New York Journal of Mathematics

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1 New York Journal of Mathematics New York J. Math. 6 (000) 54. The Chromatic E -term H 0 M for p>3 Hirofumi Nakai Abstract. In this paper, study the module structure of Ext 0 BP (BP,BP /(p, BP v,v )), which is regarded as the chromatic E -term converging to the second line of the Adams-Novikov E -term for the Moore spectrum. The main difficulty here is to construct elements x(sp r /j; k) from the Miller-Ravenel-Wilson elements (x s 3,r /vj )pk H 0 M. We achieve this by developing some inductive methods of constructing x(sp r /j; k) onk. Contents. Introduction. BP -homology and Bockstein spectral sequence 4 3. Definitions of some elements 8 4. Preliminary calculations Proof of the main theorem 46 References 54. Introduction Let BP be the Brown-Peterson spectrum for a fixed prime p. As is well known, the pair of homotopy groups BP and the BP -homology BP BP forms a Hopfalgebroid ( BP,BP BP )=(Z (p) [v,v,...],bp [t,t,...]). The Adams-Novikov spectral sequence (ANSS) is one of the fundamental tools to compute the p-component of the stable homotopy groups π S X (p) for a spectrum X: E, = Ext BP BP(BP,BP X) = π X S (p). Here, for any BP BP-comodule M, Ext BP BP(BP,M) is regarded as the right derived functor of Hom BP BP(BP,M). We abbreviate Ext s BP BP(BP,M)to H s Mas usual. Received May 5, 999. Mathematics Subject Classification. 55Q45; secondary 55Q5, 55T5, 55N. Key words and phrases. Stable homotopy of spheres, Adams-Novikov spectral sequence, chromatic spectral sequence. ISSN /00

2 Hirofumi Nakai As an important example of finite spectra, we have the Smith-Toda spectrum V (n) for each prime p with BP V (n) = BP /(p, v,,v n ), although its existence is verified only for 0 n 3 and n + p so far. (Recently, Lee S. Nave has shown the non-existence of V ((p +3)/) for p>5.) Then the E -term of the ANSS for V (n) ish BP V (n). Miller, Ravenel and Wilson [] have constructed an algebraic spectral sequence converging to H BP V (n) as follows: Denote the BP BP-comodules BP /(p, v,,v n )bynn. 0 Then, define Nn m (m ) inductively on m by short exact sequences 0 Nn m vm+nn n m Nn m+ 0. We also define Mn m by Mn m = vm+nn n m. Indeed, they can be described directly as Nn m = BP /(p,,v n,vn,,vn+m ), Mn m = vn+mbp /(p,,v n,vn,,vn+m ). Splicing the above short exact sequences, we get a long exact sequence: 0 Nn 0 Mn 0 Mn Mn, called the chromatic resolution of Nn. 0 Applying H ( ) to the above long exact sequence, we obtain a spectral sequence converging to H Nn 0 with E s,t = H t Mn s and d r : Er s,t Er s+r,t r+, called the chromatic spectral sequence. The simplest example in these E -terms is the 0-th cohomology of the n-th Morava stabilizer group H 0 Mn, 0 which is isomorphic to Z/p [v n ± ]. Moreover, H t Mn 0 ( t ) has been computed by Ravenel [5]. In general the calculation of H t Mn s becomes terribly difficult as s + t increases, except for the following case: Theorem (Morava s vanishing theorem). If (p ) n and t>n, then H t Mn 0 =0. Many chromatic E -terms are computed so far (cf. []), most of them are due to Miller-Ravenel-Wilson [], Ravenel [5] and Shimomura s works. In particular, H 0 M n is computed in [, Theorem 5.0] and H 0 M n (n,p > ) is also done in [0, Theorem.]. The purpose of this paper is to prove the following theorem about the k() -module structure of unknown H 0 M for p>3. Theorem. For each Miller-Ravenel-Wilson element x s 3,r+k /vjpk H 0 M (p j and p s) and /v mpr H 0 M (p m), there exists an element x(sp r /j; k) v3 /(p, v ) which is congruent to (x s 3,r/v j )pk mod (v ), and /X,r m BP /(p, v ) congruent to /v mpr mod (v ), so that as a k() -module H 0 M = k() x(sp r /j; k)/v N(s,r,j;k) k() /v a,r X m,r } k 0, r 0,p s Zand p j a 3,r r 0 and p m },

