby J. Steenbrink [St. 76] on the cohomology of the Milnor fiber of an holomorphicfunction with an isolated singularity at the origin of C n +1 .But it is clear that we loose some information in this procedure. The point of viewwhich is to consider E itself as a left module on the (non commutative) algebra˜ A := { X ν ≥ P ν ( a ) .b ν } P ν are polynomials with complex coefficients is richer. This is evidencedby M. Saito result [Sa. 91].The aim of the first part of this article is to build in a natural way, for any properholomorphic function f : X → D of a complex manifold X , assumed to be smoothoutside its 0 − fiber X := f − (0), a regular (geometric) (a,b)-module for each de-gree p ≥
0, which represent a filtered version of the Gauss-Manin connexion of f at the origin.This result is in fact a finiteness theorem which is a first step to refine the limiteMixte Hodge structure in this situation. It is interesting to remark that no K¨ahlerassumption is used in this construction of these geometric (a,b)-modules.This obviuosly shows that (a,b)-modules are basic objects and that they are impor-tant not only in the study of local singularities of holomorphic functions but moregenerally in complex geometry . So it is interesting to have some tools in order tocompute them.This is precisely the aim of the second part of this paper. We prove a finitenessresult which gives, for a regular (a,b)-module E , an integer N ( E ), bounded bysimple numerical invariants of E , such that you may cut the asymptotic expansions(in powers of b ) of elements of E without any lost of information on the structureof the (a,b)-module E . It is well known that the formal asymptotic expansionsfor solutions of a regular differential system always converge, and also that such aninteger exists for any meromorphic connexion in one variable (see [M.91] proposi-tion 1.12). But it is important to have an effective bound for such an integer easelycomputable from simple invariants of the (a,b)-module structure of E . Here we shall complete and precise the results of the section 2 of [B.07]. The situationwe shall consider is the following : let X be a connected complex manifold ofdimension n + 1 and f : X → C a non constant holomorphic function suchthat { x ∈ X/ df = 0 } ⊂ f − (0). We introduce the following complexes of sheavessupported by X := f − (0)1. The formal completion ”in f ” ( ˆΩ • , d • ) of the usual holomorphic de Rhamcomplex of X .2. The sub-complexes ( ˆ K • , d • ) and ( ˆ I • , d • ) of ( ˆΩ • , d • ) where the subsheavesˆ K p and ˆ I p +1 are defined for each p ∈ N respectively as the kernel and theimage of the map ∧ df : ˆΩ p → ˆΩ p +1 given par exterior multiplication by df . We have the exact sequence0 → ( ˆ K • , d • ) → ( ˆΩ • , d • ) → ( ˆ I • , d • )[+1] → . (1)4ote that ˆ K and ˆ I are zero by definition.3. The natural inclusions ˆ I p ⊂ ˆ K p for all p ≥ d . This leads to an exact sequence of complexes0 → ( ˆ I • , d • ) → ( ˆ K • , d • ) → ([ ˆ K/ ˆ I ] • , d • ) → . (2)4. We have a natural inclusion f ∗ ( ˆΩ C ) ⊂ ˆ K ∩ Ker d , and this gives a sub-complex (with zero differential) of ( ˆ K • , d • ). As in [B.07], we shall consideralso the complex ( ˜ K • , d • ) quotient. So we have the exact sequence0 → f ∗ ( ˆΩ C ) → ( ˆ K • , d • ) → ( ˜ K • , d • ) → . (3)We do not make the assumption here that f = 0 is a reduced equation of X ,and we do not assume that n ≥
2, so the cohomology sheaf in degree 1 of thecomplex ( ˆ K • , d • ), which is equal to ˆ K ∩ Ker d does not co¨ıncide, in general,with f ∗ ( ˆΩ C ). So the complex ( ˜ K • , d • ) may have a non zero cohomologysheaf in degree 1.Recall now that we have on the cohomology sheaves of the following complexes( ˆ K • , d • ) , ( ˆ I • , d • ) , ([ ˆ K/ ˆ I ] • , d • ) and f ∗ ( ˆΩ C ) , ( ˜ K • , d • ) natural operations a and b with the relation a.b − b.a = b . They are defined in a na¨ıve way by a := × f and b := ∧ df ◦ d − . The definition of a makes sens obviously. Let me precise the definition of b firstin the case of H p ( ˆ K • , d • ) with p ≥ x ∈ ˆ K p ∩ Ker d write x = dξ with ξ ∈ ˆΩ p − and let b [ x ] := [ df ∧ ξ ]. The reader will check easily that this makes sens.For p = 1 we shall choose ξ ∈ ˆΩ such that ξ = 0 on the smooth part of X (set theoretically). This is possible because the condition df ∧ dξ = 0 allows sucha choice : near a smooth point of X we can choose coordinnates such f = x k and the condition on ξ means independance of x , · · · , x n . Then ξ has to be (settheoretically) locally constant on X which is locally connected. So we may killthe value of such a ξ along X .The case of the complex ( ˆ I • , d • ) will be reduced to the previous one using the nextlemma. Lemme 2.1.1
For each p ≥ there is a natural injective map ˜ b : H p ( ˆ K • , d • ) → H p ( ˆ I • , d • ) which satisfies the relation a. ˜ b = ˜ b. ( a + b ) . For p = 1 this map is bijective. roof. Let x ∈ ˆ K p ∩ Ker d and write x = dξ where ξ ∈ ˆΩ p − (with ξ = 0on X if p = 1), and set ˜ b ([ x ]) := [ df ∧ ξ ] ∈ H p ( ˆ I • , d • ). This is independant onthe choice of ξ because, for p ≥
2, adding dη to ξ does not modify the resultas [ df ∧ dη ] = 0. For p = 1 remark that our choice of ξ is unique.This is also independant of the the choice of x in [ x ] ∈ H p ( ˆ K • , d • ) because adding θ ∈ ˆ K p − to ξ does not change df ∧ ξ .Assume ˜ b ([ x ]) = 0 in H p ( ˆ I • , d • ); this means that we may find α ∈ ˆΩ p − such df ∧ ξ = df ∧ dα . But then, ξ − dα lies in ˆ K p − and x = d ( ξ − dα ) shows that[ x ] = 0. So ˜ b is injective.Assume now p ≥
2. If df ∧ η is in ˆ I p ∩ Ker d , then df ∧ dη = 0 and y := dη liesin ˆ K p ∩ Ker d and defines a class [ y ] ∈ H p ( ˆ K • , d • ) whose image by ˜ b is [ df ∧ η ].This shows the surjectivity of ˜ b for p ≥ p = 1 the map ˜ b is not surjective (see the remark below).To finish the proof let us to compute ˜ b ( a [ x ] + b [ x ]). Writing again x = dξ , we get a [ x ] + b [ x ] = [ f.dξ + df ∧ ξ ] = [ d ( f.ξ )]and so ˜ b ( a [ x ] + b [ x ]) = [ df ∧ f.ξ ] = a. ˜ b ([ x ])which concludes the proof. (cid:4) Denote by i : ( ˆ I • , d • ) → ( ˆ K • , d • ) the natural inclusion and define the action of b on H p ( ˆ I • , d • ) by b := ˜ b ◦ H p ( i ). As i is a − linear, we deduce the relation a.b − b.a = b on H p ( ˆ I • , d • ) from the relation of the previous lemma.The action of a on the complex ([ ˆ K/ ˆ I ] • , d • ) is obvious and the action of b is zero.The action of a and b on f ∗ ( ˆΩ C ) ≃ E ⊗ C X are the obvious one, where E isthe rank 1 (a,b)-module with generator e satisfying a.e = b.e (or, if you prefer, E := C [[ z ]] with a := × z, b := R z and e := 1).Remark that the natural inclusion f ∗ ( ˆΩ C ) ֒ → ( ˆ K • , d • ) is compatible with theactions of a and b . The actions of a and b on H ( ˜ K • , d • ) are simply inducedby the corresponding actions on H ( ˆ K • , d • ). Remark.
The exact sequence of complexes (1) induces for any p ≥ ∂ p : H p ( ˆ I • , d • ) → H p ( ˆ K • , d • )and a short exact sequence0 → C X → H ( ˆ I • , d • ) ∂ → H ( ˆ K • , d • ) → p ≥ ∂ p = (˜ b ) − and that for p = 1 we have ∂ ◦ ˜ b = Id . If x = dξ ∈ ˆ K p ∩ Ker d then˜ b ([ x ]) = [ df ∧ ξ ] and ∂ p [ df ∧ ξ ] = [ dξ ]. So ∂ p ◦ ˜ b = Id ∀ p ≥
0. For p ≥ f ∧ α ∈ ˆ I p ∩ Ker d we have ∂ p [ df ∧ α ] = [ dα ] and ˜ b [ dα ] = [ df ∧ α ], so ˜ b ◦ ∂ p = Id .For p = 1 we have ˜ b [ dα ] = [ df ∧ ( α − α )] where α ∈ C is such that α | X = α .This shows that in degree 1 ˜ b gives a canonical splitting of the exact sequence (@). ˜ A− structures. Let us consider now the C − algebra˜ A := { X ν ≥ P ν ( a ) .b ν } where P ν ∈ C [ z ], and the commutation relation a.b − b.a = b , assuming that leftand right multiplications by a are continuous for the b − adic topology of ˜ A .Define the following complexes of sheaves of left ˜ A− modules on X :(Ω ′• [[ b ]] , D • ) and (Ω ′′• [[ b ]] , D • ) where (4)Ω ′• [[ b ]] := + ∞ X j =0 b j .ω j with ω ∈ ˆ K p Ω ′′• [[ b ]] := + ∞ X j =0 b j .ω j with ω ∈ ˆ I p D ( + ∞ X j =0 b j .ω j ) = + ∞ X j =0 b j . ( dω j − df ∧ ω j +1 ) a. + ∞ X j =0 b j .ω j = + ∞ X j =0 b j . ( f.ω j + ( j − .ω j − ) with the convention ω − = 0 b. + ∞ X j =0 b j .ω j = + ∞ X j =1 b j .ω j − It is easy to check that D is ˜ A− linear and that D = 0. We have a naturalinclusion of complexes of left ˜ A− modules˜ i : (Ω ′′• [[ b ]] , D • ) → (Ω ′• [[ b ]] , D • ) . Remark that we have natural morphisms of complexes u : ( ˆ I • , d • ) → (Ω ′′• [[ b ]] , D • ) v : ( ˆ K • , d • ) → (Ω ′• [[ b ]] , D • )and that these morphisms are compatible with i . More precisely, this means thatwe have the commutative diagram of complexes( ˆ I • , d • ) i (cid:15) (cid:15) u / / (Ω ′′• [[ b ]] , D • ) ˜ i (cid:15) (cid:15) ( ˆ K • , d • ) v / / (Ω ′• [[ b ]] , D • )7he following theorem is a variant of theorem 2.2.1. of [B. 07]. Th´eor`eme 2.2.1
Let X be a connected complex manifold of dimension n + 1 and f : X → C a non constant holomorphic function with the following condition: { x ∈ X/ df = 0 } ⊂ f − (0) . Then the morphisms of complexes u and v introduced above are quasi-isomorphisms.Moreover, the isomorphims that they induce on the cohomology sheaves of these com-plexes are compatible with the actions of a and b . This theorem builds a natural structure of left ˜ A− modules on each of the complex( ˆ K • , d • ) , ( ˆ I • , d • ) , ([ ˆ K/ ˆ I ] • , d • ) and f ∗ ( ˆΩ C ) , ( ˜ K • , d • ) in the derived category ofbounded complexes of sheaves of C − vector spaces on X .Moreover the short exact sequences0 → ( ˆ I • , d • ) → ( ˆ K • , d • ) → ([ ˆ K/ ˆ I ] • , d • ) → → f ∗ ( ˆΩ C ) → ( ˆ K • , d • ) , ( ˆ I • , d • ) → ( ˜ K • , d • ) → . (3)are equivalent to short exact sequences of complexes of left ˜ A− modules in thederived category. Proof.
