On the extension giving the truncated Witt vectors. Torgeir Skjøtskift January 5, 2015

Transcription

1 On the extension giving the truncated Witt vectors Torgeir Skjøtskift January 5, 2015

2 Contents 1 Introduction Why do we care? Cohomology of grous Chain comlexes and homology Resolutions The integral grou ring The cohomology of a grou Grou extensions On the extension giving the truncated Witt vectors Witt vectors in general The extension underlying W 2 (A)

3 Abstract We exlore the theory of cohomology of grous and the classification of grou extensions with abelian kernel. We then look at the grou extensions that underlie the truncated Witt vectors on the truncation set {1, } where is a rime number. It turns out that we can do without the multilicative structure on the source ring A by factoring the extension s reresenting cocycle through a ma into the -fold tensor roduct of A divided out by the C -action.

4 1 Introduction Given two grous G and N, we can ask the following question: What are the essentially different ways we can exress the grou G as a quotient of some grou E by N? Answering this question amounts to classifying certain objects called extensions, which are short exact sequences of grous 1 N E G 1. Restricting ourselves to extensions with abelian kernel, i.e. where N is an abelian grou A, we solve the classification roblem by using the theory of cohomology of grous. Turning our attention to the truncated Witt vectors on a commutative ring A, we examine the extension we get from the rojection of the Witt vectors onto A by forgetting down to abelian grous. This yields an extension 0 A W A 0 which is reresented in the second cohomology grou H 2 (A, A) by a cocycle deending on the multilication in A. It turns out that this cocycle factors through a ma into the -fold tensor roduct of A divided out by the C -action, yielding another extension 0 A /C W A 0, in which the grou structure in W does not deend on the multilication in A. We discover that tensoring this extension with F yields a long exact sequence in homology in which the connecting homomorhism is equal to a certain normalized cochain A A. 1.1 Why do we care? In this section we will give some background information that lies outside the scoe of this thesis, but which may hel the reader lace it in a wider context. The motivation for this thesis is questions in equivariant homotoy theory, more recisely the equivariant structure of smash owers. If E is a sectrum, G a finite grou and S a finite G-set, let E S = E E be the smash roduct of E with itself indexed over the elements of S. The motivation lies in an attemt to understand the G-sectrum E S, and in articular its fixed oints (E S ) G under the G-action. A variant of the Redshift conjecture is that the chromatic level of (E S ) G is higher than that of E itself. In articular, if multilication by is nullhomotoic in E then it is not null-homotoic in (E ) C, where C is the cyclic grou of order. We wish to study this more closely by examining the fundamental cofiber sequence (E ) hc (E ) C E, where (E ) hc are the homotoy orbits under the C -action. If the fixed oints are going to have higher chromatic filtration than E, then this cofiber 1

5 sequence must be non-trivial. One way to understand such a cofiber sequence is to understand the boundary ma E Σ(E ) hc, which should be easier to understand since both E and (E ) hc are easier to understand than the fixed oints. Such questions arise in connection with toological Hochschild homology: If E is a commutative ring sectrum then S E S becomes a functor and we can give meaning to E X, where X is a simlicial set. In articular, THH(E) arises as E S1 where S 1 is a simlicial model for the circle. Note that this case does not require that E be commutative. Let C be the cyclic grou of order, acting on S 1 by multilication by the -th roots of unity. This leads to the following variant of the fundamental cofiber sequence above THH(E) hc THH(E) C THH(E), where THH(E) hc are the homotoy orbits and THH(E) C are the fixed oints of THH(E). We wish to understand this cofiber sequence by studying the boundary ma THH(E) Σ THH(E) hc To discover higher eriodic classes one must work with coefficients. In articular, to discover increased divisibility by, it is fair to start out with working with homotoy mod. In this thesis we study the lowest homotoy grous for the fixed oints and the boundary mas in mod. Here we rediscover the well known truncated Witt vectors for THH, while for the finite fixed oints discover new extensions of π 0 E. In the last case we get an exlicit version of the boundary ma and an oen question is how our formulas should generalize from π 0 to a ma of sectra. Concretely, letting A be an abelian grou, E = HA the Eilenberg-MacLane sectrum on A and G = C be the cyclic grou of order, the fundamental cofiber sequence becomes (HA ) hc (HA ) C HA. In this case the boundary ma is the ma : HA Σ(HA ) hc, which in mod homotoy becomes the ma π : π HA S/ π Σ(HA ) hc S/. Looking at the long exact sequence in homotoy given by smashing with the cofiber sequence S S S/, where S is the shere sectrum, we see that for i < 1, π i Σ(HA ) hc S/ is equal to zero since taking homotoy orbits reserves connectivity. Moreover, for i > 1 we have that π i HA S/ is equal to zero. Thus, the only non-zero art of π is the connecting homomorhism π 1 HA S/ π 0 (HA ) hc S/. 2