3 The Chromatic E -term H 0 M for p>3 3 where the integers a,r and a 3,r are defined in.3. and 3.3., and the integers N(s, r, j; k) are given as follows: () () (3) (4) (5) (6) (7) (8) (9) [(p 3 + p p )p k 3 ]+[p k 3 ] for r =0and p (s ), (p k+ )/(p ) for r =0,p (s ) and s N, [(p )p k ]+[p k ] for r =0and s N, (p k+ )/(p ) [p k ] for odd r, s N 0 and j = a 3,r, p k + p k ( = a,k ) for odd r, s N 0 and j a 3,r or even r and s N 0, p k+ + p k ( = a,k+ ) for r =,s N 0 and j = p, [(p 3 + p p )p k 3 ]+[p k 3 ] for r =,s N 0 and j p, [(p + p )p k ]+[p k ] for even r and s N 0, [(p 4 + p + p )p k 4 ]+[p k 4 ] for odd r 3 and j = a 3,r, and for odd r 3, s N 0 and j a 3,r, we have (0) () () (3) (4) (5) (6) (7) [(p 3 + p )p k 3 ]+[p k 3 ] for j = a 3,r, same as the case () for a 3,r p j a 3,r, p k for j = a 3,r p, same as the case (7) for j = a 3,r p +, or p (j +) and j a 3,r p, p k +(p k+ )/(p ) for p (j +) and a 3,r p j a 3,r p, [(p + p )p k ]+[p k ] for j = a 3,r p p, or p (j +), p 3 (j+ )(j + p +) and j a 3,r p 3p, [(p 4 + p 3 p +)p k 3 ]+[p k ] for j a 3,r p p and p (j +), p k+ + p k for j a 3,r p p and p (j + p +), Here [x] is the greatest integer which does not exceed x, and N 0 = ap p a }, N = (ap p )p r + p a, r :odd }. In Section we shall review BP-theory and the Bockstein spectral sequence and recall the structure of the chromatic E -term H 0 M computed in []. Then we change the F p -module basis of H 0 M and state the method of getting the structure of H 0 M (originally due to Miller-Ravenel-Wilson). It is enough to read this section for an idea of the theoretical part. In Section 3 we give the fundamental elements u n,k, w n,k and X 3,r, construct the new element /X,r, and give the differentials on these elements (some of them are introduced in [] and [8]). In Section 4 we

4 4 Hirofumi Nakai set up the elements X 0 (v j,xs 3,r) and X(v j,xs 3,r), each of which is congruent to X s 3,r/v j mod (v ), and compute the differentials. We also introduce some inductive methods of constructing x(k) for large k. This is the hardest part of this paper. Using these results, we construct the series of elements x(sp r /j; k) and complete the proof of the main theorem in Section 5. We can deduce some applications to H BP V (0) from this result. These will appear in the forthcoming paper [3]. Acknowledgment. I wish to thank Professor Katsumi Shimomura for leading me to this field, and giving guidance and much helpful advice. I also would like to thank Professor Mikiya Masuda and Professor Zen-ichi Yosimura for their carefully reading the draft version, and for their encouragement. I could not have finished these hard calculations without stimulation from them.. BP -homology and Bockstein spectral sequence In this section we review several basic facts in BP-theory and explain how to determine the structure of H 0 M... Summary of BP -homology and related maps. For a fixed prime p, there is a spectrum BP called the Brown-Peterson spectrum, which is characterized by BP = Z (p) [v,v,,v n, ], BP BP = BP [t,t,,t n, ], where v n = t n =(p n ). The pair (A, Γ)=(BP,BP BP) form a Hopf algebroid with the following structure maps: η L (resp.η R ):A Γ (left (resp. right) unit), c :Γ Γ (conjugation), ɛ :Γ A (augmentation), :Γ Γ A Γ (coproduct). Given a BP BP-comodule M and its BP BP-comodule structure φ : M M A Γ, we use the following notation as usual: H M = Ext Γ(A, M) = H (Ω M,d), where the cobar complex (Ω M,d) is the double graded Z (p) -module with Ω n M = M A Γ A A Γ (n factors of Γ), d n (m γ γ n )=φ(m) γ γ n + ( ) k m γ (γ k ) γ n +( ) n+ m γ γ n. In this paper we compute only the 0-th differential d 0 = η R η L : M M A Γ. By definition, d 0 satisfies d 0 (xy) =d 0 (x)y+η R (x)d 0 (y) for any x, y A. This formula is frequently used for computations in this paper. When d 0 (x) z mod (p, J) for an element z Γ and an ideal J A, we also have d 0 (x p ) z p mod (p, J p ).