We have to prove that for any p ≥ H p ( u ) and H p ( v )are bijective and compatible with the actions of a and b . The case of H p ( v ) ishandled (at least for n ≥ f reduced) in prop. 2.3.1. of [B.07]. To seekcompletness and for the convenience of the reader we shall treat here the case of H p ( u ).First we shall prove the injectivity of H p ( u ). Let α = df ∧ β ∈ ˆ I p ∩ Ker d andassume that we can find U = P + ∞ j =0 b j .u j ∈ Ω ′′ p − [[ b ]] with α = DU . Then wehave the following relations u = df ∧ ζ , α = du − df ∧ u and du j = df ∧ u j +1 ∀ j ≥ . For j ≥ du j ] = b [ du j +1 ] in H p ( ˆ K • , d • ); using corollary 2.2. of [B.07]which gives the b − separation of H p ( ˆ K • , d • ), this implies [ du j ] = 0 , ∀ j ≥ H p ( ˆ K • , d • ). For instance we can find β ∈ ˆ K p − such that du = dβ . Now, by deRham, we can write u = β + dξ for p ≥
2, where ξ ∈ ˆΩ p − . Then we concludethat α = − df ∧ d ( ξ + ζ ) and [ α ] = 0 in H p ( ˆ I • , d • ).For p = 1 we have u = 0 , du = 0 so [ α ] = [ − df ∧ dξ ] = 0 in H ( ˆ I • , d • ).We shall show now that the image of H p ( u ) is dense in H p (Ω ′′• [[ b ]] , D • ) for its b − adic topology. Let Ω := P + ∞ j =0 b j .ω j ∈ Ω ′′ p [[ b ]] such that D Ω = 0. The followingrelations holds dω j = df ∧ ω j +1 ∀ j ≥ ω ∈ ˆ I p . The corollary 2.2. of [B.07]again allows to find β j ∈ ˆ K p − for any j ≥ dω j = dβ j . Fix N ∈ N ∗ .We have D ( N X j =0 b j .ω j ) = b N .dω N = D ( b N .β N )8nd Ω N := P Nj =0 b j .ω j − b N .β N is D − closed and in Ω ′′ p [[ b ]]. As Ω − Ω N liesin b N . H p (Ω ′′• [[ b ]] , D • ), the sequence (Ω N ) N ≥ converges to Ω in H p (Ω ′′• [[ b ]] , D • )for its b − adic topology. Let us show that each Ω N is in the image of H p ( u ).Write Ω N := P Nj =0 b j .w j . The condition D Ω N = 0 implies dw N = 0 and dw N − = df ∧ w N = 0. If we write w N = dv N we obtain d ( w N − + df ∧ v N ) = 0and Ω N − D ( b N .v N ) is of degree N − b . For N = 1 we are left with w + b.w − ( − df ∧ v + b.dv ) = w + df ∧ v which is in ˆ I p ∩ Ker d because dw = df ∧ dv .To conclude it is enough to know the following two factsi) The fact that H p ( ˆ I • , d • ) is complete for its b − adic topology.ii) The fact that Im ( H p ( u )) ∩ b N . H p (Ω ′′• [[ b ]] , D • ) ⊂ Im ( H p ( u ) ◦ b N ) ∀ N ≥ H p ( u ) assuming i) and ii).For any [Ω] ∈ H p (Ω ′′• [[ b ]] , D • ) we know that there exists a sequence ( α N ) N ≥ in H p ( ˆ I • , d • ) with Ω − H p ( u )( α N ) ∈ b N . H p (Ω ′′• [[ b ]] , D • ). Now the property ii)implies that we may choose the sequence ( α N ) N ≥ such that [ α N +1 ] − [ α N ] liesin b N . H p ( ˆ I • , d • ). So the property i) implies that the Cauchy sequence ([ α N ]) N ≥ converges to [ α ] ∈ H p ( ˆ I • , d • ). Then the continuity of H p ( u ) for the b − adictopologies coming from its b − linearity, implies H p ( u )([ α ]) = [Ω].The compatibility with a and b of the maps H p ( u ) and H p ( v ) is an easyexercice.Let us now prove properties i) and ii).The property i) is a direct consequence of the completion of H p ( ˆ K • , d • ) for its b − adic topology given by the corollary 2.2. of [B.07] and the b − linear isomorphism˜ b between H p ( ˆ K • , d • ) and H p ( ˆ I • , d • ) constructed in the lemma 2.1.1. above.To prove ii) let α ∈ ˆ I p ∩ Ker d and N ≥ α = b N . Ω + DU where Ω ∈ Ω ′′ p [[ b ]] satisfies D Ω = 0 and where U ∈ Ω ′′ p − [[ b ]]. With obviousnotations we have α = du − df ∧ u · · · du j − df ∧ u j +1 ∀ j ∈ [1 , N − · · · ω + du N − df ∧ u N +1 which implies D ( u + b.u + · · · + b N .u N ) = α + b N .du N and the fact that du N lies in ˆ I p ∩ Ker d . So we conclude that [ α ] + b N . [ du N ] is in the kernel of H p ( u )which is 0. Then [ α ] ∈ b N . H p ( ˆ I • , d • ). (cid:4) emark. The map β : (Ω ′ [[ b ]] • , D • ) → (Ω ′′ [[ b ]] • , D • )defined by β (Ω) = b. Ω commutes to the differentials and with the action of b . Itinduces the isomorphism ˜ b of the lemma 2.1.1 on the cohomology sheaves. So it isa quasi-isomorphism of complexes of C [[ b ]] − modules.To prove this fact, it is enough to verify that the diagram( ˆ K • , d • ) ˜ b (cid:15) (cid:15) v / / (Ω ′ [[ b ]] • , D • ) β (cid:15) (cid:15) ( ˆ I • , d • ) u / / (Ω ′′ [[ b ]] • , D • )induces commutative diagrams on the cohomology sheaves.But this is clear because if α = dξ lies in ˆ K p ∩ Ker d we have D ( b.ξ ) = b.dξ − df ∧ ξ so H p ( β ) ◦ H p ( v )([ α ]) = H p ( u ) ◦ H p (˜ b )([ α ]) in H p (Ω ′′ [[ b ]] • , D • ) . (cid:4) Let us recall some basic definitions on the left modules over the algebra ˜ A . D´efinition 2.3.1 An (a,b)-module is a left ˜ A− module which is free and of finiterank on the commutative sub-algebra C [[ b ]] of ˜ A .An (a,b)-module E is1. local when ∃ N ∈ N such that a N .E ⊂ b.E ;2. simple pole when a.E ⊂ b.E ;3. regular when it is contained in a simple pole (a,b)-module;4. geometric when it is contained in a simple pole (a,b)-module E ♯ such thatthe minimal polynomial of the action of b − .a on E ♯ (cid:14) b.E ♯ has its roots in Q + ∗ . We shall give more details and examples of (a,b)-modules in the section 3.Now let E be any left ˜ A− module, and define B ( E ) as the b − torsion of E , thatis to say B ( E ) := { x ∈ E / ∃ N b N .x = 0 } . Define A ( E ) as the a − torsion of E andˆ A ( E ) := { x ∈ E / C [[ b ]] .x ⊂ A ( E ) } . Remark that B ( E ) and ˆ A ( E ) are sub- ˜ A− modules of E but that A ( E ) is notstable by b . D´efinition 2.3.2
A left ˜ A− module E is small when the following conditionshold . E is a finite type C [[ b ]] − module ;2. B ( E ) ⊂ ˆ A ( E ) ;3. ∃ N / a N . ˆ A ( E ) = 0 ; Recall that for E small we have always the equality B ( E ) = ˆ A ( E ) and that thiscomplex vector space is finitie dimensional. The quotient E/B ( E ) is an (a,b)-module called the associate (a,b)-module to E .Conversely, any left ˜ A− module E such that B ( E ) is a finite dimensional C − vector space and such that E/B ( E ) is an (a,b)-module is small.The following easy criterium to be small will be used later : Lemme 2.3.3
A left ˜ A− module E is small if and only if the following conditionshold :1. ∃ N / a N . ˆ A ( E ) = 0 ;2. B ( E ) ⊂ ˆ A ( E ) ;3. ∩ m ≥ b m .E ⊂ ˆ A ( E ) ;4. Ker b and
Coker b are finite dimensional complex vector spaces.
As the condition 3) in the previous lemma has been omitted in [B.07] (but this doesnot affect the results of this article because this lemma was used only in a case wherethis condition 3) was satisfied, thanks to proposition 2.2.1. of loc. cit. ), we shallgive the (easy) proof.
Proof.
First the conditions 1) to 4) are obviously necessary. Conversely, assumethat E satisfies these four conditions. Then condition 2) implies that the actionof b on ˆ A ( E ) (cid:14) B ( E ) is injective. But the condition 1) implies that b N = 0 onˆ A ( E ) (see [B.06] ). So we conclude that ˆ A ( E ) = B ( E ) ⊂ Ker b N which is afinite dimensional complex vector space using condition 4) and an easy induction.Now E/B ( E ) is a C [[ b ]] − module which is separated for its b − adic topology.The finitness of Coker b now shows that it is a free finite type C [[ b ]] − moduleconcluding the proof. (cid:4) D´efinition 2.3.4
We shall say that a left ˜ A− module E is geometric when E is small and when it associated (a,b)-module E/B ( E ) is geometric. The main result of this section is the following theorem, which shows that the Gauss-Manin connexion of a proper holomorphic function produces geometric ˜ A− modulesassociated to vanishing cycles and nearby cycles.11 h´eor`eme 2.3.5 Let X be a connected complex manifold of dimension n + 1 where n ∈ N , and let f : X → D be an non constant proper holomorphic functionto an open disc D in C with center . Let us assume that df is nowherevanishing outside of X := f − (0) .Then the ˜ A− modules H j ( X, ( ˆ K • , d • )) and H j ( X, ( ˆ I • , d • )) are geometric for any j ≥ . In the proof we shall use the C ∞ version of the complex ( ˆ K • , d • ). We define K p ∞ as the kernel of ∧ df : C ∞ ,p → C ∞ ,p +1 where C ∞ ,j denote the sheaf of C ∞ − forms on X of degree j, let ˆ K p ∞ be its formal f − completion and ( ˆ K •∞ , d • ) thecorresponding de Rham complex.The next lemma is proved in [B.07] (lemma 6.1.1.) Lemme 2.3.6
The natural inclusion ( ˆ K • , d • ) ֒ → ( ˆ K •∞ , d • ) is a quasi-isomorphism. Remark.