6 This ma is actually the connecting homomorhism Tor 1 (F, A) F A /C, for which we discover a concrete formula in Section 3. 2 Cohomology of grous We will start by defining a set of basic objects that will be of use later. Most of the contents of this section is taken from [1, ch. I-III] while exanding on those arts requiring additional details. The goal of this section is to be able to give a definition of the cohomology of a grou G with coefficients in a ZG-module M. 2.1 Chain comlexes and homology Let R be a ring. A chain comlex C over R is a graded R-module C = (C n ) together with an endomorhism d of C, called the differential of C, with degree 1 satisfying d 2 = 0. If d instead has degree +1, we write C = (C n ) and say that C is a cochain comlex. If M is a module, we will let M denote the chain comlex 0 0 M, concentrated in degree 0. If C is a chain comlex, we define the homology of C to be the graded R- module H(C) = (H n (C)) = ker d/ im d. If C is a cochain comlex, we write H(C) = (H n (C)) and call it the cohomology of C. A graded module homomorhism f : C C of degree 0 between two chain comlexes is called a chain ma if d f = fd. A chain ma f : C C whose induced ma H(f) : H(C) H(C ) is an isomorhism, is called a weak equivalence. In articular, homotoy equivalences of chain comlexes are weak equivalences. 2.2 Resolutions Definition Given an R-module M, a resolution ε : F M of M over R is an exact sequence d F 2 d 2 1 ε F1 F0 M 0 of R-modules. A free resolution is a resolution where all the F i are free. Since every exact sequence is a chain comlex, we can view a resolution in terms of chain comlexes by regarding ε as a chain ma from the chain comlex F = (F n ) n 0 to M, where M is the chain comlex concentrated in degree 0: d 2 d F 2 1 F 1 F M 0 Proosition The exactness condition on the sequence F 1 F 0 M 0 holds if and only if ε is a weak equivalence ε 3

7 Proof. If the sequence is exact, then H i (ε) : H i (F ) H i (M) is the isomorhism 0 0 for i > 0 and H 0 (F ) H 0 (M) is the isomorhism F 0 / ker ε M. Conversely, if ε is a weak equivalence then H i (ε) is an isomorhism for all i which imlies that the sequence is exact. The imlication can be seen in the case i = 0 from the diagram ε F 0 M. H 0 (F ) Proosition Every module has a free resolution. Proof. Let M be a module and let d 1 be the zero ma M 0 and let F 1 = M. For each i 0, let F i be the free R-module on ker d i 1 and let d i : F i F i 1 be the ma sending each generator to its corresonding element. Proosition A module M over a rincial ideal domain R admits a resolution 0 F 1 F 0 M 0. Proof. By Proosition 2.2.3, M has a free resolution F M. Since R is a rincial ideal domain, submodules of free R-modules are free. Thus F 1 is equal to the kernel of the ma F 0 M and hence F 2 = 0. Since free modules are direct summands of themselves, free modules are rojective and it follows from that every module has a rojective resolution. Note that a free R-module M admits the free resolution 0 M 1 M The integral grou ring Let G be a grou and let M be the free Z-module on the set G. As a set, M consists of the functions f : G Z having finite suort, with sum and scalar multilication defined ointwise. Consider the ma µ : M M M sending (f, g) to x uv=x f(u)g(v) = u f(u)g(u 1 x). Proosition The ma µ is Z-bilinear and associative, giving a roduct on M. Proof. Let f, g M and x G. Let µ(f, g) be denoted by fg. To rove that µ is Z-bilinear, we must show that (nf + mf )g = nfg + mf g: (nf + mf )g(x) = (nf + mf )(u)g(v) uv=x = (nf(u) + mf (u))g(v) uv=x = n f(u)g(v) + m f (u)g(v) uv=x = nfg(x) + mf g(x). uv=x 4

8 The calculation for the second coordinate is similar. Next, let h be another element of M. To rove associativity, we need to show that (fg)h = f(gh): (fg)h(x) = fg(u)h(u 1 x) = f(v)g(v 1 u)h(u 1 x) u u v = f(v)gh(v 1 x) v = f(gh)(x). Let 1 G denote the unit element of G. The unit with resect to µ is then the function 1 : G Z maing 1 G to 1 and every other element to 0. Definition The ring given by endowing M with the roduct µ is called the integral grou ring of G and is denoted ZG. The functors G ZG and R R, where R is the multilicative grou of units of R, form an adjoint air This gives a bijection Gr Z Rng. ( ) Gr(G, R ) = Rng(ZG, R) which is natural in G and R. Letting R = End(A) be the ring of endomorhisms of A then since End(A) is equal to Aut(A), the grou of automorhisms of A, we get a bijection Gr(G, Aut(A)) = Rng(ZG, End(A)), giving that there is a one-to-one corresondence between ZG-module structures on A and G-actions on A. We will usually not distinguish between these two concets. 2.4 The cohomology of a grou We are now ready to define the cohomology of a grou G with coefficients in a ZG-module M. Let ε : F Z and η : P M be rojective resolutions of resectively Z and M over ZG, and let C and C be chain comlexes over a ring R with differentials d and d resectively. There is a chain comlex Hom R (C, C ) defined in degree n as the R-module of graded module homomorhisms from C to C with differential D defined by D(f) = d f + ( 1) n+1 fd. Observe that Hom R (C, C ) n = q Z Hom R(C q, C q+n). If C were concentrated in degree 0, we would get that Hom R (C, C ) n = q Z Hom R (C q, C q+n) = Hom R (C n, C 0), 5

