Traces in oriented homology theories, II of algebraic varieties

Transcription

1 Traces in oriented homology theories, II of algebraic varieties K. Pimenov General part 1.1 Introduction The present preprint is the second one in a short series of preprints. The main aim of this series is to give a proof of the only non-trivial result on algebraic cobordism stated by V.Voevodsky in [V, Th.3.22]. In this preprint we introduce the notion of an oriented homology theory on algebraic varieties following the outline of [PS], [P1] and [Pa]. The main result of the preprint is Theorem It states that for a given oriented homology theory there exists a family of pull-back operators, one operator for each projective morphism of smooth varieties. This family of operators satisfies a list of properties and is called a trace structure on the theory. The operators are called trace operators. It is proved that the trace structure is unique provided that it is normalized in a natural way. So in this preprint we consider a field k and the category of pairs (X, U) with a smooth variety X over k and its open subset U. By a homology theory we mean a covariant functor A from this category to the category of abelian groups endowed with a functor transformation : A(X, U) A(U) and satisfying the localization, Nisnevich excision and homotopy invariance properties (1.3.1). We consider four structures a homology theory A can be equipped with: an orientation on A, a Thom structure on A, a Chern structure on A and a trace structure. The first part of the preprint has already published as [Pim]. It concerns the relation of the first three structures to each other. Namely we prove that there are natural bijections between the first three of the structures. A heart of these section is Projective Bundle Theorem We give a rather geometric proof for it. The main theorem of this section is theorem which uses a blow up technique developed in the proof of Theorem In contrast with the paper [Ne] we work just with a homology theory A rather than with a pair (A, A ) of (co)homology. We use an extra structure on A called Chern structure (see 2.1.1) and do not use a (co)homological multiplication at all. A Chern structure on a multiplicative pair in a sense of [Ne] gives obviously a Chern structure in our sense on the homology theory A : e Z pim (L) = c Nen(L) : A Z (X) AZ (X). the author is supported by grant PD of Russian Government 1

2 A construction of the trace structure is given in the second part of the preprint. Section led on thoroughly a work [P1]. Subsection 4.1 delights some new aspects which are appeared in homological context. In this section we prove that for an oriented theory homology groups are modules over the Lazard ring. We do not prove in this text that there is a bijection between trace structures and orientations since our definition of orientation uses supports but there are no supports when we discuss trace structure. However we prove uniqueness theorem in subsection 5.3. In Appendix we show a backward relation between Trace structure and Chern prestructure. Besides we extend uniqueness theorem to a pretheory context. Given an oriented commutative ring T -spectrum E one can form as a cohomology theory E so a homology theory E. For a smooth projective variety X of dimension d with a structural morphism f : X pt the class [X] := f! (1) E 2d,d (X) is called the fundamental class of X. It is proved in [PY, Thm.2.2] that the cap product with this class [X] : E p,q (X) E 2d p,d q (X) is an isomorphism (the Poincare duality isomorphism). Our theorem allows us to deduce some functorial properties of this fundamental class to complete the proof of Voevodsky Theorem. The first part of the preprint was written during the visit of the author to the Universität Bielefeld. The author is very grateful to H.Abels for shown hospitality. The author thanks I.Panin very much for useful discussion on the subject of the present preprint. 1.2 Terminology and notation Let k be a field. The term variety is used in this text to mean a reduced quasi-projective scheme over k. If X is a variety and U X is a Zariski open then Z := X U is considered as a closed subscheme with a unique structure of the reduced scheme, so Z is considered as a closed subvariety of X. We fix the following notation: Ab - the category of abelian groups; Sm - the category of smooth varieties; SmOp - the category of pairs (X, U) with smooth X and open U in X. Morphisms are morphisms of pairs. We identify the category Sm with a full subcategory of SmOp assigning to a variety X the pair (X, ); pt = Spec(k); For a smooth X and an effective divisor D X we write L(D) for a line bundle over X whose sheaf of sections is the sheaf L X (D) (see [Har, Ch.II, 6, 6.13]). P(V ) = P roj(s (V )) - the space of lines in a finite dimensional k-vector space V ; L V = O V ( 1) - the tautological line bundle over P(V ); 1 X - the trivial rank one bundle over X, often we will write 1 for 1 X ; 2

3 P(E) - the space of lines in a vector bundle E; L E = O E ( 1) - the tautological line bundle on P(E); E 0 - the complement to the zero section of E; E - the vector bundle dual to E; z : X E - the zero section of a vector bundle E; 1.3 Homology theories Definition. A homology theory is a covariant functor SmOp A Ab together with a functor morphism : A (X, U) A (U) satisfying the following properties 1. Localization: the sequence A (U) j A(X) i A (X, U) P A (U) j A(X) is exact for each pair P = (X, U) SmOp, where j : U X and i : (X, ) (X, U) are the natural inclusions; 2. Excision: the operator A (X, U ) A (X, U) induced by a morphism e : (X, U ) (X, U) is an isomorphism, if the morphism e is etale and for Z = X U, Z = X U one has e 1 (Z) = Z and e : Z Z is an isomorphism; 3. Homotopy invariance: the operator A (X A 1 ) A (X) induced by the projection X A 1 X is an isomorphism. The operator P is called the boundary operator and is written usually as. Let us stress that the functor takes values in the category of abelian groups rather than in the category of graded abelian groups. The subscript in A is used only to stress that we work with a covariant functor. A morphism of homology theories ϕ : (A, A ) (B, B ) is a functor transformation ϕ : A B commuting with the boundary morphisms in the sense that for every pair P = (X, U) SmOp one has P B ϕ P = ϕ U P A. We write also A Z (X) for A (X, U), where Z = X U, and call the group A Z (X) homology of X with the support on Z. The operator is called the support restriction operator for the pair (X, U). A (X) i A Z (X) (1) Below in the text we will often denote a support restriction operator appearing in various contexts by β. 1.4 General properties of homology theories We specify here certain properties of an arbitrary homology theory A which are useful below in the text. 3

4 The localization property implies that A (X) = A (X, X) = 0. Therefore A ( ) = A ( ) = If any two of morphisms (X, U) (Y, V ), X Y, U V, defined by a morphism f : (X, U) (Y, V ), induce isomorphisms on the level of A then the third of these morphisms induces an isomorphism on the level of A Localization sequence for a triple. Let T Y X be closed subsets of a smooth variety X. Let : A T (X) A (X T ) be the boundary map for the pair (X, X T ). Consider the support restriction map e : A (X T ) A Y T (X T ) and set Y,T = e : A T T (X) AY (X T ). We claim that the sequence... A Y (X) α A T (X) A Y T (X T ) β A Y (X) α A T (X)... with the obvious mappings α and β is a complex and moreover it is exact. We call this sequence the localization sequence for the triple (X, X T, X Y ). If Y = X, then this sequence coincides with the localization sequence for the pair (X, X T ) Mayer-Vietoris sequence. If X = U 1 U 2 is a union of two open subsets U 1 and U 2 and if Y is a closed subset in X, then set T i = Y U i, Y i = Y U i, U 12 = U 1 U 2 Y 12 = U 12 Y. Consider the morphism of the localization sequences for the triples X Y T 1 and U 2 Y 2 T 1 induced by the inclusion of the triples (U 2, U 2 T 1, U 2 Y 2 ) (X, X T 1, X Y ) A Y 12 (U 2 T 2 ) β 2 A Y 2 (U 2 ) A T 1 (U 2 ) β 1 α 2 γ A Y 1 (X T 1) α 1 A Y (X) e A T 1 (X) A Y 12 (U 2 T 2 ) A Y 1 (X T 1). The map γ is an isomorphism by the excision property. Also by excision property we may identify A Y 12 (U 2 T 2 ) with A Y 12 (U 12 ) and A Y 1 (X T 1) with A Y 1 (U 1). Set d = γ 1 e : A Y (X) A Y 12 (U 12 ). We claim that the sequence... A Y (X) d A Y 12 (U 12 ) (β 1, β 2 ) A Y 1 (U 1) A Y 2 (U 2) α 1+α 2 A Y (X)... is exact and call this sequence the Mayer-Vietoris sequence of the open covering X = U 1 U 2. The proof of the exactness is straightforward and we skip it. The Mayer-Vietoris sequence is natural in the following sence.if f : X X is a morphism and X = U 1 U 2 is a Zariski covering of X such that f(u i) U i and if Y is a closed subset in X containing f 1 (Y ), then the mappings f : A Y (X ) A Y (X), f : A Y i (U i ) AY i (U i), f : A Y 12 (U 12 ) AY 12 (U 12 ) form a morphism of the corresponding Mayer-Vietoris sequences. 4

5 Let i r : X r X 1 X 2 be the natural inclusion (r = 1, 2). Let Y r X r be a closed subset for (r = 1, 2) Then the induced map A Y 1 (X 1) A Y 2 (X 2) A Y 1 Y 2 (X 1 X 2 ) is an isomorphism. Proof. This follows from the Mayer-Vietoris property and the fact that A ( ) = Strong homotopy invarance. Let p : T X be an affine bundle (i.e., a torsor under a vector bundle). Let Z X be a closed subset and let S = p 1 (Z). Then the natural map p : A S (T ) AZ (X) is an isomorphism. If s : X T is a section then the induced operator s : A Z (X) A S (T ) is an isomorphism as well. Proof. First consider the case Z = X. Then S = T and we have to check that the map p : A (T ) A (X) is an isomorphism. Choose a finite Zariski open covering X = U i such that T i = p 1 (U i ) is isomorphic to the trivial vector bundle over each U i and then use the morphism of the Mayer-Vietoris sequences and the homotopy invariance property of the homology theory A. To prove the general case consider the localization sequences for the pairs (X, X Z) and (T, T S). The morphism p : (T, T S) (X, X Z) form a morphism of these two long exact sequences. The 5-Lemma completes the proof Deformation to the normal cone. The deformation to the normal cone is a well-known construction (for example, see [Fu]). Since the construction and its property (3) play an important role in what follows we give here some details. Let i : Y X be a closed imbedding of smooth varieties with the normal bundle N. There exists a smooth variety X t together with a smooth morphism p t : X t A 1 and a closed imbedding i t : Y A 1 X t such that the map p t i t coincides with the projection Y A 1 A 1 and the fiber of p t over 1 A 1 is canonically isomorphic to X and the base change of i t by means of the imbedding 1 A 1 coincides with the imbedding i : Y X; the fiber of p t over 0 A 1 is canonically isomorphic to N and the base change of i t by means of the imbedding 0 A 1 coincides with the zero section Y N. Thus we have the diagram (N, N Y ) i 0 (X t, X t Y A 1 ) i 1 (X, X Y ) (2) Here and everywhere below in the text we identify a variety with its image under the zero section of any vector bundle over this variety. Recall a construction of X t, p t and i t. For that take X t to be the blow-up of X A1 with the center Y {0}. Set X t = X t X where X is the the proper preimage of X {0} under the blow-up map. Let σ : X t X A 1 be the restriction of the blow-up map σ : X t X A1 to X t and set p t to be the composition of σ and the projection X A 1 A 1. The proper preimage of Y A 1 under the blow-up map is mapped isomorphically to Y A 1 under the blow-up map. Thus the inverse isomorphism gives the desired imbedding i t : Y A 1 X t (observe that i t (Y A 1 ) does not cross X). 5

6 It s not difficult to check that the imbedding i t satisfies the mentioned two properties ( the preimage of X 0 under the map σ consists of two irreducible components: the proper preimage of X and the exceptional divizor P(N 1). Their intersection is P(N) and i t (Y A 1 ) crosses P(N 1) along P(1) = the zero section of the normal bundle N ). We claim that the diagram (2) consists of isomorphisms on the level of A. We will not here give the proof of this theorem because the proof is straightforward analogous to one given in cohomological context in [Pa] (c.f. Theorem 2.2.8) Theorem. The following diagram consists of isomorphisms A Y (N) i0 A Y A1 (X t ) i1 A Y (X). (3) Moreover for each closed subset Z Y the following diagram consists of isomorphisms as well A Z (N) i0 A Z A1 (X t ) i1 A Z (X). (4) Corollary. Let j 0 : P(1 N) X t be the imbedding of the exceptional divisor into X t and let j 1 = e t i 1 : X X t, where e t : X t X t is the open inclusion. Then the mapping is an isomorphism. Proof. Consider the commutative diagram A P(1) (P(1 N)) j0 A Y A1 (X t ) (5) A Y (N) e i 0 A Y A1 (X t ) e t A P(1) (P(1 N)) j 0 A Y A1 (X t ) where the vertical arrows are the natural mappings. The vertical arrows are isomorphisms by the excision property. The operator i 0 is an isomorphism by the first item of Theorem Thus the operator j 0 is an isomorphism Let X be a smooth variety and let L be a line bundle over X. Let E = 1 L and let ī L : X = P(L) P(E) be the closed imbedding induced by the direct summand L of E. Let A (P(E)) β A P(1) (P(E)) be the support restriction operator and let ī L : A (P(L)) A (P(E)) be the natural mapping. We claim that the following sequence 0 A (P(L)) īl A (P(E)) β A P(1) (P(E)) 0. (6) is exact. To prove this consider U = P(E) P(1) with the open inclusion j : U P(E) and observe that U becomes a line bundle over X by means of the linear projection q : U P(L) = X (the line bundle is isomorphic to L ). The obvious inclusion i L : 6

7 P(L) U is just the zero section of this line bundle, ī L = j i L and the natural operator i L : A (P(L)) A (U) is an isomorphism (the inverse to the one q ). Now consider the pair (P(E), U). By the localization property the following sequence... A (U) j A (P(E)) β A P(1) (P(E))... is exact. If P(E) p X is the natural projection then the operator i L p : A (P(E)) A (U) splits j. This implies the injectivity of j and the surjectivity of β in the long sequence above. To proof that the sequence (6) is short exact it remains to recall that the operator i L is an isomorpism and ī L = j i L We use here the notation from Let e t : X t X t be the open inclusion and let p : P(1 N) Y be the projection and let s : Y P(1 N) be the section of the projection identifying Y with the subvariety P(1) in P(1 N). The following commutative diagram will be repeatedly used below in the text j 0 P(1 N) X j 1 t X s I i t Y k 0 Y A 1 k 1 Y, where I t = e t i t and j 0 is the inclusion of the exceptional divisor and j 1 = e t i 1 and k 0, k 1 are the closed imbedding given by y (y, 0) and y (y, 1) respectively Lemma (Useful lemma). Under the notation of let j t : V t = X t Y A 1 X be the inclusion. If the support restriction operator A (P(1 N)) A P(1) (P(1 N)) is surjective then Im(j 0 ) + Im(jt ) = A (X t ), in the other words the operator j 0 jt : A (P(1 N)) A (V t ) A (X t ) is an epimorphism. In particular this holds if Y is a divisor on X. Proof. Consider the commutative diagram A (P(1 N)) β j 0 A (X t) β t A P(1) (P(1 N)) j 0 A Y A1 (X t ). where β and β t are the support restriction operators. The bottom operator j 0 is an isomorphism by Corollary The map β is surjective by the very assumption (if Y is a divisor in X then β is surjective by ). Since the composition β t j 0 coincides with the one j 0 β it is surjective as well. The localization sequence for the pair (X t, V t) shows that Im(j t) = Ker(β t). The Lemma follows. 7

8 Let i : P(V ) P(W ) and j : P(V ) P(W ) be two linear imbeddings that is imbeddings induced by linear imbeddings V into W. If the dimension of V is strictly less than the dimension of W, then i = j : A (P(V )) A (P(W )). In fact, in this case there exists a linear automorphism φ of W which has the determinant 1 and such that j = φ i. Since φ is a composite of elementary matrices and each elementary matrix induces the identity automorphism A (P(W )) (by the homotopy invariance of A ) one gets the relation φ = id. Therefore j = φ i = i. 2 Chern and Thom structures on A 2.1 Ghern structure for line bundles In this section A is a homology theory. If X is a smooth variety we write 1 X for the trivial rank one bundle over X. Often we will just write 1 for 1 X if it is clear from a context what the variety X is Definition. A Chern structure on A is an assignment which associate to each smooth X, closed subset Z X and each line bundle L/X a homomorphism e Z (L) : A Z (X) A Z (X) satisfying the following properties 1. functoriality: Let ϕ : (X, U ) (X, U) be a morphism of varieties, Z = X U, Z = X U and L be a linear bundle on X. Then the following diagram commutes A Z (X ) ϕ A Z (X) e Z (ϕ L) A Z (X ) ϕ e Z (L) A Z (X). Chern homomorphisms commutes with the boundary operator from long localization sequences. e Z (L 1 ) = e Z (L 2 ) for isomorphic line bundles L 1 and L 2 ; 2. nondegeneracy: the operator (p, p e) : A (X P 1 ) A (X) A (X) is an isomorphism, where O( 1) is the tautological line bundle on P 1, e = e(q O( 1)), q : X P 1 P 1 and p : X P 1 X are the projections; 3. vanishing: e Z (1 X ) vanishes for any pair Z X. The homomorphisms e Z (L) will be called Chern homomorphisms of the line bundle L. (It will be proved below in that these homomorphisms are nilpotent). if Z = X the superscript in e Z (L) will be omitted. 8