As the sheaves ˆ K •∞ are fine, so we have a natural isomorphism H p ( X, ( ˆ K • , d • )) ≃ H p (cid:0) Γ( X, ˆ K •∞ ) , d • (cid:1) . Let us denote by X the generic fiber of f . Then X is a smooth compact complexmanifold of dimension n and the restriction of f to f − ( D ∗ ) is a locally trivial C ∞ bundle with typical fiber X on D ∗ = D \ { } , if the disc D is small enougharound 0. Fix now γ ∈ H p ( X , C ) and let ( γ s ) s ∈ D ∗ the corresponding multivaluedhorizontal family of p − cycles γ s ∈ H p ( X s , C ). Then for ω ∈ Γ( X, ˆ K p ∞ ∩ Ker d )define the multivalued holomorphic function F ω ( s ) := Z γ s ωdf . Let now Ξ := X α ∈ Q ∩ ] − , , j ∈ [0 ,n ] C [[ s ]] .s α . ( Logs ) j j ! . This is an ˜ A− modules with a acting as multiplication by s and b as theprimitive in s without constant. Now if ˆ F ω is the asymptotic expansion at 0 of F ω , it is an element in Ξ, and we obtain in this way an ˜ A− linear map Int : H p ( X, ( ˆ K • , d • )) → H p ( X , C ) ⊗ C Ξ . To simplify notations, let E := H p ( X, ( ˆ K • , d • )). Now using Grothendieck theorem[G.66], there exists N ∈ N such that Int ( ω ) ≡
0, implies a N . [ ω ] = 0 in E .12s the converse is clear we conclude that ˆ A ( E ) = Ker ( Int ). It is also clear that B ( E ) ⊂ Ker ( Int ) because Ξ has no b − torsion. So we conclude that E satisfiesproperties 1) and 2) of the lemma 2.3.3.The property 3) is also true because of the regularity of the Gauss-Manin connexionof f . End of the proof of theorem 2.3.5.
To show that E := H p ( X, ( ˆ K • , d • )) issmall, it is enough to prove that E satisfies the condition 4) of the lemma 2.3.3.Consider now the long exact sequence of hypercohomology of the exact sequence ofcomplexes 0 → ( ˆ I • , d • ) → ( ˆ K • , d • ) → ([ ˆ K/ ˆ I ] • , d • ) → . It contains the exact sequence H p − ( X, ([ ˆ K (cid:14) ˆ I ] • , d • )) → H p ( X, ( ˆ I • , d • )) H p ( i ) → H p ( X, ( ˆ K • , d • )) → H p ( X, ([ ˆ K (cid:14) ˆ I ] • , d • ))and we know that b is induced on the complex of ˜ A− modules quasi-isomorphic to( ˆ K • , d • ) by the composition i ◦ ˜ b where ˜ b is a quasi-isomorphism of complexesof C [[ b ]] − modules. This implies that the kernel and the cokernel of H p ( i ) areisomorphic (as C − vector spaces) to Ker b and
Coker b respectively. Now toprove that E satisfies condition 4) of the lemma 2.3.3 it is enough to prove finitedimensionality for the vector spaces H j ( X, ([ ˆ K (cid:14) ˆ I ] • , d • )) for all j ≥ K (cid:14) ˆ I ] j ≃ [ Ker df (cid:14)
Im df ] j are coherent on X and supported in X . The spectral sequence E p,q := H q (cid:0) H p ( X, [ ˆ K (cid:14) ˆ I ] • ) , d • (cid:1) which converges to H j ( X, ([ ˆ K (cid:14) ˆ I ] • , d • )), is a bounded complex of finite dimensionalvector spaces by Cartan-Serre. This gives the desired finite dimensionality.To conclude the proof, we want to show that E/B ( E ) is geometric. But this is aneasy consequence of the regularity of the Gauss-Manin connexion of f and of theMonodromy theorem, which are already incoded in the definition of Ξ : the injec-tivity on E/B ( E ) of the ˜ A − linear map
Int implies that
E/B ( E ) is geometric.Remark now that the piece of exact sequence above gives also the fact that H p ( X, ( ˆ I • , d • ))is geometric, because it is an exact sequence of ˜ A− modules. (cid:4) First recall in a more na¨ıve way the definition of an (a,b)-module.
D´efinition 3.1.1 An (a,b)-module E is a free finite type C [[ b ]] − module with a C − linear endomorphism a : E → E which is continuous for the b − adic topologyof E and satisfies a.b − b.a = b .The rank of E , denote by rank ( E ) , will be the rank of E as a C [[ b ]] − module. emarks.
1. Let ( e , · · · , e k ) a C [[ b ]] − basis of a free finite type C [[ b ]] − module. Thenchoosing arbitrarily elements ( ε , · · · , ε k ) and defining a.e j = ε j ∀ j ∈ [1 , k ]gives an (a,b)-module: the commutation relation implies that ∀ n ∈ N wehave a.b n = b n .a + n.b n +1 so a is defined on P kj =1 C [ b ] .e j . The continuityassumption gives its (unique) extension.2. There is a natural (a,b)-module associated to every algebraic linear differentialsystem (see [B.95] p.42) Q ( z ) . dFdz = M ( z ) .F ( z ) , Q ∈ C [ z ] , M ∈ End ( C n ) ⊗ C C [ z ] . In the sequel of this article we shall mainly consider regular (a,b)-modules (seedefinition recalled below). To try to convince the reader that the ”general” (a,b)-module structure is interesting, let me quote the following result, which is quiteelementary in the regular case, but which is not so easy in general.
Th´eor`eme 3.1.2 ([B.95] th.1bis p.31) Let E be an (a,b)-module. Then the kerneland cokernel of ”a” are finite dimensional. This result implies a general finiteness theorem for extensions of (a,b)-modules (see[B.95] and also section 1.3).
D´efinition 3.1.3
We shall say that an (a,b)-module E has a simple pole whenthe inclusion a.E ⊂ b.E is satisfied. This terminology comes from the terminology of meromorphic connexions (see forinstance [D.70]).
Example.
For any λ ∈ C define the simple pole rank 1 (a,b)-module E λ as E := C [[ b ]] .e λ where ” a ” is defined by the relation a.e λ = λ.b.e λ . (cid:3) As an introduction to our second theorem, the reader may solve the following exerciceby direct computation.
Exercice.
For any S ∈ C [[ b ]] show that the simple pole (a,b)-module definedby E := C [[ b ]] .e S and a.e S = b.S ( b ) .e S is isomorphic to E λ with λ = S (0)(hint: begin by looking for α ∈ C such that ( a − S (0) .b )( e + α .b.e ) ∈ b .E ). (cid:3) For a simple pole (a,b)-module, the linear map b − .a : E → E is well definedand induces an endomorphism f := b − .a : E/b.E → E/b.E . For any λ ∈ C weshall denote by λ min the smallest eigenvalue of f which is in λ + Z . Then for14 = λ min − k with k ∈ N ∗ the bijectivity of the map f − λ on E/b.E implieseasily its bijectivity on E (see the exercice above). It gives then the equality( a − λ.b ) .E = b.E. Using this remark, it is not difficult to prove the following result from [B.93] (prop.1.3.p.11) that we shall use later on.
Proposition 3.1.4
Let E be a simple pole (a,b)-module, and let λ ∈ C and κ ∈ N such that λ − κ ≤ λ min . If y ∈ E satisfies ( a − λ.b ) .y ∈ b κ +2 .E then thereexists an unique ˜ y ∈ E such that ( a − λ.b ) . ˜ y = 0 and ˜ y − y ∈ b κ +1 .E . An easy consequence of this proposition is that for an eigenvalue λ of f such that λ = λ min there always exists a non zero x ∈ E such that ( a − λ.b ) .x = 0. Thisgives an embedding of E λ in E . Remark also that if E is a non zero simple pole(a,b)-module, such a λ always exists. This leads to a rather precise description aof ”general” simple pole (a,b)-module (see [B.93] th. 1.1 p.15). D´efinition 3.1.5
An (a.b)-module E is regular when its saturation by b − .a in E [ b − ] is finitely generated on C [[ b ]] . We shall denote E ♯ this saturation. It is a simple pole (a,b)-module and it is thesmallest simple pole (a,b)-module containing E in the sense that for any (a,b)-linear morphism j : E → F where F is a simple pole (a,b)-module, there existsa unique (a,b)-linear extension j ♯ : E ♯ → F of j .It is easy to show that a regular (a,b)-module of rank 1 is isomorphic to some E λ for some λ ∈ C . The classification of rank 2 regular (a,b)-module is not so obvious.We recall it here for a later use Proposition 3.1.6 (see [B.93] prop.2.4 p. 34) The list of rank 2 regular (a,b)-modules is, up to isomorphism, the following :1. E λ ⊕ E µ for ( λ, µ ) ∈ C / S .2. For any λ ∈ C and any n ∈ N let E λ ( n ) be the simple pole (a,b)-modulewith basis ( x, y ) such that a.x = ( λ + n ) .b.x + b n +1 .y and a.y = λ.b.y.
3. For any ( λ, µ ) ∈ C / S let E λ,µ the rank 2 regular (a,b)-module with basis ( y, t ) such that a.y = µ.b.y and a.t = y + ( λ − .b.t. . For any λ ∈ C , any n ∈ N ∗ and any α ∈ C ∗ let E λ,λ − n ( α ) be the rank 2regular (a,b)-module with basis ( y, t ) such that a.y = ( λ − n ) .b.y and a.t = y + ( λ − b.t + α.b n .y Note that the first two cases are simple pole (a,b)-modules.The saturation by b − .a in case 3. is generated by b − .y and t as a C [[ b ]] − module.It is isomorphic to E λ − ⊕ E µ − for λ = µ and to E λ − (0) for λ = µ .The saturation by b − .a in case 4. is generated by b − .y and t as a C [[ b ]] − module.It is isomorphic to E λ − n − ( n ) for any non zero value of α .To conclude this first section, let me recall also the theorem of existence of Jordan-H¨older sequences for regular (a,b)-module, which will be usefull in the induction inthe proof of our result . Th´eor`eme 3.1.7 (see [B.93] th. 2.1 p.30) For any regular rank k (a,b)-module E there exists a sequence of sub-(a,b)-modules E ⊂ E ⊂ · · · ⊂ E k − ⊂ E k = E such that for any j ∈ [1 , k ] the quotient E j /E j − is isomorphic to E λ j . Moreoverwe may choose for E any normal rank 1 sub-(a,b)-module of E .The number α ( E ) := P kj =1 λ j is independant of the choice of the Jordan-H¨oldersequence. It is given by the following formula α ( E ) = trace (cid:0) b − .a : E ♯ /b.E ♯ → E ♯ /b.E ♯ (cid:1) + dim C ( E ♯ /E ) . D´efinition 3.2.1
Let E be a regular (a,b)-module. We define the regularityorder of E as the smallest integer k ∈ N such that the inclusion a k +1 .E ⊂ k X j =0 a j .b k − j +1 .E (reg.) is valid. We shall note this integer or ( E ) .We define also the index δ ( E ) of E as the smallest integer m ∈ N such that E ♯ ⊂ b − m .E . Remarks. i) The (a,b)-module E has a simple pole if an only iff or ( E ) = 0. normal means E ∩ b.E = b.E , so that E/E is again free on C [[ b ]]. b − .a ) k +1 .E ⊂ Φ k ( E ) := P kj =0 ( b − .a ) j .E and this implies that Φ k ( E ) is stable by b − .a . So Φ k ( E ) is a simple pole(a,b)-module contained in b − k .E ⊂ E [ b − ]. This implies clearly the regularityof E .For k = or ( E ) we have E ♯ = Φ k ( E ) ⊂ b − k .E . So we have δ ( E ) ≤ or ( E ).iii) As the quotient b − k .E/E is a finite dimensional C − vector space, the quotient E ♯ /E is always a finite dimensional C − vector space. (cid:3) The remark iii) shows that for a regular (a,b)-module E there always exists asimple pole sub-(a,b)-module of E which is a finite codimensional vector space in E . This comes from the fact that for k = δ ( E ) we have b k .E ♯ ⊂ E and that b k .E ♯ has a simple pole. Example.