9 which would be natural to view as the degree n art of a cochain comlex defined in degree n by Hom R (C, C ) n = Hom R (C, C ) n = Hom R (C n, C 0). In this case, the differential δ = (D n : Hom R (C, C ) n Hom R (C, C ) n+1 ) would take the form δ(f) = ( 1) n+1 fd. (2.4.1) Definition The cohomology H (G, M) of G with coefficients in M is the homology of the cochain comlex Hom ZG (F, M) where M is concentrated in degree 0. We will now construct a concrete resolution F of Z over ZG, allowing us to comute the cohomology of a grou according to Definition Proosition Let X be a simlicial abelian grou having the face mas d i : X n X n 1 for 0 i n. The sequence X n n Xn 1 X 1 1 X0, where n (x) = n i=0 ( 1)i d i (x), is a chain comlex of abelian grous. Proof. Since the face mas are grou homomorhisms, so are the n s. Left to show is that n 1 n = 0. We have n 1 n n 1 n n 1 n (x) = ( 1) i d i ( 1) j d j (x) = ( 1) i+j d i d j (x). i=0 j=0 i=0 j=0 The simlicial identity d i d j = d j 1 d i holds for i < j and hence this double sum slits into the following three sums i<j( 1) i+j d j 1 d i (x) + ( 1) i+j d i d j (x) + ( 1) i+j d i d j (x). i=j j<i Observing that the terms indexed by the i s and j s satisfying i = j + 1 cancel the terms indexed by the i s and j s satisfying i = j, we have: n 1 n (x) = i+1<j( 1) i+j d j 1 d i (x) + ( 1) i+j d i d j (x) j<i = ( 1) j+i+1 d i d j (x) + j<i j<i( 1) i+j d i d j (x) = 0. Definition If X is a simlicial abelian grou then the Moore comlex MX is the chain comlex given by Proosition Let EG be the simlicial set given in degree n by G n+1. The face and degeneracy mas in EG are defined resectively by 6

10 d i (g 0,..., g n ) = (g 0,..., ĝ i,..., g n ) s i (g 0,..., g n ) = (g 0,..., g i, g i,..., g n ). The grou G acts freely on each degree EG n of EG by g(g 0,..., g n ) = (gg 0,..., gg n ), making EG a free simlicial G-set. Extending this action linearly, ZEG becomes a free simlicial ZG-module by [1, I.3.1], each degree ZEG n having basis the set of reresentatives of G-orbits of elements (g 0,..., g n ). The Moore comlex MZEG becomes a chain comlex of free ZG-modules and since EG is contractible, we obtain weak equivalences MZEG MZ Z. Definition The standard resolution of Z over ZG is the weak equivalence MZEG Z given by comosing the two weak equivalences above. Each G-orbit in ZEG n is reresented by an element (1, g 1, g 2,..., g n ), which can be written in the form [h 1 h n ] = (1, h 1, h 1 h 2,..., h 1 h n ) by letting h 1 = g 1 and h i = g 1 i 1 g i for 1 < i n. In this basis, the face mas are given by g 1 [g 2 g n ] i = 0 d i [g 1 g n ] = [g 1 g i g i+1 g n ] 0 < i < n [g 1 g n 1 ] i = n. Definition The bar resolution of Z over ZG is the standard resolution where each degree of ZEG is given the basis described above. If F Z is the standard resolution, there is for each n a submodule D n of F n generated by the elements (g 0,..., g n ) such that g i = g i+1 for some 0 i < n. Orbits of such elements are reresented by elements [h 1 h n ] where h i = 1 for some 0 < i n. Definition The normalized bar resolution of Z over ZG is the resolution defined in degree n by F n /D n and is generated by elements [h 1 h n ] such that h i 1 for all i. Since any two rojective resolutions of a module are homotoy equivalent, and homotoy equivalences are weak equivalences, the cohomology H (G, M) does not deend on the choice of resolutions F and P. Thus we may choose F to be the bar resolution, and we get Hom ZG (F, M) n = ZG-mod(MZEG n, M) = ZG-mod(Z[G n+1 ], M) = G-set(G n+1, UM) = Set(G n, U UM), 7

11 where the two isomorhisms are given resectively by the adjoint airs and G-set Set Z[ ] U G U G-mod, G-set. Exlicitly, the isomorhism Set(G n, U X) = G-set(G n+1, X) is given by sending a function f : G n U X to the G-set homomorhism (g 0,..., g n ) g 0 f(g 1 0 g 1,..., g 1 0 g n), with the inverse sending a G-set homomorhism f : G n+1 X to the function (g 1,..., g n ) f (1, g 1,..., g n ). Definition If F is the bar resolution, then Hom ZG (F, M) is called the standard comlex and is denoted C (G, M). Taking F instead to be the normalized bar resolution, we get the subcomlex C N (G, M) of C (G, M) consisting in degree n of those functions f : G n M satisfying f(g 1,..., g n ) = 0 whenever at least one g i = 1. Definition The cochain comlex CN (G, M) is called the normalized standard comlex. A function f : G n M is said to be normalized if it is a cochain in CN n (G, M). 2.5 Grou extensions Definition Let G and N be grous. An extension of G by N is a short exact sequence 1 N E G 1 ( ) of grous. Another extension 1 N E G 1 is equivalent to ( ) if there is a homomorhism E E of grous such that the following diagram commutes: E 1 N E G 1 By the Five lemma, any equivalence of extensions is necessarily an isomorhism. Definition An extension 1 N E G 1 is said to be a slit extension if it is slit as a short exact sequence of grous. 8

