NOTES ON THE MILNOR CONJECTURES DANIEL DUGGER Contents Introduction 1 1. The Milnor conjectures 1 2. Proof of the conjecture on the norm residue symbol 8 3. Proof of the conjecture on quadratic forms 15 4. Quadratic forms and the Adams spectral sequence 18 Appendix A. Some examples of the Milnor conjectures 23 Appendix B. More on the motivic Adams spectral sequence 26 References 29 Introduction These are four lectures concerning the two Milnor conjectures and their proof* *s. Voevodsky's proof of the norm residue symbol conjecture_which is now eight years old_came with an explosion of ideas. The aim of these notes is to help make this explosion a little more accessible to topologists. My intention here is not to * *give a completely rigorous presentation of this material, but just to discuss some of * *the ideas and point the reader in directions where he can learn more. I've tried to* * make the lectures accessible to topologists with no specialized knowledge in this ar* *ea, at least to the extent that such a person can come away with a general sense of how homotopy theory enters into the picture. Let me apologize for two aspects of these notes. Foremost, they reflect only my own limited understanding of this material. Secondly, I have made certain expository decisions about which parts of the proofs to present in detail and w* *hich parts to keep in a "black boxä_ nd the reader may well be disappointed in my choices. I hope that in spite of these shortcomings the notes are still useful. Sections 1, 2, and 3 each depend heavily on the previous one. Section 4 could almost be read independently of 2 and 3, except for the need of Remark 2.10. 1. The Milnor conjectures The Milnor conjectures are two purely algebraic statements in the theory of fields, having to do with the classification of quadratic forms. In this sectio* *n we'll review the basic theory and summarize the conjectures. Appendix A contains some supplementary material, where several examples are discussed. ____________ Date: August 26, 2004. 1 2 DANIEL DUGGER 1.1. Background. Let F be a field. In some sense our goal is to completely clas* *sify symmetric bilinear forms over F . To give such a form (-, -) on F nis the same as giving a symmetric n x n matrix A, where aij= (ei, ej). Two matrices A1 and A2 represent the same form up to a change of basis if and only if A1 = P A2P T for some invertible matrix P . The main classical theorem on this topic says th* *at if char(F ) 6= 2 then every symmetric bilinear form can be diagonalized by a chang* *e of basis. The question remains to decide when two given diagonal~matrices~D1~and~D2 represent equivalent bilinear forms. For instance, do 20 011and 30 01represent the same form over Q? To pursue this question one looks for invariants. The most obvious of these is the rank of the matrix A. This is in fact the unique invariant when the field * *is algebraically closed. For suppose a form is represented by a diagonal matrix D, and let ~ be a nonzero scalar. Construct a new basis by replacing the ith basis element ei by ~ei. The matrix of the form with respect to this new basis is the same as D, but with the ith diagonal entry multiplied by ~2. The conclusion is * *that multiplying the entries of D by squares does not change the isomorphism class of the underlying form. This leads immediately to the classical theorem saying that if every element of F is a square (which we'll write as F = F 2) then a symmetr* *ic bilinear form is completely classified by its rank. We now restrict to nondegenerate forms, in which case the matrix A is non- singular. The element det(A) 2 F *is not quite an invariant of the bilinear form, since after a change of basis the determinant of the new matrix will be det(P ) det(A) det(P T) = det(P )2det(A). However, the determinant is a well- defined invariant if we regard it as an element of F *=(F *)2.~ Since~22_3is~no* *t~a square in Q, for instance, this tells us that the matrices 20 011and 30 01don* *'t represent isomorphic forms over Q. The rank and determinant are by far the simplest invariants to write down, but they are not very strong. They don't even suffice to distinguish forms over R. This case is actually a good example to look at. For a1, . .,.an 2 R*, let denote the form on Rn defined by (ei, ej) = ffii,jai. Since ever* *y el- ement of R is either a square or the negative of a square, it follows that every nondegenerate real form is isomorphic to an where each ai2 {1, -1* *}. When are two such forms isomorphic? Of course one knows the answer, but let's think through it. The Witt Cancellation Theorem (true over any field) says that if ~= then ~* *=. So our problem reduces to deciding whether the n-dimensional forms <1, 1, . .,.* *1> and <-1, . .,.-1> are isomorphic. When n is odd the determinant distinguishes them, but when n is even it doesn't. Of course the thing to say is that the ass* *oci- ated quadratic form takes only positive values in the first case, and only nega* *tive values in the second_but this is not exactly an `algebraic' way of distinguishi* *ng the forms, in that it uses the ordering on R in an essential way. By the end of* * this section we will indeed have purely algebraic invariants we can use here. 1.2. The Grothendieck-Witt ring. In a moment we'll return to the problem of finding invariants more sophisticated than the rank and determinant, but first * *we need a little more machinery. From now on char(F ) 6= 2. By a quadratic space NOTES ON THE MILNOR CONJECTURES 3 I mean a pair (V, ~) consisting of a finite-dimensional vector space and a non- degenerate bilinear form ~. To systemize their study one defines the Grothendie* *ck- Witt ring GW (F ). This is the free abelian group generated by isomorphism clas* *ses of pairs (V, ~), with the usual relation identifying the direct sum of quadrati* *c spaces with the sum in the group. The multiplication is given by tensor product of vec* *tor spaces. The classical theory of bilinear forms allows us to give a complete descrip- tion of the abelian group GW (F ) in terms of generators and relations. Recall that denotes the n-dimensional space F nwith (ei, ej) = ffiijai. * * So = + . .+. in GW (F ). The fact that every symmetric bili* *n- ear form is diagonalizable tells us that GW (F ) is generated by the elements <* *a> for a 2 F *, and we have already observed the relation = for any a, b 2 F* * *. As an easy exercise, one can also give a complete description for when two-dimensi* *onal forms are isomorphic: one must be able to pass from one to the other via the two relations (1.3) = and = where in the second we assume a, b 2 F *and a + b 6= 0. As an example, working over Q we have <3, -2> = <12, -2> = <10, -240> = <90, -15>. To completely determine all relations in GW (F ), one shows that if two forms and are isomorphic then there is a chain of isomorp* *hic diagonal forms connecting one to the other, where each link of the chain differ* *s in exactly two elements. Thus, (1.3) is a complete set of relations for GW (F ). T* *he reader may consult [S1, 2.9.4] for complete details here. The multiplication in GW (F ) can be described compactly by X . = . i,j 1.4. The Witt ring. The Witt ring W (F ) is the quotient of GW (F ) by the ideal generated by the so-called `hyperbolic plane' <1, -1>. Historically W (F ) was * *studied long before GW (F ), probably because it can be defined without formally adjoin* *ing additive inverses as was done for GW (F ). One can check that the forms and <1, -1> are isomorphic, and therefore if one regards hyperbolic forms as be* *ing zero then and <-a1, . .,.-an> are additive inverses. So W (F ) ca* *n be described as a set of equivalence classes of quadratic spaces, and doesn't requ* *ire working with `virtual' objects. Because ~=<1, -1> for any a, it follows that the ideal <1, -1> is p* *recisely the additive subgroup of GW (F ) generated by <1, -1>. As an abelian group, it * *is just a copy of Z. So we have the exact sequence 0 ! Z ! GW (F ) ! W (F ) ! 0. Let GI(F ) be the kernel of the dimension function dim :GW (F ) ! Z, usu- ally called the augmentation ideal. Let I(F ) be the image of the composite GI(F ) ,! GW (F ) i W (F ); one can check that I(F ) consists precisely of equi* *va- lence classes of even-dimensional quadratic spaces. Note that I is additively g* *en- erated by forms <1, a>, and therefore In is additively generated by n-fold prod* *ucts <1, a1><1, a2> . .<.1, an>. The dimension function gives an isomorphism W=I ! Z=2. The determinant gives us a group homomorphism GW (F ) ! F *=(F *)2, but it does not extend to 4 DANIEL DUGGER the Witt ring because det<1, -1> = -1. One defines the discriminant of n(n-1)_ to be (-1) 2 . (a1. .a.n), and with this definition the discriminant gives a* * map of sets W (F ) ! F *=(F *)2. It is not a homomorphism, but if we restrict to I(* *F ) ! F *=(F *)2 then it is a homomorphism. As the discriminant of <1, a><1, b> is a * *square, the elements of I2 all map to 1. So we get an induced map I=I2 ! F *=(F *)2, wh* *ich is obviously surjective. It is actually an isomorphism_to see this, note that <-1, y> = <-x, xy, -y, y2> = <1, -x, -y, xy> and so <1, xy> (mod I2). It follows inductively that <1, 1, . .,.1, a1a2. .a.2n> (mod I2). So if is a form whose dis* *crim- inant is a square, it is equivalent mod I2 to either <1, 1, . .,.1> = 2n<1> (if* * n is even) or <1, 1, . .,.1, -1> = (2n - 2)<1> (if n is odd). In the former ca* *se 2n<1> = 2<1> . n<1> 2 I2, and in the latter case (2n - 2)<1> = 2<1> . (n - 1)<1* *> 2 I2. In either case we have 2 I2, and this proves injectivity. The examples in the previous paragraph are very special, but they suggest why one might hope for `higher' invariants which give isomorphisms between the grou* *ps In=In+1 and something more explicitly defined in terms of the field F . This is* * what the Milnor conjecture is about. Remark 1.5. For future reference, note that 2<1> = <1, 1> 2 I, and therefore the groups In=In+1 are Z=2-vector spaces. Also observe that GI(F ) does not interse* *ct the kernel of GW (F ) ! W (F ), and so GI(F ) ! I(F ) is an isomorphism. It fol* *lows that (GI)n=(GI)n+1 ~=In=In+1, for all n. 1.6. More invariants. Recall that the Brauer group Br(F ) is a set of equiva- lence classes of central, simple F -algebras, with the group structure coming f* *rom tensor product. The inverse of such an algebra is its opposite algebra, where t* *he order of multiplication has been reversed. From a quadratic space (V, ~) one can construct the associated Clifford algeb* *ra C(~): this is the quotient of the tensor algebra TF (V ) by the relations gener* *ated by v v = ~(v, v). Clifford algebras are Z=2-graded by tensor length. If ~ is even-dimensional then C(~) is a central simple algebra, and if ~ is odd-dimensi* *onal then the even part C0(q) is a central simple algebra. So we get an invariant of quadratic spaces taking its values in Br(F ) (see [S1, 9.2.12] for more detail)* *. This is usually called the Clifford invariant, or sometimes the Witt invariant. Since any Clifford algebra is isomorphic to its opposite, the invariant always produc* *es a 2-torsion class. Now we need to recall some Galois cohomology. Let ~Fbe a separable closure of F , and let G = Gal(F~=F ). Consider the short exact sequence of G-modules 0 ! Z=2 ! ~F *! ~F *! 