The inequality δ ( E ) ≤ or ( E ) may be strict for or ( E ) ≥
2. Forinstance the (a,b)-module of rank 3 with C [[ b ]] − basis e , e , e with a.e = e , a.e = b.e , a.e = 0 has index 1 and regularity order 2 : an easycomputation gives that a C [[ b ]] − basis for E ♯ is given by e , b − .e , b − .e , andthat a C [[ b ]] − basis for E + b − .a.E is given by e , b − .e , e . (cid:3) D´efinition 3.2.2
Let E be a regular (a,b)-module. The biggest simple polesub-(a,b)-module of E exists and has finite C − codimension in E . We shallnote it E b . In general, for k = δ ( E ) the inclusion b k .E ♯ ⊂ E b is strict. For instance this isthe case for E λ,µ ⊕ E ν . Lemme 3.2.3
Let E be a regular (a,b)-module. The smallest integer m such wehave b m .E ⊂ E b is equal to δ ( E ) . Proof.
Let k := δ ( E ). Then b k .E ♯ is a simple pole sub-(a,b)-module of E .So we have b k .E ⊂ b k .E ♯ ⊂ E b . Conversely, an inclusion b m .E ⊂ E b gives E ⊂ b − m .E b . As b − m .E b has a simple pole this implies E ♯ ⊂ b − m .E b ⊂ b − m .E . So δ ( E ) ≤ m . (cid:4) Examples.
In the case 3 of the proposition 3.1.6 E b is generated as a C [[ b ]] − moduleby y and b.t , so E b = b.E ♯ .In case 4 we have also E b = b.E ♯ . Lemme 3.2.4
Let E be a regular (a,b)-module. For any exact sequence of (a,b)-modules → E ′ → E π → E ′′ → we have or ( E ′′ ) ≤ or ( E ) ≤ rank ( E ′ ) + or ( E ′′ ) .As a consequence, the order of regularity of E is at most rank ( E ) − for anyregular non zero (a,b)-module. roof. The inequality or ( E ′′ ) ≤ or ( E ) is trivial because an inequality a k +1 .E ⊂ k X j =0 a j .b k − j +1 .E implies the same for E ′′ and, by definition, the best such integer k is the order ofregularity.The crucial case is when E ′ is of rank 1 . So we may assume that E ′ ≃ E λ forsome λ ∈ C (see 3.1.7 or [B.93] prop.2.2 p.23). Let k = or ( E ′′ ). Then the inclusion a k +1 .E ′′ ⊂ k X j =0 a j .b k − j +1 .E ′′ (1)implies that a k +1 .E ⊂ k X j =0 a j .b k − j +1 .E + b l .E λ (2)for some l ∈ N . In fact we can take for l the smallest integer such that thegenerator e λ of E λ (defined up to C ∗ by the relation a.e λ = λ.b.e λ ) satisfies b l .e λ ∈ Ψ k = P kj =0 a j .b k − j +1 .E .Remark that this integer l ≥ b k +1 .e λ ∈ Ψ k . Moreover, asΨ k is a C [[ b ]] − submodule of E , b l .e λ ∈ Ψ k implies b l .E λ ⊂ Ψ k .Now, thanks to (2) we have a k +2 .E ⊂ k X j =0 a j +1 .b k +1 − j .E + a.b l .E λ (3)which gives a k +2 .E ⊂ k +1 X j =0 a j .b k − j +2 .E (4)because a.b l .E λ = b.b l .E λ ⊂ b. Ψ k .This proves that or ( E ) is at most k + 1 = or ( E ′′ ) + rank ( E ′ ).Assume now that our inequality is proved for E ′ of rank p − ∗ ) with rank ( E ′ ) equal p ≥
2. Let E λ ⊂ E ′ be a normal rank 1sub-(a,b)-module of E ′ (see 3.1.7 or [B.93] prop.2.2 p.23 for a proof of the existenceof such sub-(a,b)-module) and consider the exact sequence of (a,b)-modules (usingthe fact that E λ is also normal in E ; see lemma 2.5 of [B.93])0 → E ′ /E λ → E/E λ → E ′′ → or ( E ) ≤ or ( E/E λ ) + 1 ≤ p − or ( E ′′ ) + 1 = p + or ( E ′′ ) . Now using an easy induction (or a Jordan-H¨older sequence for E ) we obtain or ( E ) ≤ rank ( E ) − E . (cid:4) emark. In the situation of the previous lemma we have δ ( E ′ ) ≤ δ ( E ). Thisis a consequence of the obvious inclusion ( E ′ ) ♯ ⊂ E ′ [ b − ] ∩ E ♯ : assume that x ∈ E ′ [ b − ] ∩ E ♯ ; then, for k := δ ( E ) we have b k .x ∈ E ′ [ b − ] ∩ E so that b N + k .x ∈ E ′ for N large enough. As E/E ′ has no b − torsion, we conclude that b k .x ∈ E ′ . So our initial inclusion implies δ ( E ′ ) ≤ k . (cid:3) In this section we consider more carefully the associative and unitary C − algebra˜ A := (cid:8) ∞ X P n ( a ) .b n with P n ∈ C [ z ] (cid:9) with the commutation relation a.b − b.a = b , and such that the left and rightmultiplications by a are continuous for the b − adic topology of ˜ A . The right structure as a commuting left-structure on ˜ A . There exits an unique C − linear (bijective) map θ : ˜ A → ˜ A with the followingpropertiesi) θ (1) = 1 , θ ( a ) = a, θ ( b ) = − b ;ii) θ ( x.y ) = θ ( y ) .θ ( x ) ∀ x, y ∈ ˜ A .iii) θ is continuous for the b − adic topology of ˜ A The uniqueness is an easy consequence of iii) and the fact that the conditions i)and ii) implies θ ( b p .a q ) = ( − p .a q .b p ∀ p, q ∈ N . Existence is then clear from theexplicit formula deduced from this remark.We define a new structure of left ˜ A− module on ˜ A , called the θ − structure anddenote by x ∗ (cid:3) , by the formula x ∗ y = y.θ ( x ) . It is easy to see that this new left-structure on ˜ A commutes with the ordinary oneand that with this θ − structure ˜ A is still free of rank one as a left ˜ A− module. D´efinition 3.3.1
Let E be a (left) ˜ A− module. On the C − vector space Hom ˜ A ( E, ˜ A ) we define a left ˜ A− module structure using the θ − structure on ˜ A . Explicitely thismeans that for ϕ ∈ Hom ˜ A ( E, ˜ A ) and x ∈ ˜ A we let ∀ e ∈ E ( x.ϕ )( e ) := x ∗ ϕ ( e ) = ϕ ( e ) .θ ( x ) . We obtain in this way a left ˜ A− module that we shall still denote Hom ˜ A ( E, ˜ A ) . remark that for each k ∈ N b k . ˜ A = ˜ A .b k .
19t is clear that E → Hom ˜ A ( E, ˜ A ) is a contravariant functor which is left exactin the category of left ˜ A− modules. As every finite type left ˜ A− module has aresolution of length ≤ Ext i ˜ A ( E, ˜ A ) , i ∈ [0 ,
2] the right derived functors of this functor.They are finite type left ˜ A− modules when E is finitely generated because ˜ A isleft noetherian (see [B.95] prop.2 p.26).Any (a,b)-module is a left ˜ A− module. They are characterized by the existence ofspecial simple resolutions. Lemme 3.3.2
Let M be a ( p, p ) matrix with entries in the ring C [[ b ]] . Thenthe left ˜ A− linear map Id p .a − M : ˜ A p → ˜ A p given by t X := ( x , · · · , x p ) → t X. ( Id p .a − M ) is injective. Its cokernel is the (a,b)-module E given as follows : E has a C [[ b ]] base e := ( e , · · · , e p ) and a is defined by the two conditions1. a.e := M ( b ) .e ;2. the left action of a is continuous for the b − adic topology of E .Any (a,b)-module is obtained in this way and so, as a ˜ A− left-module, has a reso-lution of the form → ˜ A p t (cid:3) . ( Id p .a − M ) −→ ˜ A p → E → . (@) Proof.
First remark that for x ∈ ˜ A the condition x.a ∈ b. ˜ A implies x ∈ b. ˜ A .Now let us prove, by induction on n ≥
1, that, for any ( p, p ) matrix M withentries in C [[ b ]] the condition t X. ( Id p .a − M ) = 0 implies t X ∈ b n . ˜ A p .For n = 1 this comes from the previous remark. Let assume that the assertionis proved for n ≥ X ∈ ˜ A p such that t X. ( Id p .a − M ) = 0.Using the induction hypothesis we can find Y ∈ ˜ A p such that X = b n .Y . Now weobtain, using a.b n = b n .a + n.b n +1 and the fact that ˜ A has no zero divisor, therelation t Y ( Id p .a − ( M + n.Id p .b )) = 0and using again our initial remark we conclude that Y ∈ b. ˜ A p so X ∈ b n +1 . ˜ A p .So such an X is in ∩ n ≥ b n . ˜ A p = (0).The other assertions of the lemma are obvious. (cid:4) We recall now a construction given in [B.95] which allows to compute more easilythe vector spaces
Ext i ˜ A ( E, F ) when
E, F are (a,b)-modules
D´efinition 3.3.3
Let
E, F two (a,b)-modules. Then the C [[ b ]] − module Hom b ( E, F ) is again a free and finitely generated C [[ b ]] − module. Define on it an (a,b)-modulestructure in the following way. . First change the sign of the action of b . So S ( b ) ∈ C [[ b ]] will act as ˇ S ( b ) = S ( − b ) .2. Define a using the linear map Λ :
Hom b ( E, F ) → Hom b ( E, F ) given by Λ( ϕ )( e ) = ϕ ( a.e ) − a.ϕ ( e ) .We shall denote Hom a,b ( E, F ) the corresponding (a,b)-module. The verification that Λ( ϕ ) is C [[ b ]] − linear and that Λ . ˇ b − ˇ b. Λ = ˇ b are easy (andmay be found in [B.95] p.31). Remark. In loc. cit. we defined the (a,b)-module structure on Hom a,b ( E, F )with opposite signs for a and b . The present convention is better because it fitswith the usual definition of the formal adjoint of a differential operator : z ∗ = z and ( ∂/∂z ) ∗ = − ∂/∂z . (cid:3) The following lemma is also proved in loc.cit.