12 Restricting ourselves to those extensions where N is an abelian grou A, there is to each extension an associated action of G on A: Proosition An extension of a grou G by an abelian grou A gives rise to an action of G on A, giving A the structure of a ZG-module. Proof. Consider the extension 0 A i E π G 1. By exactness, A is isomorhic to the kernel of π, which means that A is embedded as a normal subgrou of E. E acts on A by conjugation, which restricts to the trivial action on A. We get an induced action of the quotient grou E/A = G on A, given exlicitly by i(ga) = gi(a) g 1 for some g π 1 (g). Viewing the action as a grou homomorhism G Aut(A) from G to the grou of automorhisms of A, we obtain the corresonding ZG-module structure by the bijection Gr(G, Aut(A)) = Rng(ZG, End(A)). Definition For a fixed ZG-module structure on A, E(G, A) is the set of equivalence classes of extensions of G by A giving rise to the given grou action of G on A. Definition Given an action of G on A, the set A G together with the multilication (a, g)(b, h) = (a + gb, gh) is called the semi-direct roduct of G and A relative to the given action of G on A and is denoted A G. The unit element of A G is (0, 1) and inverses are defined by (a, g) 1 = (g 1 ( a), g 1 ). Note that we get an equality A G = A G exactly when the G-action on A is trivial, in articular when E is abelian. The canonical inclusion and rojection gives us a slit extension 0 A incl A G roj G 1. (2.5.1) Definition Given an action of G on A, the extension (2.5.1) is called the canonical slit extension of G by A Since slittings s : G A G of the extension (2.5.1) are of the form s(g) = (a, g) for some a in A, with a deending on g, and since s(gh) = s(g)s(h) = (a, g)(b, h) = (a + gb, gh), the slittings of the extension (2.5.1) are in one to one corresondence with functions d : G A satisfying d(gh) = d(g) + gd(h). The canonical slit extension (2.5.1) gives rise to the given action of G on A. To see this, let g G. Then g π 1 (g) is of the form (a, g) for some a A 9

13 and so if b A, the the induced G-action on A is given by i(g b) = gi(b) g 1 = (a, g)(b, 1)(a, g) 1 = (a + gb, g)(g 1 ( a), g 1 ) = (a + gb a, 1) = (gb, 1) = i(gb), showing that the induced action is is indeed the action we started out with. Now, let 0 A i E π G 1 (2.5.2) be another extension giving rise to the same action of G on A. Proosition The extension (2.5.2) slits if and only if it is equivalent to the canonical slit extension of G by A. Proof. Assume that the extension slits and let s : G E be a slitting. If e is an element of E then e = e(sπe) 1 (sπe) where e(sπe) 1 ker π and sπe im s. Moreover, if e is in both the the kernel of π and image of s, then πe = 1 = πsg = g so that e = 1. By exactness, ker π is equal to im i which is isomorhic to A, and since πs is a bijection, s is an injection so that im s is isomorhic to G. Thus, we get a bijection A G E of sets by (a, g) i(a)s(g). The unique grou structure on the set A G making this bijection an isomorhism is calculated by noting that i(ga) = gi(a) g 1 imlies i(ga) g = gi(a), so that i(a)s(g)i(b)s(h) = i(a)i(gb)s(g)s(h) = i(a + gb)s(gh), giving the multilication (a, g)(b, h) = (a+gb, gh) on A G, which is just A G. Since the diagram 0 A incl i A G E = roj π G 1 commutes, we have the required equivalence. Conversely, assuming we have such an equivalence, we get a slitting s by letting s(g) = φs (g) where φ is the isomorhism A G E given by the equivalence and s a slitting of (2.5.1). From this roosition, we see that for a given ZG-module structure A, there is, u to equivalence, only one slit extension of G by A giving rise to the given action of G on A. 10

14 Given the extension (2.5.2), we may always choose a normalized set-theoretic section of π, i.e. a function s : G E such that πs = id G and such that s(1) = 1. Since π(s(g)s(h)) = πs(g)πs(h) = gh = πs(gh) we have that s(g)s(h)s(gh) 1 is in i(a). This gives a function f : G G A defined by i(f(g, h)) = s(g)s(h)s(gh) 1. Note that if s is normalized, then this imlies that f too is normalized, i.e. f(g, 1) = f(1, g) = 0. (2.5.3) This is true since i(f(1, g)) = s(g)s(1)s(g) 1 = s(g)s(g) 1 = 1 and i(f(g, 1)) = s(1)s(g)s(g) 1 = 1. Since f is identically 1 if and only if s is a homomorhism, f measures the failure of s being a slitting. Letting φ be the bijection φ : A G E, defined by φ(a, g) = i(a)s(g), we have that i(a)s(g)i(b)s(h) = i(a)i(gb)s(g)s(h) = i(a + gb)i(f(g, h))s(gh) = i(a + gb + f(g, h))s(gh). Thus, the grou law on the set A G making φ an isomorhism is defined by (a, g)(b, h) = (a + gb + f(g, h), gh). Letting E f denote the set A G together with this grou structure, we get an extension 0 A incl roj E f G 1, (2.5.4) which is equivalent to (2.5.2), seen again from the commutativity of the diagram 0 A incl E f = roj G 1. i E Moreover, the G-action on A induced by the extension (2.5.4), is the given G-action: incl(g a) = g incl(a) g 1 = (b, g)(a, 1)(b, g) 1 π = (b + ga + f(g, 1), g)( g 1 f(g, g 1 ) g 1 b, g 1 ) = (b + ga f(g, g 1 ) b + f(g, g 1 ), 1) = (ga, 1) = incl(ga). 11