9 2.1.2 Lemma. Let L be a tautological line bundle O( 1) over X P 1. Then one has the following relations: e(l) 2 = e(l)e(l ) = 0; e(l ) = e(l) Hom(A (X P 1 ), A (X P 1 )). Proof. To prove the first assertion we show that for any two line bundle L 1 and L 2 over P 1 one has e(q L 2 )e(q L 1 ) = 0 where q : X P 1 P 1 is the projection. Fix two points {0}, { } P 1 and consider a commutative diagram A (X P 1 ) β A X {0} γ (X P 1 ) A X {0} (X (P 1 { })) e(q L 1 ) e X {0} (q L 1 ) e X {0} (q L 1 ) A (X P 1 ) β A X {0} (X P 1 ) γ A X {0} (X (P 1 { })) where β is a support restriction and γ is an excision isomorphism. The right vertical arrow vanishes because q L 1 X A 1 is a trivial line bundle. Therefore the middle vertical arrow vanishes. We get β e(q L 1 ) = 0. By the sequence 0 A (X) s A (X P 1 ) β A X {0} (X P 1 ) 0 is exact. It implies that Im(e(q L 1 )) Im(s ). Since the line bundle s q L 2 is trivial one has e(q L 2 )e(q L 1 )(A (X P 1 )) (e(q L 2 )s )(A (X)) = (s e(s q L 2 ))(A (X)) = 0 and the first part of the lemma is proved. In the proof of the second part we assume for simplicity that X = pt. Let i 1, i 2 : P 1 P 1 P 1 be closed imbeddings, p 1, p 2 : P 1 P 1 P 1 be corresponding projections and : P 1 P 1 P 1 be a diagonal imbedding. Also let s : pt P 1 P 1 be any closed imbedding and P : P 1 P 1 pt be a projection. The projection P 1 pt will be denoted by a small letter p. We claim the following relation holds: = i 1 + i2 s p. (7) By property 2 there is an isomorphism A (P 1 P 1 ) A (pt) A (pt) A (pt) A (pt) given by (P, P e(p 1L), P e(p 2L), P e(p 1L)e(p 2L)). In order to check the relation 7 we must check four equations: P = P (i 1 + i2 s p ) P e(p 1 L) = P e(p 1 L)(i1 + i2 s p ) P e(p 2L) = P e(p 2L)(i 1 + i 2 s p ) P e(p 1 L)e(p 2 L) = P e(p 1 L)e(p 2 L)(i1 + i2 s p ). The first one is obvious. To prove the other three ones it is enough to observe that e(p r L) = e(l); e(p r L)ir = ir e(l); e(p r L)ik = 0 for k r. Then both hands in the second and third equations are equal to p e(l) and both hands in the fourth equation are zero. 9

10 Now we are ready to complete the proof of this lemma. Consider a line bundle M = p 1L p 2L over P 1 P 1. Then i 1M = L; i 2M = L ; M = 1 P 1. By equation (7) one has P e(m) = P e(m)(i 1 + i2 s p ). Computing left and right hand we get 0 = p (e(l) + e(l )). To prove the relation u = v Hom(A (P 1 ), A (P 1 )) it is sufficient by property 2 in Definition to prove that p u = p v & p e(l)u = p e(l)v. The first one in our case (u = e(l), v = e(l )) is checked just above. In the second one both sides of the relation vanish because e(l) 2 = 0 and e(l)e(l ) = 0 by the first assertion of this Lemma Definition. If one has a Chern structure (L, X, Z) e Z (L) on a homology theory A then the assignment (L, X, Z) (e ) Z (L) = e Z (L ) will be called a dual Chern structure with respect to {e}. We must check that all properties of Chern structure are valid. The only point to prove is the nondegeneracy. It is hold because by Lemma e (O( 1)) = e(o(1)) = e(o( 1) Definition. One says that A is endowed with a Thom structure if for each smooth variety X, closed subset Z X and each line bundle L/X it is chosen and fixed a homomorphism th Z (L) : A Z (L) AZ (X) satisfying the following properties 1. functoriality: Let Z X and Z X be closed subsets, L/X be a line bundle over X; ϕ : X X be a morphism such that ϕ 1 (Z) Z. Then the following diagram commutes A Z th Z (L ) (L ) A Z (X ) ϕ L A Z (L) th Z (L) ϕ A Z (X), where L = L X X is a line bundle over X and ϕ L : L L is a morphism of line bundles induced by ϕ. Homomorphisms th Z (L) commutes with from a long localization sequence. 2. If τ : L 1 L 2 is an isomorphism of line bundles over X then for any closed subset Z X one has th Z (L 2 ) τ = th Z (L 1 ) Hom(A Z (L 1), A Z (X)). 3. nondegeneracy: th(1) : A X (X A 1 ) A (X) is an isomorphism ( here X is identified with X {0} ) Remark. Chern and Thom homomorphisms commutes with from the Mayer- Vietoris sequence Lemma. Any homomorphism th Z (L) is an isomorphism. 10

11 Proof. The usual arguments using Mayer-Vietoris and long exact sequence for a triple. Now we are going to describe a one-to-one correspondence between Chern and Thom structures on A Lemma. Suppose that a homology theory theory A is equipped with a Chern structure. Then the following assertions hold 1. Let e = e(o P n( 1)) and β : A (P n ) A Pn 1 (P n ) be a support restriction. Then there exists a unique homomorphism ê : A Pn 1 (P n ) A (P n ) such that ê β = e. 2. Let L/X be a line bundle, Z X be a closed subset. Let p : P(1 L) X be a projection, s : X P(1 L) be a closed imbedding identifying X with P(1), s L : X P(1 L) be a closed imbedding identifying X with P(L) and β : A p 1 (Z) (P(1 L)) A s (Z) (P(1 L)) be the support restriction. Consider the line bundle M = O 1 L (1) p L over P(1 L). Then s M = L, s L M = 1 X; there exists a unique homomorphism α Z (L) : A s (Z) (P(1 L)) A Z (X) such that α Z (L) β = p e(m). Proof. To prove the first statement consider a long localization sequence for pair (P n, P n P n 1 ). As A (P n P n 1 ) = A (A n ) = A (pt) by homotopic invariance each third arrow in this sequence j : A (P n P n 1 ) A (P n ) is mono. Therefore the long sequence splits into short exact sequences 0 A (A n ) j A (P n ) β A Pn 1 (P n ) 0. (8) As j O P n( 1) is trivial then e j = j e(j O P n( 1)) = 0. Hence e factors through the factorgroup A Pn 1 (P n ) as required. The proof of the second statement is similar the first one. First of all we compute s M and s L M. One has s O 1 L ( 1) = 1 X and s L O 1 L( 1) = L. Therefore s M = 1 X s p L = L and s L M = L L = 1 X. Consider a long localization sequence for triple P(1 L) p 1 (Z) s (Z):... A p 1 (Z) s (Z) (P(1 L) P(1)) j A p 1 (Z) (P(1 L)) β A s (Z) (P(1 L))... where A p 1 (Z) s (Z) (P(1 L) s (Z)) = A p 1 (Z) s (Z) (P(1 L) P(1)) by the excision property. Since P(1 L) P(1) is a line bundle over P(L) (isomorphic to L ) then s L : A Z (X) (P(1 L) P(1)) and (p P(1 L) P(1) ) are inverse to each other isomorphisms. Hence j in the long sequences has splitting (s L ) 1 p and this sequence splits into short exact ones: A p 1 (Z) s (Z) 0 A Z (X) sl A p 1 (Z) (P(1 L)) β A s (Z) (P(1 L)) 0. (9) 11

12 Since s L M is trivial then the homomorpfism ep 1 (Z) (M) factors through β: e p 1 (Z) (M) = ê p 1 (Z) (M) β. We can set α Z (L) = p êp 1 (Z) (M). (10) Remark. Notation used in the previous lemma will be used below in this section without any further explanation Lemma. Let γ : A Z (L) = As (Z) (P(1 L) P(L)) A s (Z) (P(1 L)) be the excision isomorphism. Suppose that A is endowed with a Chern structure. Let α Z (L) = α Z (L) γ where a homomorphism α Z (L) is constructed in Lemma Then homomorphisms α Z (L) define a Thom structure on a homology theory A. Proof. The functiorial properties of such constructed α Z (L) are obvious consequences from the analogous properties of Chern homomorphisms. The only item we need to check is nondegeneracy. By the nondegeneracy property of Chern structure we have a commutative diagram: 0 A (X A 1 ) j A (X P 1 ) β A X {0} (P 1 ) 0 0 A (X) id i A (X P 1 ) p ζ α(1) A (X) 0 where ζ = e(o(1)) : A (X P 1 ) A (X P 1 ) is the Chern homomorphism. The top row is exact in the same way as sequence (8). The bottom row is exact by the nondegeneracy of the dual Chern structure (c.f. Definition 2.1.3). As two of vertical arrows are isomorphisms the third one is an isomorphism as well Remark. The bar-notation introduced above will be used below without any further explanation Lemma. Let {th Z (L)} be a Thom structure on a homology theory A. Define a homomorphism c Z (L) as the composition [ ] A Z s (X) A p 1 (Z) (P(1 L)) β A s (Z) (P(1 L)) thz (L) A Z (X). Then the assignment (L, X, Z) c Z (L) is a Chern structure on A. Moreover, if we begin with a Chern structure {e Z (L)} and take a Thom structure to be the one corresponding to the Chern structure {e Z (L)} by the construction from lemma then for any (L, X, Z) one has c Z (L) = e Z (L). Proof. Functorial properties are obvious. 12

13 To proof the nondegeneracy property for a just defined Chern structure it is sufficient to prove nondegeneracy of the dual Chern structure. The last will follow from the fact that the sequence 0 A (X) A (X P 1 ) p c(o(1)) A (X) 0 is exact. For simplicity of notation we proceed only the case X = pt. Consider P 1 P 2 and a point { } P 2 P 1. Then P 2 { } is turned out to be a line bundle L over P 1 which is isomorphic to O(1). Then P(1 L) is isomorphic to the blow up P 2 { } of the P 2 at the point { }. The exceptional divisor D is identified with P(L) under this isomorphism. We will write P 2 for P 2 { } and σ : P 2 P 2 for the blowing up. By the excision property one has A P(1) (P(1 L) D) = A P(1) (P(1 L)) and A P1 (P2 { }) = A P1 (P2 ). Since σ : P(1 L) D P 2 { } is an isomorphism then the operator A P(1) (P(1 L)) σ A P1 (P 2 ) is an isomorphism. Take a projective line l on P 2 crossing the point. Let {0} = P 1 l and l P 2 be the strict transform of l. It is easy to see that l is just a fibre of projective bundle P(1 L) over a point {0}. Then closed imbedding i : l P(1 L) can be regarded as a morphism of the projective bundles l/{0} and P(1 L)/P 1. Let i 0 : P 1 P 2, τ : P 1 l be arbitrary linear embeddings. By the functoriality of the Thom isomorphisms we have a commutative diagram A (P 1 τ ) β 0 {0} id A (l) A (l) A {0} th(1) (l) A ({0}) id σ i σ i i i A (P 1 ) i 0 A (P 2 ) β A P1 (P2 ) σ 1 A P(1) (P(1 L)) th(l) A (P 1 ). The left hand side square commutes by The bottom row composition map is the homomorphism c(l) by the very definition of c(l). Since the diagram commutes one has the relation p c(l) = β 0 th(1). Replacing in the short exact sequence of type 8. 0 A (pt) A (P 1 ) β A {0} (P 1 ) 0 the term A {0} (P 1 ) by the one A (pt) we get the short exact sequence 0 A (pt) A (P 1 ) th(1) β A (pt) 0 Since p c(l) = th(1) β the nondegeneracy property of the c(l) follows. It remains to prove the relation c(l) = e(l) for any line bundle L over X. 13

14 As in the proof of the second part of Lemma consider the commutative diagram A Z (X) e Z (L) A Z (X) s p A 1 (Z) (P(1 L)) A s (Z) (P(1 L)) e p 1 (Z) (M) α Z (L) s p A 1 (Z) (P(1 L)) β p A Z (X). The right hand side square commutes by the very definition of the operator α Z (L) from Lemma The left hand side square commutes by the functoriality of the Chern classes and the relation s M = L. Then c Z (L) = α Z (L) β s = p s e Z (L) = e Z (L). 2.2 Projective bundle theorem. The nearest aim is to compute the homology of a projective bundle Theorem (Projective Bundle Homology). Let A be a homology theory equipped with a Chern structure (L, X, Z) e Z (L) on A. Let X be a smooth variety and let E/X be a vector bundle with rk(e) = n + 1. For the endomorphism e = e(o E ( 1)) : A (P(E)) A (P(E)) we have an isomorphism (p, p e,..., p e n ) : A (P(E)) A (X) A (X) A (X) where p : P(E) X is a projection. Moreover, for the trivial rank n + 1 bundle E we have e n+1 = 0. In addition, all the assertions hold if the endomorphism e is replaced by the one e(o E (1)) : A (P(E)) A (P(E)). Proof. This variant of the proof is based on an unpublished notes of I.Panin. Let P n 1 P n be a hyperplane and 0 P n P n 1 be a point. Recall that the projection P n {0} P n 1 with the center 0 makes P n {0} into a line bundle over P n 1 which is isomorphic to O P n 1(1). Let us denote this bundle by L below. Projective bundle P(1 L) is isomorphic to the blow up P n {0} of the P n at the point {0}. We will write P n for the P n {0} and σ : P n P n for the blowing up. If P(1 L) is identified with P n then σ(p(1)) = P n 1 P n. Let us regard our homology theory A to be equipped with Thom structure corresponding given Chern structure by Lemma We will write below in the proof e k for e(o P k( 1)) Lemma. Let L 1 and L 2 be two line bundle over P(1 L), and i : P 1 P(1 L) be a closed imbedding of any fiber. Then the following three conditions are equivalent: 1. L 1 = L2 ; 2. s L 1 = s L 2 and i L 1 = i L 2 ; 14

15 3. s L L 1 = s L L 2 and i L 1 = i L 2. Proof. This lemma is an easy consequence of a general projective bundle Picar group computation. Recall that the sequence 0 P ic(x) p P ic(p(e)) i P ic(p n ) = Z 0 is exact for any projective bundle p : E X. Here i : P n X is any fibre of this bundle. If E = P(1 L) then s and s L be various splitting of this sequence. The lemma follows from this remark directly Lemma. Let M = O 1 L (1) p L be a line bundle over P(1 L). Then the diagram below commutes. A (P(1 L)) β A P(1) (P(1 L)) be(m ) A (P(1 L)) σ A (P n ) β cσ σ A Pn 1 (P n ) be(opn ( 1)) A (P n ) Proof. The left square commutes by natural reason. Homomorphisms β and β are taken from sequences (8) and (9). Therefore they both are onto. This implies that one has to check the relation σ ê(m) β = ê n βσ in order to prove that right square commutes. Last relation is equivalent to σ e(m ) = e n σ. It follows from the functoriality of Chern homomorphism and the fact that σ O P n( 1) = O 1 L ( 1) p L. Last equality is easy to prove using lemma 2.2.2: s L σ O P n( 1) = σ O P n( 1) { } is trivial and s L M is trivial by Lemma 2.1.7; restricting both side of the desired relation onto a fibre we see that in both cases we get the line bundle L = O P n 1( 1). So two bundles over P(1 L)) are isomorphic to each other and lemma follows. Before the general case we prove Theorem for the trivial bundle E = X A n+1. For simplicity of notation we procceed only the case X = pt. In this case P(E) = P n and we proceed the proof by the induction on the integer n. By induction assumption there is an isomorphism γ n 1 = ((p n 1 ), (p n 1 ) e n 1,..., (p n 1 ) e n 1 n 1 ) : A (P n 1 ) A (pt) A (pt) and e n n 1 = 0. Let u : A (P n ) A (P n 1 ) be a composition map [ ] A (P n ) (p,p en,...,p en 1 n ) A (pt) A (pt) γ 1 n 1 A (P n 1 ). Since i e n 1 = e n i Hom(A (P n 1 ), A (P n )) where i : P n 1 P n is a linear embedding then ui = id A (P n 1 ). Under this notation if one prove that is an isomorphism then the theorem follows. (p, u e n ) : A (P n ) A (pt) A (P n 1 ) (11) 15

16 2.2.4 Lemma. Under induction hypothesis there is a following short exact sequence: 0 A (P n 1 ) i A (P n ) β A { } (P n ) 0. Proof. Given a point { } not lying on P n 1 consider a long exact sequence for the pair (P n, P n { }):... A (P n { }) j A (P n ) β A { } (P n )... Since P n { } can be regarded as a linear bundle on P n 1 then zero section natural map z : A (P n 1 ) A (P n { }) is an isomorphism by strong homotopic invariance Moreover, since j z = i then the homomorphism A (P n ) z u A (P n { }) is the splitting for j. Hence, j is mono and the long exact sequence splits into short exact sequences: 0 A (P n { }) j A (P n ) β A { } (P n ) 0. We can write A (P n 1 ) for A (P n { }) and so we get a desired sequence Lemma. Under induction hypothesis one has: 1. The image of the operator e n : A (P n ) A (P n ) lies in the image of i : A (P n 1 ) A (P n ) and e n+1 n = The image of an operator e(m) : A (P(1 L)) A (P(1 L)) lies in the image of s : A (P n 1 ) A (P(1 L)) where s : P n 1 P(1 L) is a closed imbedding identifying P n 1 with P(1) as in Lemma The following relations hold where P : P(1 L) P n 1 is a projection. Proof. Consider a diagram s P e(m) = e(m); i ue n = e n (12) A (P n ) β A { } (P n ) j A { } (A n ) A (P n 1 ) e n i A (P n ) β e { } n j e { } n A { } (P n ) A { } (A n ) It is commutative from functoriality of Chern homomorphism. Show that the right vertical arrow vanishes. Since the bundle O P n( 1) becomes trivial being restricted to A n = P n P n 1 j P n. A { } (A n ) e{ } (j O P n ( 1)) A { } (A n ) then the right vertical arrow vanishes by property 3 in definition The middle vertical arrow is null as well because j is an excision isomorphism. Therefore the diagram shows that β e n = e { } n β = 0. Since Ker(β) = Im(i ) by Lemma then Im(e n ) Ker(β) = Im(i ) as required. 16