Lemme 3.3.4
Let
E, F two (a,b)-modules. Then there is a functorial isomorphismof C − vector spaces H i (cid:16) Hom a,b ( E, F ) a → Hom a,b ( E, F ) (cid:17) → Ext i ˜ A ( E, F ) ∀ i ≥ . Here the map a of the complex Hom a,b ( E, F ) a → Hom a,b ( E, F )) is equal tothe Λ defined above which is, by definition, the operator ′′ a ′′ of the (a,b)-module Hom a,b ( E, F ) . Now the following corollary of the lemma 3.3.2 gives that the two natural ways ofdefining the dual of an (a,b)-module give the same answer.
Corollaire 3.3.5
Let E an (a,b)-module. There is a functorial isomorphism of(a,b)-modules between the following two (a,b)-modules constructed as follows :1. Ext A ( E, ˜ A ) with the ˜ A− structure defined by the θ − structure of ˜ A .2. Hom a,b ( E, E ) where E := ˜ A (cid:14) ˜ A .a . Proof.
Using a free resolution (@) of E deduced from a C [[ b ]] − basis e := ( e , · · · , e p ) we obtain, by the previous lemma, an exact sequence0 → ˜ A p ( Id p .a − t M ) . (cid:3) −→ ˜ A p → Ext A ( E, ˜ A ) → . (@@)of left ˜ A− modules where ˜ A p is endowed with its θ − structure. Writing the sameexact sequence with the ordinary left-module structure of ˜ A p gives0 → ˜ A p t (cid:3) . ( Id p .a − t ˇ M ) −→ ˜ A p → Ext A ( E, ˜ A ) → . (@@ bis)21here t ˇ M ( b ) := t M ( − b ).Denote by e ∗ := ( e ∗ , · · · , e ∗ p ) the dual basis of Hom C [[ b ]] ( E, E ). By definition ofthe action of a on Hom a,b ( E, E ) we get, if ω is the class of 1 in E :( a.e ∗ i )( e j ) = e ∗ i ( a.e j ) − a.e ∗ i ( e j ) = e ∗ i ( p X h =1 m j,h .e h ) − a.δ i.j .ω = ˇ m j,i .ω because a.ω = 0 in E , and the definition of the action of b on Hom a,b ( E, E ).So we have a.e ∗ = t ˇ M .e ∗ concluding the proof. (cid:4) D´efinition 3.3.6
For any (a,b)-module E the dual of E , denoted by E ∗ , is the(a,b)-module Ext A ( E, ˜ A ) ≃ Hom a,b ( E, E ) . Of course, for any ˜ A− linear map f : E → F between two (a,b)-modules we havean ˜ A− linear ”dual” map f ∗ : F ∗ → E ∗ .It is an easy consequence of our previous description of Ext A ( E, ˜ A ) that we havea functorial isomorphism ( E ∗ ) ∗ → E . Examples.
1. For each λ ∈ C we have ( E λ ) ∗ ≃ E − λ .2. For ( λ, µ ) ∈ C we have E ∗ λ,µ ≃ E − µ +1 , − λ +1 .3. Let E be the rank two simple pole (a,b)-module E (0) defined by a.e = b.e + b.e and a.e = b.e . Then its dual is isomorphic to E − (0).It is also an elementary exercice to show the following isomorphisms : E (0) ≃ C [[ z ]] ⊕ C [[ z ]] .Logz and E − (0) ≃ C [[ z ]] 1 z ⊕ C [[ z ]] . Logzz with a := × z and b := R z . Proposition 3.3.7
For any exact sequence of (a,b)-modules → E ′ u → E v → E ′′ → we have an exact sequence of (a,b)-modules → ( E ′′ ) ∗ v ∗ → E ∗ u ∗ → ( E ′ ) ∗ → . If E is a simple pˆole (a,b)-module, E ∗ has a simple pole.For any regular (a,b)-module E its dual E ∗ is regular. Moreover, if E b and E ♯ are respectively the biggest simple pole submodule of E and the saturation of E by b − .a in E [ b − ] , we have ( E ♯ ) ∗ ≃ ( E ∗ ) b and ( E b ) ∗ ≃ ( E ∗ ) ♯ . roof. The first assertion is a direct consequence of the vanishing of
Ext i ˜ A ( E, ˜ A )for i = 0 ,
2, for any (a,b)-module and the long exact sequence for the ”Ext”.The condition that E has a simple pole is equivalent to the fact that for anychoosen basis e of E the matrix M has its coefficients in b. ˜ A = ˜ A .b . Then thisremains true for t ˇ M .To prove the regularity of E ∗ when E is regular, we shall use induction onthe rank of E . The rank 1 case is obvious because we have a simple pole in thiscase. Assume that the assertion is true for rank < p and consider a rank = p regular (a,b)-module E . Using the theorem 3.1.7 we have an exact sequence of(a,b)-modules 0 → E λ → E → F → F is regular of rank p −
1. This gives a short exact sequence0 → F ∗ → E ∗ → E − λ → F ∗ and of E − λ implies the regularity of E ∗ .Now the inclusions E b ⊂ E ⊂ E ♯ gives exact sequences0 → Ext A ( E/E b , ˜ A ) → E ∗ → ( E b ) ∗ → Ext A ( E/E b , ˜ A ) → → Ext A ( E ♯ /E, ˜ A ) → ( E ♯ ) ∗ → E ∗ → Ext A ( E ♯ /E, ˜ A ) → Ext A ( V, ˜ A ) = 0 for any ˜ A− module whichis a finite dimensional vector space, and also the finiteness (as a vector space) of Ext A ( V, ˜ A ). This implies that we have, for any regular (a,b)-module, the inclusions E ∗ ⊂ ( E b ) ∗ and ( E ♯ ) ∗ ⊂ E ∗ . They imply, thanks to the fact that ( E b ) ∗ and ( E ♯ ) ∗ have simple poles,( E ∗ ) ♯ ⊂ ( E b ) ∗ and ( E ♯ ) ∗ ⊂ ( E ∗ ) b . But the inclusion ( E ∗ ) b ⊂ E ∗ gives E = ( E ∗ ) ∗ ⊂ (( E ∗ ) b ) ∗ ⊂ (( E ♯ ) ∗ ) ∗ = E ♯ and the minimality of E ♯ gives (( E ∗ ) b ) ∗ = E ♯ because (( E ∗ ) b ) ∗ has a simplepole and contains E . Dualizing again gives ( E ♯ ) ∗ ≃ ( E ∗ ) b . The last equality isobtained in a similar way from E ∗ ⊂ ( E ∗ ) ♯ . (cid:4) Lemme 3.3.8
Let V be an ˜ A− module of finite dimension over C . Then wehave Ext i ˜ A ( V, ˜ A ) = 0 for i = 0 , and Ext A ( V, ˜ A ) is again a ˜ A− module (viathe θ − structure of ˜ A ) which is a finite dimensional vector space. Moreover it hasthe same dimension than V and there is a canonical ˜ A− module isomorphism Ext A ( Ext A ( V, ˜ A ) , ˜ A ) ≃ V. roof. We begin by proving the first assertion of the lemma for the special case V λ := ˜ A (cid:14) ˜ A . ( a − λ ) + ˜ A .b for any λ ∈ C . Let us show that we have the freeresolution 0 → ˜ A α → ˜ A β → ˜ A → V λ → α ( x ) := ( x.b, − x. ( a − b − λ )) , β ( u, v ) := u. ( a − λ ) + v.b . The map α is clearly injective and β ( α ( x )) = x. ( b.a − λ.b − ( a − b − λ ) .b ) = 0. If we have β ( u, v ) = 0 then u ∈ ˜ A .b ; let u = x.b . Then we get x. ( a − b − λ ) .b + v.b = 0 and so v = − x. ( a − b − λ ) . This gives the exactness of our resolution.Now the
Ext i ˜ A ( V λ , ˜ A ) are given by the cohomology of the complex0 → ˜ A β ∗ → ˜ A α ∗ → ˜ A → . The map β ∗ ( x ) = (( a − λ ) .x, b.x ) and α ∗ ( u, v ) = b.u − ( a − b − λ ) .v are ˜ A− linear forthe θ − structure of ˜ A . Clearly β ∗ is injective and α ∗ ( β ∗ ( x )) ≡
0. If α ∗ ( u, v ) = 0set v = b.y and conclude that u = ( a − λ ) .y . This gives the vanishing of the Ext i for i = 0 ,
1. The
Ext is the cokernel of β ∗ which is easily seen to be isomorphicto V λ .Consider now any finite dimensional ˜ A− module V over C . We make an inductionon dim C ( V ) to prove the vanishing of the Ext i for i = 0 , Ext .The dim V = 1 case is clear because reduced to the case V = V λ for some λ ∈ C .Assume that the case dim V = p is proved, for p ≥ V with dim V = p + 1. Then Ker b is not { } and is stable by a . Let λ ∈ C an eigenvalue of a acting on Ker b . Then a eigenvector generates in V a sub-˜ A− module isomorphic to V λ .The exact sequence of ˜ A− modules0 → V λ → V → W → W := V (cid:14) V λ has dimension p allows us to conclude, looking at the longexact sequence of Ext .The last assertion follows from the remark that we produce a free resolution of Ext A ( V, ˜ A ) by taking Hom ˜ A ( − , ˜ A ) of a free (length two, see [B.97]) resolution of V because of the already proved vanishing of the Ext i for i = 0 ,
1. Doing this againgives back the initial resolution (remark that we use here that the θ ◦ θ − structureon Hom ˜ A ( Hom ˜ A ( ˜ A , ˜ A ) , ˜ A ) is the usual left structure on ˜ A ). (cid:4) Corollaire 3.3.9
For a simple pole (a,b) module E denote by S ( E ) the spectrumof b − .a acting on E/b.E . Then we have S ( E ∗ ) = − S ( E ) . roof. We make an induction on the rank of E . In rank 1 the result is clearbecause we have E ≃ E λ for some λ ∈ C , and S ( E λ ) = { λ } . But we know that E ∗ λ = E − λ .Assume the assertion proved for any rank p ≥ E with rank p + 1. Using theorem 3.1.7, there exists λ ∈ C and anexact sequence (a,b)-modules 0 → E λ → E → F → rank ( F ) = p and where F has a simple pole (because a quotient of asimple pole (a,b)-module has a simple pole !). The exact sequence of vector spaces0 → E λ /b.E λ → E/b.E → F/b.F → S ( E ) = S ( F ) ∪ { λ } . Now proposition 3.3.7 gives the exact sequence0 → F ∗ → E ∗ → E − λ → S ( E ∗ ) = S ( F ∗ ) ∪{− λ } . The induction hypothesis S ( F ∗ ) = − S ( F ) allows to conclude. (cid:4) Lemme 3.3.10
For any pair of (a,b)-modules E and F there is a canonical iso-morphism of vector spaces D : Ext A ( E, F ) → Ext A ( F ∗ , E ∗ ) associated to the correspondance between 1-extensions (i.e. short exact sequences) (0 → F → G → E → D → (0 → E ∗ → G ∗ → F ∗ → . Proof.
We have a obvious isomorphism of C [[ b ]] − modules I : Hom b ( E, F ) → Hom b ( Hom b ( F, E ) , Hom b ( E, E )) ≃ Hom b ( F ∗ , E ∗ )because E ≃ C [[ b ]] as a C [[ b ]] − module. But recall that Ext A ( E, F ) (resp.