15 Thus, given the function f defined above we can, u to equivalence, reconstruct the extension (2.5.2). Proosition Let A be a ZG-module and let f : G G A be a function satisfying (2.5.3). The binary oeration (a, g)(b, h) = (a + bg + f(g, h), gh) defines a grou structure on A G if and only if gf(h, k) f(gh, k) + f(g, hk) f(g, h) = 0. (2.5.5) Moreover, if f satisfies (2.5.5), the canonical inclusion and rojection gives an extension 0 A incl roj (A G) f G 1 (2.5.6) inducing the given G-action on A. This extension also has the roerty that if s : G A G is the canonical cross-section g (0, g), then the function f : G G A defined by incl(f (g, h)) = s (g)s (h)s (gh) 1, is the function f. Proof. Assume that the given oeration defines a grou structure, then associativity holds only if is equal to [(a, g)(b, h)](c, k) = (a + gb + f(g, h) + ghc + f(gh, k), ghk) (a, g)[(b, h)(c, k)] = (a + gb + ghc + gf(h, k) + f(g, hk), ghk). This holds only if f satisfies (2.5.5). Conversely, if f satisfies the given condition, then associativity obviously holds. Now assume f satisfies the condition (2.5.5). The canonical inclusion and rojection define homomorhisms in the sequence (2.5.6) since and incl(a + b) = (a + b, 1) = (a + b + f(1, 1), 1) = incl(a) incl(b) roj((a, g)(b, h)) = roj(a + bg + f(g, h), gh) = gh = roj(a, g) roj(b, h). Thus, by the injectivity of incl and surjectivity of roj, the sequence (2.5.6) is an extension, inducing the G- action on A defined by incl(g a) = g incl(a) g 1 = (b, g)(a, 1)(b, g) 1 = (b + ga + f(g, 1), g)( g 1 f(g, g 1 ) g 1 b, g 1 ) = (b + ga f(g, g 1 ) b + f(g, g 1, 1) = (ga, 1) = incl(ga), which is the given G-action. Lastly, incl(f (g, h)) = (0, g)(0, h)(0, gh) 1 showing that f is equal to f. = (f(g, h), gh)( (gh) 1 f(gh, (gh) 1 ), (gh) 1 ) = (f(g, h) f(gh, (gh) 1 ) + f(gh, (gh) 1 ), 1) = (f(g, h), 1) = incl(f(g, h)), 12

16 Observe that functions G G A satisfying (2.5.3) corresond to 2-cochains in the normalized chain comlex CN (G, A), and that (2.5.5) holds if and only if f is a cocycle. Let s : G E be a normalized section of π in (2.5.2). If s is any other normalized section, then due to the bijection φ : A G E, the element s (g) in E can be written i(a)s(g) for some a A, and so s must be of the form g i(c(g))s(g) for some function c : G A satisfying c(1) = 0. The function f : G G A defined by i(f (g, h)) = s (g)s (h)s (gh) 1 then has the values i(f (g, h)) = i(c(g))s(g)i(c(h))s(h)(i(c(gh)s(gh)) 1 = i(c(g))i(gc(h))s(g)s(h)s(gh) 1 i(c(gh)) 1 = i(c(g) + gc(h) + f(g, h))i(c(gh)) 1 = i(c(g) + gc(h) + f(g, h) c(gh)), and so f (g, h) = f(g, h) + c(g) + gc(h) c(gh) = f(g, h) + (δc)(g, h). Thus, the change of section of π in (2.5.2) is reflected by modifying f by a coboundary in CN 2 (G, A), and we have the following theorem [1, IV.3.12]: Theorem For a fixed ZG-module structure on A, there is a bijection E(G, A) = H 2 (G, A) given by sending an extension 0 A i E π G 1 to the function defined by f : G G A i(f(g, h)) = s(g)s(h)s(gh) 1 where s is any normalized set-theoretic section of the extension. The inverse is given by sending a function f : G G A to the extension 0 A incl (A G) f roj G 1. 3 On the extension giving the truncated Witt vectors For the remainder of this thesis, let A be a commutative ring and a rime number. 3.1 Witt vectors in general Definition A set S of ositive integers is a truncation set if n S and d n imlies d S. 13

17 A truncation set S gives rise to the Witt ring W S (A), defined as a set by W S (A) = A S. The ghost ma w : W S (A) A S is given in the n-th coordinate by da n/d d. a w n (a) = d n By [2, roosition 2], there is a unique ring structure on W S (A) such that the ghost ma is a natural transformation of functors from rings to rings. This unique ring structure is obtained exlicitly by forcing w to be a homomorhism of rings. Definition For n a non-negative integer, W n (A) is the Witt ring on the truncation set S n = {1,, 2,..., n 1 }. The vectors in W n (A) and A Sn are indexed over the set {0,..., n 1} of the owers of in S n 3.2 The extension underlying W 2 (A) Let W = W 2 (A). The ghost ma w : W A A is given by (a 0, a 1 ) (a 0, a 0 + a 1). If a = (a 0, a 1 ) and b = (b 0, b 1 ) are two vectors in W, the sum a + b is given by solving the equation w(a + b) = w(a) + w(b) for a + b. We have w(a) + w(b) = (a 0 + b 0, a 0 + b 0 + (a 1 + b 1 )) which gives the following equations: w(a + b) = w(s) = (s 0, s 0 + s 1), s 0 = a 0 + b 0, 1 s 1 = ( i ) a 0 b i 0. Since 1 i <, the binomial coefficients ( ) i =! i!( i)! are all divisible by, and hence we can solve the second equation above for s 1, giving s 1 = (a 0 + b 0 (a 0 + b 0 ) )/. This forces the following grou law on W a + b = (a 0 + b 0, a 1 + b 1 + (a 0 + b 0 (a 0 + b 0 ) )/). Letting f : A A A be the ma (x, y) (x + y (x + y) )/, this grou law takes on the familiar form a + b = (a 0 + b 0, a 1 + b 1 + f(a 0, b 0 )). The identity element in W is (0, 0) and the additive inverse of a is { ( a 0, a 1 ) > 2 a = ( a 0, a 2 0 a 1 ) = 2. 14