0, where the second map is squaring. Hilbert's Theorem 90 implies that H1(G; ~F)*= 0, which means that the induced long exact sequence in Galois cohomology splits up into 0 ! H0(G; Z=2) ! F *-2! F *! H1(G; Z=2) ! 0 and 0 ! H2(G; Z=2) ! H2(G; ~F)*-2!H2(G; ~F)*. The group H2(G; ~F)*is known to be isomorphic to Br(F ), so we have H0(G; Z=2) = Z=2, H1(G; Z=2) = F *=(F *)2, and the 2-torsion in the Brauer group is precisely H2(G; Z=2). From now on we will write H*(F ; Z=2) = H*(G; Z=2). NOTES ON THE MILNOR CONJECTURES 5 At this point we have the rank map e0: W (F ) ! Z=2 = H0(F ; Z=2), which gives an isomorphism W=I ! Z=2. We have the discriminant e1: I(F ) ! F *=(F *)2 = H1(F ; Z=2) which gives an isomorphism I=I2 ! F *=(F *)2, and we have the Cliff* *ord invariant e2: I2 ! H2(F ; Z=2). With a little work one can check that e2 is a homomorphism, and it kills I3. The question of whether I2=I3 ! H2(F ; Z=2) is an isomorphism is difficult, and wasn't proven until the early 80s by Merkurjev [M ] (neither surjectivity nor injectivity is obvious). The maps e0, e1, e2 are* * usually called the classical invariants of quadratic forms. The above isomorphisms can be rephrased as follows. The ideal I consists of a* *ll elements where e0 = 0; I2 consists of all elements such that e0 = 0 and e1 = 1;* * and by Merkujev's theorem I3 is precisely the set of elements for which e0, e1, and* * e2 are all trivial. Quadratic forms will be completely classified by these invaria* *nts if I3 = 0, but unfortunately this is usually not the case. This brings us to the s* *earch for higher invariants. One early result along these lines is due to Delzant [De* * ], who defined Stiefel-Whitney invariants with values in Galois cohomology. Unfortunat* *ely these are not the `right' invariants, as they do not lead to complete classific* *ations for elements in In, n 3. 1.7. Milnor's work. At this point we find ourselves looking at the two rings GrIW (F ) and H*(F ; Z=2), and we have maps between them in dimensions 0, 1, and 2. I think Milnor, inspired by his work on algebraic K-theory, wrote down t* *he best ring he could find which would map to both rings above. In [Mi ] he defined what is now called `Milnor K-theory' as KM*(F ) = TZ(F *)= where TZ(V ) denotes the tensor algebra over Z on the abelian group V . The gra* *ding comes from the grading on the tensor algebra, in terms of word length. I will w* *rite {a1, . .,.an} for the element a1 . . .an 2 KMn(F ). Note that when dealing with KM*(F ) one must be careful not to confuse the addition_which comes from multiplication in F *_with the multiplication. So for instance {a}+{b} = {ab} but {a}.{b} = {a, b}. This is in contrast to the operat* *ions in GW (F ), where one has + = and = . Unfortunately it is very easy to get these confused. Note that {a2} = 2{a}, and more generally {a2, b1, . .,.bn} = 2{a, b1, . .,.bn}. Remark 1.8. From a modern perspective the name `K-theory' applied to KM*(F ) is somewhat of a misnomer; one should not take it too seriously. The constructi* *on turns out to be more closely tied to algebraic cycles than to algebraic K-theor* *y, and so I personally like the term `Milnor cycle groups'. I doubt this will ever* * catch on, however. Milnor produced two ring homomorphisms j :KM*(F )=2 ! H*(F ; Z=2) and :KM*(F )=2 ! GrIW (F ). To define the map , note first that we have already established an isomorphism F *=(F *)2 ! I=I2 sending {a} to = - <1> (this is the inverse of the discriminant). This tells us what does to element* *s in degree 1. Since these elements generate KM*(F ) multiplicatively, to construct * * it suffices to verify that the appropriate relations are satisfied in the image. * *So we firstineed tojcheckithatj 0 = -<1> . <1-a>-<1> = --<1-a>+<1> = -, 6 DANIEL DUGGER but this follows directly from the second relation in (1.3). We also must check* * that 2{a} maps to 0, but 2{a} = {a2} 7! - <1> and the latter vanishes by the fi* *rst relation in (1.3). For future reference, note that ({a}) is equal to both and <-a, 1> in I=I2, since this group is 2-torsion. Defining j is similar. We have already noticed that there is a natural iso- morphism H1(F ; Z=2) ~= F *=(F *)2, and so it is clear where the element {a} in KM1(F ) = F *must be sent. The verification that a [ (1 - a) = 0 in H2(F ; Z=2)* * is in [Mi , 6.1]. Milnor observed that both j and were isomorphisms in all the cases he could compute. The claim that j is an isomorphism is nowadays known as the Milnor conjecture, and was proven by Voevodsky in 1996 [V1 ]. The claim that is an isomorphism goes under the name Milnor's conjecture on quadratic forms. For characteristic 0 it was proven in 1996 by Orlov, Vishik, and Voevodsky [OVV ],* * who deduced it as a consequence of the work in [V1 ]. I believe the proof now works in characteristic p, based on the improved results of [V3 ]. A second proof, a* *lso in characteristic 0, was outlined by Morel [M2 ] using the motivic Adams spectr* *al sequence, and again depended on results from [V1 ]; unfortunately complete deta* *ils of Morel's proof have yet to appear. It is interesting that the conjecture on quadratic forms doesn't have an inde- pendent proof, and is the less primary of the two. Note that both KM*(F )=2 and GW (F ) can be completely described in terms of generators and relations (altho* *ugh the latter does not quite imply that we know all the relations in GrIW (F ), wh* *ich is largely the problem). The map is easily seen to be surjective, and so the * *only question is injectivity. Given this, it is in some ways surprising that the con* *jecture is as hard as it is. Remark 1.9. The map j is called the norm residue symbol, and can be de- fined for primes other than 2. The Bloch-Kato conjecture is the statement that j :KMi(F )=l ! Hi(F ; ~li) is an isomorphism for l a prime different from char(* *F ). This is a direct generalization of the Milnor conjecture to the case of odd pri* *mes. A proof was released by Voevodsky in 2003 [V5 ] (although certain auxiliary resul* *ts re- quired for the proof remain unwritten). I'm not sure anyone has ever considered* * an odd-primary analog of Milnor's conjecture on quadratic forms_what could replace the Grothendieck-Witt ring here? At this point it might be useful to think through the Milnor conjectures in a* * few concrete examples. For these we refer the reader to Appendix A. Let's at least * *note here that through the work of Milnor, Bass, and Tate (cf. [Mi ]) the conjectur* *es could be verified for all finite fields and for all finite extensions of Q (in * *fact for all global fields). Finally, let's briefly return to the classification of forms over R. We saw earlier that this reduces to proving that the n-dimensional forms <1, 1, . .,.1> and <-1, -1, . .,.-1> are not isomorphic. Can we now do this algebraically? If they were isomorphic, they would represent the same element of W (R). It would follow that (2n)<1> = 0 in W (R). Can this happen? The isomorphisms Z=2[a] ~=H*(Z=2; Z=2) ~=KM*(R)=2 ~=GrI W (R) show that GrIW (R) is a poly- nomial algebra on the class <-1, -1> (the generator a corresponds to the genera* *tor -1 of R*=(R*)2, and (-1) = <-1, -1>). It follows that 2k<1> = <-1, -1>k is a generator for the group Ik=Ik+1 ~= Z=2. If m = 2ir where r is odd, then m<1> = 2i<1> . r<1>. Since r<1> is the generator for W=I and 2i<1> is a generat* *or NOTES ON THE MILNOR CONJECTURES 7 for Ii=Ii+1, it follows that m<1> is also a generator for Ii=Ii+1. In particula* *r, m<1> is nonzero. So we have proven via algebraic methods (although in this case also somewhat pathological ones) that <1, 1, . .,.1> 6~=<-1, -1, . .,.-1>. 1.10. Further background reading. There are several good expository papers on the theory of quadratic forms, for example [Pf1] and [S2]. The book [S1] is * *a very thorough and readable resource as well. For the Milnor conjectures themselves t* *here is [Pf2], which in particular gives several applications of the conjectures; it* * also gives detailed references to original papers. The beginning sections of [AEJ ] offer * *a nice survey concerning the search for `higher' invariants of quadratic forms. It's w* *orth pointing out that after Milnor's work definitions of e3, e4, and e5 were eventu* *ally given_with a lot of hard work_but this was the state of the art until 1996. Fin* *ally, the introduction of [V3 ] gives a history of work on the Milnor conjecture. 