17 This fact implies that e n+1 n = 0. Indeed, let e(a) = i (b) for some b A (P n 1 ). Since e n n 1 = 0 then e n+1 n (a) = e n n e(a) = e n n i (b) = i e n n 1(b) = 0. In order to proof the second part of the lemma we consider the commutative diagram: A (P(1 L)) β A P(L) (P(1 L)) j A P(L) (P(1 L) P(1)) A (P(1)) i e(m) A (P(1 L)) β e P(L) (M) j e P(L) (M) A P(L) (P(1 L)) A P(L) (P(1 L) P(1)). As above j is an excision isomorphism. The projection P : P(1 L) P(1)X is a homotopic equivalence with inverse s L. Since s L M = 1 X by lemma then j M = P s L M is a trivial line bundle. Hence, ep(l) (M) = 0. From exact sequence (9) one has Ker(β) = Im(s ). Since β e(m) = e P(L) (M) β = 0 then Im(e(M)) Ker(β) = Im(s ) Remark. One can write M for M and write ê(m ) for e(m) in the second and the third assertion of the lemma because their images coincide. We are ready now to complete the proof of Projective Bundle Theorem Denote by α the Thom isomorphism th (L) in the dual Thom structure, i.e. α = P ê(m ) = P ê(o 1 L ( 1) p L ) Hom(A P(1) (P(1 L)), A (P n 1 )) (see formula (10)). Here hat-notation is borrowed from Lemma By the short exact sequence (8) 0 A (A n ) j A (P n ) β A Pn 1 (P n ) 0 one has an isomorphism A (P n ) (p, β ) A (pt) A Pn 1 (P n ). The isomorphism of the second summand A Pn 1 (P n ) with A (P n 1 ) is given by formula α σ 1 where σ : A P(1) (P(1 L)) A Pn 1 (P n ) is an isomorphism. By equation (11) it is sufficient to prove that α σ 1 β = u e n. Applying i to both sides of desired equality one has to prove i α σ 1 β = i u e n. (13) Recall that α = P ê(m ). Therefore i α = σ s P ê(m ) = σ ê(m ) by Lemma Since σ ê(m ) σ 1 = ê n by Lemma then the left hand of equality (13) is equal to ê n β = e n. But the right hand of this equality is equal to e n by relation (12). Now we have finished the proof for the case of the trivial vector bundle E. The general case can be proceeded in the usual way by the Mayer-Vietoris argument. 17

18 2.2.7 Lemma. Let Z be a closed subset in X, L = q O P n( 1) be a line bundle over X P n. Then there is an isomorphism ( p, p e Z Pn (L),..., p (e Z Pn (L)) n) : A Z Pn (X P n ) A Z (X) A Z (X). Proof. The proof is a straightforward consequence from Theorem 2.2.1, the long localization sequence for the pair (X P n, (X Z) P n ) and five lemma Corollary. Let E = M N be vector bundles of constant rank. Then there is an exact sequence 0 A (P(N)) i A (P(E)) β A P(M) (P(E)) 0. Proof. Consider a long exact sequence for the pair (P(E), P(E) P(M)). As P(E) P(M) can be regarded as vector bundle over P(N) we may replace A (P(E) P(M)) by A (P(N)): A (P(N)) i A (P(E)) β A P(M) (P(E))... By projective bundle theorem i : A (P(N)) A (P(E)) has an obvious splitting. Indeed, let rkn = n, ξ E = e(o E ( 1)) and ξ N = e(o N ( 1)). Since i ξ N = ξ E i then the homomorphism (p E, p E ξ E,..., p E ξ n 1 E ) : A (P(E)) A (pt) A (pt) = A (P(N)) is a desired splitting Lemma. For each line bundle L over a smooth X and closed subset Z X the endomorphism e Z (L) of A Z (X) is nilpotent. Proof. Recall that the assertion holds in the case L = O P n(1) by Lemma applied to the dual Chern structure. To prove the general case recall that by Claim from [Pa] one can find for any smooth variety X a diagram of the form X p X f P(V ) (14) with a torsor under a vector bundle p : X X and a morphism f : X P(V ) such that the line bundles L = p (L) and f (O V (1)) over X are isomorphic. By the strong homotopy invariance we can replace X by X and regard that L = f O V (1). The homomorphism e(o V (1)) A(P(V )) is nilpotent as just mentioned above. Thus the homomorphism e X = e(o X P n(1)) is nilpotent as well. (id, f) Let g : X X P n. Then L = g O X P n(1) and the following diagram commutes: 18

19 A (X) g A (X P n ) e(l) A (X) e X g A (X P n ). Since pg = id X the natural homomorphism g is injective. Therefore it is sufficient to check that g e(l) n+1 = 0. As g e(l) n+1 = e n+1 X g = 0 the Lemma is proved Remark. Define a group A Z P (X P ) as an injective limit of the groups A Z Pn (X P n ). Then by Remark to Lemma and Lemma one has that e Z Pn n (A Z Pn (X P n )) = i n 1,n (A Z Pn 1 (X P n 1 )) where e n = e(o X P n( 1)) and i n 1,n : X P n 1 X P n is a linear imbedding. Particularly e Z P (q L) is a surjection where L is a tautological line bundle over P and q : X P P is a projection Lemma. Let A be a homology theory equipped with the Thom structure {th Z (L)} and let {c Z (L)} be the Chern structure corresponding to this Thom structure as define in Lemma Let us take the Thom structure {α Z (L)} corresponding to the Chern structure {c Z (L)} by the construction from lemma Then for any (L, X, Z) one has α Z (L) = th Z (L). Proof. In the proof of the lemma we will use the notation from lemma Since α Z (L) is defined by formula (10) one has to check a relation p ĉp 1 (Z) (M) = th Z (L) (15) First of all to prove a relation α Z (L) = th Z (L) we assume the homomorphism c Z (L) to be a surjection. Consider a diagram: A Z (X) c Z (L) A Z (X). A s (Z) β s id (P(1 L)) thz (L) A Z (X) where β : A p 1 (Z) (P(1 L)) A s (Z) (P(1 L)) is a support restriction. The diagram commutes by the very construction of the homomorphism c Z (L) from Lemma Since c Z (L) is a surjection by assumption and th Z (L) is an isomorphism then β s is a surjection. Therefore equality (15) follows from p ĉp 1 (Z) (M) β s = th Z (L) β s. Since ĉ p 1 (Z) (M)β = c p 1 (Z) (M) and s M = L by Lemma then the left hand is equal to p c p 1 (Z) (M)s = p s c Z (s M) = c Z (L). Right hand is equal to c Z (L) from the diagram above. Lemma and following remark imply that the lemma holds in the (universal) case: for the line bundle L = O( 1) over X P. 19

20 The general case can be be reduced to the universal one as follows. By Jouanolou trick we can find a diagram of the form (14). Therefore one can regard X to be an affine variety. Let f : X P be a morphism such that L = f O P ( 1). Set g : X (id,f) X P. Denoting O X P ( 1) by L one has L = g L. Let G : P(1 L) P(1 L ) be an induced morphism of projective bundles. For a closed subset Z X denote Z P X P by Z. Recall that M = O 1 L ( 1) p L is a line bundle over P(1 L) and M = O 1 L ( 1) P L is a line bundle over P(1 L ). Then one has a commutative diagram: A p 1 (Z) (P(1 L)) thz (L) β A Z (X) s A p 1 (Z) (P(1 L)) p A Z (X). G A P 1 (Z ) (P(1 L )) thz (L ) β A Z g (X P ) G S A P 1 (Z ) (P(1 L )) g P A Z (X P ) One has to check the relation th Z (L) β = p c p 1 (Z) (M). Since g is mono then it is sufficient to check a relation g th Z (L) β = g p c p 1 (Z) (M). The right hand is equal to P G c p 1 (Z) (M) = P c P 1 (Z ) (M )G because G M = M. Since for the universal case the lemma holds then P c P 1 (Z ) (M ) = th Z (L ) β. Therefore the right hand is equal to th Z (L ) β G. Since the diagram commutes the last is equal to the left hand. The lemma follows Definition. The Chern structure on a homology theory A is said to be COM- MUTATIVE if for any smooth variety X, closed subset Z X and line bundles L 1 /X and L 2 /X the Chern homomorphisms e Z (L 1 ) and e Z (L 2 ) commute with each other. 2.3 Chern classes Definition. Let A be a homology theory. We say that A is endowed with a Chern class theory if for any vector bundle E/X and any closed subset Z X there are given homomorphisms c Z i (E) : A Z (X) A Z (X) such that 1. c Z i depends only on an isomorphism class of E; For any morphism f : X X and vector bundle E/X the following diagram is commutative: A (X ) cz i (E ) A (X ) f f A (X) cz i (E) A (X) where E = E X X is a pullback vector bundle over X ; Z X is a closed subset and Z = f 1 (Z); 20

21 2. c Z 0 (E) = id A Z (X); the restriction of the assignment (L, X, Z) c Z 1 (L) to line bundles is a Chern structure on A ; 3. Any two homomorphisms c Z i (E) and cz j (F ) commute with each other; 4. Cartan formula: For each short exact sequence of vector bundles 0 E 1 E E 2 0 we have c Z r (E) = cz r (E 1)c Z 0 (E 2) + c Z r 1 (E 1)c Z 1 (E 2) c Z 0 (E 1)c Z r (E 2); 5. Vanishing property: c Z i (E) = 0 for i > rke Theorem. Let A be endowed with a commutative Chern structure (L, X, Z) e Z (L). Then there exists a unique Chern class theory on A such that for each line bundle L one has c Z 1 (L) = e Z (L). Moreover the Chern class homomorphisms c Z i (E) are nilpotent for i > 0. Proof. For simplicity of notation we proceed only the absolute case. First of all we prove the uniqueness assertion. If there are two assignements E/X c i(e) and E/X c i (E) satisfying the required properties. Then they coincide on line bundles by the properties 2 and 5. Therefore they coincide on direct sums of line bundles by the Cartan formula (4). Thus they coincide on all vector bundles by the splitting principle [PY, Prop.3.5]. It remains to construct a Chern classes theory. Let X be a smooth variety and E/X be a vector bundle with rke = n. Set ξ = e(o E ( 1)) Hom(A (P(E), A (P(E))). Let γ = (p, p ξ,..., p ξ n 1 ) : A (P(E)) A (X) is an isomorphism from n Theorem Then the homomorphism p ξ n γ : A (X) A (X) can be written uniquely as a row (( 1) n 1 c n, ( 1) n 2 c n 1,..., c 1 ) where c i End(A (X)). By definition the homomorphism c k : A (X) A (X) is an k-th Chern class homomorphism for the vector bundle E. One can say the same in a slightly different form: c 1,..., c n are uniquely determined homomorphism such that p ξ n = c 1 p ξ n 1 c 2 p ξ n ( 1) n 1 c n p. (16) In order to define Chern class homomorphism with supports one has to apply Projective Bundle Theorem with supports Claim. Homomorphisms c i (E) satisfy the theorem. The rest of the proof is devoted to the proof of this Claim. The property c 0 (E) = 1 holds by the very definition. To prove the property c 1 (L) = e(l) for a line bundle L observe that P(L) = X and O L ( 1) = L over X. Thus ξ = e(l) Hom(A (X), A (X)) and the relation (16) shows that c 1 (L) = e(l). Now prove the property 1 of Definition A vector bundle isomorphism ϕ : E E induces an isomorphism Φ : P(E) P(E ) of the projective bundles and a line bundle isomorphism Φ (O E ( 1)) O E ( 1) over P(E). Therefore Φ ξ = ξ Φ ; p = p Φ. (17) 21

22 By the formula (16) one has equation p ξ n = n ( 1) k 1 c k (E) p ξ n k k=1. Using relations (17) one can derive. p (ξ ) n Φ = n ( 1) k 1 c k (E) p (ξ ) n k Φ k=1 Since Φ is an isomorphism then one gets p (ξ ) n = n ( 1) k 1 c k (E) p (ξ ) n k. Therefore by very definition of Chern classes given by (16) c k (E ) = c k (E). The proof of the other funktoriality assertion is essentialy the same. Let f : X X be a morfism, E/X be a vector bundle and E = f E be the induced vector bundle over X. Denote by F : P(E ) P(E). The following relations hold: f p = p F ; ξ F = F ξ ; f p = p F. (18) One need to show that f c k (E ) = c k (E) f. One has the following chain of relations f p (ξ ) n = p p ξ n F = ( = k=1 n ( 1) k c k (E) p ξ n k )F = k=1 n ( 1) k c k (E)f p (ξ ) n k. k=1 From the one hand the homomorphism f p (ξ ) n (γ ) 1 : n : A (X ) A (X) can be given by the row (( 1) n f c n (E ), ( 1) n 1 f c n 1 (E ),..., f c 1 (E )). For the rest of the proof we need in the following Claim. For a rank r vector bundle E set c t (E) = id + c 1 (E)t + + c r (E)t r End(A (X))[t]. Let T be a smooth variety and let E = r i=1 L i for certain line bundles L i over T. Then one has r c t (E) = c t (L i ). Proof. Let ξ E = e(o E ( 1)) where O E ( 1) is the tautological line bundle on P(E). First of all we shall prove the relation i=1 r (ξ e(p E L i)) = 0 (19) i=1 where p E : P(E) T is a projection. Since all first Chern class homomorphisms commute the formula above has sense. 22

23 To prove the very last relation set F = L 1 L r 1. The canonical projection P(E) P(F ) P(L r ) makes P(E) P(F ) into a vector bundle over X = P(L r ) with the zero section P(L r ). Therefore the natural mapping A (P(L r )) A (P(E) P(F )) is an isomorphism. Let j : P(E) P(F ) P(E) be an open inclusion. Since O E ( 1) P(Lr) = L r then the bundles j O E ( 1) and j p E L r coincide. One has a commutative diagram 0 A (P(F )) i A (P(E)) β A P(Lr) (P(E)) 0 0 A (P(F )) ξ F e(p F Lr) i A (P(E)) ξ E e(p E Lr) ξ P(Lr) E e P(Lr) (p E Lr) β A P(Lr) (P(E)) 0 where strings are the short exact sequences of type The right vertical arrow is null by following reasons. A P(Lr) (P(E)) = A P(Lr) (P(E) P(F )) by exscision property; but the bundles O( 1) and p L r coincide being restricted to P(E) P(F ). So we get that the image of ξ E e(p E L r) lies in the image of i. Since ( r 1 i=1 ξ F e(p F L i)) = 0 by induction hypothesis then r 1 (ξ E e(p EL i ))) i = i (ξ F e(p F L i )) = 0 r 1 ( i=1 and the relation (19) is proved. We need to check that c k (E) is just a symmetric polynomial σ k (e(l 1 ),..., e(l r )). Expanding brackets in equation (19) one has ξ n σ 1 ξn ( 1) n σ n = 0 where σ k = σ k(e(p L 1 ),..., e(p L r )). Since p σ k = σ kp then p ξ n = σ 1 p ξ n 1 σ 2 p ξ n ( 1) n+1 p. Comparing last formula with the formula (16) one completes the proof. i=1 Claim and splitting principle [PY, Prop.3.5] allow us to complete the proof of Claim Let E and F be a vector bundles over X. Prove that the homomorfhisms c k (E) and c l (F ) commute. Indeed, by the splitting principle [PY, Prop.3.5] there exists a smooth variety T and a morphism r : T X such that each the vector bundle r E and r F is a sum of line bundles and r : A (T ) A (X) is a split surjection. The homomorphisms c k (r E) and c l (r F ) commute as symmetric polynomes of line bundles chern classes, which are commute because Chern structure is commutative. Then c k (E)c l (F )r = r c k (r E)c l (r F ) = r c l (r F )c k (r E) = c l (F )c k (E)r. Last implies commutativity of higher chern classes because r is onto. Cartan formula and nilpotence are easily can be proved essentially in the same way. Below we will need Proposition Proposition. Namely, let Y be a smooth variety and let E be a vector bundles over Y of constant rank n. Let p : P(E) Y and q : P(1 E) Y be the projections and s : Y P(1 E) be the closed imbedding identifying Y with P(1). Then 23

24 1. the homomorphism c n (O E (1) p E) is null; 2. s q c n (O E (1) q E) = c n (O E (1) q E). Proof. Define a rank n 1 vector bundle Q via the short exact sequence 0 O E ( 1) p E Q 0. Tensoring with line bundle O E (1) and applying Cartan formula we get c n (O E (1) p E) = c 1 (O)c n 1 (Q) = 0. The second part of this proposition is a straightforward extension of the third assertion of Lemma Orienting a theory In this subsection A is a homology theory. Two theorems in this subsection shows how one can construct an orientation using a commutative Chern structure (or a Thom structure) on A and how one can construct a commutative Chern structure (or a Thom structure) using an orientation. Let us recall that for a vector bundle E over a variety X we identify X with z(x) where z : X E is the zero section Definition. An orientation on the theory A is a rule assigning to each smooth variety X, to each its closed subset Z and to each vector bundle E/X an isomorphism which satisfies the following properties th Z (E) : A Z (E) A Z (X) 1. invariance: for each vector bundle isomorphism ϕ : E F the diagram commutes A Z (E) ϕ th Z (E) A Z (X) id A Z (F ) th Z (F ) A Z (X) 2. base change: for each morphism f : (X, X Z ) (X, X Z) with closed subsets Z X and Z X and for each vector bundle E/X and for its pull-back E over X and for the projection g : E = E X X E the diagram commutes A Z (E ) g th Z (E ) A Z (X ) f A Z (E) th Z (E) A Z (X) 24