18 Since the ghost ma restricts to the identity on A in its zeroth coordinate, the rojection W A from W to A sending (a, b) to a is a surjection of rings. The kernel of this rojection is I = {(0, a) a A}. Forgetting down to abelian grous, we get an extension 0 I incl W roj A 0 (3.2.1) of A by I. Since W is abelian, the associated ZA-module structure on I is trivial. Observing that the ma I A sending (0, a) to a is an isomorhism of grous, and that interchanging the two coordinates in W yields an isomorhism W = (A A) f, we see that f reresents the class of this extension in H 2 (A, A). Indeed, by alying the coboundary ma δ : C 2 (A, A) C 3 (A, A) described in (2.4.1) to the cochain f we find that δf(a 1, a 2, a 3 ) = (f(a 2, a 3 ) f(a 1 + a 2, a 3 ) + f(a 1, a 2 + a 3 ) f(a 1, a 2 )) = a 2 + a 3 (a 2 + a 3 ) a 1 + (a 2 + a 3 ) (a 1 + (a 2 + a 3 )) + a 1 + a 2 (a 1 + a 2 ) = 0, showing that f is a cocycle. Moreover, the equations f(a, 0) = a a = 0 f(0, b) = b b = 0 + (a 1 + a 2 ) + a 3 ((a 1 + a 2 ) + a 3 ) hold for any a, b in A, showing that f is normalized. The following diagram then shows that f reresents the class of (3.2.1). 0 I W A 0 = 0 A W A 0 0 A (A A) f A 0. Examle Let A = F be the field of characteristic. As we have seen, the sum a + b of two vectors a and b in W is determined by the value of the function f : F F F on the vector (a 0, b 0 ). Examining the case = 2, we see that a + b = (a 0 + b 0, a 1 + b 1 a 0 b 0 ). For the vector (1, 0) in W, we have that 2(1, 0) = (1, 0)+(1, 0) = (2, 1) = (0, 1), which is not zero in F 2 F 2. This imlies that W cannot be isomorhic to = 15

19 F 2 F 2, and since W is of order 4, then by the classification of finitely generated abelian grous, we must have W = Z/4Z. In fact, if is any rime number and n a non-negative integer, then [3, corollary 6.8] characterizes W n (F ) by W n (F ) = Z/ n Z. If acts injectively on H 2 (A, A), the extension (3.2.1) is a slit extension. To see this, let g : A A be the -th ower ma a a. Then δg(a, b) = ag(b) g(a + b) + g(a) = a + b (a + b) = f(a, b), so the class of f is zero in H 2 (A, A), imlying that the class of f is zero if acts injectively. Note that g is a normalized cochain in C 1 (A, A). Since A is commutative, the multilication ma A A factors through the rojection A A /C of A onto its C -orbits, and thus corresonds uniquely to the linear ma µ : A /C A defined by sending an element a 1 a to the roduct a 1 a. Let X be the set of surjective functions {1,..., } {0, 1}, and let the function f : A A A /C be given by f (a, b) = [x] X/C [a x1 a x ], where a 0 = a and a 1 = b. In words, f (a, b) is the sum of all the orbits of tensors consisting of both a s and b s. Lemma Let a b. For each 1 i <, the number of summands in f (a, b) whose reresenting tensors consist of exactly i a s is ( i) /. This gives a total of (2 2)/ summands. Proof. Let 1 i <. There are ( i) functions in X that hit 0 exactly i times. Since is rime, C is a simle grou, and since each orbit [x] in X is in one to one corresondence with the quotient grou C /C x (where C x is the isotroy subgrou of x), we have that each orbit must be of order either or 1. Suose [x] has order 1. Then for a generator t of C, we must have t j x = x for each j = 0,..., 1. Since t j x(i) = x(i + j mod ), this means that x is not surjective, a contradiction. Thus, every orbit of X has elements, and so the total number of orbits of tensors consisting of exactly i a s is equal to ( i) /. 16

20 Using the Binomial theorem, we see that the total number of summands is 1 ( ) / = 1 1 ( i i ( = 1 i=0 ) 1 i 1 i ( i = (1 + 1) 2 = 2 2. ) 1 i 1 i 2 ) Proosition The diagram below commutes. A A f A f µ A /C Proof. Need to show that µf = f. If a, b A, then µf (a, b) = µ [a x1 a x ] [x] X/C = [x] X/C µ[a x1 a x ] = a x1 a x [x] X/C 1 ( = i) ai b i = a + b ( i=0 i = a + b (a + b) = f(a, b). ) a i b i Since x 0 = 0, we have for any a, b in A f (a, 0) = f (0, b) = 0, showing that f is normalized. To show that f is also a cocycle, we need the following lemma. 17