8 DANIEL DUGGER 2.Proof of the conjecture on the norm residue symbol This section outlines Voevodsky's proof of the Milnor conjecture on the norm residue symbol [V1 , V3]. Detailed, step-by-step summaries have been given in [* *M1 ] and [Su]. My intention here is not to give a complete, mathematically rigorous presentation, but rather just to give the flavor of what is involved. Several steps in the proof involve manipulations with motivic cohomology based on techniques that were developed in [VSF ]. I have avoided giving any details * *about these steps, in an attempt to help the exposition. Most of these details are no* *t hard to understand, however_there are only a few basic techniques to keep track of, * *and one can read about them in [VSF ] or [MVW ]. But I hope that by keeping some of this stuff in a black box the overall structure of the argument will become cle* *arer. 2.1. Initial observations. The aim is to show that j :KM*(F )=2 ! H*(F ; Z=2) is an isomorphism. To do this, one of the first things one might try to figure * *out is what kind of extra structure KM*(F )=2 and H*(F ; Z=2) have in common. For instance, they are both covariant functors in F , and the covariance is compati* *ble with the norm residue symbol. It turns out they both have transfer maps for finite separable extensions (which, for those who like to think geometrically, * *are the analogs of covering spaces). That is, if j :F ,! F 0is a separable extensio* *n of degree n then there is a map j!:KM*(F 0) ! KM*(F ) such that j!j* is multiplica* *tion by n, and similarly for H*(F ; Z=2). (Note that the construction of transfer ma* *ps for Milnor K-theory is not at all trivial_some ideas were given in [BT , Sec. 5* *.9], but the full construction is due to Kato [Ka , Sec. 1.7]). It follows that if n* * is odd then KM*(F )=2 ! H*(F ; Z=2) is a retract of the map KM*(F 0)=2 ! H*(F 0; Z=2). So if one had a counterexample to the Milnor conjecture, field extensions of al* *l odd degrees would still be counterexamples. This is often referred to as "the trans* *fer argument". Another observation is that both functors can be extended to rings other than fields, and if R is a discrete valuation ring then both functors have a `locali* *zation sequence' relating their values on R, the residue field, and the quotient field* *. I will not go into details here, but if F is a field of characteristic p then by using* * the Witt vectors over F and the corresponding localization sequence, one can reduce the Milnor conjecture to the case of characteristic 0 fields. The argument is i* *n [V1 , Lemma 5.2]. In Voevodsky's updated proof of the Milnor conjecture [V3 ] this st* *ep is not necessary, but I think it's useful to realize that the Milnor conjecture* * is not hard because of `crazy' things that might happen in characteristic p_it is hard even in characteristic 0. 2.2. A first look at the proof. The proof goes by induction. We assume the norm residue map j :KM*(F )=2 ! H*(F ; Z=2) is an isomorphism for all fields F and all * < n, and then prove it is also an isomorphism for * = n. The basic th* *eme of the proof, which goes back to Merkurjev, involves two steps: (1)Verify that jn is an isomorphism for certain `big' fields_in our case, those which have no extensions of odd degree and also satisfy Kn(F ) = 2Kn(F ) (so that one must prove Hn(F ; Z=2) = 0). Notice that when n = 1 the condition K1 = 2K1 says that F = F 2. (2)Prove that if F were a field for which jn is not an isomorphism then one cou* *ld expand F to make a `bigger' counterexample, and could keep doing this until NOTES ON THE MILNOR CONJECTURES 9 you're in the range covered by step (1). This would show that no such F could exist. In more detail one shows that for any {a1, . .,.an} 2 Kn(F ) one can con- struct an extension F ,! F 0with the property that {a1, . .,.an} 2 2Kn(F 0) and jn :Kn(F 0)=2 ! Hn(F 0; Z=2) still fails to be an isomorphism. By doing this over and over and taking a big colimit, one gets a counterexample where KMn = 2KMn. Neither of the above two steps is trivial, but step (1) involves nothing very fancy_it is a calculation in Galois cohomology which takes a few pages, but is * *not especially hard. See [V3 , Section 5]. Step (2) is the more subtle and intere* *sting step. Note that if a_= {a1, . .,.an} =22KMn(F ) then none of the ai's can be in* * F 2. There are several ways one can extend F to a field F 0such that a_2 2KMn(F 0): * *one can adjoin a square root of any ai, for instance. The problem is to find such an extension where you have enough control over the horizontal maps in the diagram KMn(F )=2______//KMn(F 0)=2 jF || |jF0| fflffl| fflffl| Hn(F ; Z=2)___//_Hn(F 0; Z=2) to show that if jF fails to be an isomorphism then so does jF0. The selection of the `right' F 0is delicate. We will alter our language at this point, because we will want to bring more geometry into the picture. Any finitely-generated separable extension F ,! F 0 is the function field of a smooth F -variety. A splitting variety for an eleme* *nt a_2 KMn(F ) is a smooth variety X, of finite type over F , with the property th* *at a_2 2KMn(F (X)). Here F (X) denotes the function field of X. As we just re- marked, there are many such varieties: X = SpecF [u]=(u2 - a1) is an example. The particular choice we'll be interested in is more complicated. Given b1, . .,.bk 2 F , let qb_be the quadratic form in 2k variables correspo* *nding to the element <1, -b1> <1, -b2> . . .<1, -bk> 2 GW (F ). For example, qb1,b2(x1, . .,.x4) = x21- b1x22- b2x23+ b1b2x24. Such q's are ca* *lled Pfister forms, and they have a central role in the modern theory of quadratic forms (see [S1, Chapter 4], for instance). For a1, . .,.an 2 F , define Qa_to be the projective quadric in P2n-1given by* * the equation qa1,...,an-1(x0, . .,.x[2n-1-1]) - anx22n-1= 0. In [V3 ] these are called norm quadrics. A routine argument [V3 , Prop. 4.1] sh* *ows that Qa_is a splitting variety for a_. The reason for choosing to study this pa* *rticular splitting variety will not be clear until later; isolating this object is one o* *f the key aspects of the proof. The name of the game will be to understand enough about the difference be- tween KMn(F )=2 and KMn(F (Qa_))=2 (as well as the corresponding Galois cohomol- ogy groups) to show that KMn(F (Qa_))=2 ! Hn(F (Qa_); Z=2) still fails to be an isomorphism. Voevodsky's argument uses motivic cohomology_of the quadrics Qa_ and other objects_to `bridge the gap' between KMn(F )=2 and KMn(F (Qa_))=2. 10 DANIEL DUGGER 2.3. Motivic cohomology enters the picture. Motivic cohomology is a bi- graded functor X 7! Hp,q(X; Z) defined on the category of smooth F -schemes. Actually it is defined for all simplicial smooth schemes, as well as for more g* *eneral objects. One of the lessons of the last ten years is that one can set up a mod* *el category which contains all these objects, and then a homotopy theorist can deal with them in much the same ways he deals with ordinary topological spaces. From now on I will do this implicitly (without ever referring to the machinery invol* *ved). The coefficient groups Hp,q(SpecF ; Z) vanish for q < 0 and for p > q 0. Fo* *r us an important point is that the groups Hn,n(SpecF ; Z) are canonically isomorphic to KMn(F ). Proving this is not simple! An account is given in [MVW , Lecture * *5]. Finally, we note that one can talk about motivic cohomology with finite coeffic* *ients Hp,q(X; Z=n), related to integral cohomology via the exact sequence . .!.Hp,q(X; Z) xn-!Hp,q(X; Z) ! Hp,q(X; Z=n) ! Hp+1,q(X; Z) ! . . . The sequence shows Hn,n(SpecF ; Z=2) ~=KMn(F )=2 and Hp,q(SpecF ; Z=2) = 0 for p > q 0. Now, there is also an analagous theory Hp,qL(X; Z) which is called Lichten- baum (or 'etale) motivic cohomology. There is a natural transformation Hp,q(X; Z) ! Hp,qL(X; Z). The theory H*,*Lis the closest theory to H*,*which satisfies descent for the 'etale topology (essentially meaning that when E ! B * *is an 'etale map there is a spectral sequence starting with H*,*L(E) and converging to H*,*L(B)). The relation between H*,*and H*,*Lis formally analagous to that between a cohomology theory and a certain Bousfield localization of it. It is k* *nown that Hp,qL(X; Z=n) is canonically isomorphic to 'etale cohomology Hpet(X; ~nq),* * if n is prime to char(F ). From this it follows that Hp,qL(SpecF ; Z=2) is the Ga* *lois cohomology group Hp(F ; Z=2), for all q. At this point we can re-phrase the Mil* *nor conjecture as the statement that the maps Hp,p(SpecF ; Z=2) ! Hp,pL(SpecF ; Z=2) are isomorphisms. There are other conjectures about the relation between H*,*and H*,*Las well. A conjecture of Lichtenbaum says that Hp,q(X; Z) ! Hp,qL(X; Z) should be an isomorphism whenever p q + 1. Note that this would imply a corresponding statement for Z=n-coefficients, and in particular would imply the Milnor conjec* *ture. Also, since one knows Hn+1,n(SpecF ; Z) = 0 Lichtenbaum's conjecture would im- ply that Hn+1,nL(SpecF ; Z) also vanishes. This latter statement was conjectur* *ed independently by both Beilinson and Lichtenbaum, and is known as a the Gener- alized Hilbert's Theorem 90 (the case n = 1 is a translation of the statement that H1Gal(F ; ~F)*= 0, which follows from the classical Hilbert's Theorem 90). By knowing enough about how to work with motivic cohomology, Voevodsky was able to prove the following relation among these conjectures (as well as ot* *her relations which we won't need): Proposition 2.4. Fix an n 0. Assume that Hk+1,kL(SpecF ; Z(2)) = 0 for all fields F and all 0 k n. Then for any smooth simplicial scheme X over a field F , the maps Hp,q(X; Z=2) ! Hp,qL(X; Z=2) are isomorphisms when q 0 and p q n; and they are monomorphisms for p - 1 = q n. In particular, applying this when p = q and X = SpecF verifies the Milnor conjecture in dimensions n. It's worth pointing out that the proof uses nothing special about the prime 2, and so the statement is valid for all other primes as well. NOTES ON THE MILNOR CONJECTURES 11 For us, the importance of the above proposition is two-fold. First, it says t* *hat to prove the Milnor conjecture one only has to worry about the vanishing of one se* *t of groups (the Hn+1,nL's) rather than two sets (the kernel and cokernel of j). Sec* *ondly, inductively assuming that the Generalized Hilbert's Theorem 90 holds up through dimension n is going to give us a lot more to work with than inductively assumi* *ng the Milnor conjecture up through dimension n. Instead of just knowing stuff abo* *ut Hn,n of fields, we know stuff about Hp,qof any smooth simplicial scheme. The need for this extra information is a key feature of the proof. 