Ext A ( F ∗ , E ∗ )) is the cokernel of the C − linear map ” a ” defined on Hom b ( E, F )by the formula ( a.ϕ )( x ) = ϕ ( a.x ) − a.ϕ ( x )So it is enough to check that the isomorphism I commutes with ” a ” in order toget an isomorphism between the cokernels of ” a ” in these two spaces.Let ϕ ∈ Hom b ( E, F ) and ξ ∈ F ∗ . Then I ( ϕ )( ξ ) = ϕ ◦ ξ . So, for x ∈ E we have(using Λ to avoid too many ” a ”)Λ( I ( ϕ )( ξ ) = I ( ϕ )( a.ξ ) − a. ( I ( ϕ )( ξ ))Λ( I ( ϕ )( ξ )( x ) = ( ϕ ◦ ξ )( a.x ) − a.ξ ( ϕ ( x )) − (cid:0) ξ ( ϕ ( a.x )) − a.ξ ( ϕ ( x )) (cid:1) = (cid:2) (Λ( ϕ )) ◦ ξ (cid:3) ( x ) = I (Λ( ϕ ))( x ) . but be carefull with the b → ˇ b !
25o Λ ◦ I = I ◦ Λ. The map I gives an isomorphism of complexes Hom a,b ( E, F ) Λ / / I (cid:15) (cid:15) Hom a,b ( E, F ) I (cid:15) (cid:15) Hom a,b ( F ∗ , E ∗ ) Λ / / Hom a,b ( F ∗ , E ∗ )and this conclude the proof, using lemma 3.3.4. (cid:4) For an (a,b)-module E and an integer m ∈ N it is clear that b m .E is again an(a,b)-module. This can be generalize for any m ∈ C . D´efinition 3.3.11
For any (a,b)-module E and any complex number m ∈ C define the (a,b)-module b m .E as follows : as an C [[ b ]] − module we let b m .E ≃ E ≃ C [[ b ]] rank ( E ) ; the operator a is defined as a + m.b . Precisely, this means that if ( e , · · · , e k ) is a C [[ b ]] − basis of E such that wehave a.e = M ( b ) .e where M ∈ End ( C p ) ⊗ C C [[ b ]], the (a,b)-module b m .E admita basis, denote by ( b m .e , · · · , b m .e k ), such that the operator a is defined by therelation a. ( b m .e ) := ( M ( b ) + m.b.Id k ) . ( b m .e ).Remark that for m ∈ N this notation is compatible with the preexisting one,because of the relation a.b m = b m . ( a + m.b ).For any m ∈ N there exists a canonical (a,b)-morphism b m .E → E which is an isomorphism of b m .E on Im ( b m : E → E ). But remark that the map b m : E → E is not a − linear (but the image is stable by a ).For any m ∈ N there is also a canonical (a,b)-morphism E → b − m .E which induces an isomorphism of E on Im ( b m : b − m .E → b − m .E ). So we maywrite, via this canonical identification, b m . ( b − m .E ) = E .It is easy to see that for any m, m ′ ∈ C we have a natural isomorphism b m ′ . ( b m .E ) ≃ b m + m ′ .E and also b .E ≃ E. Remark.
It is easy to show that for any m ∈ C there exists an unique C − algebraautomorphism η m : ˜ A → ˜ A such that η (1) = 1 , η ( b ) = b and η ( a ) = a + m.b. Using this automorphism, one can define a left ˜ A− module b m .F for any left˜ A− module F and any m ∈ C . This is, of course compatible with our definitionin the context of (a,b)-modules. (cid:3) The behaviour of the correspondance E → b m .E by duality is given by the followingeasy lemma; the proof is left as an exercice.26 emme 3.3.12 For any (a,b)-module E and any m ∈ C there is natural (a,b)-isomorphism ( b m .E ) ∗ → b − m .E ∗ . The following corollary of the lemma 3.2.3 and the proposition 3.3.7 allows to showthat duality preserves the index.
Lemme 3.3.13
Let E be a regular (a,b)-module. Then we have δ ( E ∗ ) = δ ( E ) . Proof.
By definition δ ( E ) is the smallest integer k ∈ N such that E ♯ ⊂ b − k .E .Now E ♯ ⊂ b − m .E implies by duality that b m .E ∗ ⊂ ( E ∗ ) b . So, by lemma 3.2.3, wehave m ≥ δ ( E ∗ ). This proves that δ ( E ) ≤ δ ( E ∗ ) and we obtain the equality bysymetry. (cid:4) Remark.
Duality does not preserve the order of regularity : in the example givenbefore the definition 3.2.2 we have or ( E ) = 2 and or ( E ∗ ) = 1. (cid:3) Let us conclude this section by an easy exercice.
Exercice.
For any (a,b)-modules
E, F and any λ ∈ C there are natural(a,b)-isomorphisms1. b λ .E µ ≃ E λ + µ .2. b λ .Hom a,b ( E, F ) ≃ Hom a,b ( b − λ .E, F ) ≃ Hom a,b ( E, b λ .F ).3. Then deduce from the previous isomorphisms that Hom a,b ( E, E λ ) ≃ b − λ .E ∗ ,and Ext A ( E, E λ ) ≃ E ∗ / ( a + λ.b ) .E ∗ . Notation.
For a complex number λ we shall note by ˜ λ is class in C (cid:14) Z . Weshall order elements in each class modulo Z by its natural order on real parts. (cid:3) D´efinition 3.4.1
Let E be a regular (a,b)-module and let ˜ λ ∈ C (cid:14) Z . We definethe following complex numbers : ˜ λ min ( E ) := inf { λ ∈ ˜ λ/ ∃ a non zero morphism E λ → E } ˜ λ max ( E ) = sup { λ ∈ ˜ λ/ ∃ a non zero morphism E → E λ } L ˜ λ ( E ) = ˜ λ max ( E ) − ˜ λ min ( E ) ∈ Z L ( E ) = sup { ˜ λ ∈ C / Z / L ˜ λ ( E ) } ith the following conventions : inf {∅} = + ∞ , sup {∅} = −∞ and − ∞ − λ = −∞ ∀ λ ∈ ] − ∞ , + ∞ ]+ ∞ − λ = + ∞ ∀ λ ∈ [ −∞ , + ∞ [ . We shall call L ( E ) the width of E . Remarks.
1. A non zero morphism E λ → E is necessarily injective. Either its image is anormal submodule in E or there exists an integer k ≥ E λ − k → E whose image is normal an contains the image of the previous one.2. In a dual way, a non zero morphism E → E λ has an image equal to b k .E λ ≃ E λ + k , where k ∈ N .3. A non zero morphism E λ → E µ implies that λ lies in µ + N . It is possiblethat for some E we have ˜ λ max ( E ) < ˜ λ min ( E ). For instance this is the casefor the rank 2 regular (a,b)-module E λ,µ from 3.1.6. So the width of a regularbut not simple pole (a,b)-module is not necessarily a non negative integer.4. Let E and F be regular (a,b)-modules. If there is a surjective morphism E → F then for all ˜ λ ∈ C (cid:14) Z we have ˜ λ max ( E ) ≥ ˜ λ max ( F ).If there is an injective morphism E ′ → E then for all ˜ λ ∈ C (cid:14) Z we have˜ λ min ( E ) ≤ ˜ λ min ( E ′ ).5. Every submodule of E isomorphic to E λ is contained in E b . So we have˜ λ min ( E ) = ˜ λ min ( E b ), for every regular (a,b)-module E and every ˜ λ ∈ C (cid:14) Z .6. In a dual way, every morphism E → E λ extends uniquely to a morphism E ♯ → E λ with the same image. So for every regular (a,b)-module E andevery ˜ λ ∈ C (cid:14) Z , we get ˜ λ max ( E ) = ˜ λ max ( E ♯ ). (cid:3) Lemme 3.4.2
1. Let E a simple pole (a,b)-module and let S ( E ) denotes thespectrum of the linear map b − .a : E/b.E → E/b.E , we have ˜ λ min ( E ) = inf { λ ∈ S ( E ) ∩ ˜ λ } and ˜ λ max ( E ) = sup { λ ∈ S ( E ) ∩ ˜ λ } (@)
2. For any regular (a,b)-module E we have ] ( − λ ) max ( E ∗ ) = − ˜ λ min ( E ) ] ( − λ ) min ( E ∗ ) = − ˜ λ max ( E ) . This implies L − ˜ λ ( E ∗ ) = L ˜ λ ( E ) ∀ ˜ λ ∈ C / Z , and so L ( E ∗ ) = L ( E ) .3. For any regular (a,b)-module E and any ˜ λ ∈ C (cid:14) Z we have equivalencebetween ˜ λ min ( E ) = + ∞ and ˜ λ max ( E ) = −∞ . roof. Let E be a simple pole (a,b)-module. We have already seen (in propo-sition 3.1.4) that if λ ∈ S ( E ) is minimal in its class modulo 1, there exists a nonzero x ∈ E such that a.x = λ.b.x . This implies that ˜ λ min ≤ inf { λ ∈ S ( E ) ∩ ˜ λ } .But the opposite inequality is obvious, so the first part of (@) is proved.Using corollary 3.3.9 and the result already obtained for E ∗ gives ] ( − λ ) min ( E ∗ ) = inf {− λ ∈ S ( E ∗ ) ∩ ] ( − λ ) } = − sup { λ ∈ S ( E ) ∩ ˜ λ } . So for µ = sup { λ ∈ S ( E ) ∩ ˜ λ } we have an exact sequence of (a,b)-modules0 → E − µ → E ∗ → F → E → E µ . This implies ˜ λ max ≥ µ . As, again, theopposite inequality is obvious, the second part of (@) is proved.Let us prove now the relations in 2.Remark first that these equalities are true for a simple pole (a,b)-module becauseof (@) and corollary 3.3.9.For any regular (a,b)-module E we know that˜ λ min ( E ) = ˜ λ min ( E b ) = inf { λ ∈ S ( E b ) ∩ ˜ λ } and ] ( − λ ) max ( E ∗ ) = ] ( − λ ) max (( E ∗ ) ♯ ) . But we have ] ( − λ ) max (( E ∗ ) ♯ ) = sup {− λ ∈ S (( E ∗ ) ♯ ) ∩ ] ( − λ ) } = − inf { λ ∈ S (( E ∗ ) ♯ ) ∗ ∩ ˜ λ } because ( E ∗ ) ♯ has a simple pole, using corollary 3.3.9. So we obtain ] ( − λ ) max ( E ∗ ) = − ˜ λ min ( E b ) = − ˜ λ min ( E )because ( E ∗ ) ♯ ) ∗ = E b (see proposition 3.3.7).The second relation is analoguous.The equivalence in 3 is obvious in the simple pole case using (@).The general case is an easy consequence using E b , E ♯ : if ˜ λ min ( E ) = + ∞ so is˜ λ min ( E ♯ ) because E ⊂ E ♯ . Then ˜ λ max ( E ♯ ) = −∞ and so is ˜ λ max ( E ). Theconverse is analoguous using E b . (cid:4) Remarks.