5 The Chromatic E -term H 0 M for p>3 5 See also [, Observation 5.8]. To compute d 0, we summarize some known results about η R. Ravenel [4] has shown the following congruence of formal group laws: F t i η R (v pi j ) F v i t pi j mod p. i,j 0 i,j 0 In general η R (v n ) v n +v n t pn v p n t mod I n for the invariant prime ideal I n =(p, v,,v n ). Then, for instance, direct calculation shows that (..) η R (v) i (v + v t p )i iv p vi t i(i ) v p+ v i t p+ mod p, v p+. More precisely, η R (v 3 ) and η R (v 4 ) satisfy the following congruences (cf. [6](4.3.)): (..) η R (v 3 ) v 3 + v t p vp t + v t p v v p t p mod p, v, 3 η R (v 4 ) v 4 + v 3 t p3 vp 3 t + v t p vp + (..3) tp3 mod p, v,v p+... Bockstein spectral sequence. In this paper will deduce the k() -module structure of H 0 M from H 0 M, which has already been computed in []. By definitions of comodules M and M, we obtain the short exact sequence 0 M /v M v M 0. Applying H ( ) to this sequence, we get the long exact sequence 0 H 0 M /v H 0 M v H 0 M δ H M /v. Regarding this long sequence as an exact couple, we get a Bockstein type spectral sequence in the usual way, leading from H M to H M. But as in [] we compute H 0 M directly by making use of the following lemma. Lemma.. (cf. [], Remark 3.). Assume that there exists a k() -submodule B t of H t M for each t<n, such that the following sequence is exact: 0 H 0 M /v B 0 v B 0 δ H M /v /v B N v B N δ H N M, where δ : B t H t+ M is the restriction of the coboundary map δ : H t M H t+ M. Then the inclusion map i t : B t H t M is an isomorphism between k() -modules for each t<n. Sketch of the proof. Because H t M is a v -torsion module, we can filter B t and H t M as P t (m) =x B t v m x=0} and Q t (m) =x H t M v m x=0}. Assume that the inclusion i k is an isomorphism for k t (the t = 0 case is obvious), and consider the following commutative ladder diagram: /v B t δ H t M δ Pt (m) Pt (m ) H t+ M = i t i t i t H t M δ H t M /v v Qt (m) δ Qt (m ) H t+ M. Using the Five Lemma, we can show that P t (m) = Q t (m) (m ) by induction on m. v

6 6 Hirofumi Nakai We shall construct B 0 satisfying the above lemma to determine the k() -module structure of H 0 M. In order to construct a k() -module basis of B 0, it is natural to push each element of H 0 M to H 0 M and to divide it by v as many times as possible. So we need to review the module structure of H 0 M..3. H 0 M and changing its basis. We first recall some notations defined in [] to write down a k() -module basis of H 0 M. Hereafter we assume that p>. Definition.3. ([], (5.) and (5.3)). Define integers a 3,k by a 3,0 =, a 3, = p, a 3,t = pa 3,t, a 3,t+ = pa 3,t +p for t and elements x 3,k v3 BP by x 3,0 = v 3, x 3, = v p 3 vp v 3 v 4, x 3,t = x p 3,t, x 3,t+ = x p 3,t va3,t+ p v (p )pt + 3 for t. Using these notations, Miller-Ravenel-Wilson ([, Theorem 5.0]) have shown the following: Theorem.3.. As a k() -module, H 0 M = k() x m 3,k /v a 3,k k 0, p m Z } F p /v i i }. In this paper we will consider an analogue to Miller-Ravenel-Wilson construction of the elements x m 3,k v 3 BP ([, Section 5]): The elements x 3,k have been defined inductively on k with x 3,0 = v 3, and each of them has the relation x 3,k v pk 3 mod (p, v N ) for a small enough integer N. Motivated by this, we shall construct elements x(k) v3 BP /(p, v ) inductively on k with x(0) = x s 3,r/v j so that x(k) (x s 3,r/v