2.5. ~Cech complexes. We only need one more piece of machinery before returning to the proof of the Milnor conjecture. This piece is hard to motivate, and its introduction is one of the more ingenious aspects of the proof. The reader will* * just have to wait and see how it arises in section 2.6 (see also Remark 3.10). Let X be any scheme. The ~Cech complex ~CX is the simplicial scheme with (C~X)n = X x X x . .x.X (n + 1 factors) and the obvious face and degeneracies. This simplicial scheme can be regarded as augmented by the map X ! SpecF . For a topological space the realization of the associated ~Cech complex is al* *ways contractible_in fact, choosing any point of X allows one to write down a contra* *ct- ing homotopy for the simplicial space ~CX. But in algebraic geometry the scheme X may not have rational points; i.e., there may not exist any maps SpecF ! X at all! If X does have a rational point then the same trick lets one write down* * a contracting homotopy, and therefore ~CX behaves as if it were SpecF in all com- putations. (More formally, ~CX is homotopy equivalent to SpecF in the motivic homotopy category). Working in the motivic homotopy category, one finds that for any smooth scheme Y the set of homotopy classes [Y, ~CX] is either empty or a singleton. The latt* *er holds precisely if Y admits a Zariski cover {Uff} such that there exist scheme * *maps Uff! X (not necessarily compatible on the intersections). The object C~X has no `higher homotopy information', only this very simple discrete information ab* *out whether or not certain maps exist. One should think of ~CX as very close to bei* *ng contractible. I point out again that in topology there is always at least one m* *ap between nonempty spaces, and so ~CX is not very interesting. If E ! B is an 'etale cover, then there is a spectral sequence whose input is H*,*L(E; Z) and which converges to H*,*L(B; Z) (this is the 'etale descent prop* *erty). In particular, if X is a smooth scheme and we let F 0= F (X), X0= X xF F 0, then X0 ! X and SpecF 0! SpecF are both 'etale covers. The scheme X0 necessarily has a rational point over F 0, so ~CX0 and SpecF 0look the same to HL. The 'eta* *le descent property then shows that C~X and SpecF also look the same: in other words, the maps Hp,qL(SpecF ; Z) ! Hp,qL(C~X; Z) are all isomorphisms (and the same for finite coefficients). This is not true for H*,*in place of H*,*L. One * *might paraphrase all this by saying that in the 'etale world ~CX is contractible, jus* *t as it is in topology. 2.6. The proof. Now I am going to give a complete summary of the proof as it appears in [V1 , V3 ]. Instead of proving the Milnor conjecture in its original form one instead concentrates on the more manageable conjecture that Hi+1,iL(SpecF ; Z(2)) = 0 for all i and all fields F . One assumes this has be* *en proven in the range 0 i < n, and then shows that it also follows for i = n. 12 DANIEL DUGGER Suppose that F is a field with Hn+1,nL(F ; Z(2)) 6= 0. The transfer argument * *shows that any extension field of odd degree would still be a counterexample, so we c* *an assume F has no extensions of odd degree. One checks via some Galois cohomology computations_see [V3 , section 5]_that if such a field has KMn(F ) = 2KMn(F ) then Hn+1,nL(SpecF ; Z(2)) = 0. So our counterexample cannot have KMn(F ) = 2KMn(F ). By the reasoning from section 2.2, it will suffice to show that for e* *very a1, . .,.an 2 F the field F (Qa_) is still a counterexample. We will in fact sh* *ow that Hn+1,nL(F ; Z(2)) ! Hn+1,nL(F (Qa_); Z(2)) is injective. Suppose u is in the kernel of the above map, and consider the diagram Hn+1,nL(SpecF ; Z(2))_//_Hn+1,nL(SpecF (Qa_); Z(2)) ~=|| fflffl| Hn+1,n(C~Qa_; Z(2))___//_Hn+1,nL(C~Qa_; Z(2)). Let u0 denote the image of u in Hn+1,nL(C~Qa_; Z(2)). One can show (after some extensive manipulations with motivic cohomology) that the hypothesis on u impli* *es that u0 is the image of an element in Hn+1,n(C~Qa_; Z(2)). It will therefore be sufficient to show that this group is zero. Let C~ be defined by the cofiber sequence (C~Qa_)+ ! (SpecF )+ ! C~. This means ~H*,*(C~) fits in an exact sequence ! Hp-1,q(C~Qa_) ! ~Hp,q(C~) ! Hp,q(SpecF ) ! Hp,q(C~Qa_) ! ~Hp+1,q(C~) ! . . . So the reduced motivic cohomology of ~Cdetects the `difference' between the mo- tivic cohomology of C~Qa_and SpecF . The fact that Hi,n(SpecF ; Z) = 0 for i > n shows that Hn+1,n(C~Qa_; Z(2)) ~= H~n+2,n(C~; Z(2)). Since Qa_has a ra- tional point (and therefore C~Qa_is contractible) over a degree 2 extension of * *F , it follows from the transfer argument that the above group is killed by 2. To show that the group is zero it is therefore sufficient to prove that the image * *of ~Hn+2,n(C~; Z(2)) ! H~n+2,n(C~; Z=2) is zero. This is the same as the image of ~Hn+2,n(C~; Z) ! ~Hn+2,n(C~; Z=2), which I'll denote by ~Hn+2,nint(C~; Z=2). So far most of what we have done is formal; but now we come to the crux of the argument. For any smooth scheme X one has cohomology operations acting on H*,*(X; Z=2). In particular, one can produce analogs of the Steenrod operations: the Bockstein acts with bi-degree (1, 0), and Sq2i acts with bi-degree (2i, 2i-* *1). From these one defines the Milnor Qi's, which have bi-degree (2i+1- 1, 2i- 1). * *In ordinary topology these are defined inductively by Q0 = fi and Qi = [Qi-1, Sq2i* *], whereas motivically one has to add some extra terms to this commutator (these arise because the motivic cohomology of a point is nontrivial). One shows that QiO Qi = 0, and that Qi = fiq + qfi for a certain operation q. It follows from the latter formula that Qi maps elements in ~Hintto elements in ~Hint. All of t* *hese facts also work in ordinary topology, it's just that the proofs here are a litt* *le more complex. The next result is [V3 , Cor. 3.8]. It is the first of two main ingredients n* *eeded to complete the proof. NOTES ON THE MILNOR CONJECTURES 13 Proposition 2.7. Let X be a smooth quadric in P2n, and let C~X be defined by the cofiber sequence (C~X)+ ! (SpecF )+ ! ~CX. Then for i n, every element of ~H*,*(C~X; Z=2) that is killed by Qi is also in the image of Qi. This is a purely `topological' result, in that its proof uses no algebraic ge* *ometry. It follows from the most basic properties of the Steenrod operations, motivic c* *o- homology (like Thom isomorphism), and elementary facts about the characteristic numbers of quadrics. The argument is purely homotopy-theoretic. The second main result we will need is where all the algebraic geometry enters the picture. Voevodsky deduces it from results of Rost, who showed that the mot* *ive of Qa_splits off a certain direct summand. See [V3 , Th. 4.9]. Proposition 2.8. H~2n,2n-1(C~; Z(2)) = 0. Using the above two propositions we can complete the proof of the Milnor con- jecture. In order to draw a concrete picture, let us just assume n = 4 for the moment. We are trying to show that ~H6,4int(C~; Z=2) = 0. Consider the diagram Hp,q(SpecF ; Z=2)____//Hp,q(C~Qa_; Z=2) | | | | fflffl| ~= fflffl| Hp,qL(SpecF ; Z=2)__//_Hp,qL(C~Qa_; Z=2). Our inductive assumption together with Proposition 2.4 implies that the vertical maps are isomorphisms for p q n-1, and monomorphisms for p-1 = q n-1. So the top horizontal map is an isomorphism in the first range and a monomorphi* *sm in the second. The long exact sequence in motivic cohomology then shows that ~Hp,q(C~; Z=2) = 0 for p q n - 1. This is where our induction hypothesis has gotten us. The following diagram depicts what we now know about ~Hp,q(C~; Z=2) (the group marked ?? is ~H6,4, the one we care about): | ||| | | | | | | | | | | | | | | | | | | |||q| | | | | | | | | | | | | | | | | |___|___|__|___|___|___|___|___|___|__|___|___|___|___|___|__|___|___|_|| | ||6|| | | | | | | | | | | | | | | *| * | |___|___|__|___|___|___|___|___|___|__|___|___|___|___|___|__|___|___|_|| | ||| | | | | | | | | | | | | | | | | | | ||| | | | | | | | | | | | | | | | | | |___|___|__|___|___|___|___|___|___|__|___|___|___|___|___|__|___|___|_|| | ||| | | | | | | | | | | | | Q2| | | | | |___|___|__|___|___|___|___|___|___|__|___|___|___|___|___|__|___|___|_|| | ||| | | | | | | | | | | | | | | | | | | ||| | | | | | | | | 1| * | | | | | | | | |___|___|__|___|___|___|___|___|___|__|___|___|___|___|___|__|___|___|_||Q1* | ||| | | | | | | | | | | | | | | | | | | ||| | | | | Q1| 1|?? | | | | | | | | | | | |___|___|__|___|___|___|___|___|___|__|___|___|___|___|___|__|___|___|_|| | |||0|0 | 0 |0 | | |Q2 | | | | | | | | | | | |___|___|__|___|___|___|___|___|___|__|___|___|___|___|___|__|___|___|_|| | ||| | | | | | | | | | | | | | | | | | | |||0|0 | 0 | | | | | | | | | | | | | | | |___|___|__|___|___|___|___|___|___|__|___|___|___|___|___|__|___|___|_|| | ||| | | | | | | | | | | | | | | | | | | |||0|0 | | | | | | | | | | | | | | | | |___|___|__|___|___|___|___|___|___|__|___|___|___|___|___|__|___|___|_|| | |||0| | | | | | | | | | | | | | | | | |___|___|__|___|___|___|___|___|___|__|___|___|___|___|___|__|___|___|_||_-_oep | ||| | | | | | | | | | | | | | | | | | | ||| | | | | | | | | | | | | | | | | | |___|___|__|___|___|___|___|___|___|__|___|___|___|___|___|__|___|___|_||? 14 DANIEL DUGGER At this point Proposition 2.7 shows that Q1: H6,4! H9,5is injective, and that Q2: H9,5! H16,8is injective. Since the Qi's take integral elements to integral elements, we have an inclusion Q2Q1: ~H6,4int(C~; Z=2) ,! ~H16,8int(C~; Z=2). But it follows directly from Proposition 2.8 that ~H16,8int(C~; Z=2) = 0, and s* *o we are done. The argument for general n follows exactly this pattern: one uses the composi* *te of the operations Q1, Q2, . .,.Qn-2, but everything else is the same. 2.9. Summary. Here is a list of some of the key elements of the proof: (1)The re-interpretation of the Milnor conjecture as a comparison of different bi-graded motivic cohomology theories. An extensive knowledge about such theories allows one to deduce statements for any smooth simplicial scheme from statements only about fields (cf. Proposition 2.4). (2)Choice of the splitting variety Qa_(needed for Propositions 2.7 and 2.8). (3)The introduction and use of ~Cech complexes. (4)The construction of Steenrod operations on motivic cohomology and develop- ment of their basic properties, leading to the proof of Proposition 2.7. (5)The `geometric' results of Rost on motives of quadrics, which lead to Propos* *i- tion 2.8. 