25 3. for each vector bundles p : E X and q : F X the following diagram commutes A Z (E F ) th Z (p F ) A Z (E) th Z (q E) th Z (E) A Z (F ) th Z (F ) A Z (X) and both compositions coincide with the operator th Z (E F ). The operators th Z (E) are called Thom isomorphisms. The theory A is called orientable if there exists an orientation of A. The theory A is called oriented if an orientation is chosen and fixed Lemma. If an assignment (E, X, Z) th Z (E) is an orientation on A, then its restriction to line bundles is a Thom structure on A. If two orientations coincide on each line bundle then they coincide. Proof. The first assertion is obvious. To prove the second assertion consider two orientations th( ) and th ( ) which coincide on line bundles. To prove that for a vector bundle E one has the relation th Z (E) = (th ) Z (E) one may assume by the splitting principle [PY, Prop.3.5] that E = L i is a direct sum of line bundles. Since for each line bundle L one has th Z (L) = (th ) Z (L) the property 3 in Definition shows that th Z (E) = (th ) Z (E) Theorem. Given a commutative Chern structure (L, X, Z) e Z (L) on A and corresponding Thom structure (L, X, Z) α Z (L) there exists an orientation (X, Z, E) th Z (E) on A such that the following properties hold 1. for each smooth variety X, each line bundle L/X and each closed subset Z X one has α Z (L) = th Z (L); 2. for each smooth X, closed subset Z X and each line bundle L/X one has e Z (L) = th Z (L) z where z : (X, X Z) (L, L Z) is the zero section. Moreover the required orientation is uniquely determined both by the property 1 and by the property Theorem. If (X, Z, E) th Z (E) is an orientation on A then the assignment (L, X, Z) th Z (L) z is a COMMUTATIVE Chern structure on A, the assignment (L, X, Z) th Z (L) is a Thom structure on A and so constructed Chern and Thom structures correspond to each other. Moreover the construction of an orientation by means of a Chern (or a Thom) structure given by Theorem and the construction of a Chern and a Thom structure by means of an orientation are inverse of each other. Proof of Theorem Let E/X be a rank n vector bundle, Z X be a closed subset and let F = E 1. Recall that the sequence is exact: 0 A p 1 E (Z) (P(E)) i A p 1 F (Z) (P(F )) β A s (Z) (P(F )) 0. 25

26 Since c p 1 F (Z) n (O F (1) p F E)i = i c p 1 E (Z) n (O E (1) p EE) = 0 (Z) by Proposition then c p 1 F n (O F (1) p E) factors through support restriction β: Set c p 1 F (Z) n (O F (1) p F E) = ĉ n p 1 F (Z) (O F (1) p F E) β. th Z (E) = p F p ĉ 1 F (Z) n (O F (1) p E) : A s (Z) (P(F )) A Z (X). (20) and define the homomorphism th Z (E) : A Z (E) AZ (X) as follows th Z (E) = th Z (E) j. (21) (here p : P(F ) X is the projection and j : E = P(F ) P(1) P(F ) is the open inclusion). To show that the assignment (X, Z, E) th Z (E) is an orientation it remains to check the properties from Definition The second and the first property follows immediately from the functoriality of the Chern classes. To prove that the operator th Z (E) is an isomorphism it is enough to check that th(e) is an isomorphism. For that consider the commutative diagram with exact rows 0 A (P(E)) A (P(F )) A P(1) (P(F )) 0 γ (th(e) β, α) th(e) 0 A(X) n A(X) A(X) n A(X) 0 where γ = (p E, pe ζ E,..., p E ζn 1 E ), ζ E = e(o E (1)) and α = (p F, pf ζ F,..., p F ζn 1 F ), ζ F = e(o F (1)). The map γ is an isomorphism by the projective bundle theorem. Thus to prove that the right vertical arrow is an isomorphism it suffices to check that the map (th(e) β, α) is an isomorphism. Using the Mayer-Vietoris property and Remark one may assume that the bundle E is the trivial rank n bundle. In this case one has c n (O F (1) n ) = ζf n. Thus the map (th(e) β, α) = (p F c n (O F (1) p F E), α) coincides in this case with the map ( p F ζn F, (pf,..., pf ζn 1 F ) ) and it is an isomorphism by the projective bundle theorem. Basically the property 3 in Definition follows from the Cartan formula for Chern classes. But to give a detailed prove one needs certain preliminaries. For simplicity we write the proof below without supports. It will be an easy exercise to give a complete proof in usual way. Let E = E 1 E 2 be a vector bundle over a smooth variety X and let F r = E r 1 (r = 1, 2) and let F = E 1 and let P : P(F ) X be the projection. We will identify E with the open subset P(F ) P(E) of P(F ) and identify E i with the open subset P(F r ) P(E i ) of P(F r ). Let P(F i ) be the subvariety in P(F ) defined by the direct summund F r of F. Let i r : P(F i ) P(F ) be the closed embedding. 26

27 2.4.5 Lemma. Let M and N be vector bundles over X and p : P(M) X and P(N) be corresponding projective bundles. Then the natural map q : P(M N) P(N) P(M) makes P(M N) P(N) into a vector bundle over P(M) which is isomorphic to p N O M (1) Lemma. Under condition of lemma denote by E the bundle p N O M (1) over P(M). Then the total space of projective bundle P(1 E) is naturally isomorphic to the blowing up P(M N)P(N). We apply the above lemmas to the bundles M = F 2 and N = E 1. The blowing up P(F ) P(E1 ) will be denoted below without indices. The bundle p E 1 O F2 (1) over P(F 2 ) will be denoted by T. We use a letter q for the projection q : P(1 T ) P(F 2 ) Lemma. Let σ : P(F ) P(F ) be a blowing up. Then σ O F (1) = q O F2 (1) O 1 T (1). It is enough to show that the composition map A X (E) thx (p 2 E 1) A X (E 2 ) th(e 2) A (X) coincide with th(e) Lemma. Let γ : A X (E) A P(1) ( P(F )) and γ 2 : A X (E 2 ) A P(1) (P(F 2 )) be excision isomorphisms. Then the following diagram commutes A X (E) bγ th X (p 2 E 1) A X (E 2 ) γ 2 A P(1) ( P(F )) th P(1) (T ) A P(1) (P(F 2 )). In order to proof this lemma it is suffice to compute γ 2 T = γ 2 p E 1 γ 2 O F 2 (1) = p 2 E Lemma. Let α Z 1 = cz n 1 (σ (P E 1 O F (1))) be the Chern homomorphism. Then the following diagram commutes: β 1 A ( P(F P(F )) A 1 ) ( P(F )) α 1 P(F α 1 ) A ( P(F )) 1 β 1 A P(F 1 ) ( P(F )) β 2 A P(1) th P(1) (T ) ( P(F )) eq A P(1) (P(F 2 )). In the above diagram the strict transform of P(F 1 ) P(F ) under blowing up is denoted also by P(F 1 ). It is easy to see that σ restricted to this strict transform is an isomorphism. By Lemma σ (P E 1 O F (1)) = (q p E 1 q O F2 (1)) O 1 T (1) = q T O 1 T (1). Since q 1 (P(1)) = P(F 1 ) then commutativity of the right square is provided by the very definition of Thom homomorphism Lemma. Let α 2 = c n2 (σ (P E 2 O F (1))) and s : P(F 2 ) P(F ) is a closed imbedding identifying P(F 2 ) with P(1 P(F2 )) P(1 T ) = P(F ). Then the following diagram commutes: 27

28 A (P(F 2 )) s c n2 (p E 2 O F2 (1)) A (P(F 2 )) s A ( P(F )) α 2 A ( P(F )). It is enough to check that s σ (P E 2 O F (1)) = p E 2 O F2 (1)). The relation P σ s = p is obvious consequence of relations σ s = i 2 where i 2 : P(F 2 ) P(F ) is a natural inclusion and P i = p. By Lemma one has s σ O F (1) = s q O F2 (1) s O 1 T (1) = O F2 (1) 1 P(F2 ) Then the lemma follows from the functoriality of Chern homomorphisms Lemma. Let γ : E P(F ) be an open inclusion identifying E with P(1 E) P(E). Then the following diagram commutes: β A ( P(F σ )) A (P(F )) A P(1) (P(F )) A X (E) α 2 α 1 c n1 +n 2 (P E O F (1)) th(e) th(e) A ( P(F )) σ A (P(F )) P A (X) γ id A (X). Indeed the left square is commutative by Cartan formula: c n1 +n 2 (P E O F (1)) = c n1 (P E 1 O F (1)) c n2 (P E 2 O F (1)) and functoriality of Chern. The other two squares are commutative by the very construction of Thom homomorphism Lemma. The support restriction operator β : A ( P(F )) A P(1) ( P(F )) is a surjection. Consider a diagram where the rows is restrictions snd vertical arrows are blow up: bβ A ( P(F )) A P(F 2) ( P(F )) A P(1) ( P(F )) σ σ σ A (P(F )) β A P(F 2) (P(F )) β c β A P(1) (P(F )). The map β is onto by Corollary apllied to the bundle 1 T over P(F 2 ). The map β is onto by the same reason. Two right vertical arrows are isomorphisms by exsicion property. Since β = β β = (σ ) 1 β σ β the lemma follows Lemma. Under notation introduced above one has the following relation: th(e) β σ = th(e 2 ) th P(1) (T ) β. Lemma implies that th(e) β σ = P σ α 2 α 1. Since s q α 1 = α 1 by proposition then the right hand is equal to P σ (α 2 s )q α 1 = P σ s c n2 (p E 2 O F2 (1))q α 1. The last equality uses Lemma By obvious geometrical reason P σ s = p. By the very definition of Thom p c n2 (p E 2 O F2 (1)) = th(e 2 ) β 0 where β 0 : A (P(F 2 )) A P(1) (P(F 2 )) is the support restriction. Therefore one can prolong the (22) 28

29 chain of relations as p c n2 (p E 2 O F2 (1)) q α 1 = th(e 2 ) β 0 q α 1 = th(e 2 ) q β 1 α 1. The last equality is provided by the commutativity of the square A ( P(F )) β 1 q A (P(F 2 )) β 0 A P(F 1) (P(F )) eq A P(1) (P(F 2 )). We only need to check now that q β 1 α 1 = th P(1) (T ) β. This follows from Lemma and obvious relation β = β 2 β 1. Now we are ready to finish the proof of the multiplicativity of the Thom isomorphisms. By Lemma it suffices to check that th(e) ( γ ) 1 β = th(e 2 ) th X (p 2 E 1) ( γ ) 1 β. (23) By Lemma th(p 2 E 1) ( γ ) 1 = (γ 2) 1 th P(1) (T ). Therefore the right hand side of Formula (23) is equal to th(e 2 ) (γ 2) 1 th P(1) (T ) β = th(e 2 ) th P(1) (T ) β. The last expression is equal to th(e) β σ by Lemma Since th(e) = th(e) (γ ) 1 all we need to check is (γ ) 1 β σ = ( γ ) 1 β. This is the case that is easy to see from the diagram (22) and obvious relations γ ( γ ) 1 = σ ; β = β. β Thus the assignment (X, Z, E) th Z (E) is indeed an orientation on A. We still have to check that the orientation satisfies the requirements 1 and 2 of Theorem The first requirement holds obvious by the construction (21) of the Thom isomorphisms. The requirement 2 follows from the first one and Lemma To complete the proof of Theorem it remains to prove the uniqueness of the orientation. To prove the uniqueness of the orientation satisfying the property 1 take two orientations ω and ω on A satisfying the property 1. To check that they coincide it suffices by Lemma to check that their restrictions to line bundles coincide. This is the case by the requirement 1. Thus ω = ω. Now prove the uniqueness of the orientation satisfying the requirement 2. Let L e(l) be a Chern structure and let ω and ω be two orientations satisfying the requirement 2. We will show that ω = ω. The restriction of these orientations to line bundles are Thom structures on A. By the same Lemma the orientations coincide if the mentioned Thom structures on A coincide. To prove that the two Thom structures on A coincide it suffices by Lemma to check that the two corresponding Chern structures L e ω (L) and L e ω (L) coincide. But that is the case. The proof of the relation ω = ω is completed. Proof of Theorem The restriction of an orientation to line bundles is a Thom structure. By Lemma the assignment L e(l) = th(l) z is a Chern structure. The only we need to prove is the commutativity of the this Chern structure. 29

30 Let L 1 and L 2 be line bundles over X, Z X is a closed subset. We need to check e Z (L 1 )e Z (L 2 ) = e Z (L 2 )e Z (L 1 ). Consider a rank 2 bundle E = L 1 L 2. Let z E : X E, z 1 : X L 1 and u 2 : L 1 E be a zero sections of respective bundles. Let us compute the composition A Z (X) se A Z (E) thz (E) A Z (X) in two different ways. For the first one, th Z (E)s E = th Z (L)th Z (p 1L 2 )s E where p 1 : L 1 X is the projection. Consider a diagram z 1 A Z (X) A Z (L 1) A Z (E) e Z (p 1 L 2) th Z (p 1 L 2) e Z (L 2 ) A Z (X) s 1 A Z (L) u 2 id A Z (L). The right hand side square is commutative by the construction of Chern classes (cf. Lemma ). The left hand side square is commutative by the functoriality of the Chern classes. Recall that th Z (L 1 )z 1 = e Z (L 1 ) by Lemma As u 2 s 1 = s E we get from the diagram above the chain of relations th Z (L)th Z (p 1 L 2)s E = th Z (L 1 )s 1 e Z (L 2 ) = e Z (L 1 )e Z (L 2 ). The commutativity of the Chern homomorphisms is proved because switching of L 1 and L 2 does not change the very left term in the last relation. Now verify that the correspondences between orientations and Chern structures in the two theorems are inverse to each other. We know that the correspondences between Thom structures and Chern structures are inverse to each other. By Theorem an orientation is uniquely defined by corresponding Thom structure. Moreover, the correspondences between orientation being restricted to line bundles and Chern structure defined by Theorem are the same as the correspondences between Thom structures and Chern structures described in Lemmas and Since the last correspondences are inverse to each other then the correspondences described by Theorem are inverse to each other. The theorem is proved. 3 Gysin Structures 3.1 Gysin structure definition and the main theorem Let A be a homology theory. The main aim of this subsection is to prove Theorem Let us recall a notion Definition. Let i : Y X be a closed imbedding of smooth varieties. Consider a 30

31 Cartesian square ( in the category of schemes ) Ỹ ϕ Y ĩ X ϕ i X consisting of smooth varieties. This square is called transversal if the canonical morphism ϕ (N) Ñ is an isomorphism (here N and Ñ are the normal bundle to Y in X and to Ỹ in X) Definition. A Gysin structure on A is an assignment which associate to each closed imbedding i : Y X of smooth varieties an operator i! : A (X) A (Y ) satisfying the following properties (1) functoriality: composition property: (i j)! = i! j! (24) for any closed imbeddings Z j Y and Y i X of smooth varieties; normalization: for any smooth variety X one has (id X )! = id A (X); (2) Base change: for the transversal square from (3.1.1) the following diagram commutes A ( X) ϕ A (X) ĩ! A (Ỹ ) ϕ i! A (Y ) (25) (3) localization: Let i : S T be a closed imbedding of smooth varieties and let j : T S T be the open inclusion. Then the sequence is exact. A (T S) j A(T ) i! A(S) Definition. Let A be a homology theory equipped with a commutative Chern structure (X, Z, L) e Z (L). One says that a Gysin structure f f! on A is subjected to the Chern structure if for any line bundle L/X and its zero section z the Chern operator e(l) coincides with the operator z! z : A (X) A (X) Theorem. Let A be a homology theory equipped with a commutative Chern structure. Then there exists a unique Gysin structure on A subjected to the given Chern structure. The proof of the theorem occupies Subsections 3.2 to

32 3.1.5 Corollary. Let A be a homology theory equipped with a commutative Chern structure. Let f f! be a Gysin structure on A subjected to the Chern structure. Then the Gysin operators commute with the Chern operators, i.e. for any closed imbedding f : Y X and a line bundle L over X one has e(f L) f! = f! e(l). Proof. Denote a bundle f L by M. Let z L : X L and z M : Y M be zero sections of these bundles. Let F : M L be a bundle map covering f. It is enough to prove a following diagram to be commutative: A (X) z L A (L) f! F! z! L A (X) f! A (Y ) z M A (M) z! M A (Y ). The left hand side square commutes by Base Change property 25. The right hand side square commutes by composition property Construction of Gysin homomorphism Below we give a construction of Gysin homomorphisms. The deformation to the normal cone (see 1.4.7) plays an important role here (the deformation to the normal cone construction is a perfect substitute of the turbular neighbourhood in the differential topology). For a closed imbedding s : Y X of smooth varieties and closed subset Z Y define an operator s th Z : AZ (X) AZ (Y ) (26) as the composition A Z (X) i1 A Z A1 (X t ) (i0 ) 1 A Z (N) thz (N) A Z (Y ) where the notation for N = N X/Y, i 0 and i1 are taken from (1.4.7) and thz (N) is the Thom operator corresponding to the orientation ω (2.4.1). The operator s th Z is an isomorphism. Define the Gysin operator s! : A (X) A (Y ) (27) as the composition A(X) β A Y sth (X) A(X) where β is the support extension operator for the pair (X, X Y ) (see 1.3.1). The following properties of the Gysin operators will be proved below before the theorem and are useful when proving this theorem Additivity: Let j 1 and j 2 be the natural imbeddings of smooth varieties Y 1 and Y 2 to Y = Y 1 Y 2. For a closed imbedding i : Y X one has where i r is the composition i j r. i! = j 1 i! 1 + j2 i! 2 (28) 32

33 Normalization. Let z : X E be a zero section of the vector bundle. Then an operator z th : A Y (E) A (Y ) coincides with an operator th(e). Particularly, if E = L is a line bundle over Y then an operator z! z : A (Y ) A (Y ) coincides with e(l) Smooth divisor case: for a smooth divisor s : D X one has s s! = e(l(d)) (29) in End(A (X)). There are more properties of the Gysin operators which are useful themself and which are useful in proving some of the properties required by definition of gysin structure Let E be a rank n vector bundle over a smooth Y and let s : Y P(1 E) be the section of the projective bundle p : P(1 E) Y identifying Y with the closed subvariety P(1) in P(1 E). Then one has s s! = c n (p (E) O E (1)). (30) Let z : Y E be the zero section of a vector bundle E/Y. Then the operator z th coincides with the Thom operator th E Y Let F/Y be a rank m vector bundle and let s : Y P(F ) be a section of the natural projection p : P(F ) Y. Consider the natural inclusion O F ( 1) p (F ) and let Q be the factor-bundle p (F )/O F ( 1). In A(P(F )) one has 3.3 Proof of Theorem s s! = c n 1 (O F (1) p s Q). (31) Identity to identity. The relation (id X )! = id A (X) is obvious because in this case Y = X and X t = Y A 1 and N = Y and th(n) = id : A (X) A (Y ) Gysin sequence exact. This is clear, because the localization sequence A (X Y ) j A (X) β A Y (X) for the pair (X, X Y ) is exact and the operator s th : A Y (X) A (Y ) is an isomorphism and s! = s th β Base change property The base change property for the Gysin maps s! follows from the one for the Thom maps s th. We will prove it now. Let N = N X/Y and Ñ = N X/ Ỹ. Since the square from is transversal the canonical map Ỹ Y N N is an isomorphism. We will identify Ỹ Y N and Ñ by means of this isomorphism and denote Φ : Ỹ Y N N the projection. The functoriality of the Thom class shows that th(n) Φ = ϕ th(ñ). Therefore the diagram AỸ (Ñ) Φ th(ñ) A (Ỹ ) ϕ (32) A Y (N) th(n) A (Y ) 33