j mod (p, v N )pk ) for a small enough N. Keeping it in mind, we now change the F p -basis of H 0 M to a p k -power basis. Lemma.3.3. As a k() -module, H 0 M } = F p (x s 3,r/v j k 0, r 0,p s Z,and p j a )pk 3,r F p /v i i }. Proof. It is sufficient to prove that any F p -module base x m 3,u/v l ( l a 3,u ) displayed in Theorem.3. can be written as a linear sum (.3.4) x m 3,u/v l = a λ x λ, λ where a λ F p and each x λ has a form (x s 3,r/v j with p s Z and p j a )pk 3,r. We shall show it by induction on mp u. Notice that x m 3,u/v l = v mpu 3 /v l + (elements with v 3 -exponents less than mp u ). When mp u =, the only such a base is v 3 /v, so it is clear. Suppose that each base x a 3,b /vc ( ap b <mp u ) can be expressed as (.3.4) and l = dp e with p d. Because l a 3,u <p u+, we may assume that e u. Define y = x m 3,u/v l ( ) x m 3,u e /v d p e H 0 M. Since the maximal v 3 -exponent in y is less than mp u, y has a form (.3.4) by inductive assumption, hence so does x m 3,u/v. l

7 The Chromatic E -term H 0 M for p> The construction of the module B 0. Let x(sp r /j; k) and y(mp r ) be elements of v 3 BP /(p, v ) satisfying x(sp r /j; k)/v =(x s 3,r/v j )pk /v and y(mp r )/v =/v v mpr in H 0 M, and N(s, r, j; k) (resp. N(m; r)) be the maximal integer such that each x(sp r /j; k)/v i (resp. y(mp r )/v i ) with i N(s, r, j; k) (resp. i N(m; r)) is a cycle of H 0 M. Proposition.4.. As a k() -module, B 0 = k() x(sp r /j; k)/v N(s,r,j;k) k() y(mp r )/v N(m;r) } k 0, r 0,p s Z and p j a3,r } r 0 and p m is isomorphic to H 0 M if it satisfies the following condition for the coboundary map δ : B 0 H M in Lemma..: ( )} ( )} δ x(sp r /j; k)/v N(s,r,j;k) δ y(mp r )/v N(m;r) is F p -linearly independent. Proof. All exactness of the sequence 0 H 0 M B 0 v B 0 δ H M is obvious, but Ker δ Im v. So we need to show only this inclusion. Separate the F p -basis of B 0 into two parts } } A = x(sp r /j; k)/v N(s,r,j;k) y(mp r )/v N(m;r) B = x(sp r /j; k)/v l l<n(s, r, j; k) } y(mp r )/v l l<n(m;r) }. Then it is obvious that δ(x λ ) 0 H M for x λ A, and that δ(y µ )=0 H M for y µ B. Thus for any element z = λ a λx λ + µ b µy µ of B 0 (a λ,b µ Z/p), we have δ(z) = λ a λδ(x λ ). The condition implies that all a λ are zero when δ(z) =0, and so v µ b µy µ /v = z. This completes the proof. We will construct the element x(sp r /j; k) from (x s 3,r/v j in Section 5, and the )pk element /X,r m from /v mpr as a candidate for y(mp r ) in Section F p -linear independence and Coker δ. When we compute the coboundaries δ : B 0 H M of x(sp r /j; k)/v N(s,r,j;k) and y(mp r )/v N(m;r), we expect each of these images to have an appropriate form so that we can judge whether the set of δ-images is F-linearly independent or not in H M for use in Proposition.4.. Though the structure of H M (p >3) has already been computed by Shimomura ([9] and []), we don t need the whole structure of H M for our purpose. We follow the same method as in the proof of [, Theorem 6. (p. 500)]. Consider the long exact sequence H 0 M δ H M3 0 /v H M v H M δ. Since the coboundaries on elements of H 0 M have already been computed in [, Proposition 5.7], we can obtain generators of Ker ( ) v H M. An easy calculation shows that Coker(δ : H 0 M H M3 0 ) is isomorphic to the following direct /v