2.10. A notable consequence. The integral motivic cohomology groups of a point Hp,q(SpecF ) are largely unknown_the exception is when q = 0, 1. How- ever, the proof of the Milnor conjecture tells us exactly what Hp,q(SpecF ; Z=2) is. First of all, independently of the Milnor conjecture it can be shown to van* *ish when p q and when q < 0. By Proposition 2.4 (noting that we now know the hypothesis to be satisfied for all n), it follows that Hp,q(SpecF ; Z=2) ! Hpet(SpecF ; ~2q) is an isomorphism when p q and q 0. As ~2q ~=~2, the 'etale cohomology groups are periodic in q; that is, H*et(SpecF ; ~2*) ~=H*Gal(F ; Z=2)[ø, ø-1 ] * *where ø has degree (0, 1). The conclusion is that H*,*(SpecF ; Z=2) ~= H*Gal(F ; Z=2)[ø], where ø is the canonical class in H0,1and the Galois cohomology is regarded as the subalgebra lying in degrees (k, k). Of course the Milnor conjecture tells us that the Gal* *ois cohomology is the same as mod 2 Milnor K-theory, and so we can also write H*,*(SpecF ; Z=2) ~= KM*(F )=2 [ø]. 2.11. Further reading. Both the original papers of Voevodsky [V1 , V3] are very readable, and remain the best sources for the proof. Summaries have also been g* *iven in [M1 ] and [Su]. A proof of the general Bloch-Kato conjecture was recently gi* *ven in [V5 ]_the proof is similar in broad outline to the 2-primary case we describ* *ed here, but with several important differences. See the introduction to [V5 ]. Of course in this section I have completely avoided discussing the two main elements of the proof, namely Propositions 2.7 and 2.8. The proof of Propositio* *n 2.7 is in [V1 , V3] and is written in a way that can be understood by most homotopy theorists. Proposition 2.8 depends on results of Rost, which seem to be largely unpublished. See [R1 , R2] for summaries. For more about why ~Cech complexes arise in the proof, see Proposition 3.9 in the next section. NOTES ON THE MILNOR CONJECTURES 15 3.Proof of the conjecture on quadratic forms In this section and the next I will discuss two proofs of Milnor's conjecture* * on quadratic forms. The first is from [OVV ], the second was announced in [M2 ]. * *Both depend on Voevodsky's proof of the norm residue conjecture. As I keep saying, I'm only going to give a vague outline of how the proofs go, but with reference* *s for where to find more information on various aspects. The present section deals wi* *th the [OVV ] proof. 3.1. Preliminaries. Recall that we are concerned with the map :KM*(F )=2 ! GrIW (F ) defined by ({a1, . .,.an}) = <1, -a1> . .<.1, -an>. The fact that I* * is additively generated by the forms <1, x> shows that is obviously surjective; * *so our task is to prove injectivity. In general, the product <1, b1> . .<.1, bn> is c* *alled an n-fold Pfister form, and denoted <>. Note that it has dimension 2n. The proof is intimately tied up with the study of such forms. Milnor proved that the map :KM2(F )=2 ! I2=I3 is an isomorphism. He used ideas of Delzant [De ] to define Stiefel-Whitney invariants for quadratic forms* *, which in dimension 2 give a map I2=I3 ! KM2(F )=2. One could explicitly check that th* *is was an inverse to . Unfortunately, this last statement generally fails in lar* *ger dimensions; the Stiefel-Whitney invariants don't carry enough information. See [Mi , 4.1, 4.2]. 3.2. The Orlov-Vishik-Voevodsky proof. We first need to recall some results about Pfister forms proven in the 70's. The first is an easy corollary of the s* *o-called Main Theorem of Arason-Pfister (cf. [S1, 4.5.6]). For a proof, see [EL , pp. 19* *2-193]. Proposition 3.3 (Elman-Lam). <> <> (mod In+1) if and only if <> = <> in GW (F ). Combining the result for n = 2 with Milnor's theorem that KM2(F ) ! I2=I3 is an isomorphism, we get the following (note that the minus signs are there becau* *se ({a1, . .,.an}) = <<-a1, . .,.-an>>): Corollary 3.4. <> = <> in GW (F ) if and only if {-a1, -a2} = {-b1, -b2} in KM*(F )=2. Say that two n-fold Pfister forms A = <> and B = <>* * are simply-p-equivalent if there are two indices i, j where <> = <>* * and ak = bk for all k =2{i, j}. The forms A and B are chain-p-equivalent if there i* *s a chain of forms starting with A and ending with B in which every link of the cha* *in is a simple-p-equivalence. Note that it follows immediately from the previous coro* *llary that if A and B are chain-p-equivalent then {-a1, . .,.-an} = {-b1, . .,.-bn}. The following result is [EL , Main Theorem 3.2]: Proposition 3.5. Let A = <> and B = <>. The following are equivalent: (a)A and B are chain-p-equivalent. (b){-a1, . .,.-an} = {-b1, . .,.-bn} in KM*(F )=2. (c)A B (mod In+1). (d)A = B in GW (F ). Note that (a) ) (b) ) (c) is trivial, and (c) ) (d) was mentioned above. So t* *he new content is in (d) ) (a). I will not give the proof, but refer the reader to* * [S1, 4.1.2]. The result below is a restatement of (c) ) (b): 16 DANIEL DUGGER Corollary 3.6. The equality ({a1, . .,.an}) = ({b1, . .,.bn}) can only occur * *if {a1, . .,.an} = {b1, . .,.bn}. Unfortunately the above corollary does not show that is injective, as a typ* *ical element x 2 KM*(F )=2 is a sum of terms {a1, . .,.an}. A term {a1, . .,.an} is called a pure symbol, whereas a general x 2 KM*(F ) is just a symbol. The key ingredient needed from [OVV ] is the following: Proposition 3.7. If x 2 KM*(F )=2 is a nonzero element then there is a field extension F ,! F 0such that the image of x in KM*(F 0)=2 is a nonzero pure symb* *ol. It is easy to see that the previous two results prove the injectivity of . * *If x 2 KMn(F )=2 is a nonzero element in the kernel of , then by passing to F 0we find a nonzero pure symbol which is also in the kernel. Corollary 3.6 shows thi* *s to be impossible, however. We are therefore reduced to proving Proposition 3.7. If we write x = a_1+. .+* *.a_k, where each a_iis a pure symbol, then we know we can make a_ivanish by passing to the function field F (Qa_i) (where Qa_iis the splitting variety produced in the* * last section). Our goal will be to show that a_iis the only term that vanishes: Proposition 3.8 (Orlov-Vishik-Voevodsky). If a_= {a1, . .,.an} is nonzero in KMn(F )=2, then the kernel of KMn(F )=2 ! KMn(F (Qa_))=2 is precisely Z=2 (gene* *r- ated by a_). Granting this for the moment, let i be the largest index for which x is nonze* *ro in KMn(F 0)=2, where F 0= F (Qa_1x . .x.Qa_i). Since x will become zero over F 0(Qa_i+1), the above result says that x = a_i+1in KMn(F 0)=2. This is precis* *ely what we wanted. So finally we have reduced to the same kind of problem we tackled in the last section, namely controlling the map KMn(F )=2 ! KMn(F (Qa_))=2. The techniques needed to prove Proposition 3.8 are exactly the same as those from the last sec* *tion. There is a again a homotopical ingredient and a geometric ingredient. Proposition 3.9. If X is a smooth scheme over F , then for every n 0 there is an exact sequence of the form 0 ! Hn,n-1(C~X; Z=2) ! Hn,n(SpecF ; Z=2) ! Hn,n(SpecF (X); Z=2). Recall that Hn,n(SpecE; Z=2) ~=KMn(E)=2 for any field E. So the above se- quence is giving us control over the kernel of KM*(F )=2 ! KM*(F (Qa_))=2. The proof uses the conclusion from Proposition 2.4 (which is known by Voevodsky's proof of the Milnor conjecture) and some standard manipulations with motivic cohomology. See [OVV , Prop. 2.3]. Remark 3.10. In some sense Proposition 3.9 explains why ~Cech complexes are destined to come up in the proofs of these conjectures. If the above proposition is thought of as a `homotopical' part of the proof, * *the geometric part is the following. It is deduced using Rost's results on the moti* *ve of Qa_; see [OVV , Prop. 2.5]. Proposition 3.11. There is a surjection Z=2 ! H2n-1,2n-1-1(C~Qa_; Z=2). The previous two results immediately yield a proof of 3.8. By Proposition 3.9* * we must show that Hn,n-1(C~Qa_; Z=2) ~=Z=2 (and we know the group is nontrivial). NOTES ON THE MILNOR CONJECTURES 17 But we saw in the last section that Hn,n-1(C~Qa_; Z=2) ~= Hn+1,n-1(C~Qa_; Z=2), where C~Qa_is the homotopy cofiber of (C~Qa_)+ ! (SpecF )+ . We also saw that the operation Qn-2 . .Q.2Q1 gives a monomorphism Hn+1,n-1(C~Qa_; Z=2) ,! H2n-1,2n-1-1(C~Qa_; Z=2). But now we are done, since by 3.11 the latter group h* *as at most two elements. This completes the proof of the injectivity of . 18 DANIEL DUGGER 4.Quadratic forms and the Adams spectral sequence In [M2 ] Morel announced a proof of the quadratic form conjecture over charac* *ter- istic zero fields, using the motivic Adams spectral sequence. The approach depe* *nds on having computed the motivic Steenrod algebra, but I'm not sure what the stat* *us of this is_certainly no written account is presently available. Despite this fr* *ustrat- ing point, Morel's proof is very exciting; while it uses Voevodsky's computation of H*,*(SpecF ; Z=2)_see Remark 2.10_it somehow avoids using any other deep results about quadratic forms! So I'd like to attempt a sketch. The arguments below take place in the motivic stable homotopy category. All the reader needs to know as background is that it formally behaves much as the usual stable homotopy category, and that there is a bigraded family of spheres * *Sp,q. The suspension (in the triangulated category sense) of Sp,qis Sp+1,q, and S2,1is the suspension spectrum of the variety P1. 4.1. Outline. We have our maps n :KMn(F )=2 ! In=In+1, and need to prove that they are injective. We will see that the Adams spectral sequence machinery gives us, more or less for free, maps sn :In=In+1 ! KMn(F )=(2, J) where J is a subgroup of boundaries from the spectral sequence. The composite sn n is the natural projection, and so the whole game is to show that J is zero. That is, one needs to prove the vanishing of a line of differentials. Using the multipli* *cative structure of the spectral sequence and the algebra of the E2-term, this reduces* * just to proving that the differentials on a certain `generic' element vanish. This a* *llows one to reduce to the case of the prime field Q, then to R, and ultimately to a * *purely topological problem. 