34 commutes. Since the square from is transversal the map ϕ id : X A 1 X A 1 gives rise to a commutative diagram Ỹ ϕ Y z Ñ ĩ0 X ĩ 1 t X Φ ϕ t ϕ z N i 0 X t i 1 X in which rows coincide with the deformation to the normal cone diagrams for the pairs ( X, Ỹ ) and (X, Y ). The commutativity of the last diagram shows that the diagram consisting of natural mappings commutes AỸ (Ñ) Φ ĩ 0 AỸ A1 ( X t ) ϕ t ĩ 1 AỸ ( X) ϕ A Y (N) i 0 A Y A1 (X t ) i 1 A Y (X) Gluing this diagram with the commutative diagram (32) we get the commutativity of the diagram A (Ỹ ) s th AỸ ( X) ϕ ϕ A (Y ) s th A Y (X). The commutativity of the diagram (25) is proved. The property follows. To complete the proof of the theorem it remains to verify the functoriality of the Gysin operators that is the relation (i j)! = i! j!. This property is proved in the very end of the Section after certain auxiliary results Normalization Property. We prove here the normalization property It suffices to prove that the composition A Y (E) (i 1 ) A Y A1 (E t ) (i 0 ) 1 A Y (E) is the identity map. For that we construct a morphism q : E t E such that the map q makes E t into a line bundle over E and q i 1 = id and q i 0 = id and q 1 (Y ) = i t (Y A 1 ) in E t, 34

35 where i t : Y A 1 E t is the imbedding from the deformation to the normal cone construction In this case one gets q 1 (Y ) = i t (Y A 1 ) E t. Therefore the relations i A 0 = (qa ) 1 = i A 1 holds by the strong homotopy invariance property and we are done. It remains to construct the desired morphism q : E t E. Let F/Y be a vector bundle and let F be the blow-up of F at the zero section. The variety F coincides with the total space of the line bundle O F ( 1) over P(F ). Let q F : F P(F ) be projection of the line bundle to its base P(F ). If F = E 1 for a vector bundle E over Y then one has the following commutative diagram E F F E = E t P(1) A 1 = Y A 1 q E q F q pr P(E) P(F ) P(F ) P(E) = E P(1) = Y. in which all the vertical arrows are the projections of the line bundles to their bases. The projection q has two sections s 0 and s 1. The section s 0 is the zero section and the section s 1 is given by x (x, 1). Observe that the variety P(F ) P(E) coincides with E, the variety E t coincides with the variety F E, the imbedding i 1 : E E t coincides with the section s 1 : E F E. The normal bundle N = N E/Y to Y in E coincides with bundle E itself and the imbedding i 0 : N E t coincides with the section s 0 : E F E. Finally the variety Y A 1 coincides with P(1) A 1 and the imbedding Y A 1 E t coincides with the imbedding P(1) A 1 F E. Thus the map q : E t E makes E t into a line bundle over E and satisfies the relations q i 0 = id = q i 1, q 1 (Y ) = i t (Y A 1 ). Thus q is the desired map. The property is proved Proof of the relation (30). Set F = 1 E. Take the zero section z of the vector bundle E, the section s of the projective bundle P(F ) which identifies Y with the closed subvariety P(1) in P(F ) and consider the commutative diagram A Y (E) e z th A (Y ) id A P(1) (P(F )) s th A (Y ) where e : E P(1 E) = P(F ) is the open imbedding. As it was already proved z th = th(e). By the very construction of the Thom isomorfism 21 and 20 one has th(e) β = p c n (O F (1) p (E))) where β : A (P(F )) A P(1) (P(F )). Since the map e is an isomorphism and z th = th(e) = e th(e) one gets a relation s th = th(e). Therefore s s! = s s th β = s th(e)β = s p c n (O F (1) p (E)) = c n (O F (1) p (E)) as desired. The property is proved. 35

36 Section of a projective bundle Let F/Y be a rank m vector bundle and let s : Y P(F ) be a section of the natural projection p : P(F ) Y. Consider the natural inclusion O F ( 1) p (F ) and let Q be the factor-bundle p (F )/O F ( 1) and let E = s (Q). We have to prove the relation s s! = c m 1 (O F (1) p (E)) in End(A (P(F ))). Set L = s (O F ( 1)). Then one has the obvious exact sequence of vector bundles on X 0 L F E 0. If M is a line bundle over X then replacing F by F M we does not change as the variety P(F ) so the vector bundle O F (1) p (E). Thus one may assume that L is the trivial line bundle. By the splitting principle [PY, Prop.3.5] one may assume further that F = 1 E. In this case the relation s s! = c m 1 (O F (1) p (E)) is just the relation (30) which is proved just above (n = m 1). The relation (3.2.6) is proved Smooth divisor case For a smooth divisor i : D X one has to check the relation i i! = e(l(d)) in End(A (X)) (see 1.1 for notation). For that consider the commutative diagram j 0 P(1 N) X j 1 t X s I i t D k 0 D A 1 k 1 D, from Since both squares in this diagram are transversal the following diagram commutes A (P(1 N)) j 0 A (X t) j 1 A (X) s! (I t)! i! A (D) k 0 A (D A 1 ) k 1 A (D) s (I t) i A (P(1 N)) j 0 A (X t ) j 1 A (X), Now consider the line bundle L t = L(D A 1 ) over X t (see the notation in 1.1). One can show that its restriction j0 (L t) to P(1 N) is isomorphic to the line bundle O 1 N (1) p (N) and its restiction j1(l t ) to X is isomorphic to L(D). Since s s! = e(o 1 N (1) p (N)) then one has the relation e(l t )j 0 = j0 s s! = (I t ) k 0s!. Since the upper left square is transversal then k 0s! = (I t )! j 0. Therefore (e(l t) (I t ) (I t )! )j 0 = 0. In the notation of lemma jt (L t ) is a trivial vector bundle and e(l t )j t = 0. By localization property (I t )! j to = 0. Therefore (e(l t) (I t ) (I t )! )j t = 0. Appliing useful lemma we get e(l t ) (I t ) (I t )! = 0. Since e(l t )j 1 = j e(l(d)) 1 and (I t ) (I t )! j 1 = (I t ) k i 1! = j i 1 i! then one has j (e(l(d) 1 i i! ) = 0. The composition σ j 1 is identity. Therefore j 1 is mono and the proof is completed. 36

37 Proof of the additivity property (28). The proof is strictly the same as in the cohomological case. The proof of this property will be given just after three preliminary Lemmas 3.3.9, and We beging with the following particular case Lemma. Let s = s 1 s 2 : S = S 1 S 2 T 1 T 2 = T be a closed imbedding of smooth varieties. Let i m : S 1 S and j m : T m T be the open inclusions and let r m = j m s m : S m T. Then s! = i 1 r! 1 + i2 r! 2 To prove this Lemma consider the isomorphism (j 1, j2 ) : A (T 1 ) A (T 2 ) A (T ). To prove the Lemma it suffices to check the following relations s! j 1 = (i1 r! 1 + i2 r! 2 ) j1 (33) The transversal square s! j 2 = (i 1 r! 1 + i 2 r! 2) j 2 (34) T 1 j 1 s 1 T s i S 1 1 S prove relations s! j 1 = i 1 s! 1. The transversal square j 1 T 1 T s 1 r 1 S 1 id S 1 proves the relations r! 1 j 1 = s! 1. The transversal square T T 1 j 1 r 2 S 2 proves the relation r! 2 j= 0. Now combining the last three relations one gets the chain of relations (i 1 r! 1 + i2 r! 2 ) j1 = i1 (r! 1 j1 ) = i1 s! 1 = s! j 1. which proves the relation (33). The relation (34) is proved in a similar way. The Lemma is proved. Now suppose we are given with varieties V and U = U 1 U 2 and a transversal square of the form v T 1 T 2 V s=s 1 s 2 t u=u S 1 S 1 u 2 2 U1 U 2 37

38 Lemma. Let k m : U m U be the open inclusion and let t m = t k m : U m V. Then t! v = [k 1 t! 1 + k2 t! 2 ] v To prove this Lemma observe that the following squares are transversal (m = 1, 2) T 1 T 2 r m v V t m S m u m Um Thus for t! m v = u m r! m for m = 1, 2 and one gets the following chain of relations using Lemma in the last step. [k 1 t! 1 + k 2 t! 2] v = k 1 u 1 r! 1 + k 2 u 2 r! 2 = u [i 1 r! 1 + i 2 r! 2] = u s! where i 1 : S 1 S and i 2 : S 2 S is inclusions. Finally u s! = t! v by the transversality of the diagram considered in the Lemma. The Lemma follows. Now consider the deformation to the normal cone diagram from the subsection j 0 P(1 N) X j 1 t X s i i t Y k 0 Y A 1 k 1 Y. Since Y = Y 1 Y 2 one see that Y A 1 = Y 1 A 1 Y 2 A 1 and P(1 N) = P(1 N 1 ) P(1 N 2 ) where N i is the normal bundle of X to Y i. Let l m : Y m A 1 Y A 1 be the inclusion for m = 1, 2 and let i m t = i t l m. Since the left hand side square in the very last diagram is transversal Lemma gives the following relation (i t )! j 0 = [l1 (i1 t )! + l 2 (i2 t )! ] j 0. (35) To complete the proof of the Proposition we need in the following Lemma. Let V t = X t Y A1 and let j t : V X t (i t )! j t = 0 and (i m t )! j t = 0 for m = 1, 2. be the open inclusion. Then In fact, the first relation follows from the localization property. The remaning relations for m = 1, 2 follows from the relation (i m t )! (jt 1) = 0 where jt 1 : X t Y 1 A 1 X t. Now we are ready to complete the proof of the additivity relation (28). By Lemma one has [l 1 (i1 t )! +l 2 (i2 t )! i! t ] jt By relation (35) one has [l1 (i1 t )! +l 2 (i2 t )! i! t ] j0 Thus [l 1 (i1 t )! + l 2 (i2 t )! i! t ] = 0 by Lemma Therefore l 1 (i1 t )! + l 2 (i2 t )! = i! t. Since the right hand side square from the very last diagram is transversal one gets the relation k 1 i! = i! t j 1. 38

39 Since k 1 = k 11 k 12 : Y 1 Y 2 Y 1 A 1 Y 2 A 1 the squares i (m) t X t j 1 X i m Y m A 1 k 1m Ym. are transversal. Thus one has relations (i m t )! j 1 = k1m for m = 1, 2. Recall that u m : Y m Y for m = 1, 2 are the inclusions and i m = i u m. Combining last relations one gets a chain of relations i! m k 1 i! = i! t j1 = [l1 (i1 t )! + l 2 (i2 t )! ] j 1 = l1 k11 i! 1 + l2 k12 i! 2 = which proves the following one k 1 u1 i! 1 + k1 u2 i! 2 k 1 i! = k 1 [u 1 i! 1 + u 2 i! 2]. Since the operator k 1 is an isomorphism the additivity relation (28) follows Proof of the relation (24). We beginning with proving the following generalization of the smooth divisor case relation (29). Namely, Let Y = n i=1 D i be a transversal intersection of n smooth divisors D r (r = 1, 2,..., n) in X. Denoting i : Y X the closed imbedding one has i i! = c n ( n r=1l r ), (36) where L r = L(D r ) for (r = 1, 2,..., n). First of all we proceed the case Y = by the induction on the integer n. To prove that c n ( n r=1 L r) = 0 when n i=1 D i = (37) consider a diagram A (X D 1 ) j A (X) A D 1 (X n i=2d i ) c n 1 ( n r=2 Lr) c n 1 ( n r=2 Lr) c D 1 n 1 ( n r=2 Lr) j A (X D 1 ) A (X) A D 1 (X n i=2d i ). The right vertical arrow is null by the inductional assumption. Hence, the image of the map c n 1 ( n r=2 L r) lies in the image of the map j. Since c 1 (L 1 )j = j c 1 (j L 1 ) = 0 then c n ( n r=1l r ) = c 1 (L 1 )c n 1 ( n r=2l r ) = 0 and the relation (37) is proved. 39

40 To prove the relation (36) consider the commutative diagram j 0 P(1 N) X j 1 t X s I i t Y k 0 Y A 1 k 1 Y from for the imbedding i : Y X. Observe that the normal bundle N = N X/Y to Y in X is the direct sum of the line bundles L r Y. Let (D r ) t be the proper preimage of the divisor D r A 1 under the blow-up map σ : X t X A 1. Clearly the subvariety Y A 1 in X t is a complete intersection of the smooth divisors (D r) t. Furthermore the divisor (D r) t intersect in X t the divisor P(1 N) transversally and their intersection coincides with the smooth divisor P r = P(1 E r ) in P(1 N) where E r is a direct summund of N which is the direct sum of all L s Y except of the exactly L r Y. Clearly the line bundle L(P r ) on P(1 N) is isomorphic to the line bundle p (L r Y ) O 1 N (1). Thus c n (p (N) O 1 N (1)) coincides with the product of homomorphisms c 1 (p (L r Y ) O 1 N (1)). Now consider the line bundle (L r ) t = L((D r ) t ) over X t (see the notation in 1.1). The line bundle j0((l r ) t ) over P(1 N) is isomorphic to the line bundle L(P r ) and thus it is isomorphic to the line bundle O 1 N (1) p (L r Y ). The line bundle j1 ((L r) t ) over X is isomorphic to L r. Therefore one has the relation (c 1 ((L r ) t ))j 0 = j0 c 1(O 1 N (1) p (L r Y )) and the relation (c 1 ((L r ) t ))j 1 = j c 1 1 (L r ). Recall that by (30) s s! = c n (O 1 N (1) p (N) and thus j 0s n s! = j 0 c 1 (O 1 N (1) p (L r Y )) = n c 1 ((L r ) t )j 0. r=1 As an intermidiate step in the proof of the relation (36) we need to check the one in End(A (X t)). r=1 n c 1 ((L r ) t ) = (I t ) (I t )! (38) r=1 To prove this relation it is suffices by lemma to check that r=1 n c 1 ((L r ) t )j 0 = (I t ) (I t )! j 0 and n c 1 ((L r ) t )j t = (I t) (I t )! j t where j t : V t = X t Y A1 X t is an open inclusion. The first of the above relations follows from the equality (I t ) (I t )! j 0 = (I t ) k s 0! = j s 0 s!. The second of the above relations is the consequense of the localisation sequence wich implies (I t )! j t = 0 and the relation 37 which implies c n( n (L r ) t )j t = 0 because n r=1(d r ) t V t =. Now the relation (38) is proved. The chain of relations n n c 1 (L r ) = ( c 1 ((L r ) t ))j 1 == (I t ) (I t )! j 1 = (I t ) k i 1! = j i 1 i! j 1 r=1 r=1 40 r=1 r=1

41 and injectivity of j 1 completed the proof of the relation (36). The next step in the proof of the composition property is to prove the following relation. Let Y be a smooth variety and let E and F be two vector bundles on Y of constant ranks n and m respectively.let M = 1 E F. Let l : P(1 E) P(1 E F ) be the obvious closed imbedding and let p : P(1 E F ) Y be the projection. Then one has the relation in End(A (P(1 E F ))) l l! = c m (p (F ) O M (1)). (39) In fact, by Splitting principle [PY, Prop.3.5] one may assume that the vector bundle F splits in a direct sum of line bundles F = m r=1 N r. Let F r is a direct summund of F which is the direct sum of all line bundles N s except of N r. Let D r = P(1 E F r ) be the divisor on P(1 E F. Clearly the subvariety P(1 E) in P(1 E F ) is the transversal intersection of the smooth divisors D r. And furthermore the line bundle L(D r ) on P(1 E F is isomorphic to the line bundle p (N r ) O M (1). The relation (36) shows that one has a chain of relations in End(A (P(1 E F )): l l! = m c 1 (L(D r )) = r=1 m c 1 (p (N r ) O M (1)) = c m (( m r=1 p (N r )) O M (1)) = c m (p (F ) O M (1)). r=1 The relation (39) is proved. Now we are ready to prove the composition property in one particular case. Let s E : Y P(1 E) be the section of the projective bundle p E : P(1 E) Y identifying Y with the closed subvariety P(1). Let s = l s E : Y P(1 E F ). Clearly s is a section of the projection p : P(1 E F ) Y identifying Y with the closed subvariety P(1). We claim that s! = s! El!. (40) The equalities (30) and (39) gives rise to a chain of relations s s! E l! = l s E s! E l! = l [c n (p E (E) O 1 E(1))]l! = = l [c n (l (p (E) O M (1)))]l! == c n (p (E) O M (1))l l! = = c n (p (E) O M (1))c m (p (F ) O M (1)) = c m+n (p (F E) O M (1)) = s s! which proves the relation (40) by injectivity of s. Now we are ready to prove the composition property. For that consider closed imbeddings of smooth varieties j : Z Y and i : Y X and consider the diagram from for the closed imbedding j i : Z X. j 0 P(1 N X/Z ) X j 1 t X s (I J) t i j Z k 0 Z A 1 k 1 Z. 41