21 Lemma Let a, b, c be elements of A and let Y be the set of surjective functions {1,..., } {0, 1, 2}. Then the following two equations hold f (a + b, c) = f (a, c) + f (b, c) R f (a, b + c) = f (a, b) + f (a, c) R, where R = [y] Y/C [ρ y1 ρ y ], ρ 0 = a, ρ 1 = b and ρ 2 = c. Proof. In the first equality, the summands in f (a, c) are summands in f (a+b, c) since any summand [α x1 α x ] in f (a, c) is a term in the summand [β x1 β x ] in f (a + b, c). Since A is abelian, the same holds for f (b, c). The remaining summands are reresented by tensors consisting of both a s, b s and c s, and hence are of the form [ρ y1 ρ y ]. Let ϕ be the ma {0, 1, 2} {0, 1} defined by Each summand [ρ y1 ρ y ] in R is a term in the summand [β ϕy1 β ϕy ] in f (a + b, c). This shows the first equality. Finally, since f is symmetric, f (a, b + c) = f (b + c, a) = f (a, b) + f (a, c) R. Let us now calculate δf where δ : C 2 (A, A /C ) C 3 (A, A /C ) is the coboundary ma. Using Lemma 3.2.4, we have δf (a, b, c) = ( 1) 3 f d(a, b, c) = f (b, c) + f (a + b, c) f (a, b + c) + f (a, b) = f (b, c) + f (b, c) f (a, b) + f (a, b) = 0. This shows f is a normalized cocycle in C 2 (A, A /C ), and f reresents the class of the extension 0 A /C W A 0 (3.2.2) in E(A, A /C ), where W = (A /C A) f. Notice that unlike W, the grou law on W does not deend on the multilication in A. We see that the extension (3.2.2) is still slit if acts injectively on H 2 (A, A /C ). Indeed, if 18

22 g : A A /C is the cochain a [a ], then δg (a, b) = ag (b) g (a + b) + g (a) = [b ] [(a + b) ] + [a ] ( ) = [b ] [a x1 a x ] + [a ] + [b ] + [a ] x X = x X[a x1 a x ] = [a x1 a x ] [x] X/C = [a x1 a x ] [x] X/C = f (a, b). If acts injectively, this imlies that the class of f is zero in H 2 (A, A /C ). Definition Let R be a ring and let M and N be R-modules. If F M and P N are rojective resolutions of M and N over R, then Tor R (M, N) = H (F R N) = H (F R P ) = H (M R P ). Note that when R = Z, we write Tor (M, N) instead of Tor Z (M, N). Proosition For any abelian grou A, Tor i (Z, A) = 0 for i > 0 and Tor 0 (Z, A) = A. Proof. Let ε : F A be a rojective resolution of A over Z. By definition, we have Tor i (Z, A) = H i (Z F ) = H i (F ). Since F is exact at all ositive degrees, H i (F ) = 0 for i > 0. If i = 0 then, since ε is a weak equivalence, H 0 (F ) = A. Let now ε : P F denote the sequence 0 Z Z ε F 0, where is multilication with and ε is the quotient ma. Since the sequence above is exact and each P i is free, this forms a rojective resolution of F over Z. Since rojective modules are flat, tensoring P with the extension (3.2.2) yields a short exact sequence 0 P A /C P W P A 0 (3.2.3) of chain comlexes. following diagram The chain comlexes in (3.2.3) are the columns in the 0 Z A /C Z W Z A 0 0 Z A /C Z W Z A 0 19

23 Alying the Snake lemma to the diagram above yields the following long exact sequence in homology: 0 Tor 1 (F, A /C ) Tor 1 (F, W ) Tor 1 (F, A) F A /C F W F A 0. (3.2.4) By Proosition 2.2.3, A has a rojective resolution F over Z. The tensor roduct of P with F yields another short exact sequence 0 Z F Z F F F 0 (3.2.5) of chain comlexes. following diagram: The chain comlexes in (3.2.5) are the columns in the 0 Z F 1 Z F 1 F F Z F 0 Z F 0 F F 0 0 This yields another long exact sequence in homology: 0 Tor 1 (F, A) A A F A 0. (3.2.6) If is zero in A then : A A is the zero ma, and so (3.2.6) becomes 0 Tor 1 (F, A) A 0 A F A 0 giving the following isomorhisms: Moreover, if is zero in A then we have Tor 1 (F, A) = A = F A. (3.2.7) Tor 1 (F, A /C ) = A /C = F A /C. (3.2.8) Alying the isomorhisms in (3.2.7) and (3.2.8) to (3.2.4), we get the long exact sequence 0 A /C Tor 1 (F, W ) A A /C W F A 0. Note that the ma in the sequence above is the connecting homomorhism Tor 1 (F, A) F A /C in (3.2.4). Chasing the diagram given by (3.2.3), an element x 1 in A is the image of the element (0, x 1 ) in W. Alying the multilication ma to the element (0, x 1 ), we get that (x 1 ) is the first coordinate of (0, x 1 ). Definition Let (0, x 1 ),..., (0, x n ) be elements of W. Define h to be the function A n A /C satisfying the following relation in W. (0, x 1 ) + + (0, x n ) = (h(x 1,..., x n ), x x n ). 20