4.2. Basic setup. Now I'll expand on this general outline. The first step is to produce a map q :GW (F ) ! {S0,0, S0,0} where {-, -} denotes maps in the motivic stable homotopy category. Recall from Section 1.2 that one knows a complete description of GW (F ) in terms of generators and relations. For a 2 F *we let q() be the map P1 ! P1 defined in homogeneous coordinates by [x, y] ! [x, ay* *]. By writing down explicit A1-homotopies one can verify that the relations in GW * *(F ) are satisfied in {S0,0, S0,0}, and so q extends to a well-defined map of abelia* *n groups. It is actually a ring map. Further details about all this are given in [M3 ]. Now we build an Adams tower for S0,0based on the motivic cohomology spec- trum HZ=2. Set W0 = S0,0, and define W1 by the homotopy fiber sequence W1 ! S0,0! HZ=2. Then consider the map W1 ~=S0,0^ W1 ! HZ=2 ^ W1, and let W2 be the homotopy fiber. Repeat the process to define W3, W4, etc. This gives us a tower of cofibrations H ^OW2O H ^OW1O H ^ W0OO | | | | | | | | | . ._.____//_W2________//W1_______//_W0, where we have written H for HZ=2. For any Y the tower yields a filtration on {Y, S0,0} by letting Fn be the subgroup of all elements in the image of {Y, Wn} (note that there is no a priori guarantee that the filtration is Hausdorff.) Th* *e tower yields a homotopy spectral sequence whose abutment has something to do with the associated graded of the groups {S*,0^ Y, S0,0}. If the filtration is not Hausd* *orff NOTES ON THE MILNOR CONJECTURES 19 these associated graded groups may not be telling us much about {S*,0^ Y, S0,0}, but this will not matter for our application. We will be interested in the case Y = S0,0. Set Ea,b1= {Sa,0, H ^ Wb}, so that dr: Ea,br! Ea-1,b+rr. My indexing has been chosen so that the picture of the spectral sequence has Ea,b1in spot (a, b* *) on a grid, rather than at spot (b - a, a) as is more typical for the Adams spectral sequence_but the picture itself is the same in the end. Formal considerations g* *ive inclusions Fk{Sn,0, S0,0}=Fk+1{Sn,0, S0,0} ,! En,k1 (however, there is no a priori reason to believe the map is surjective). In par* *ticular, if F* is the filtration on {S0,0, S0,0} then we have inclusions Fk=Fk+1 ,! E0,k* *1. Let GI(F ) be the kernel of the mod 2 dimension function dim:GW (F ) ! Z=2. The powers GI(F )n define a filtration on GW (F ). One can check that q maps GI1 into F1. Since the Adams filtration Fn on ß0,0(S0,0) will be multiplicative, on* *e finds that q maps GIn into Fn. So we get maps (GI)n=(GI)n+1 ! Fn=Fn+1 ! E0,n1. In a moment I'll say more about what the Adams spectral sequence looks like in this case, but first let's relate GI to what we really care about. One easi* *ly checks that GI = GI Z, where the Z is the subgroup generated by <1, 1> = 2<1>. So GIn = GIn Z, where the Z is generated by 2n<1>. It follows that GIn=GIn+1 ~=[GIn=GIn+1] Z=2. Finally, recall from Remark 1.5 that the natural map GI ! I is an isomorphism. Putting everything together, we have produced invariants [In=In+1] Z=2 ! E0,n1. 4.3. Analysis of the spectral sequence. So far the discussion has been mostly formal. We have produced a spectral sequence, but not said anything concrete about it. The usefulness of the above invariants hinges on what E0,n1looks like* *. If things work as in ordinary topology, then the E2 term will turn out to be Ea,b2= ExtbH**H( b+a,0H**, H**) where I've again written H = HZ=2 and k,0denotes a grading shift on the bi-graded module H**. So we need to know the algebra H**H, but unfortunately there is no published source for this calculation. In [* *V2 ] Voevodsky defines Steenrod operations and shows that they satisfy analogs of the usual Adem relations; he doesn't show that these generate all of H**H, though. However, let's assume we knew this_so we are assuming H**H is the algebra Voevodsky denotes A** and calls the motivic Steenrod algebra [V2 , Section 11]. The form of H**H is very close to that of the usual Steenrod algebra, and so one has a chance at doing some of the Ext computations. In fact, it is not very hard. Some hints about this are given in Appendix B, but for now let me just te* *ll you the important points: (1)Ep,q2= 0 if p < 0. (2)E0,02= Z=2. (3)For n 1, E0,n2= Hn,n Z=2. The inclusion nHn,n ,! nE0,n2is a ring homomorphism, where the domain is regarded as a subring of H**. Most of these computations make essential use of Remark 2.10, and therefore de- pend on Voevodsky's proof of the norm residue conjecture. Also note the connect* *ion between (3) and Milnor K-theory, given by the isomorphism Hn,n~=KMn(F )=2. The above two facts show that everything in E0,n2is a permanent cycle and thus E0,n1= (Z=2 KMn(F )=2)=J where J is the subgroup of all boundaries. Recall that 20 DANIEL DUGGER one has maps KMn(F )=2 -n!In=In+1 ! E0,n1~=[KMn(F )=2 Z=2]=J. The composition can be checked to be the obvious one. To prove that n is in- jective, we need to prove that J = 0. That is, we need to prove the vanishing of all differentials landing in E0,*(which necessarily come from E1,*). As for * *the computation of the E1,*column, here are the additional facts we need: (4)E1,02= 0. (5)E1,12= H0,1 H2,2~=Z=2 H2,2. (6)The images of the two maps E0,12 E1,n-12! E1,n2 E1,n-12 E0,12! E1,n2 generate E1,n2as an abelian group. (7)The composite H1,1 H2,2,! E0,12 E1,12! E1,22is zero. Again, let me say that none of these computations is particularly difficult, * *and the reader can find some hints in Appendix B. Portions of columns 0 and 1 of our E2-term are shown below: | |____________|_____________|____________|_____________|| ||6 | | | | ||4,4 | | | | |H| Z=2 | ?? | | | |____________|_____________|____________|_____________|| || | | | | ||3,3 | | | | |H| Z=2 | ?? | | | |____________|_____________|____________|_____________|| || | | | | ||2,2 | | | | |H| Z=2 | ?? | | | |____________|_____________|____________|_____________|| || | | | | ||1,1 | 2,2 | | | |H| Z=2 | H Z=2 | | | |____________|_____________|____________|_____________|| || | | | | || | | | | || Z=2 | 0 | | | |____________|_____________|____________|_____________|o|_- Remark 4.4. If one only looks at the Z=2's appearing in the above diagram, the picture looks just like the ordinary topological Adams spectral sequence. The Z* *=2's in our 0th column indeed turn out to be "hn0's", just as in topology. The Z=2 in E1,12is a little more complicated, though_it doesn't just come from Sq2, like t* *he usual h1 does (see Appendix B for what it does come from). We need to prove that all the differentials leaving the E1,*column vanish. By fact (6) and the multiplicative structure of the spectral sequence, it is suffi* *cient to prove that all differentials leaving E1,12vanish (starting with d2: E1,12! E0,3* *2). We will do this in several steps. The following result basically shows that, just as in ordinary topology, all * *the Z=2's in column 0 survive to E1 . Lemma 4.5. The image of dr: E1,1r! E0,r+1rlies in the subgroup Hr+1,r+1, for every r 2. NOTES ON THE MILNOR CONJECTURES 21 Proof.Suppose there is an element x 2 E1,1rsuch that dr(x) does not lie in Hr+1* *,r+1 (or rather its image in Er). We can write x = a_+ y where a_2 H2,2= KM2(F )=2 and y 2 H0,1~= Z=2. In expressing a_as a sum of pure symbols, one notes that only a finite number of elements of F are involved. By naturality of the spectr* *al sequence, we can therefore assume F is a finitely-generated extension of Q. But now we can choose an embedding F ,! C, and again use naturality. The groups KMn(C)=2 are all zero, and therefore our assumption implies that over C * *we have E0,r+1r+1= 0 (in other words, the Z=2 in E0,r+12dies in the spectral seque* *nce). But there is a `topological realization map' from our spectral sequence over C * *to the usual Adams spectral sequence in topology, where we know that none of the Z=2's in E0,*ever die. Remark 4.6. There is also a purely algebraic proof of the above result. One reduces via naturality to the case of algebraically closed fields, where all th* *e Hn,n's are zero. Then one shows that the Z=2's in the 0th column form a polynomial algebra, and that the composite Z=2 Z=2 ,! E1,12 E0,12! E1,22is zero (just as in ordinary topology). The fact that the spectral sequences is multiplicative t* *akes care of the rest. Lemma 4.7. For a_2 H2,2one has dr(a_) = 0, for every r. Proof.It follows from facts (3) and (7), together with the multiplicative struc* *ture of the spectral sequence, that everything in the image of dr: H2,2! Hr+1,r+1is killed by H1,1. This is the key to the proof. Let z = dr(a_). Consider the naturality of the spectral sequence for the map j :F ! F (t). It follows from the previous paragraph that j(z) = dr(ja_) is kil* *led by F (t)*. In particular, {t} . j(z) = 0 in KMr+2(F (t))=2. But by [Mi , Lem. 2.1]* * there is a map @t:KMr+2(F (t))=2 ! KMr+1(F )=2 with the property that @t({t} . j(z)) * *= z. So we conclude that z = 0, as desired. Proposition 4.8. All differentials leaving E1,1are zero. Proof.Recall E1,12~=H0,1 H2,2~=Z=2 H2,2. By the previous lemma we are reduced to analyzing the maps dr:H0,1! Hr+1,r+1. Since H0,1(Q) ! H0,1(F ) is an isomorphism, it suffices to prove the result in the case F = Q. Now use naturality with respect to the field extension Q ,! R. The maps KMn(Q)=2 ! KMn(R)=2 are isomorphisms for n 3 (see Appendix A), so now we've reduced to F = R. But here we can again use a `topological realization' m* *ap to compare our Adams spectral sequence to the corresponding one in the context * *of Z=2-equivariant homotopy theory. This map is readily seen to be an isomorphism on the E0,*column: the point is that the Z=2-equivariant cohomology groups Hn,n are isomorphic to the corresponding mod 2 motivic cohomology groups over R (see [Du , 2.8, 2.11], for instance). We are essentially seeing a reflection of the * *fact that GW (R) may be identified with the Burnside ring of Z=2, which coincides with {S0,0, S0,0} in the Z=2-equivariant stable homotopy category. In any case, we a* *re finally reduced to showing the vanishing of certain differentials in a topologi* *cal Adams spectral sequence: the paper [LZ ] seems to essentially do this (but I ha* *ven't thought about this part carefully_I'm relying on remarks from [M2 ]). This completes Morel's proof of the quadratic form conjecture for characteris* *tic zero fields (modulo the identification of H**H, which we assumed). 