42 The proper preimage Y t of Y A 1 under the blow-up map σ : X t X A1 and the projective bundle P(1 N Y/Z ) fit in the this diagram making a big commutative diagram with transversal squares P(1 N X/Z ) l i 0 X i 1 t X i l t P(1 N Y/Z ) s Y i Y 0 Y t J t i Y 1 Y j Z k 0 Z A 1 k 1 Z. in which the bottom and the middle rows form the deformation to the normal cone diagram for the closed imbedding j : Z Y and l s Y = s, l t J t = (I J) t. Using the transversality of the left hand side squares in the last diagram one gets a chain of relations k s 0! Y l! = (J t )! (i Y ) 0 l! = (J t )! i 0 (l t )! and We already proved the relation Thus one has the relation k 0 s! = ((I J) t )! i 0 = (l t J t )! i 0 s! Y l! = s! : A (P(1 N X/Z )) A (Z). [(l t J t )! (J t )! (l t )! ]i 0 = 0 Observe that for V t = X t Z A 1 and the open inclusion j t : V t X t the compositions (l t J t )! j t and (J t)! (l t )! j t vanish. In fact, the first composition vanishes because the Gysin sequence for the closed imbedding Z A 1 X is exact. To prove that the second composition vanishes let V Y t = Y Vt Y Y t j t m t = l t jt Y. The relation be the open inclusion and let m t : V Y t (l t )! j t = (jy t ) (m t )! t Z A1 and let jt Y : V t be the closed inclusion such that together with the vanishing of the composition (J t )! (jt Y ) (look at the Gysin sequence for the closed imbedding J t : Z A 1 Y t ) prove the relation (J t)! (l t )! j t = 0. Now the useful Lemma implies the relation (l t J t )! (J t )! (l t )! = 0. The trasversality of the two right hand side squares from the last diagram gives rise to the chain of relations k 1 (i j)! = (l t J t )! i 1 = (J t )! ) ((l t )! i 1 ) = (J t )! ((i Y ) 1 i! ) = k 1 (j! i! ). 42

43 The operator k 1 is an isomorphism by the homotopy invariance property. The required relation (i j)! = j! i! follows. The functoriality of the Gysin operators is checked. So the proof of Theorem is completed. 3.4 Uniqueness of Gysin structure In the previous section we constructed a Gysin structure on an oriented homology theory A. In this section we prove that any Gysin structure subjected to the Chern structure on A coincide with the one constructed in Subsection 3.2. We need to prove equations (30) and (29) for any Gysin structure. Earlier we have proved them only for the Gysin structure constructed in the previous section. However for arbitrary Gysin structure we need other arguments. The main proving tool is the Useful lemma which is recalled below Lemma (Useful lemma). Let A be an orientable homology theory. Then for the diagram (see for notation ) j 0 P(1 N) X j 1 t X s I i t Y k 0 Y A 1 k 1 Y, and open inclusion j t : X t Y A1 X one has Im(j 0 ) + Im(j t ) = A (X t). Proof. It follows from Lemma , because the surjectivity condition on the restriction operator is satisfied by Lemma. Let A be an oriented homology theory with the corresponding Chern structure (X, Z, L) e Z (L). Let f f! be a Gysin structure subjected to the Chern structure. Let s : Y P(1 L) be a zero section of projective bundle identifying Y with P(1). Then where p : P(1 L) Y is the natural projection. s s! = e(p L O 1 L (1)) (41) Corollary. Under the assumptions of Lemma let s : D X be a smooth divisor and L(D) be the corresponding line bundle over the smooth X. Then one has s s! = e(l(d)). 43

44 The proof of the corollary is exactly the same as in the subsection It uses the formula (41) instead of the formula (30). Proof. The proof of Lemma will proceed in several steps. Step 1. Three auxiliary relations. First of all we show that s! s = e(l). Consider a transversal square id Y Y z j L s P(1 L). One has s! s = s! j z = (id) z! z = e(l). Then we claim that s! L s = 0 where s L : X P(1 L) is a closed imbedding identifying X with P(L). To prove this consider an open embedding j : P(1 L) P(L) P(1 L) and j L : P(1 L) P(1) P(1 L). Note that restriction of p on P(1 L) P(L) gives us original line bundle L X, but the restriction of p on P(1 L) P(1) gives us a dual line bundle L X. Further consider a diagram j A (P(1 L) P(L)) A (P(1 L) s z A (Y ) s! L A (Y ) Since s! L j = 0 by Gysin localization then s! L s = s! L j z = 0. Remark that we get the relation e(1) = 0 in the particular case L = 1 because e(l) = s! s = s! L s = 0. Then we prove a relation s p e(p L O(1)) = e(p L O(1)) (42) Recall that s L O(1) = L. Since p s = id one needs to prove Ime(p L O(1)) Ims. This is the case because Kers! L = Ims from the above diagram and by lemma s! L e(p L O(1)) = e(s L p L s L O(1))s! L = e(l L )s! L = 0. Step 2. Prove the lemma in the case Y = X P and L = O(1). By projective bundle theorem the image of ξ n : A (P n ) A (P n ) coincides with the image of i n 1,n : A (P n 1 ) A (P n ). Therefore e(l ) : A (P ) A (P ) is a surjection. Consider a map s L s : A (Y ) A (Y ) A (P(1 L)). It is a surjection since the composition map A (Y ) A (Y ) sl s A (P(1 L) (p,p ζ) A (Y ) A (Y ) ( ) ( ) p s is expressed by matrix L p s 1 1 p ζs L = p ζs e(l. ) 0 From the chain of relations e(p L O(1))s = s e(s p L s O(1)) = s e(l 1) = s e(l) = s s! s and e(p L O(1))s L = s L e(s L p L s L O(1)) = sl e(l L ) = 0 = s L s! s one may conclude that s! s (s s L ) = e(p L O(1))(s s L ). A desired equation is followed from the surjectivity of s s L.. 44

45 Step 3. Now we are ready to prove the general case. Let L X is an arbitrary line bundle. By the Jouanolou trick one can regard X as affine variety and take an f : X P such that L = f O P (1). Consider X = X P and a line bundle L X induced from O P (1) by projection X P P. All previous considerations are valid for X and L as well as for P an O P (1). Particularly, our lemma holds for the closed embedding S : X P(1 L ). Let g : X X = X P be a closed inclusion defined via g = (id, f). Then L = g L, p L O 1 L (1) = G (P L O 1 L (1)) and one gets a commutative diagram. A (P(1 L)) s! A (X) s A (P(1 L)) p A (X) G g G g A (P(1 L )) S! A (X ) S A (P(1 L )) P A (X ). Since G s s! = S S! G = e(p L O 1 L (1))G = G e(g (P L O 1 L (1))) = G e(p L O 1 L (1)) then g p s s! = P G s s! = P G e(p L O 1 L (1)) = g p e(p L O 1 L (1)). Since g is injective one has p s s! = p e(p L O 1 L (1)). Therefore s s! = s p s s! = s p e(p L O 1 L (1)) = e(p L O 1 L (1)) by equation (42). Lemma is proved. Now we are ready to complete the proof of Theorem In fact it remains to prove only the following assertion Theorem. Let A be an oriented homology theory with the corresponding Chern structure (X, Z, L) e Z (L). Let f f! 1 and f f! 2 be two Gysin structures which are both subjected to the given Chern structure. Then for any closed embedding f : Y X one has f! 1 = f! 2. Proof. The proof follows the proof of [PY, Thm.2.4]. We shall consequently extend a class O of morphisms f which are satisfied to relation f! 1 = f! 2 holds Lemma. Let f : Y X be a closed imbedding of smooth varieties. Consider the useful lemma diagram j 0 P(1 N) X j 1 t X s I t f k Y 0 Y A 1 k 1 Y. Suppose morphism s belongs to the class O. Then f belongs to O too. Proof. One has I! 1 t j 0 = k s 0! 1 = k s 0! 2 = I! 2 t j. 0 Since I tj! t = 0 by Gysin localization then by useful lemma we conclude that I t belongs to O. Further k 1f! 1 = I! 1 t j 1 = I! 2 t j 1 = k1 f! 2 and by injectivity of k 1 the lemma follows. 45

46 For any line bundle L/Y the section s : Y P(1 L) of the projection p : P(1 L) Y belongs to O. In fact, let D be the divisor s(y ) on P(1 L) then by corollary one has a chain of relations s s! 1 = e(l(d)) = s s! 2. Since s is injective one gets the required relation s! 1 = s! 2 and s belongs to O. Then the lemma lemma implies that for any smooth divisor i : D X morphism i belongs to O. Let E be a vector bundle over a smooth variety Y. Let s : Y P(1 E) be the zero section of the projective bundle P(1 E). We claim that the morphism s belongs to O. In fact, by the splitting principle [PY, Prop.3.5] we may assume that E = L 1 L r is a direct sum of line bundles. The map s is a composition of smooth divisor inclusions i k : P(1 L 1 L k 1 ) P(1 L 1 L k ) each of them belongs to the class O. The class O is closed under the composition by the property (24). Therefore the morphism s belongs to O too. The claim is proved. Let f : Y X be a closed imbedding with the normal bundle N = N X/Y. The zero section s : Y P(1 N) belongs to the class O. Thus f belongs to O by lemma

47 4 Quillen map 4.1 A formal group law. Let A be a homology theory endowed with a commutative Chern structure. The main object of this section is the variety X P n. The projective bundle isomorphism (p, p ζ,..., p ζ n ) : A (X P n ) A (X) A (X) (43) with ζ = e(o(1)) will be repeatedly used in this Section. We intent to simulate a cohomology ring of X which is absent in our subject. The first step on the way is to define a ring U A (X). For a variety Y we will denote by U A (Y ) a subring in End Z (A (Y )), generated by Chern homomorphisms e(l) : A (Y ) A (Y ) for all line bundles L/Y. Since the Chern structure is commutative the ring U A (X) is commutative too. The group A (X) is carried out an U A (X) module structure. The assignment X U A (X) is not a functor as there are no reasonable way to define an induced map. However, in a special case we can define one Lemma. There exists a unique ring inclusion p U : U(X) U(X P n ) given by p U (e(l)) = e(p (L)). Proof. It must be checked that all relations holding in U(X) holds in U(X P n ). Let f(x 1,..., X n ) be a polynomial with integral coefficients such that f(e(l 1 ),..., e(l n )) = 0 for line bundles L 1,..., L n over X. For any a A (X P n ) one has p (ζ k e(p L i )(a)) = p (e(p L i )(ζ k (a))) = e(l i )(p (ζ k (a))). Therefore for any k = 0... n one has a chain of relations p (ζ k (f(e(p L 1 ),..., e(p L n ))(a))) = f(e(l 1 ),..., e(l n ))(p ζ k (a)) = 0. Whence f(e(p L 1 ),..., e(p L n ))(a) = 0 as required. The injectivity is obvious. Identifying A (X P n ) with A (X) A (X) by isomorphism (43) the operator e(p L) is identified with a scalar matrix which all components are equal to e(l). So the map U(X) U(X P n ) is simply an assignment α α id A (X). In other words e(p L) acts on the direct sum A (X P n ) = A (X) A (X) by a scalar matrix which has e(l) on its diagonal. So the map U(X) U(X P n ) is simply the assignment α α id A (X). Below we will identify U(X) with its image in U(X P n ) Remark. A natural map p : A (X P n ) A (X) is an U(X)-module homomorphism. the isomorphism (43) is an U(X)-module isomorphism. One has U(X P n ) End U(X P n )(A (X P n )) End U(X) (A (X P n )). Denote a noncommutative ring End U(X) (A (X)) by Λ(X). Obviously U(X) Λ(X). Using (43) identify End U(X) (A (X P n )) with the matrix algebra M n+1 (Λ(X)). Clearly Λ(X P n ) End U(X) (A (X P n )) = M n+1 (Λ(X)). We will identify Λ(X) with the subring of scalar matrices in M n+1 (Λ(X)). Under this identification A (X P n ) becomes a 47

48 Λ(X)-module such that isomorphism (43) is a Λ(X)-module isomorphism. The ring Λ(X) is an upper bound for nonexisting cohomology ring A (X). A ring R(X) pretending to replace cohomology will be constructed below right after Remark It will be between U(X) and Λ(X) Lemma. Λ(X P n ) Λ(X)[ζ] and U(X)[ζ] U(X P n ) Λ(X)[ζ]. Proof. The image of Λ(X P n ) in M n+1 (Λ(X)) contains only those matrices which commute with a matrix representing ζ. Recall that ζ = e(o(1)) is represented by Jordan cell with unities over a diagonal. Since every matrix commuting with Jordan cell is a polinomial of this Jordan cell one has Λ(X P n ) Λ(X)[ζ]. Since U(X P n ) is the center of Λ(X P n ) it contains the center of Λ(X)[ζ] which is equal to U(X)[ζ] Corollary. For a variety Y P n any element from U(Y P n ) can be uniquely written as a polynomial a 0 + a 1 ζ + a 2 ζ a n ζ n where a 0 U(Y ) and other a i Λ(Y ). Proof. By Lemma one has a desired expression e(m) = a 0 + a 1 ζ + + a n ζ n where a i Λ(Y ). To see that a 0 lies in U(Y ) consider an imbedding i : Y Y P n. One has a chain of relations p e(m)i = p i e(i M) = e(i M) U(Y ). Since ζ i = 0 one has p e(m)i = p a 0 i = a 0 and the Lemma follows Corollary. Let L be a line bundle over Y P n P m. Then there is a unique family of elements a kl Λ(Y ) such that e(l) = a kl ζ k nζ l m with a kl Λ(Y ). Proof. U(Y P n P m ) Λ(Y P n P m ) Λ(Y P n )[ζ m ] Λ(Y )[ζ n, ζ m ]. In the next two lemmas we shall prove some sort of functoriality for the coefficients a i s from Corollary It should be mentioned that for the projection p : Y P n Y and linear imbedding i : Y P n 1 Y P n the natural maps p and i are homomorphisms of free Λ(Y )-modules. Consider a variety Y P n P m. Let p : Y P n Y and P : Y P n P m Y P m be natural projections. Denote by ζ m a Chern class e(o Y P m(1)) and by ζ m a Chern class e(p O Y P m(1)). In the same way let ζ n be a Chern class e(o Y P n(1)). In these notations one has Lemma. Let L be a line bundle over Y P m and L = P L be a line bundle over (Y P n ) P m. Let e(l) = a i ζ i m and e(l ) = a i(ζ m) i with a i Λ(Y ) and a i Λ(Y P n ) be expansions as in Lemma Using Lemma consider the elements a i s as elements of the ring Λ(Y )[ζ n]. Consider the elements a i s as elements of the same ring Λ(Y )[ζ n ] by means of the inclusion Λ(Y ) Λ(Y )[ζ n ]. Then for all indexes i one has a i = a i. Proof. Since e(l ) = e(p L) in Λ(Y P m )[ζ n ] the element e(l ) belongs to the subring Λ(Y P m ). Therefore in expansion e(l ) = b kl ζn(ζ k m) l coefficients b kl vanish for all nonzero k and one can rewrite this expansion as e(l ) = b l (ζ m )l. So we have two expansions: the last one where b l Λ(Y ) and the one e(l ) = a iζy i P where n a i 48

49 Λ(Y P n ). Since coefficients in both expansions can be considered as elements of Λ(Y )[ζ n ] these expansions coincide. We need only to check that b l = a l. That is the case because al ζ l m = e(l) = e(l)p i = P e(l )i = P b l ζ l mi = b l P (ζ m) l i = b l ζ l m. In the above chain of relations the symbol i was used for a natural inclusion Y P m Y P n P m. In this chain of relations the equality before the last holds because P is a Λ(X)-module homomorphism Lemma. Let i : Y P n 1 Y P n be a linear imbedding over Y, L be a a line bundle over Y P n. If e(l) = n k=0 a kζn k then e(i L) = n 1 k=0 a kζn 1 k. Proof. One has a chain of relations: i e(i L) = e(l)i = a k ζ k n i = a k i ζ k n 1 = i ak ζ k n 1. Since i is mono the lemma follows. Consider a line bundle E = p 1O(1) p 2O(1) over Y P P where p 1,2 : Y P P P are two projections. Then there exists a unique formal power series G Y (t 1, t 2 ) = g ij t i t j whose coefficients g ij Λ(Y ) are such that the Chern class e(e) is equal to G Y (ζ 1, ζ 2 ) where ζ i = e(p i O(1)) Remark. By definition the coefficients of G Y P n(t 1, t 2 ) belong to Λ(Y P n ). By lemma we can consider them as ones beloning to the ring Λ(Y ). Then G Y P n(t 1, t 2 ) = G Y (t 1, t 2 ) Definition. Define a ring R(Y ) as a subring in Λ(Y ) generated over U(Y ) by all coefficients g ij Lemma. The ring R(Y ) is a commutative ring; For any two line bundles L 1, L 2 /Y one has G Y (e(l 1 ), e(l 2 )) = e(l 1 L 2 ); Proof. First of all we claim that all coefficients of series G Y (t 1, t 2 ) commute with each other. For that it suffices to check that series G(t 1, t 2 ) and G(t 3, t 4 ) commute as elements of Λ[[t 1, t 2, t 3, t 4 ]]. The last is the case because G(ζ 1, ζ 2 ) and G(ζ 3, ζ 4 ) are two Chern classes for the variety Y P P P P which commute with each other. The first assertion of the lemma follows. In order to prove the second use Jouanalou trick. Replacing Y by an affine bundle over it we may assume that L 1 = f1 O(1) and L 2 = f2 O(1) for certain morphisms f i : Y P. Consider a morphism F : Y (id,f 1,f 2 ) Y P P. Then L 1 L 2 = F (p 1 O(1) p 2 O(1)) and one has F e(l 1 L 2 ) = G Y (ζ 1, ζ 2 )F. Let p : Y P P Y be the projection. Then e(l 1 L 2 ) = p F e(l 1 L 2 ) = p G Y (ζ 1, ζ 2 )F = G Y (e(l 1 ), e(l 2 )) because p ζ i F = p e(p i O(1))F = p F e(f p i O(1)) = e(f i O(1)) = e(l i) for i = 1, Theorem. The series G Y [t 1, t 2 ] is a formal group law over the ring R(Y ). 49