24 To be able to define the connecting homomorhism exlicitly, we need a concrete formula for the function h from Definition Lemma Let x = (x 1,..., x n ) be a sequence in A n and let Z be the set of non-constant functions {1,..., } {1,..., n}. The following equality holds h(x) = [x z1 x z ]. [z] Z/C Proof. Let X be the sum n x i and observe that [x z1 x z ] = [X x 1 x n ]. z Z Since each C -orbit in Z has exactly elements, this imlies that [x z1 x z ] = [X x 1 x n ]. [z] Z/C For the case n = 1, we have h(x) = 0 = [(x 1 ) x 1 ]. Assume the statement holds for some n = k > 1. By definition of h and the grou law in W, we get k+1 (0, x i ) = k (0, x i ) + (0, x k+1 ) = (h(x), X) + (0, x k+1 ) = (h(x) + f (X, x k+1 ), X + x k+1 ) Multilying the first coordinate in the sum above by yields h(x) + f (X, x k+1 ) = [X x 1 x k ] [(X + x k+1 ) X x k+1 ] = [(X + x k+1 ) x 1 x k+1 ], which is the statement for n = k + 1, comleting the roof. Proosition If Z is the set of non-constant functions {1,..., } {1,..., n} then h(x) = [x z1 x z ]. [z] Z/C Proof. Let B = ZA be the free abelian grou on A and let g : A n A /C be defined by g(x) = [x z1 x z ]. [z] Z/C We will rove the Proosition by showing that h g = 0. Since B = A Z we have B = ( A Z) = A (Z ) = A Z = Z[A ] 21

25 so that B is isomorhic to the free abelian grou on A. Since Z[ ] is the left adjoint to the forgetful functor from abelian grous to sets, it takes colimits to colimits. Thus, B /C = Z[A ]/C = Z[A /C ] so that B /C is the free abelian grou on A /C. The naturality of η = h g now yields the commutative diagram B n A n η B η A B /C A /C. The horizontal mas are given by sending each generator to its corresonding element. Since B /C is torsion free, Lemma imlies that η B is zero and since the to horizontal ma in the diagram above is surjective, we have that η A must also be zero. The sum n(0, x) in W is given by the last roosition as (h(x,..., x), nx), where h(x,..., x) is equal to the sum [z] Z/C [x ]. Since there are exactly n n functions in Z, there are (n n)/ orbits in Z/C. Thus, ( ) n(0, x) = n n [x ], nx. This shows that (x) = [x ]. If is zero in A then since 2, we have = = 1 1 = 1, showing that the connecting homomorhism : A A /C is equal to the ma g. Observe that is zero in A, this imlies that is not zero in W since (0, 1) in W would then be equal to (g (1), 0) which is not zero. What we have discovered is that in the case of an extension of A by A /C reresented by the 2-cocycle f, the connecting homomorhism coincides with a articular given 1-cochain g when is zero in A. An effort to generalize this discovery rovokes the following question: Question. Given a module M and a cocycle ϕ in CN 2 (A, M) satisfying ϕ = δγ for some cochain γ : A M in CN 1 (A, M), will tensoring the resolution P with extensions 0 M E A 0 reresented by ϕ always give long exact 22

26 sequences 0 Tor 1 (F, M) Tor 1 (F, E) Tor 1 (F, A) F M F E F A 0 in which the connecting homomorhism is equal to γ? Observe that the question above assumes a riori that times the reresenting cocycle ϕ is equal to the coboundary of some cochain γ, while the assumtion that is zero in A and M is removed. To answer the question, let us again construct the connecting homomorhism by chasing the diagram 0 M E A 0 0 M E A 0 We start by icking an element a in A in the kernel of, i.e. such that a = 0. This element is the image of the element (0, a) in E. Alying the multilication ma to (0, a), we have that (a) is the first coordinate of (0, a). Recall that the grou oeration in E is determined by the cocycle ϕ in the following way: (m, a) + (n, b) = (m + n + ϕ(a, b), a + b). Calculating (0, a) according to this grou law then gives (0, a) = (ϕ(a, a) + ϕ(2a, a) + + ϕ(( 1)a, a), a) ( 1 ) = ϕ(ia, a), a. Multilying the first coordinate of (0, a) by, we get ( 1 ) 1 ϕ(ia, a) = ϕ(ia, a). By assumtion, ϕ(ia, a) = δγ(ia, a). Substituting this in the sum above yields 1 1 ϕ(ia, a) = δγ(ia, a) For examle if = 2 we have 1 = γ(a) γ(ia + a) + γ(ia) 1 = γ(a) γ((i + 1)a) + γ(ia). 2ϕ(a, a) = γ(a) γ(2a) + γ(a) = 2γ(a). 23

27 In general, the terms γ((j + 1)a) and γ(ia) cancel for j = i 1 whenever 1 < i <. The remaining terms in the sum above are then ( 1 1 ) γ(a) γ((i + 1)a) + γ(ia) = γ(a) γ(a) + γ(a) = ( 1)γ(a) + γ(a) = γ(a), which shows that (a) = γ(a). By modifying the roof of Proosition we see that = γ and we have roved the following theorem. Theorem Let M be a module, A an abelian grou and ϕ a cocycle in H 2 (A, M) reresenting an extension 0 M E A 0. If ϕ has the roerty that ϕ = δγ for some cochain γ in CN 1 (A, M), then the connecting homomorhism in the long exact sequence is equal to γ. 0 Tor 1 (F, M) Tor 1 (F, E) Tor 1 (F, A) F M F E F A 0 24

28 References [1] Kenneth S. Brown. Cohomology of Grous. Sringer-Verlag, [2] Lars Hesselholt. Lecture notes on Witt vectors. htt:// nagoya-u.ac.j/~larsh/aers/s03/wittsurvey.df, [3] Jean-Pierre Serre. Local Fields. Sringer-Verlag,