8 8 Hirofumi Nakai sum as a F p -vector space: } (.5.) F p v spr 3 h 0 either s N 0 and even r 0, or p s and odd r } F p v spr 3 h either s N 0 and odd r, or p s and even r 0 } F p h } F p v tp 3 h t Z K(3) ζ 3 }, where N 0 = s Z p s + and p s + }, and h i is the cohomology class of t pi. Let S be the set of elements A k = v b k 3 h a k /v c k 0 k n, 0 a k, c c n,v b k 3 h a k Coker δ }, and consider the following condition on S: (.5.) (a i,b i,c i ) (a j,b j,c j ) for any two elements with i j. For a linear sum A = n k= λ ka k, assume that A =0inH M and c = c = = c m for some m n. Because H M is a v -torsion module, we can consider the multiplication by v c, so that v c A = m k= λ k(v b k 3 h a k /v ) = 0 in Ker v ( = Coker δ). By the condition (.5.), the set v b k 3 h a k } is linearly independent in Coker δ and thus all λ k ( λ k m) should be zero. Iterating this, we conclude that all coefficients λ k are zero and S is linearly independent in H M. 3. Definitions of some elements In this section we introduce the elements u n,k, w n,k, X 3,r and /X,r. We will use these to define many elements in Section Moreira s element u n,k and Shimomura s element w n,k. Definition 3... As is done in [, 6(4)] or [8,.8], we define the element u n,k vn BP by the following recursive formula: u n,0 = vn and v n+i u pi n,j = 0 for k. i+j=k Remark 3... In [] u n,k is defined only for k n and denoted as u n+k. By definition u n, = vn p v n+, u n, = v p p n v p+ n+ v p n v n+, u n,3 v p3 p p n v p +p+ n+ + v p3 p n (v p n+ v n+ + v n+ v p n+ ) v p3 n v n+3 mod (p), and so on. Computing the right unit η R on u n,n, Moreira has shown that the K(n) -module base ζ n of H M 0 n is homologous to ζ p n: Proposition 3..3 ([], Theorem 6...). η R (u n,n ) η L (u n,n ) ζ n ζ p n mod I n.

9 The Chromatic E -term H 0 M for p>3 9 We now recall the elements w n,k introduced by Shimomura [8]. Define elements T j (j 0) by T 0 = and j i=0 t iη R (v pi j i ) j i= v it pi j i mod (p) for j, and an element e n (x) vn mod I n. BP BP/I n for x v n BP by the congruence η R (x) e n (x) Definition 3..4 (cf. [8],.0). Define elements w n,k vn BP BP (n ) inductively on k by w n,0 = 0 and w n,k = Remark By definition, w n, = vn t pn, w n, vn p v n+ t pn + vn p k j= (t pn e n (u pj pn n,k j )Tj for k. )+vn t pn + t pn +p n mod p, v pn. In general, as an element of M A Γ w,k =( ) k )/(p ) k v(p 3 t p v (pk )/(p ) + (elements killed by v (pk )/(p ) ). We note that w, is similar to ζ = v p v 3 t p + v p (tp +p tp )+v t.in fact w n,n (n ) is related to ζ n by the following congruence (cf. [8, 4.8]): ζ n w n,n i<j n u pj i n,n j t k c(t pk i k j j k ) pn i mod I n. In particular we have ζ w, v (t t p+ )modi for n =. Using the notation w n,k, Shimomura [8] has proved the following proposition, which is a generalization of Moreira s result: Proposition 3..6 ([8], Proposition.). For n, η R (u n,k ) u n,i t pi j w p n,k v n w n,k+ η R (vn ) mod I n +(v p n ). i+j=k 3.. Shimomura s element X 3,r. The elements x n,r vn BP (n and r 0) have been defined in [, (5.)] to express a basis of H 0 Mn (x 3,r was listed in.3.). As is shown in [, Proposition 5.7], the differentials d 0 : v n v n BP A Γmod(p, v,v +a3,r ) are given by BP d 0 (x 3,0 ) v t p and d 0 (x 3,r ) v a3,r x p tpδ(r) 3,r for r, where δ(r) = 0 for odd r and for even r. However, we need to calculate d 0 (x 3,r ) mod (p, v,v) t with t>+a 3,r for our computation. So we use the following elements X 3,r defined by Shimomura [8] instead of x 3,r.