1. If E has a simple pole, we have L ˜ λ ( E ) ≥ L ˜ λ ( E ) = −∞ for any ˜ λ in C / Z . So L ( E ) is always ≥ λ min and ˜ λ max for any ˜ λ ∈ C / Z .For the remaining cases we can compute these numbers using the fact that wealready know the corresponding E b and E ♯ and the remark 5 and 6 beforethe preceeding lemma. (cid:3) roposition 3.4.3 Let E be a regular (a,b)-module and let ˜ λ ∈ C (cid:14) Z . Assumethat λ = ˜ λ min ( E ) < + ∞ . Consider an exact sequence of (a,b)-modules → E λ → E π → F → . Then we have for all ˜ µ ∈ C / Z the inequality L ˜ µ ( F ) ≤ L ˜ µ ( E ) + 1 . (i) Proof.
As ˜ µ max ( F ) ≤ ˜ µ max ( E ) for any µ ∈ C it is enough to prove that wehave ˜ µ min ( E ) ≤ ˜ µ min ( F ) + 1 for all ˜ µ ∈ C / Z .Let begin by the case of ˜ µ = ˜ λ . We want to show the inequality˜ λ min ( F ) ≥ λ − E λ − d ֒ → F with d ≥
0. The rank 2 (a,b)-module G := π − ( E λ − d ) is containedin E , so λ = ˜ λ min ( G ). We have the exact sequence of (a,b)-modules0 → E λ → π − ( E λ − d ) π → E λ − d → . Now let us compare G with the list in proposition 3.1.6.If G is in case 1, we have E λ − d ⊂ G so d = 0 because λ = ˜ λ min ( G ).If G is in case 2, we have λ − d = λ + n with n ∈ N , so d = 0.If G is in case 3, we have G ≃ E λ,λ + k with k ∈ N . Then the theorem 3.1.7 gives2 λ − d = 2 λ + k − d = 1 − k ≤ G is in case 4, we have G ≃ E λ,λ + n ( α ). Again theorem 3.1.7 gives 2 λ − d =2 λ + n − d = 1 − n ≤ n ∈ N ∗ . So d = 0.We conclude that we always have d ≤ µ = ˜ λ let us prove now the following inequality :˜ µ min ( F ) ≤ ˜ µ min ( E ) ≤ ˜ µ min ( F ) + 1 . (iii)Consider an injective morphism E µ → E with µ = ˜ µ min ( E ). The restriction of π to E µ is injective and so it gives ˜ µ min ( E ) ≥ ˜ µ min ( F ). Assume now that wehave an injective morphism E µ ′ ֒ → F with µ ′ = ˜ µ min ( F ), and consider the rank2 (a,b)-module π − ( E µ ′ ). Using the proposition 3.1.6 where only cases 1 or 3 arepossible now, it can be easily check that (iii) is satisfied. (cid:4) Remarks.
1. In the situation of the previous proposition we have either ˜ λ min ( E ) ≥ ˜ λ max ( E )or ˜ λ max ( E ) = ˜ λ max ( F ) : Assume that we have λ < λ ′ := ˜ λ max ( E ). Thenthere exists a surjective morphism q : E → E λ ′ , and, as the restriction of q to E λ is zero, the map q can be factorized and gives a surjective morphism˜ q : F → E λ ′ . So we get ˜ λ max ( E ) ≤ ˜ λ max ( F ), and the desired equality thanksto the preceeding lemma. 30. We shall use later that in the situation of the previous proposition we havethe inequality ˜ λ max ( F ) ≤ λ + L ( E ) . (cid:3) Corollaire 3.4.4
In the situation of the previous proposition we have the inequality L ( E ) + rank ( E ) ≥ L ( F ) + rank ( F ) . So this integer is always positive for any nonzero regular (a,b)-module. Proof.
As the rank 1 case is obious, an easy induction on the rank of E usingthe propositions 3.1.4 and 3.4.3 gives the proof. (cid:4) Examples.
1. The (a,b)-module J k ( λ ) := ˜ A (cid:14) ˜ A . ( a − ( λ + k − .b )( a − ( λ + k − .b ) · · · ( a − λ.b )which has rank k , satisfies λ max = λ and λ min = λ + k −
1. So its width is L ( J k ( λ )) = − k + 1.To understand easily the (a,b)-module J k ( λ ) the reader may use the followingalternative definition of it : there is a C [[ b ]] − basis ( e , · · · , e k ) in which theaction of ” a ” is given by a.e = e + λ.b.e , a.e = e + ( λ + 1) .b.e , · · · , a.e k = ( λ + k − .b.e k .
2. The rank 2 (a,b)-module E λ ⊕ E λ + n has width n . This shows that, despitethe fact that the width is always bigger than − rank ( E ) + 1, the width maybe arbitrarily big, even for a rank 2 regular (a,b)-module. (cid:3) Lemme 4.1.1
Let E be a regular (a,b)-module of index δ ( E ) = k . For N ≥ k +1 the quotient map q N : E → E (cid:14) b N .E induces a bijection between simple pole sub-(a,b)-modules F containing b k .E ♯ and sub ˜ A− modules F ⊂ E (cid:14) b N .E satisfyingthe following two conditionsi) a. F ⊂ b. F ;ii) b k .E ♯ (cid:14) b N .E ⊂ F . Proof.
It is clear that if F is a simple pole sub-(a,b)-module of E containing b k .E ♯ the image F := q N ( F ) is a ˜ A− submodule of E (cid:14) b N .E such that i) and ii)are fullfilled. Conversely, if a ˜ A− submodule F satisfies i) and ii), let F := q − N ( F ).Of course, F is a sub-(a,b)-module of E and contains b k .E ♯ . The only point to seeis that F has a simple pole. If x ∈ F then a.q N ( x ) ∈ b. F so a.x ∈ b.F + b N .E . As N ≥ k +1 we may write a.x = b.y + b.z with y ∈ F and z ∈ b N − .E ⊂ b k .E ♯ ⊂ F .This completes the proof. (cid:4) .31 emarks.
1. we may replace b k .E ♯ by b k .E in the second condition imposed on F and F : if a simple pole (a,b)-submodule F contains b k .E it contains b k .E ♯ bydefinition of E ♯ . This allows to avoid the use of E ♯ in the previous lemma.2. The biggest F satisfying i) and ii) corresponds to E b . So we may recover E b from the quotient E (cid:14) b N .E for N ≥ δ ( E ) + 1. (cid:3) Corollaire 4.1.2
Let E be a regular (a,b)-module of order of regularity k . Fix N ≥ k + 1 and assume that we has an isomorphism of ˜ A− modules ϕ : E (cid:14) b N .E → E ′ (cid:14) b N .E ′ where E ′ is an (a,b)-module. Then E ′ is regular and has order of regularity k .Moreover we have the equality ϕ ( E b (cid:14) b N .E ) = ( E ′ ) b (cid:14) b N .E ′ . Proof. As k is the order of regularity of E we have a k +1 .E ⊂ P kj =0 a j .b k − j +1 E .The inequality N ≥ k + 1 gives a k +1 .E (cid:14) b N .E ⊂ P kj =0 a j .b k − j +1 E (cid:14) b N .E , and thisis also true for E ′ (cid:14) b N .E ′ , and implies a k +1 .E ′ ⊂ P kj =0 a j .b k − j +1 E ′ . So the orderof regularity of E ′ is at most k . We conclude that it is exactly k by symetry.The last stament comes from the second remark above, as or ( E ) ≥ δ ( E ). (cid:4) Lemme 4.2.1
Let E be an (a,b)-module et fix a complex number λ . There exists N ( E, λ ) ∈ N such that for any N ≥ N ( E, λ ) we have the following inclusion : b N .E ⊂ ( a − λ.b ) .E. Proof.
With the b − adic topology, E is a Frechet space. The C − linear map a − λ.b : E → E is continuous. The finiteness theorem of [B.95], theorem 1.bis p.31gives that the kernel and cokernel of this map are finite dimensional vector spaces.So the subspace ( a − λ.b ) .E is closed in E . This statement corresponds to theequality ∩ N ≥ (cid:2) ( a − λ.b ) .E + b N .E (cid:3) = ( a − λ.b ) .E (@)But the images of the subspaces b N .E in the finite dimensional vector space E (cid:14) ( a − λ.b ) .E is a decreasing sequence. So it is stationnary, and, as the intersectionis { } thanks to (@), the result follows. (cid:4) Proposition 4.2.2
Let F be an (a,b)-module and λ a complex number. Considera short exact sequence of (a,b)-modules → E λ α → E β → F → here E λ := ˜ A (cid:14) ˜ A . ( a − λ.b ) . Then, for any N ≥ N ( F ∗ , − λ ) , the extension (@@) is uniquely determined by the following extension of ˜ A− modules which are finitedimensional vectors spaces → E λ (cid:14) b N .E λ α → E (cid:14) b N .E β → F (cid:14) b N .F → N ) obtained from (@@) by ”quotient by b N ”. Comments.
This statement needs some more explanations. Denote by K N thekernel of the obvious map (forget ”a”) ob N : Ext A ( F/b N .F, E λ /b N .E λ ) → Ext b ( F/b N .F, E λ /b N .E λ )where Ext b ( − , − ) is a short notation for Ext C [[ b ]] ( − , − ). The short exact sequencecorrespondance (@@) → (@@ N ) gives a map δ N : Ext A ( F, E λ ) → Ext A ( F/b N .F, E λ /b N .E λ )whose range lies in K N , because the C [[ b ]] − exact sequence (@@) is split as F is C [[ b ]] − free, and so is the exact sequence (@@ N ). The precise signification ofthe previous proposition is that for N ≥ N ( F ∗ , − λ ) the map δ N is a C − linearisomorphism between the vector spaces Ext A ( F, E λ ) and K N . (cid:3) Proof.