22 DANIEL DUGGER Remark 4.9. We restricted to characteristic zero fields because the identificat* *ion of H**H has never been claimed in characteristic p. If we make the wild guess t* *hat in positive characteristic H**H still has the same form, most of the argument g* *oes through verbatim. There are two exceptions, where we used topological realizati* *on functors. The first place was to show that the image of the dr's didn't touch t* *he Z=2's in E0,*2, but Remark 4.6 mentioned that this could be done another way. T* *he second place we used topological realization was at the final stage of the argu* *ment, to analyze the differentials dr: H0,1! Hr+1,r+1. As before, this reduces to the case of a prime field. But for F a finite field one has KMn(F ) = 0 for n 2, * *so for prime fields there is in fact nothing to check. In summary, the same general argument would work in characteristic p if one knew that H**H had the same form. 4.10. Further reading. There is very little completed literature on the subjects discussed in this section. Several documents are available on Morel's website, * *how- ever; the draft [M5 ] is particularly relevant, although it only slightly expan* *ds on [M2 ]. For information on the motivic Steenrod algebra, see [V2 ]. Finally, M* *orel recently released another proof of Milnor's quadratic form conjecture, using ve* *ry different methods. See [M4 ]. NOTES ON THE MILNOR CONJECTURES 23 Appendix A. Some examples of the Milnor conjectures This is a supplement to Section 1. We examine the Milnor conjectures in the cases of certain special fields F . (a) F is algebraically closed. Since F = F 2, every nondegenerate form is isomo* *rphic to one of the form <1, 1, . .,.1>. So GW (F ) ~=Z, and W (F ) ~=Z=2 with I(F ) * *= 0. Thus, GrIW (F ) ~=Z=2. The absolute Galois group is trivial, so H*(F ; Z=2) = Z=2. Finally, the fact that F = F 2implies that KM*(F )=2 = 0 for * 1. This is because the generators all lie in KM1(F ), and if a = x2 then {a} = {x2} = 2{x}* * = 0 2 KM1(F )=2. (b) F = F 2. This case is suggested by the previous one. We only need to check that the hypothesis implies H*(F ; Z=2) = 0 for * 1. Strangely, I haven't been able to find an easy proof of this. (c) F = R. In this case we know forms are classified by their rank and signatur* *e, and it follows that GW (R) is the free abelian group generated by <1> and <-1>. Also, <-1>2 = <1>. So GW (R) ~=Z[x]=(x2 - 1), and W (R) ~=Z with I(R) = 2Z. Hence GrIW (R) ~=Z=2[a]. The absolute Galois group of R is Z=2, so H*(R; Z=2) = H*(Z=2; Z=2) = Z=2[a]. Finally we consider KM*(R)=2. The group KM1(R)=2 = R*=(R*)2 ~={1, -1} (the set consisting of 1 and -1). A similar calculation, based on the fact that eve* *ry element of R is a square up to sign, shows that KMi(R)=2 ~=Z=2 for every i, with the nonzero element being {-1, -1, . .,.-1}. So KM*(R)=2 ~=Z=2[a] as well. (d) F = Fq, q odd. Here F *~=Z=(q - 1) and so KM1=2 = F *=(F *)2 ~=Z=2. If g is the generator, then {g, g, . .,.g} generates KMn=2 (but may be zero). In fac* *t one can show (cf. [Mi , Ex. 1.5]) that {g, g} = 0 in KM2, from which it follows t* *hat KM* = 0 for * 2. So KM*(F )=2 ~=Z=2 Z=2, in degrees 0 and 1. For a finite field the absolute Galois group is ^Z, the profinite completion * *of Z. The Galois cohomology H*(^Z; Z=2) is just the mod 2 cohomology of BZ ' S1; so it is Z=2 Z=2, with the generators in degrees 0 and 1. Again noting that F *=(F *)2 ~=Z=2, it follows that the Grothendieck-Witt gro* *up is generated by <1> and . A simple counting argument (cf. [S1, Lem. 2.3.7* *]) shows that every element of F*qis a sum of two squares. Writing g = a2 + b2 one finds that <1, 1> = = = = . That is, 2(<1> - ) = 0. It follows that GW (F ) = Z Z=2, with corresponding generators <1> and <1> - . The computation of the Witt group depends on whether or not -1 is a square; since F *= Z=(q - 1) and -1 has order 2, then -1 is a square precisely when 4|(q - 1). So if q 1(mod 4) then <1> = <-1> and W (F ) ~=Z=2 Z=2; in this case I(F ) = (<1> - ) ~=Z=2. If q 3(mod 4) then = <-1> and we have W (F ) ~=Z=4 with I(F ) = (2). In either case GrIW (F ) ~=Z=2 Z=2. 24 DANIEL DUGGER Remark A.1. Although Milnor's quadratic form conjecture says that GrIW (F ) depends only on the absolute Galois group of F , this example makes it clear th* *at the same cannot be said for W (F ) itself. (e) F = Q. This case is considerably harder, so we will only make a few observa- tions. Note that as an abelian group one has i j Q* ~=Z=2 x pZ , by the fundamental theorem of arithmetic; the direct sum is over the set of all primes. Here the isomorphism sends a fraction q to its sign (in the Z=2 factor) together with the list of exponents in the prime factorization of q. So KM1(Q)=* *2 ~= Z=2 ( pZ=2). As the above isomorphism may suggest, to go further it becomes convenient to work with one completion at a time. The case F = R has already been discussed, so what is left is the p-adics. We will return to F = Q after discussing them. (f) F = Qp. We will concentrate on the case where p is odd; the case p = 2 is similar, and can be left to the reader. We know KM1(Qp)=2 ~= H1(Qp; Z=2) ~= Q*p=(Q*p)2. A little thought (cf. [S1, 5.6.2]) shows this group is Z=2 Z=2,* * with elements represented by 1, g, p, and pg, where 1 < g < p is any integer which generates the multiplicative group F*p. By [Se, Section II.5.2] one has H2(Qp; * *Z=2) ~= Z=2 and Hi(Qp; Z=2) = 0 for i 3. The fact that KM1(Qp)=2 only has four elements tells us that KM*(Qp)=2 can't * *be too big. By finding the appropriate relations to write down, Calvin Moore proved that KM*(Qp)=2 = 0 for * 3 [Mi , Ex. 1.7], and that KM2(Qp)=2 = Z=2. This is an exercise for the reader. The group GW (Qp) will be generated by the four elements <1>, ,

, and . The theory again depends on whether or not -1 is a square, which is when p 1(mod 4). When p 1(mod 4) one has <1> = <-1> and so = <-x> for any x. As a result = = <1, -1> = <1, 1>, and similarly = = <1, 1>. One finds that GW (Qp) = Z (Z=2)3 with corresponding generators <1>, <1> -

, <1> - , and <1> - . Since <1, -1> = 2<1>, W (Qp) = (Z=2)4 with the same generators. I is generated by <1, p>, <1, g>, and <1, pg>; I2 is gener* *ated by <1, p, g, pg>, and I3 = 0. So GrIW = Z=2 (Z=2 Z=2) Z=2. Note that this is the first example we've seen where I2 6= 2I. When p 3(mod 4) we can take g = -1. One has <1, 1> = <-1, -1> by the same reasoning as for Fp (-1 is the sum of two squares), and so = <-p, -p>. N* *ote that = = <1, -1, -1, 1> = <1, 1, 1, 1> and so 4(<1> -

) = 0. Also, = = <1, -1, -p> and <1, 1, 1> = <1, -1, -1>. So 3(<1> -

) = <-1> - <-p>. Of course GW (Qp) is generated by <1>, <1> - <-1* *>, <1> -

, and <1> - <-p>, and the previous computation shows the last generator is not needed. So we have a surjective map Z Z=2 Z=4 ! GW (Qp) sending the standard generators to <1>, <1> - <-1>, and <1> -

. This is readily chec* *ked to be injective once one knows that <1, 1> 6~=. If these forms were isomo* *rphic it would follow by reduction mod some power of p that <1, 1> was isotropic over NOTES ON THE MILNOR CONJECTURES 25 some Fpe; that is, we would have <1, 1> ~=<1, -1>. But we've already computed GW (Fpe), and know this is not the case. The Witt ring is W (Qp) ~=Z=4 Z=4 with generators <1> and <1> -

. The ideal I is generated by 2<1> and <1> -

; I2 is generated by 2(<1> -

); I3* * = 0. Again we have GrIW ~=Z=2 (Z=2 Z=2) Z=2. (g) Return to F = Q. Our understanding of the higher Milnor K-groups of Q is based on passing to the various completions Qp and R. A computation of Bass and Tate [Mi , Lem. A.1] gives an exact sequence i j 0 ! KM2(Q)=2 ! KM2(R)=2 pKM2(Qp)=2 ! Z=2 ! 0, and we already know KM2(Qp)=2 ~=KM2(R)=2 ~=Z=2. A computation of Tate [Mi , Th. A.2, Ex. 1.8] shows that for * 3 one has KM*(Q)=2 ~= pKM*(Qp)=2 KM*(R)=2 ~=0 Z=2. To compute H*(Q; Z=2) we again work one completion at a time. A theorem of Tate [Se, Section II.6.3, Th. B] says that for i 3 one has Y Hi(Q; Z=2) ~=Hi(R; Z=2) x Hi(Qp; Z=2) ~=Hi(R; Z=2) ~=Z=2. p Our computationQof Q*=(Q*)2 ~=H1(Q; Z=2) shows that the map H1(Q; Z=2) ! H1(R; Z=2)x p H1(Qp; Z=2) is injective. More of Tate's work [Se, Sec. II.6.3, * *Th. A] identifies the dual of the kernel with the kernel of H2(Q; Z=2) ! H2(R; Z=2)* * x ( pH2(Qp; Z=2))_thus, this latter map is also injective. Using this, [Se, Sec. * *II.6.3, Th. C] gives a short exact sequence 0 ! H2(Q; Z=2) ! H2(R; Z=2) ( pH2(Qp; Z=2)) ! Z=2 ! 0. As we have already remarked that H2(Qp; Z=2) = H2(R; Z=2) = Z=2, this com- pletes the calculation of H*(Q; Z=2). The method for computing the Witt group W (Q) proceeds similarly by working one prime at a time. See [S1, Section 5.3]. One has an isomorphism of groups W (Q) ~=Z ( pW (Fp)) [S1, Thm. 5.3.4]. With enough trouble one can compute GrIW (Q), but we will leave this for the reader to consider. Remark A.2. Note that the verification of the Milnor conjectures for F = Q tells us exactly how to classify quadratic forms over Q by invariants. First one needs the invariants over R (which are just rank and signature), and then one needs the invariants over each Qp_but for Qp one has I3 = 0, and so p-adic forms are classified by the three classical invariants e0, e1, and e2. These observation* *s are essentially the content of the classical Hasse-Minkowski theorem. The method we've used above, of working one completion at a time, works for a* *ll global fields; this is due to Tate for Galois cohomology, and Bass and Tate for* * KM*. In this way one verifies the Milnor conjecture for this class of fields [Mi , L* *emma 6.2]. Note in particular that the class includes all finite extensions of Q. 26 DANIEL DUGGER Appendix B. More on the motivic Adams spectral sequence This final section is a supplement to Section 4. I will give some hints on co* *mput- ing the E2-term of the motivic Adams spectral sequence, for the reader who would like to try this at home. The computations are not hard, but there are several * *small issues that are worth mentioning. B.1. Setting things up. H**H is the algebra of operations on mod 2 motivic cohomology. We will write this as A from