10 30 Hirofumi Nakai Definition 3... cf. [8, (3.3)]] Define elements X 3,r v 3 BP by X 3,0 = v 3, X 3, = X p 3,0 +vp vp 3 u 3,, X 3, = X p +p 3, vp X p 3, u 3,, X 3,3 = X p 3, +v+a3,3 X p 3, u, + v b(3) X p p 3, u 3,3, X 3,4 = X p 3,3 vb(4) X p p 3, (u, u 3,3 ), X 3,r = X p 3,r + v+a3,r X p 3,r u, v b(r) X p p 3,r (u, u 3,3 ) for odd r 5, X 3,r = X p 3,r vb(r) X p p 3,r ( u, u 3,3 ) for even r 6. with b(0)=,b() = p + and b(r) =a 3,r + a 3,r +=(p +p+)p r for r. Remark 3... By definition, it is obvious that X 3,r x 3,r addition, they satisfy the congruences mod (v +a3,r ). In X 3,i+ X p 3,i for i. X 3, v p 3 mod v p, X 3, X p 3, mod v p +p, mod v a3,i+ p, X 3,i+ X p 3,i+ mod v b(i+) p p Then d 0 (X 3,r ) is computed as follows. Let s be an integer with p s. Proposition Mod (p, v,v b(r)+ ), d 0 (X3,r) s may be expressed as follows. For small values of r, sv v3w s 3, ( sv v3 s t p ), for r =0 d 0 (X3,r) s = sv p vsp 3 t sv p+ v sp 3 w 3,, for r = sv a3, X sp 3, t p svb() v sp p 3 (u,0 t w 3,3 ), for r =. For r =3, d 0 (X3,r) s =sv a3,3 X sp 3, t sv b(3) X sp p 3, w, + u,0 t w 3,3 + u 3,0 (t p t t 3 )}. For r 4 and even, d 0 (X3,r) s =sv a3,r X sp 3,r tp svb(r) X sp p 3,r ( ζ + u,0 t ζ 3 ). For r 5 and odd, d 0 (X3,r) s =sv a3,r X sp 3,r t sv b(r) X sp p ( ) 3,r ζ + w, ζ 3. Proof. Originally, this is proved in [8, Proposition 3.]. Notice that it is sufficient to prove the s = case because d 0 (X3,r) s sx3,r s d 0(X 3,r ). For r =0,d 0 (X 3,0 ) v t p mod p, v,v by (..). For r =, Proposition 3..6 shows that d 0 (v p vp 3 u 3,) v p vp 3 t v p tp vp+ v p 3 w 3, mod (p, v,v p+ ). Summing this with d 0 (X p 3,0 ) vp tp3 gives d 0 (X 3, ). For r =, we have d 0 ( v p +p v p p 3 u 3, ) v p +p v p p 3 (u 3, t p +u 3,0t w p 3, v v 3 w 3,3)

11 The Chromatic E -term H 0 M for p>3 3 mod (p, v,v p +p+ ). On the other hand, using the congruence v p p 3 X p 3, + v p p vp 3 u 3, we have d 0 (X p 3, ) vp X p 3, tp + +p vp v p p 3 (u 3, t p wp 3, ). Again, summing the terms gives d 0 (X 3, ). For r =3,wehave d 0 (v p3 +p X p 3, u,) v +a3,3 X p 3, (u,0t w p, ) v b(3) X p p 3, (e (u, )t p + u,0t + u 3,0 t p t w 3,3 ), d 0 (v b(3) X p p 3, u 3,3 ) v b(3) X p p 3, ( i=0 u 3,i t pi 3 i wp 3,3 ). mod (p, v,v b(3)+ ). On the other hand, by the congruences X p p 3, X p 3, v p +p X p p 3, u 3, and v p3 p p 3 X p p 3, + v p p Xp 3, u 3,,wehave d 0 (X p 3, ) vp3 Xp 3, tp vb(3) X p p 3, (u p,0 tp wp 3,3 + Using a congruence for w, of Remark 3..5, we obtain d 0 (X 3,3 ). For r =4,wehave d 0 ( v b(4) X