50 Proof. The ring R is commutative by The commutativity relation G Y (t 1, t 2 ) = G Y (t 2, t 1 ) is obvious. The associativity relation G Y (t 1, G Y (t 2, t 3 )) = G Y (G Y (t 1, t 2 ), t 3 ) follows from the second assertion of Lemma applied to the bundles L 1 = p 1O(1) and L 2 = p 2 O(1) p 3 O(1) over Y P P P. In fact, one has the following relations G Y (ζ 1, G Y (ζ 2, ζ 3 )) = G Y (e(p 1 O(1)), e(p 2 O(1) p 3 O(1))) = e(p 1 O(1) (p 1 O(1) p 1 O(1))). Since G Y (ζ 1, 0) = e(p 1O(1) 1) = ζ 1 and G Y (0, ζ 2 ) = ζ 2 one has G Y (t 1, t 2 ) = t 1 +t 2 +. The theorem follows. Denote an inverse series for the formal group law by I Y (t). Recall that it is a unique series such that G Y (t, I Y (t)) = 0. It exists for any formal group law. If ξ = e(o( 1)) over Y P then ξ = I Y (ζ) Corollary. For any line bundle T over Y P k all the coefficients a i of an expansion of e(t ) = a i ζ i are in the ring R(Y ). Proof. Define series G k Y (t 1,..., t k ) by induction: G 1 Y (t) = t, G2 Y (t 1, t 2 ) = G Y (t 1, t 2 ) and G k Y (t 1, t 2,..., t k ) = G Y (t 1, G k 1 Y (t 2,..., t k )). Obviously one has the relation G k Y (e(l 1),..., e(l k )) = e(l 1 L k ). Any line bundle over Y P k has the form T = p L O(±n) = p L O(±1) O(±1). Then either e(t ) = G n+1 Y (e(l), ζ,..., ζ) or e(t ) = G n+1 Y (e(l), ξ,..., ξ). In both cases the e(t ) has an expansion by degrees of ζ with coefficients in R(Y ) Remark. The third assertion holds for any line bundle over Y P... P Lemma. R(Y P n ) = R(Y )[ζ n ] Proof. The lemma asserts that two subrings of Λ(Y )[ζ n ] coincide. To see the inclusion R(Y P n ) R(Y )[ζ n ] one only need to check that U(Y P n ) R(Y )[ζ n ] and to show that coefficients of G Y P n(t 1, t 2 ) are in R(Y ). This is the case by and To see the inverse inclusion it is enough to recall that U(Y ) U(Y P n ) and ζ n U(Y P n ) Corollary. R(Y P ) = R(Y )[[t]] where the variable t is identified with ζ. Recall that A (Y ) is equipped with a structure of R(Y )-module. Moreover, the projective bundle isomorphism (43) is R(Y )-module isomorphism. The last lemma allows us to describe the structure of A (Y P n ) as an R(Y P n )-module. Recall that ζ acts on A (Y P n ) = A (Y ) shifting summands to the left on one step. It is convenient to write formally this direct sum as A (Y ) 1 A (Y ) t 1 A (Y ) t 2 A (Y ) t n = A (Y ) Z t n Z[t]/(t). Then the action of ζ coincides with the multiplication by t. Since R(Y P n ) = R(Y )[ζ n ] = R(Y )[t]/(t n+1 ) = R(Y ) Z Z[t]/(t n+1 ) one has the following corollary Corollary. The isomorphism is an R(Y )[t]/(t n+1 )-module isomorphism. A (Y P n ) = A (Y ) Z t n Z[t]/(t) 50

51 Consider a module t Z[t]/(t) which is a standard indecomposable injective module over the ring Z[[t]]. Such module will be denoted below as Z(t). Increasing n in the last corollary one gets the following one Corollary. A (Y P ) and A (Y ) Z Z(t) are isomorphic as R(Y )[[t]]-modules. Let Laz be the Lazard ring, that is the ring of coefficients of the universal commutative formal group law of dimension one. Let ϕ X : Laz R(X) be a unique homomorphism defined by the universal property. Consider A (X) as an Laz-module through this homomorphism Lemma. For any morphism f : Y X the natural map f : A (Y ) A (X) is an Laz-module homomorphism. Proof. First of all recall that the projective bundle isomorphisms (43) is natural with respect to base change, that is for any morphism f : Y X the diagram commutes: A (Y P n ) (p,p ζ Y,...,p ζy n ) A (Y ) F f A (X P n ) (p,p ζ X,...,p ζ n X ) A (X), where F = f id : Y P n X P n. The group A (Y P P ) is equipped as with the R(Y )-module structure so with a Z[[t 1, t 2 ]]-module structure (t i acts as the Chern homomorphism ζ i = e(o(1))). The natural map F : A (Y P P ) A (X P P ) induced by F is an Z[[t 1, t 2 ]]-module homomorphism by the functoriality of the Chern homomorphisms. Recall that the Lazard ring is generated by coefficients of the universal formal group law G(t 1, t 2 ) = g ij t i t j. It is sufficient to prove that the action of g ij commutes with f. Consider a diagram A (Y P P ) = A (Y ) Z(t 1, t 2 ) F =f 1 A (X P P ) = A (X) Z(t 1, t 2 ) G Y (t 1,t 2 ) G X (t 1,t 2 ) A (Y ) Z(t 1, t 2 ) = A (Y P P ) F =f 1 A (X) Z(t 1, t 2 ) = A (X P P ). It commutes by the functoriality of Chern classes applied to the line bundles p 1 O(1) p 2 O(1) over X P P and Y P P. Since F G Y (t 1, t 2 ) = G X (t 1, t 2 ) F one has (f (g Y ) ij )t i 1t j 2 = ((g X ) ij f )t i 1t j 2. Equating coefficients in the last relation we get the desired relations (g X ) ij f = f (g Y ) ij Lemma. For any closed imbedding f : Y X the Gysin map f! : A (X) A (Y ) is a Laz-module homomorphism. 51

52 Proof. We shall use Lemma By this lemma there is a commutative square: A (X P n ) ζ k X A (X P n ) F! A (Y P n ) F! ζ k Y A (Y P n ). By the base change property (25) there is a commutative square A (X P n ) p X A (X) F! A (Y P n ) f! p Y A (Y P n ). Gluing these diagrams together we get the following commutative diagram: A (Y P n ) (p,p ζ Y,...,p ζy n ) A (Y ) F! f! A (X P n ) (p,p ζ X,...,p ζx n ) A (X), Now the proof of Lemma proceeds as the proof of Lemma By Lemma the diagram A (Y ) Z(t 1, t 2 ) F! =f! 1 A (X) Z(t 1, t 2 ) G Y (t 1,t 2 ) G X (t 1,t 2 ) A (Y ) Z(t 1, t 2 ). F! =f! 1 A (X) Z(t 1, t 2 ) commutes. So one has f! ϕ X (g ij ) = ϕ Y (g ij )f!. Since coefficients g ij generates the Lazard ring the Lemma follows. 4.2 Construction of a Quillen map. Let ω = (1 + a 1 t + a 2 t )dt with a i Laz be the (unique) normalized invariant differential on the universal formal group law G(t 1, t 2 ). Set a 0 = 1. Let X be a smooth variety and p : X P n X, q : X P n P n be the projections. Set ζ = e(q (O(1))) : A (X P n ) A (X P n ). By the projective bundle theorem the map (p, p ζ,..., p ζ n ) : A (X P n ) A (X) A (X) A (X) is an isomorphism. Define p! : A (X) A (X P n ) (44) 52

53 as follows. For an element x A (X) let p! (x) A (X P n ) be a unique element such that for any integer k with 0 k n one has p (ζ k (p! (x))) = a n k x. The operator p! is called Quillen operator. We can reformulate the above definition by means of residues Lemma. Let R = R A (X) and let h(t) R[[t]]. Consider the element h(ζ) ( R(X P n ). Then one has the relation p h(ζ)p! = Res h(t)ωx residue map. t n+1 ) R, where Res is the The proof is obvious. The Quillen operators (44) defined just above satisfy the following list of properties which will be proved below. 1. Composition property: the following diagram commutes A (X) p! m p! n A (X P n ) p! m A (X P m ) p! n A (X P n P m ). where p n : X P n X, p m : X P m X, p n : X P n P m X P m and p m : X P n P m X P n are the natural projections. 2. Base change property: Let f : Y X be a morphism of smooth varieties and let F : Y P n X P n be the base change of f by the natural projection p X : X P n X. Then one has the relation F p! Y = p! X f where p Y : Y P n Y is the natural projection. 3. Consistence with linear imbeddings: For a linear imbedding i : P m P n the following diagram commutes A (X) p! n A (X P n ) id A (X) p! m i! A (X P m ) where p n : X P n X and p m : X P m X are the natural projections. 4. Consistence with Gysin operators: For a closed imbedding of smooth varieties i : Y X the following diagram commutes A (X) p! X A (X P n ) i! A (Y ) p! Y I! A (Y P n ) where p X : X P n X and p Y : Y P n Y are the natural projections. 5. Consistence with a section of trivial projective bundle: For a section s : X X P n of the projection p : X P n X one has s! p! = id A (X). 53

54 4.3 Proof of properties of the Quillen s maps p! In this Subsection we prove properties of Quillen s operators stated in Subsection 4.2. The base change property 2 is obvious. In fact, let us take f : Y X. Since f : A (Y ) A (X) is an Laz-module homomorphism then f ϕ Y (a i ) = ϕ X (a i ) f for any coefficients a i Laz of the differential form ω(t)dt. So one has a chain of relations p X ζk X F p! Y = px F ζ k Y p! Y = f p Y ζk Y p! Y = f ϕ Y (a n k ) = ϕ X (a n k )f = p X ζk X p! X f. Since the map (p, p ζ,..., p ζ n ) : A (X P n ) A (X) A (X) A (X) is an isomorphism we get the required relation F p! Y = p! X f. To prove the property 1 observe that A (X P n P m ) is a direct sum of mn copies of A (X) and for an element b A (X P n P m ) its coordinates b kl A (X) are given by b kl = p n p m (ζmζ k n(b)), l where ζ r = e(qro(1)) and q r : X P n P m P r is the projection for r = m, n. In order to check the equality p! m p! n = p! n p! m let us compute pn p m (ζk m ζl n ( p! m p! n (x))) for arbitrary x A (X). We can rewrite it as (p n (ζ n) l )( p m ζ k m p! m(p! n(x))) = (p n (ζ n) l )(a k p! n(x)) = = a k (p n (ζ n) l p! n)(x) = a k a l x. where ζ n = e( p m q n O(1)) R(X Pn ). We can interchange a k with an action of ζ k since ϕ X P n(a k ) and ζ k commute. Since the last expression is symmetric with respect to indexes k and l the property 1 is proved. Now prove the linear imbedding property 3. It suffices to check the case n = m + 1 because of the compatibility of the Gysin homomorphisms with the composition of closed imbeddings (24). In order to prove that p! m = i! p! m+1 we need to check the equality pm (ζk m (p! m (x))) = p m (ζm(i k! p! m+1(x))). The left hand is equal to a m k (x) by the very definition of p! m. The right hand is equal to p m+1 i (ζm k (i! p! m+1 (x))) = pm+1 ζm+1 k (i i! p! m+1 (x)). Applying the normalization property i i! = ζ m+1 several times we finally get p m+1 ζm+1(p k+1! m+1(x)) = a (m+1) (k+1) (x). Whence p! m = i! p! m+1. Now prove the property 4. Proof. In order to check the equality p! Y i! = I! p! X it is enough to establish that p Y ζy k p! Y i! = p Y ζy k I! p! X. Recall that by lemma ϕ Y (a k ) i! = i! ϕ X (a k ) and by lemma I! ζ X = ζ Y I! because O Y P n(1) = I O X P n(1). Then one has p Y ζy k p! Y i! = ϕ Y (a n k )i! = i! ϕ X (a n k ) = i! p X ζx k p! X. Using the base change for Gysin (25) we see that the last term is equal to p Y I! ζx k p! X = py ζk Y I! p! X. The property is proved. The Key property 5 is proved below. 54

55 4.4 Proof of the Key property. Let X be a smooth variety and let p : X P n X be the projection and let s : X X P n be a section of the projection p. One has to prove the relation s! p! = id A (X) (45) Since s is a section of p it has a form (id X, f) for a morphism f : X P n. Let S : X P n (X P n ) P n be given by formula (s p, q) where q : X P n P n is the projection. Consider two diagrams and S X P n X P n P n p X s S p 1 X P n X P n X P n P n s X s X P n. We need to compute p s s! p!. Since the last diagram is Cartesian and transversal one has s s! p! =! S p!. The first diagram shows that! S p! =! p! 1 s. By Proposition below one has! p! 1 = id A (X). Therefore one has It remains to prove the following p s s! p! = p (! p! 1 ) s = p s = id Proposition. One has the relation! p! 1 = id A (X). Observe that for the diagonal imbedding : X P n X P n P n one has the relation! = c n (p 2 (O(1)) p 1 (Q)). (46) This relation is a particular case of the one which is proved in In fact take Y = X P n, E = 1 n+1, s =, p = p 2. To proof the property it suffices to prove p! p! = 1 R(X P n ). The left hand side equals to p c n (p 2 (O(1)) p 1 (Q))p!. We can find a series h(t) = h(ζ, t) representing an element c n (p 2(O(1)) p 1(Q)). For that consider an exact sequence of vector bundles over X P n P 0 p 2O(1) p 1O( 1) p 2O(1) 1 n+1 p 2(O(1)) p 1(Q) 0. The sequence gives the relation in R(X)[ζ][[t]] = R(X P n P ) c n+1 (p 2O(1) 1 n+1 ) = c 1 (p 2O(1) p 1O( 1))c n (p 2O(1) p 1Q). 55

56 Moreover c n+1 (p 2 O(1) 1n+1 ) = t n+1, c 1 (p 2 O(1) p 1 O( 1)) = t G X ζ, where G X (t 1, t 2 ) R(X)[[t 1, t 2 ]] is a formal group low defined in the subsection 4.1. Since the series t GX ζ has invertible the lowest term it is invertible in R(X)[ζ]((t)) and the series h(t, ζ) = is well defined. We just proved that tn+1 t GX ζ h(t, ζ 2 ) t=ζ1 =! in R(X)[ζ 1, ζ 2 ] = R(X P n P n ). (47) Therefore p! p! = p h(t, ζ)p!. By Lemma one has p h(t, ζ)p! = Res( tn+1 ω(t)dt t GX ) = ζ t n+1 Res( ω(t)dt t GX ) R(X)[ζ]. The last term is equal to 1 by [Pa, Claim 4.6.2]. ζ The property follows (compare with [Qu1, Th.1]). 5 Trace structures on a homology theory 5.1 Definition and the main theorem Here we define a notion of a trace structure on a homology theory A and claim in that an orientation of A gives rise to a unique trace structure on A subjecting certain normalization property Definition. Let A be a homology theory. A trace structure on A is a rule assigning to each projective morphism of smooth varieties f : Y X an operator f! : A (X) A (Y ) called the trace (for f) and satisfying the following properties 1. (f g)! = g! f! for any projective morphisms Z g Y and Y f X of smooth varieties; 2. for the transversal square from (3.1.1) the following diagram commutes A ( X) ϕ A (X) ĩ! A (Ỹ ) ϕ i! A (Y ) 3. for any morphism of smooth varieties f : Y X the following diagram commutes A (P n Y ) p! Y (id f) A A (P n X) p! X A (Y ) f A (X) where p Y : P n Y Y and p X : P n X X are the natural projections; 56

57 4. normalization: for any smooth variety X one has (id X )! = id A (X). 5. localization: for any closed imbedding of smooth varieties i : Y X and the inclusion j : X Y X the sequence A (X Y ) j A (X) i! A (Y ) is exact. This sequence is often called Gysin localization sequence below in the text Remark. The restriction of a trace structure to the family of all closed imbeddings is a Gysin structure Definition. Let A be a homology theory equipped with a commutative Chern structure (L, X, Z) c Z (L). One says that a trace structure f f! on A is subjected to the Chern structure if for each smooth variety X and each line bundle L/X one has the following relation in End(A (X)) z! z = c(l). (48) Theorem. Let A be a homology theory. Let (L, X, Z) c Z (L) be a commutative Chern structure on A, then there exists a unique trace structure f f! on A subjected to the Chern structure. A construction of the required trace structure is given in section 5.2. The uniqueness of the trace structure is proved in section Remark. Let A be a homology theory. Orientations of A and commutative Chern structures bijectively correspond to each other by Theorems and So for an oriented cohomology theory A with the corresponding Chern structure there exists a unique trace structure subjected to this Chern structure. 5.2 Construction of a trace structure Let A be a homology theory endowed with an orientation ω and therefore endowed with the corresponding Chern structure ( (2.4.4) and (2.4.3) ). The main aim of this subsection is to construct a trace structure required by Theorem Let f : Y X be a projective morphism of smooth varieties. One can present f as a composition of a closed imbedding i : Y X P n and the projection p : X P n X, i.e. f = p i. Define now an operator f! : A (X) A (Y ) by the formula f! = i! p! Theorem. The operator f! does not depend on the particular choice of a decomposition of the morphism f in the form f = p i, where i : Y X P n is a closed imbedding and p : X P n X is the projection. Moreover the assignment f f! is a trace structure on A required by Theorem