As a first step to realize the map δ N let us consider the following com-mutative diagramm of complex vector spaces, deduced from the exact sequences of˜ A− modules: 0 → E λ + N → E λ → E λ /b N .E λ → → b N .F → F → F/b N .F → Ext ( F/b N .F, E λ + N ) (cid:15) (cid:15) / / Ext ( F, E λ + N ) α (cid:15) (cid:15) / / Ext ( b N .F, E λ + N ) (cid:15) (cid:15) Ext ( F/b N .F, E λ ) (cid:15) (cid:15) / / Ext ( F, E λ ) β (cid:15) (cid:15) u / / Ext ( b N .F, E λ ) v (cid:15) (cid:15) Ext ( F/b N .F, E λ /b N .E λ ) γ / / Ext ( F, E λ /b N .E λ ) w / / Ext ( b N .F, E λ /b N .E λ )We have the following propreties :1. The surjectivity of the map β is consequence of the vanishing of the vectorspace Ext A ( F, E λ + N ) thanks to the proposition 3.3.7.33. the vanishing of the composition u ◦ v is consequence of lemma 3.3.4 and ofthe fact that the restriction map Hom b ( F, E λ ) → Hom b ( b N .F, E λ ) → Hom b ( b N .F, E λ /b N .E λ )is obviously zero.3. So the map w is zero and γ is surjective.4. The kernel of γ is given by the image of the injective map ∂ : Hom ˜ A ( b N .F, E λ /b N .E λ ) ֒ → Ext A ( F/b N .F, E λ /b N .E λ ) . This is a consequence of the vanishing of the map
Ext A ( F, E λ /b N .E λ ) → Ext A ( b N .F, E λ /b N .E λ ) . Let us show now that for N ≥ N ( F ∗ , − λ ) the map α is zero. Using againthe isomorphisms given by the lemma 3.3.4, α is induced by the obvious map Hom b ( F, b N .E λ ) → Hom b ( F, E λ ), whose image is b N .Hom b ( F, E λ ). Denote respec-tively by G and H the (a,b)-modules given by Hom b ( F, b N .E λ ) and Hom b ( F, E λ )with the action of ” a ” defined by Λ (see 3.3.4). Then we have the following com-mutative diagramm G i / / (cid:15) (cid:15) H (cid:15) (cid:15) G/a.G / / ≃ (cid:15) (cid:15) H/a.H ≃ (cid:15) (cid:15) Ext A ( F, b N .E λ ) α / / Ext A ( F, E λ )and the image of i is b N .H . So the map α will be zero as soon as b N .H ⊂ a.H and this is fullfilled for N ≥ N ( H,
0) = N ( F ∗ , − λ ). This last equality coming fromthe isomorphisms H/a.H ≃ Ext A ( F, E λ ) ≃ Ext A ( E − λ , F ∗ ) ≃ F ∗ / ( a + λ.b ) .F ∗ see the exercice concluding section 3.3.Consider now the commutative diagramm0 (cid:15) (cid:15) K Ni (cid:15) (cid:15) Ext A ( F, E λ ) β (cid:15) (cid:15) ˆ δ N o o δ N t t iiiiiiiiiiiiiiiii / / Hom ˜ A ( b N .F, E λ /b N .E λ ) ∂ / / ob N (cid:15) (cid:15) Ext A ( F/b N .F, E λ /b N .E λ ) γ / / ob N (cid:15) (cid:15) Ext A ( F, E λ /b N .E λ ) Hom b ( b N .F, E λ /b N .E λ ) ≃ / / Ext b ( F/b N .F, E λ /b N .E λ )34he surjectivity of β implies that the map i ◦ γ is surjective ( we know that theextensions in the image of δ N comes from K N , so δ N factors in ˆ δ N ◦ i ).We have i ( K N ) ∩ Im ( ∂ N ) = (0) because ob N is injective on Im ( ∂ N ).So i induces an isomorphism of vector spaces from K N to Ext A ( F/b N .F, E λ /b N .E λ ) /Im ( ∂ N ) γ ≃ Ext A ( F, E λ /b N .E λ ) β − ≃ Ext A ( F, E λ ) . This completes the proof . (cid:4)
We shall need some bound for the integer N ( F ∗ , − λ ) which appears in the previousproposition for the proof of our theorem. Lemme 4.2.3
Let G be a regular (a,b)-module and let µ ∈ C . A sufficientcondition on N ∈ N in order to have the inclusion b N .G ⊂ ( a − µ.b ) .G is N ≥ µ − ˜ µ min ( G ) + δ ( G ) + 2 . Proof.
As we know that ˜ µ min ( G b ) = ˜ µ min ( G ) , for M ∈ N , the assumption M > µ − ˜ µ min ( G ) implies that ( a − ( µ − M ) .b ) .G b = b.G b (see the remarkbefore proposition 3.1.4). By definition of the index of G we have b δ ( G ) .G ⊂ G b .Combining both gives b M + δ ( G )+1 .G ⊂ b M . ( a − ( µ − M ) .b ) .G = ( a − µ.b ) .b M .G ⊂ ( a − µ.b ) .G. Now let N = M + δ ( G ) + 1 ; a sufficient condition on the integer N is now N ≥ µ − ˜ µ min ( G ) + δ ( G ) + 2 . (cid:4) Corollaire 4.2.4
A sufficient condition for N ≥ N ( F ∗ , − λ ) in the situation ofprop. 4.2.2 in the regular case is that N ≥ or ( E ) + L ( E ) + rank ( E ) + 1 . Remark that the inequality L ( E ) + rank ( E ) ≥ E implies that we have or ( E ) + L ( E ) + rank ( E ) + 1 ≥ or ( E ) + 2. Proof.
We apply the previous lemma with F ∗ = G and µ = − λ = − ˜ λ min ( E ).The conclusion comes now from the following facts :1. − ] ( − λ ) min ( F ∗ ) = ˜ λ max ( F ) ≤ λ + L ( E ) this last inequality is proved in 3.4.3.2. δ ( F ∗ ) = δ ( F ) ≤ or ( F ) ≤ or ( E ) proved in 3.3.13 and 3.2.4 (cid:4) Th´eor`eme 4.3.1
Let E be a regular (a,b)-module. There exists an integer N ( E ) such that for any (a,b)-module E ′ , any integer N ≥ N ( E ) and any ˜ A− isomorphism ϕ : E/b N .E → E ′ /b N .E ′ (1) there exists an unique ˜ A− isomorphism Φ : E → E ′ inducing the given ϕ .Moreover the choice N ( E ) = N ( E ) := or ( E ) + L ( E ) + rank ( E ) + 1 is possible. emarks.
1. It is easy to see that for a rank 1 regular (a,b)-module the integer 2 is the bestpossible.2. In our final lemma 4.3.2 we show that the integer given in the theorem isoptimal for the rank k (a,b)-module J k ( λ ), (defined in the lemma), for any k ∈ N ∗ .3. For the rank 2 (a,b)-modules E λ,µ the integer given by the theorem is or ( E )+ L ( E ) + 2 + 1 = 3 is again optimal, as it can be shown in the same maner thatin our final lemma.4. For the rank 2 simple pole (a,b)-module E λ (0) the integer given by thetheorem is 3 = L ( E ) + rank ( E ) + 1 and the best possible is 2 : the actionof b − .a on E/b.E which is determined by
E/b .E characterizes this rank2 regular (a,b)-module in the classification given in proposition 3.1.6.5. For the (a,b)-module E associated to an holomorphic germ at the origine of C n +1 with an isolated singularity we have the uniform bounds or ( E ) ≤ n + 1and L ( E ) ≤ n so the choice N ( E ) = 2 n + µ + 2 is always possible, where µ is the Milnor number (equal to the rank). (cid:3) Proof.
We shall make an induction on the rank of E . So we shall assume thatthe result is proved for a rank p − E of rank p ≥
1, an (a,b)-module E ′ , an integer N ≥ N ( E ) andan ˜ A− isomorphism ϕ as in (1). From 4.1.2 we know that E ′ is then regularand has order of regularity or ( E ′ ) = or ( E ).Choose now a complex number λ which is minimal modulo Z such there existsan exact sequence of (a,b)-module ( so λ = ˜ λ min ( E ) with the terminology of § → E λ α → E β → F → . (2)This exists from theorem 3.1.7. The (a,b)-module F has rank p − N ( E ) ≥ N ( F ∗ , − λ ). So we know from 4.2.2 that the extension (2)is determined by the extension0 → E λ /b N .E λ α N → E/b N .E β N → F/b N .F → . (2 N )Now, using the ˜ A− isomorphism ϕ we obain an injective ˜ A− linear map j N : E λ /b N .E λ ֒ → E ′ /b N .E ′ . Using the proposition 3.1.4 with the fact that N ≥ or ( E ′ ) + 2 there exists a uniquenormal inclusion j : E λ ֒ → E ′ inducing j N .36efine F ′ := E ′ /j ( E λ ). Then F ′ is a rank p − → E λ j → E ′ → F ′ → N ). Using the induction hypothesis, because the inequalities or ( E ) ≥ or ( F ) from 4.1.2 and L ( E ) + rank ( E ) ≥ L ( F ) + rank ( F ) from 3.4.4implies N ( E ) ≥ N ( F ) , we have a unique isomorphism Ψ : F → F ′ compatiblewith the one induced by ϕ between F/b N .F and F ′ /b N .F ′ . Using 4.2.2, 4.2.4 andthe inequality N ( E ) ≥ N ( F ∗ , − λ ) we have an unique isomorphism of extensions0 / / E λ = (cid:15) (cid:15) α / / E Φ (cid:15) (cid:15) β / / F Ψ (cid:15) (cid:15) / / / / E λ j / / E ′ / / F ′ / / (cid:4) Lemme 4.3.2
Let E := J k ( λ ) the rank k (a,b)-module defined by the C [[ b ]] − basis e , · · · , e k and by the following relations a.e j = ( λ + j − .b.e j + e j +1 ∀ j ∈ [1 , k ] with the convention e k +1 = 0 . We have δ ( E ) = or ( E ) = k − , L ( E ) = − k + 1 . Theinteger or ( E ) + L ( E ) + rank ( E ) + 1 = k + 1 is the best possible for the theorem. Proof.
It is easy to see that the saturation E ♯ is generated by e , b − .e , · · · , b − k +1 .e k .This gives the equality δ ( E ) = or ( E ) = k − E µ ֒ → E such that e µ b.E . Then there exists( α , · · · , α k ) ∈ C k \ { } such that a. ( k X h =1 α h .e h ) = µ.b ( k X h =1 α h .e h ) + b .E. Then we obtain k X h =1 α h . (cid:0) ( λ + j − .b.e h + e h +1 (cid:1) = k X h =1 α h .µ.b.e h + b .E and so α = · · · = α k − = 0 and we conclude that µ = λ + k − J k ( λ ) ∗ = J k ( − λ − k +2) and so we have λ max = λ .So L ( E ) = − k + 1.Now we shall prove that the integer k + 1 is the best possible in the theorem 4.3.1for E = J k ( λ ) by giving a regular (a,b)-module F such that F/b k .F ≃ E/b k .E and not isomorphic to E . 37et consider the rank k (a,b)-module F defined by P kj =1 C [[ b ]] .e j with thefollowing relations a.e j = ( λ + j − .b.e j + e j +1 ∀ j ∈ [1 , k ] a.e k = ( λ + k − .b.e k + k − X h =1 α h .b k − h +1 .e h Then define, for β , · · · , β k − ∈ C , ε := e k + k − X j =1 β j .b k − j .e j . We have a.ε :=( λ + k − .b.e k + k − X h =1 α h .b k − h +1 .e h + k − X j =1 β j . (cid:2) b k − j . (cid:0) ( λ + j − .b.e j + e j +1 ) + ( k − j ) .b k − j +1 .e j (cid:3) a.ε :=( λ + k − .b.e k + k − X h =1 (cid:0) α h + β h . ( λ + k −
1) + β h − (cid:1) .b k − h +1 .e h Let now choose β , · · · , β k − such that we have α h + β h . ( λ + k −
1) + β h − = ( λ + k − β k − ) .β h ∀ h ∈ [1 , k − β = 0. We obtain the system of equations α h + β h − = β k − .β h ∀ h ∈ [1 , k − . This implies, assuming β k − = 0, that β k − is solution of the equation x k = α k − .x k − + · · · + α .x + α . Now choose α = · · · = α k − = 0 and α := ρ k with ρ ∈ ]0 , β j = ρ k − j ∀ j ∈ [1 , k − F ρ satisfies F/b k .F ≃ E/b k .E as a.e k = e k + ρ.b k .e in F ρ . But the relation a.ε = ( λ + k − ρ k ) .b.ε with ε = 0 shows that F ρ cannot be isomorphic to J k ( λ ). (cid:4) Bibliography
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