p p 3, u, ) v b(4) X p p 3, ( ζ p up,0 tp +ζ w 3,3 i<j 3 i= u 3,i t pi 3 i ). u pj i 3,3 j (t i c(t pi j i ) ))p i and d 0 (v b(4) X p p 3, u 3,3 ) v b(4) X p p 3, (ζ 3 ζ p 3 )mod(p, v,v b(4)+ ). On the other hand, by the congruences X p p 3, X p +p 3,3 +vp3 X p p 3, u, and X p3 p p 3, X p p 3, v p +p X p3 p p 3, u 3,, d 0 (X p 3,3 ) is congruent to v a3,4 X p 3,3 tp vb(4) X p p 3, u, t p +u 3,t p +wp, +up,0 tp wp 3,3 +up 3,0 (tp3 tp tp 3 )} mod (p, v,v b(4)+ ). Moreover, noticing the congruences w p 3,3 ζ 3 ζ p 3 (u 3, t p + u 3,0t 3 )+u p 3,0 tp3 tp u p 3,0 tp 3 i<j 3 i<j 3 we have d 0 (X p 3,3 ) va3,4 X p 3,3 tp vb(4) X p p 3, v b(4) X p p 3, (ζ 3 ζ p 3 i=0 u 3,i t pi 3 i, u pj i 3,3 j [t i+ c(t pi+ j i, )]p i u pj i 3,3 j [t i+ c(t pi+ j i, )]p i ( u, t p + wp, + up,0 tp ) i<j 3 u pj i 3,3 j [ i+ k j t k c(t pk j k )]p i).

12 3 Hirofumi Nakai We obtain d 0 (X 3,4 ) in the desired expression by summing the terms and using the congruences u, t p + wp, ζp ζ +u,0 t, u pj i 3,3 j ( t k c(t pk j k )p i) ζ 3 w 3,3. i<j 3 i k j The r 5 cases are proved by induction on r. By Proposition 3..3, we have congruences d 0 (v b(r) X p p 3,r u, ) v b(r) X p p 3,r (ζ ζ p ), d 0 (v b(r) X p p 3,r u 3,3 ) v b(r) X p p 3,r (ζ 3 ζ p 3 ). We also have the congruences X (p )p 3,r X p 3,r and d 0 (v +a3,r X p 3,r u,) v +a3,r X p 3,r (u,0t w p, ) vb(r) X p p 3,r t e (u, ) for odd r, and X (p )p 3,r X p 3,r +v+a3,r X p p 3,r u, for even r. Easy calculation shows the desired results for r 5. Moreover, we can obtain the following result in the same way as the proof of Proposition Proposition Mod (p, v,v b(r) ), d 0 (X3,r) s is expressed as follows. For small values of r, weget sv a3, X sp 3, t p, r =; sv a3,3 X sp 3, (t v w, ) sv b(3) X sp p 3, v ζ v t p (w, + v t ) }, r =3; sv a3,4 X sp 3,3 t p svb(4) X sp p 3, v ζ v (w,3 + t p ζ )}, r =4. For r 5 and odd, we get sv a3,r X sp 3,r (t v w, ) sv b(r) X sp p 3,r v ζ v (w,3 + t p w,)}. For r 6 and even, we get sv a3,r X sp 3,r tp svb(r) X sp p 3,r v ζ v w,3 } The element /X,r m. Here we modify x,r into X,r in the analogous way of modifying x 3,r into X 3,r, and introduce new elements /X,r. m Definition Define integers a,r by a,0 = and a,r = p r + p r for r, and elements X,r v BP by X,0 = v, X, = X p,0 +vp vp u,, X, = X p, +v+a, X p, (u, u, ), X,r = X p,r + v+a,r X p,r ( u, u, ) for r 3. Then we have: Proposition Mod (p, v +a,r ), mv v m t p + ( ) m v v m t p for r =0 d 0 (X,r) m mv p vmp (t v w, ) for r = v a,r v (mp )pr ( t v ζ ) for r.