58 Proof. For a projective morphism f : Y X we first check that the map f! does not depend on the particular choice of the decomposition of f. Let f = p i be another decomposition of f, where i : Y X P m is a closed imbedding and p : X P m X is the projection. We have to check the relation For that consider the commutative diagram i! p! = (i )! (p )!. (49) f f X Y X I p p X P m p m X P n P m p n X P n where I is a unique imbedding such that p m I = i and p n I = i. Recall that p! n p! = p! m (p )! by the property 1. Now clearly it suffices to check two relations I! p! m = (i )! I! p! n = i!. Both these relations are particular cases of the one from the following Claim Claim. Consider a commutative diagram id Y Y j T P k j with the closed imbeddings j and j. One has the following relation j! p! = j!. Proof of Claim. To prove this Claim consider the commutative diagram J p T Y P k T P k q p Y j with J = j id and consider a section s : Y Y P k of the projection q such that J s = j. The relations j! = s! J! (see (24)) and J! p! = q! j! (see 4) and s! q! = id A (Y ) give a chain of relations j! p! = s! J! p! = s! q! j! = j!. T 58

59 The Claim is proved which completes the proof of the relation (49). Thus the operator f! is well-defined. The nearest goal is to prove the relation g! f! = (f g)!. For that consider the commutative diagram j Z Y P m I X P n P m g Y p m i q m X P n f X with a smooth variety Z and a closed imbedding j and a closed imbedding i and with I = i id and with the projection q m : X P n P m X P n. By the very definition one has relations f! = i! p! n and g! = j! p! m. Thus one has a chain of relations g! f! = j! p! m i! p! n = j! I! q! m p! n = = (I j)! (q! m p! n ). Now consider a closed imbedding k : P n P m P N and the projection p N : X P N X and the commutative diagram Since it remains to prove I j Z X P f g n P m id k X P N p n q m p N X (f g)! = ((id k) I j)! p! N = (I j)! (id k)! p! N Claim. The following relation holds p n (id k)! p! N = q! m p! n. (50) Proof of Claim. First consider the projections q n : X P n P m X P n and q m : X P n P m X P m and consider the commutative diagram X P n P m P N k 2 q m,n X P n P m k 1 X P m P N k q N X P N 59

60 with the obvious projections q m,n, q N and with the closed imbeddings k 1 = (q m, k) and k 2 = (id, k). The Claim gives a chain of relations Now consider one more commutative diagram k! = k! 1 q! N = k! 2 q! m,n q! N. (51) X P n P m q n X P m X p m q n,m q N p N X P n P m P N q m,n X P m P N q m X P N By the property (1) of the Quillen map the diagram gives rise to the chain of relations Thus one gets the following relation q! m,n q! m p! N = q! m,n q N! p! m = q! n,m q! n p! m. (52) k! p! N = k! 2 q! n,m q! n p! m (53) To complete the proof of the Claim it remains to check the relation id = k! 2 q! n,m because q! n p! m = q! m p! n by the property 1 of the mappings p!. To prove the relation id = k! 2 q! n,m observe that the map k 2 is a section of the projection q n,m. Therefore this relation is just a particular case of the property 5 of the Quillen map. To complete the proof of the Theorem it remains to check the properties 2 to 4 of the trace structure and to prove the relation i i! = e(l(d)) for a smooth divisor i : D X. The property 2 coincides with the one The property 3 coincides with the property 2 of the maps p!. The property 4 is obvious. The relation i i! = e(l(d)) is exactly the property (29) of the Gysin map i!. 5.3 Uniqueness of the trace structure In this subsection we prove the uniqueness assertion of Theorem The uniqueness assertion of Theorem is an obvious consequence of Theorem and the following Lemma. Suppose that A is equipped with a commutative Chern structure (X, L) e(l). Let A is equipped with two trace structures f f! 1 and f f! 2 subjected to the Chern structure. Then for each n 0 and the projection p : X P n X one has p! 1 = p! 2. Proof. Recall the following elementary lemma. 60

61 5.3.2 Lemma. Let s k : P n k P n be a linear inclusion. Then s k s! k = ζk n, where ζ n = c(o(1)). Proof. Proceeding by induction over k. For k = 1 the lemma holds by Assume the assertion is valid for k. Then s k s! k = ζk n. For s : Pn k 1 P n k one has s s! = ζ n k. Therefore s k+1 s! k+1 = sk s s! s! k = sk ζ n k s! k = ζ ns k s! k = ζ nζn k = ζ k+1 Let : X P n X P n P n be the diagonal imbedding. We claim that the composition! : A (P n X Pn X ) A (P n X Pn X ) belongs to the ring Λ(X Pn P n ) (see for notation subsection 4.1). It follows from that! commutes with e(l) for any line bundle L over X P n P n : one has e(l)! = e( L)! =! e(l) by Lemma Since Λ(X P n P n ) Λ(X)[ζ 1, ζ 2 ] by Lemma one can regard an expansion n! = D ij ζ1ζ i 2, j where D ij Λ(X). i,j=0 n Remark. In fact It follows from equation! = tn+1 t GX ζ 2 t=ζ1! R(X)[ζ 1, ζ 2 ] Lemma. D 0k = D k0 = 0, k n; D 0n = D n0 = 1. (see (47)) that Proof. Let i = i 1 : P n X Pn X Pn X be natural inclusion. Then ζ 1i = i ζ and ζ 2 i = 0. Consider a transversal square P n X s X P n X Pn X i s P n X Lemma shows that! i = s! s = ζ n. On the other hand! i = n D ij ζ1 iζj 2 i = n D i0 i ζ i. Let P : P n X Pn X Pn X be the projection onto the first i,j=0 i=0 factor. Applying P one gets ζ n = n P D i0 i ζ i = n D i0 P i ζ i = n D i0 ζ i provided i=0 by P is Λ(X)-module homomorphism. Therefore for each i < n one has D i0 = 0 and D n0 = 1. For the rest of the proof of Theorem we shall use following notations: p k : P k X X is the projection, P k : P n X Pk X Pn X is the projection onto first factor. Note that P k! ζ = ζ 1P k! : A (P k X ) A (P k X X P k X ). Prove the theorem by induction on n. In order to prove that p! 1 n = p! 2 n one need to check i=0 p n ζk p! 1 n = p n ζk p! 2 n, k = 0,..., n (54) i=0 61

62 Since p n ζk p! n = pn sk s! k p! n = pn k p! n k then equation (54) holds for k 0 by the induction hypothesis. It remains to check that p n p! 1 n = p n p! 2 n. Consider the chain of relations Thus one has n i,j=0 id A (P n X ) = P n! P! n = n D ij P n ζj 2 P n! ζi = i,j=0 D ij P n j P! 1 n j ζi = n i,j=0 n i,j=0 n P n D ijζ1 i ζj 2 P n! = i,j=0 D ij P n j P! 2 n j ζi. D ij P n j P! n j ζi. The summands with j > 0 in both hands are equal each other by induction hypothesis. Since D i0 = 0 for i n one has Consider the transversal square P n P! 1 n ζn = P n P! 2 n ζn (55) P n X p n X P k P n X P k X P n p k P k X. Then p n P kp k! = pk P np k! = pk p! k pn. Using the last equation for k = n one can rewrite relation (55) in the form p n p! 1 n p n ζn = p n p! 2 n p n ζn. Since p n ζn is an epimorphism one has p n p! 1 n = p n p! 2 n. The induction runs. 5.4 Examples Here we give two examples of oriented homology theories. One is the motivic homology theory and another is the algebraic cobordism theory MGL Motivic Homology. Recall that the motivic homology theory could be defined as follows. For a smooth X and a closed Z in X let M Z (X) be the motive with support (see [SV, the text just below the proof of Thm.4.8]). Now H Z p,q(x) := Hom DM (Z(p)[q], M Z (X)). The theory H, is an oriented homology theory. To show this recall that for a smooth variety X the motivic cohomology group H 2,1 (X) coincides with the Chow group CH 1 (X) by [SV, Cor.3.2.1]. For a line bundle L on a smooth X and a closed Z in X consider the cap-product operator c 1 (L) : H Z, (X) HZ 2, 1 (X) 62

63 where c 1 (L) CH 1 (X) is the classical first Chern class of L. These operators define a Chern structure on the motivic homology theory H,. In fact, the only property one has to check is the non-degeneracy property 2 of This is the case because [SV, Thm.4.5] and the canonical morphism M Z P 1(X P 1 ) = M Z (X) M Z (X)(1)[2] Hom DM (Z(p 1)[q 2], M Z (X)) Hom DM (Z(p)[q], M Z (X)(1)[2]) is an isomorphism by the cancellation theorem [V2, Cor.4.10]. Therefore the theory H, is equipped with a Chern structure. Whence it is oriented and is equipped with a unique trace structure subjected to the given Chern structure The algebraic cobordism of Voevodsky. Given an oriented commutative ring T -spectrum E one can form as a cohomology theory E, so a homology theory E,. In this case the class c E 2,1 (P ) defines a Chern structure L/X c(l) E 2,1 (X) on the ring cohomology theory E, (see [P1, Def.3.2] for definitions and [S, Thm.3] for the construction and the proof). The cap-products operators c(l) : E Z, (X) E Z 2, 1(X) define a Chern structure on the homology theory E,. Theorem gives a unique trace structure on E, subjected to the given Chern structure on E,. An example of an oriented commutative ring T -spectrum is the algebraic cobordism spectrum MGL (see [V1]). To get the required element c MGL 2,1 (P ) one should do the following. Take the element th MGL 2,1 P (O( 1)) described in [P1, 3.8.7]. Consider the image of this element under the support extension operator MGL 2,1 P (O( 1)) MGL 2,1 (O( 1)) and finally apply to the image the pull-back operator z : MGL 2,1 (O( 1)) MGL 2,1 (P ). induced the zero section z. The resulting element is the required element c by [S, Thm.4]. Whence the homology theory MGL, is oriented and equipped with a unique trace structure subjected to the given Chern structure. 6 Appendix 6.1 In this section we show how one can extend the uniqueness theorem to a pretheory context. We worked above in the context of a homology theory which is given on pairs (X, U) where X is a smooth quasi-projective variety and U X is a Zarisky open subset. However the notion of a trace structure can be easily extended to the context of functors defined on smooth varieties rather than on pairs. This is done, for example, in [PY]. We call such a functor a pretheory in order to distinguish it of a theory. Recall the notion. 63

64 6.1.1 Definition. A homology pretheory is a covariant functor X A (X) from category of smooth quasi-projective varieties to Ab which satisfies two properties: Strong homotopy invariance. For any vector bundle E X and a principal homogeneous space p : T X for E the natural map p : A (T ) A (X) is an isomorphism. Additivity. Let i r : X r X 1 X2 be natural inclusions (r = 1, 2). Then the induced map A (X 1 ) A (X 2 ) A (X 1 X2 ) is an isomorphism Definition. Let A be a homology pretheory. A trace structure on A is a rule assigning to each projective morphism of smooth varieties f : Y X an operator f! : A (X) A (Y ) called the trace operator (for f) and satisfying as Axioms so the following axiom called below the projective bundle axiom: for any rank n + 1 vector bundle E X with projective bundle p : P(E) X let ζ E := z! z : A (P(E)) A (P(E)) where z is a zero section for line bundle O E (1). Then the operator is an isomorphism. (p, p ζ E,..., p ζ n E ) : A (P(E)) n A (X) (56) We can state the uniqueness theorem in the context of a pretheory A like as uniqueness theorem in the context of a theory A : Theorem. Let A be a homology pretheory and let f f! 1 and f f! 2 be two trace structures on the pretheory A. Suppose that for any smooth X and any line bundle L/X with the zero section z : X L operators z! 1 z, z! 2 z : A (X) A (X) coincide. Then for any projective morphism f : Y X one has f! 1 = f! 2. The proof is essentially the same as ones for Theorem and Lemma with several replacements. Below we describe what these replacements are. First of all one has to define a Chern prestructure on a pretheory A Definition (Chern Prestructure). Let A be a homology pretheory. A Chern prestructure on A is a rule assigning to each smooth X and a line bundle L over X an operator e(l) : A (X) A (X) satisfying the following list of axioms: 1. Functoriality. Let ϕ : X X be a morphism of varieties, and L be a line bundle on X. Then the following diagram commutes r=0 A (X ) ϕ A (X) e(ϕ L) A (X ) ϕ e(l) A (X). 64

65 2. e(l 1 ) = e(l 2 ) for isomorphic line bundles L 1 and L 2 ; 3. commutativity: e(l 1 )e(l 2 ) = e(l 2 )e(l 1 ) for any two line bundles L 1, L 2 over X. 4. vanishing: e(1 X ) vanishes for any smooth X. 5. nilpotence: for any line bundle L there exists an integer n such that e(l) n = Theorem. Let A be a homology pretheory endowed with a trace structure f f!. For any smooth X and a line bundle L/X consider an operator e(l) = z! z where z : X L is a zero section. Then an assignment (L, X) e(l) is a Chern prestructure on A. Proof. Functoriality is a straightforward consequence of the base change axiom 2. The transversality of the diagram ϕ X z L X L Φ z L L implies that e(l )ϕ = z L! zl ϕ = z L! Φ z L = ϕ z L! zl = ϕ e(l). The proof of the second item is the same. In order to prove the commutativity consider a transversal square: Z L 1 2 L1 L 2 z 2 Z 2 X z 1 L 1. Then e(l 1 )e(l 2 ) = z! 1 z1 z! 2 z2 = z! 1 Z! 2 Z1 z2 = z! 2 Z! 1 Z2 z1 z! 2 z2 z! 1 z1 = e(l 2)e(L 1 ). Vanishing. To prove that e(l) = 0 if L = 1 consider s, s L : X P(1 L) which identiies X with P(1) and P(L) respectively. Then one has s! = s! L since both s, s L : X X P 1 go through an open inclusion X A 1 X P 1. From the other hand s! L s = 0 by Gysin localization property 5 applying to the sequence A (P(1 L) P(L)) A (P(1 L)) s! L A (X). Therefore e(l) = s! s = s! L s = 0. Before proving the nilpotency property we need the following lemma: Lemma. Let f : Y X be a closed imbedding and L be a line bundle over X. Denote by L a pullback bundle f L over Y and denote an induced map L L by F. Then f! e(l) = e(l ) f!. Proof. Consider a transversal square L F L z Y f 65 z X,

66 where z, z are zero sections. Then one has f! e(l) = f! z! z = (z )! F! z = (z )! z f! = e(l )f!. To prove that e(l) is nilpotent first of all we consider the case X = P n and L = O(1). Denote e(o(1)) by ζ n as usually. We prove by induction that ζn n+1 = 0. Let s : P n 1 P n be a linear imbedding. Let j : P n P n 1 P n be an open imbedding. Since j L is a trivial bundle one has ζ n j = 0. By Lemma one has s! ζn n = ζn n 1 s! = 0 since by inductive hypothesis ζn 1 n = 0. By property 5 of trace structure the image of j coincide with the kernel of s!. Therefore Im(ζn) n Kers! = Imj. Then Im(ζ n ζn) n ζ n (Imj ) = Im(ζ n j ) = 0. The induction runs. The general case is reduced to the one just considered as it is done in Lemma The property follows. In the context of a pretheory A endowed with a trace structure one can easily prove a number of results proven above in the context of an oriented homology theory. To do this one has to reformulate Useful Lemma Lemma. Let A be a homology pretheory endowed with a trace structure. Then for the diagram (see for notation ) j 0 P(1 N) X j 1 t X s I i t Y k 0 Y A 1 k 1 Y, and the open inclusion j t : X t Y A1 X one has Im(j 0 ) + Im(jt ) = A (X t ). Proof. By Gysin localization property 5 one has Imj t = KerI t!. It remains to prove that the operator I t! j 0 is surjective. That is the case because I t! j 0 = k 0 s! where k 0 is an isomorphism and s! is surjective from relation s! p! = id A (Y ). Using this lemma one gets the following analogues of Lemma and Corollary Lemma. Let A be a homology pretheory endowed with a trace structure f f!. Let s : Y P(1 L) be a zero section of projective bundle identifying Y with P(1). Then s s! = e(p L O 1 L (1)) where p : P(1 L) Y is a natural projection Corollary. Let A be a homology pretheory endowed with a trace structure. Then for a smooth divisor s : D X on a smooth X one has s s! = e(l(d)). Proofs of these Lemma and Corollary follow the proofs of Lemma and Corollary One should use Chern prestructure just defined and Projective Bundle Axiom (56) instead of the Chern structure and Projective Bundle Theorem Since the proof of Theorem uses only the ingredients mentioned above one gets the following 66

67 Theorem. Let A be a homology pretheory endowed with two trace structures f f! 1 and f f! 2 such that the corresponding Chern prestructures coincide. Then for any closed imbedding f : Y X one has f! 1 = f! 2. In Subsection 4.1 we doid not use all the power of the Chern structure. Namely, all proofs given in that Subsection work well in the context of a pretheory A equipped with a trace structure and the corresponding Chern prestructure. This is why all the results of Subsection 5.3 also valid in the pretheory context. In particularly one can reformulate Lemma as Theorem. Suppose that a pretheory A is equipped with two trace structures f f! 1 and f f! 2 such that corresponding Chern prestructures coincide. Then for each n 0 and the projection p : X P n X one has p! 1 = p! 2. Theorem follows from Theorem and Theorem References [Fu] W. Fulton. Intersection theory. Springer-Verlag, [Har] R. Hartshorne. Algebraic geometry. Graduated texts in Mathematics 52,1977. [Ne] [P1] [PS] [Pa] [Pim] [PY] [Qu1] [S] A. Nenashev Projective bundle theorem in homology theories with Chern structure, I. Panin. Push-forwards in Oriented Cohomology Theories of Algebraic Varieties: II, I. Panin, A. Smirnov. Push-forwards in Oriented Cohomology Theories of Algebraic Varieties, I. Panin (after I.Panin, A.Smirnov). Riemann Roch Theorems for Oriented Cohomology, pp in the book J.P.C.Greenless(ed.) Axiomatic, Enriched, and Motivic Homotopy Theory. Kluwer Academic Publishers, Netherlands, K.Pimenov Traces in oriented homology theories, I.Panin, S.Yagunov. Poincare Duality for Algebraic varieties. D. Quillen. On the formal group laws of unoriented and complex cobordism theory. Bull. AMS 75(1969), A. Solynin. Chern and Thom elements in the representable cohomology theories. Preprint POMI-03/2004, 67

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