now on. There is the Bockstein fi 2 A1* *,0 and there are squaring operations Sq2i2 A2i,i. We set Sq2i+1= fiSq2i2 A2i+1,i. Finally, there is an inclusion of rings H** ! A sending an element t to the ope* *ration left-multiplication-by-t. Under our standing assumptions about A (see Section 4* *), it is free as a left H**-module with a basis consisting of the admissible seque* *nces Sqi1Sqi2. .S.qik. There are two main differences between what happens next and what happens in ordinary topology. These are: (a)The vector space H** = H**(pt), regarded as a left A-module, is nontrivial. (b)The image of H** ,! A is not central. The above two facts are connected. Let t 2 H** and let Sq denote some Steenrod operation. It is not true in general that Sq(t . x) = t . Sq(x)_instead there * *is a Cartan formula for the left-hand side [V2 , 9.7], which involves Steenrod opera* *tions on t. So the operations Sq . t and t . Sq are not the same element of A. There * *is one notable exception, which is when all the Steenrod squares vanish on t. This happens for elements in Hn,n, for dimension reasons. So we have (c)Every element of Hn,n is central in A. It is important that we can completely understand H** as an A-module. This will follow from (1) the fact that H** ~= nHn,n [ø] (see Remark 2.10); (2) all Steenrod operations vanish on Hn,n for dimension reasons; (3) all Sqi's vanish * *on ø except for Sq1, and Sq1(ø) = æ = {-1} 2 H1,1; (4) the Cartan formula. In particular we note the following two facts about H**, which are all that will be needed later (the second fact only needs Remark 2.10): (d)The map Sq2: Hn-1,n! Hn+1,n+1is zero for all n 1. (e)The map Hp,q Hi,j! Hp+i,q+jis surjective for q p 0 and j i 0. We are aiming to compute ExtaA(H**, b,0H**). In ordinary topology we could use the normalized bar construction to do this, but one has to be careful here because H**, as a left A-module, is not the quotient of A by a two-sided ideal. One way to see this is to use the fact that Sq1(ø) = æ. Under the quotient map A ! H** sending ` to `(1), Sq1 maps to zero but Sq1ø does not (it maps to æ). So instead of the normalized bar construction we must use the unnormalized on* *e. This can be extremely annoying, but for the most part it turns out not to influ* *ence the öl w-dimensional" calculations we're aiming for. It is almost certainly an * *issue when computing past column two of the Adams E2 term, though. Anyway, let Bn = A H** A H** . . .H**A H** H** (n + 1 copies of A). The final H** can be dropped off, of course, but it's usef* *ul to keep it there because the A-module structure on H** is nontrivial and enters NOTES ON THE MILNOR CONJECTURES 27 into the definition of the boundary map. If we denote the generators of Bn as x = a[`1|`2| . .|.`n]t then the differential is d(x) = (a`1)[`2| . .|.`n]t + a[`1`2|`3| . .|.`n]t + . .+.a[`1| . .|.`n-1]`n* *(t). The good news is that our coefficients have characteristic 2, and so we don't h* *ave to worry about signs. Note that Bn, as a left H**-module, is free on generators 1[`1| . .|.`n]1 where each `i is an admissible sequence of Steenrod operations * *(and we must include the possibility of the null sequence Sq0 = 1). We will often dr* *op the 1's off of either end of the bar element, for convenience. Generators of Hom A(Bn, H**) can be specified by giving a bar eleme* *nt [`1| . .|.`n] together with an element t 2 H**. This data defines a homomorphism Bn ! H** sending the generator [`1| . .|.`n] to t and all other generators of B* *n to zero. Let's denote this homomorphism by t[`1| . .|.`n]*. These elements generate Hom A(Bn, H**) as an abelian group. The last general point to make concerns the multiplicative structure in the c* *obar construction. If we were working with ExtA(k, k) where k is commutative and A is an augmented k-algebra, multiplying two of the above generators in the cobar complex just amounts to concatenating the bar elements_the labels t 2 k commute with the `'s, and so can be grouped together: e.g. t[`1| . .|.`n] . u[ff1| . .* *|.ffk] = tu[`1| . .|.`n|ff1| . .|.ffk]. In our case, the fact that H** is not central in* * A immensely complicates the product on the cobar complex: very roughly, the u has to be commuted across each `i, and in each case a resulting Cartan formula will intro* *duce new terms into the product. Luckily there is one case where these complications aren't there, which is when u 2 Hn,n_for then u is in the center of A, and the product works just as above. We record this observation for future use: (f)t[`1| . .|.`n]* . u[ff1| . .|.ffk]* = tu[`1| . .|.`n|ff1| . .|.ffk]* when u * *2 Hq,q. B.2. Computations. We are trying to compute the groups ExtaA(H**, b,0H**), and from here on everything is fairly straightforward. As an example let's look* * at b = 1. Since Hp,q6= 0 only when 0 p q, one sees that Hom A(B0, H**) = 0 and Hom A(B1, 1,0H**) ~=H0,0 H1,1. The generators for this group are elements of the form s[Sq1]* and t[Sq2]*, where s 2 H0,0and t 2 H1,1. We likewise find that Hom A(B2, 1,0H**) ~=H0,1 H0,1 H0,1 H0,1, generated by elements s[Sq1|1]*, s[1|Sq1]*, t[Sq2|1]*, and t[1|Sq2]*. A similar analysis * *shows that Hom A(Bn, 1,0H**) only has such `degenerate' terms for n 2. No degenera* *te terms like these contribute elements to Ext (at worst they can contribute relat* *ions to Ext). So the Extn's vanish for n 2. An analysis of the coboundary shows th* *at everything in dimension 1 is a cycle. So we find that 0 = Ext0(H**, 1,0H**) = Extn(H**, 1,0H**), for n 2 and Ext1(H**, 1,0H**) ~=H0,0 H1,1 with a typical element in the latter group having the form s[Sq1]*+ t[Sq2]* (wh* *ere s 2 H0,0and t 2 H1,1). In general, one sees for degree reasons that the `non-degenerate' terms in Hom A(Bn, n,0H**) all have the form t[`1| . .|.`n]* where each `i is either Sq1 28 DANIEL DUGGER or Sq2. In Hom A(Bn-1, n,0H**) one has non-degenerate terms u[`1| . .|.`n-1]* * *of the following types: (i)Each `i2 {Sq1, Sq2}, and at least one Sq2 occurs. Here u 2 Hj-1,jwhere j is the number of Sq2's. (ii)Each `i 2 {Sq1, Sq2, Sq3}, and exactly one Sq3 occurs. Here u 2 Hj+1,j+1 where j is the number of Sq2's. (iii)Each `i 2 {Sq1, Sq2, Sq2Sq1}, and exactly one Sq2Sq1 occurs. Here one has u 2 Hj+1,j+1where j is the number of Sq2's. (iv)Each `i 2 {Sq1, Sq2, Sq4}, and exactly one Sq4 occurs. Here u 2 Hj+2,j+2 where j is the number of Sq2's. To analyze the part of the boundary Bn ! Bn-1 that we care about, one only needs to know the Adem relations Sq1Sq2 = Sq3 and Sq2Sq2 = øSq3Sq1. (In fact, since Sq3Sq1 doesn't appear in any of the bar elements relevant to Hom (Bn-1, n,0H**), one may as well pretend Sq2Sq2 = 0.) From this it's easy to compute that Extn(H**, n,0H**) ~= H0,0 Hn,n where a typical element has the form s[Sq1|Sq1| . .|.Sq1]* + t[Sq2|Sq2| . .|.Sq2]*. The computation uses r* *e- mark B.1(d). Also, one sees that all elements s[Sq1|Sq2]* and s[Sq2|Sq1]* are zero in Ext2(being the coboundaries of s[Sq3]* and s[Sq2Sq1]*, respectively). U* *s- ing remark (f) from Section B.1, this completely determines n Extn(H**, nH**) as a subring of the whole Ext-algebra. The next step is to compute Ext0(H**, 1,0H*,*), Ext1(H**, 2,0H*,*), and Ext2(H**, 3,0H*,*) completely. The first group is readily seen to vanish. For the second group one has to grind out another term of the bar construction, but it's a very small term. One finds that Ext1(H**, 2,0H*,*) ~=H0,1 H2,2 where the generators have the form s[Sq2]* + (Sq1s)[Sq3]* and t[Sq4]*. To get t* *he Ext2group one will need three more Adem relations, namely Sq2Sq3 = Sq5 + Sq4Sq1, Sq2Sq4 = Sq6 + øSq5Sq1, and Sq3Sq2 = æSq3Sq1. Then the same kind of coboundary calculations (but a few more of them) show that Ext2(H**, 3,0H*,*) ~=H1,2 H2,2 where the generators are s[Sq2|Sq2]* + (Sq1s)[Sq3|Sq2]* and t[Sq1|Sq4]* = t[Sq4|Sq1]* (these last two classes are the same in Ext). It is important to note that all elements u[Sq2|Sq4]* and u[Sq4|Sq2]* are coboundaries (of u[Sq6]* and u[Sq4Sq2]*, respectively). This justifies fact (7) on page 20. To jus- tify fact (6) from that same page (for n = 2), one notices that the cycles s[Sq2|Sq2]* + (Sq1s)[Sq3|Sq2]* and t[Sq4|Sq1]* decompose as a products 2 * 1 3 * 2 * 4 * 1 * s1[Sq ] + (Sq s1)[Sq ] . (s2[Sq ] ) and t1[Sq ] . t2[Sq ] for some s1 2 H0,1, s2 2 H1,1, t1 2 H2,2, and t2 2 H0,0. This uses remarks (e) and (f) from Section B.1, together with the fact that (Sq1s1)s2 = Sq1(s1s2) for s2 2 H2,2(by the Cartan formula). The final step is to analyze the groups Extn-1(H**, n,0H**) for n 4; these complete the E1,*column of the Adams spectral sequence. One doesn't have to NOTES ON THE MILNOR CONJECTURES 29 compute them explicitly, just enough to know that every element is decomposable as a sum of products from Extn-2(H**, n-1,0H**) and Ext1(H**, 1,0H**). The calculations involve nothing more than what we've done so far, except for more sweat. It's fairly easy to write down all the cocycles made up from the cl* *asses of types (i)-(iv) listed previously. All bar elements which have a Sq4 in them * *are cocycles, for instance. But note that such a bar element will either begin or e* *nd with a Sq1 or a Sq2, so that it decomposes as a product of smaller degree cocyc* *les (this again depends on B.1(e,f)). One also finds cocycles of the form s[Sq1|Sq1| . .|.Sq3|Sq1| . .|.Sq1]* + s[Sq1|Sq1| . .|.Sq2Sq1|Sq1| . .|.Sq1]* **, but for each of these a common [Sq1]* can be pulled off of either the left or r* *ight side_again showing it to be decomposable. Certainly there are cocycles which are not decomposable, like ones of the form s[Sq2|Sq1| . .|.Sq1|Sq3]* + s[Sq2Sq1|Sq1| . .|.Sq1|Sq2]*. But this is the coboundary of s[Sq2Sq1|Sq1| . .|.Sq1|Sq3], and so vanishes in E* *xt. Anyway, I am definitely not going to give all the details. But with enough diligence one can see that all elements of Extn-1(H**, n,0H**) for n 3 do indeed decompose into products. Remark B.3. A final note about Adem relations, for those who want to try their hand at further calculations. Every formula I've seen for the motivic Adem relations_in publications or preprints_seems to either contain typos or else is just plain wrong. 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