13 The Chromatic E -term H 0 M for p>3 33 Proof. It is sufficient to prove the s = case because d 0 (X m,r) mv (m )pr d 0 (X,r ). The r = 0 case is given by (..). For r =,wehave d 0 (v p vp u,) v p d 0(u, )v p +η R(u, )d 0 (v p )} v p vp d 0(u, ) v p (vp t t p ) vp+ v p w, mod p, v p+. Summing this with d 0 (X p,0 ) gives d 0(X, ) v p vp (t v w, ). (See also [, Proposition 5.4].) For r =,wehave d 0 (v +a,r X p, (u, u, )) v p Xp, (vp t t p )+vp (wp, ζ )} mod p, v +a,. Again, summing this with d 0 (X p, ) gives For r 3, we have d 0 (X, ) v a, v p p (t v ζ ). d 0 (v +a,r X p,r (u, u, )) v a,r p+ X p,r (vp t t p )+vp (ζp ζ )} mod p, v +a,r. This gives d 0 (X,r ) inductively on r. Shimomura [7] has replaced /v mp with /x m, in computing H 0 M0 for p =. Analogously, we construct elements /X,r m as a substitute for /v mpr. Notice that our /X,r m should be different from X,r m because there is no inverse element of X,r (r ) in v BP. Definition Define elements /X m,r v BP (r 0 and p m ) by /X,0 m =/v m and /X,r m =(/X mp,r ) Xm,r (/X mp,r ) for r. For example, easy calculation shows the congruence /X m, /v mp + mv p v 3/v (m+)p+ mod (v p ). Although our /X,r m is not equal to (X,r) m = k=0 ( Xm,r) k, the above definition is justified by the congruences (X, m v mp ) 0mod(p, v p ) and (Xm,r X mp,r ) 0mod(p, v (a,r p) ) for r. Hereafter we denote Y (/X,r) m by Y/X,r m and (/X,r)/v m l by /vx l,r m for simplicity. Proposition Mod (p, v +a,r ), mv t p /vm+ + ( ) m+ v t p /vm+ for r =0 d 0 (/X,r) m mv p (t v w, )/v mp+ for r = mv a,r ( t v ζ )/v (mp+)pr for r. Proof. We prove that the all above differentials have the form d 0 (X m,r)/v mpr.

14 34 Hirofumi Nakai The case r = 0 is easy. For r, it is shown by induction on r. In fact, we have d 0 (/X,r) m =d 0 (/X mp,r ) d 0(X,r)/X m mp,r η R(X,r)d m 0 (/X mp,r ) d 0 (/X,r ) m p d 0 (X,r)/v m mpr v mpr d 0 (/X,r ) m p d 0 (X,r ) m p /v mpr d 0 (X,r)/v m mpr + d 0 (X,r ) m p /v 3mpr d 0 (X m,r)/v mpr + d 0 (X m,r ) v mpr d 0 (X m,r ) } p /v 3mpr mod p, v +a,r. By the congruence d 0 (X,r) m mv (m )pr d 0 (X,r ), the second term is trivial. Then we directly obtain the v -divisibilities of /v X m,r (= /v v mpr )inh 0 M. Corollary /v X,r m can be divided by v a,r in H 0 M, and the image of /v a,r X,r m under the coboundary map δ : H 0 M H M is mh /v m+ for r =0 δ(/v a,r X,r) m = mh 0 /v mp+ for r = mh 0 /v (mp+)pr for r. This corollary asserts that N(m; r) =a,r when we choose /X m,r to be y(mp r ) in Proposition.4.. Using.5, we easily see that δ(/v a,r X m,r) } is linearly independent. 4. Preliminary calculations In Section 5 we shall construct elements x(sp r /j; k) v 3 BP /(p, v ) and compute the differentials on them. For the sake of this, we define some elements and compute differentials in Subsection 4.. Based on these results, we describe some inductive methods of constructing the elements x(sp r /j; k) in Subsection 4.. This is the hardest part in this paper. Hereafter we assume that p>3, and that all elements are either in v 3 BP /(p) or in v 3 BP /(p, v ). We shall compute the differential d 0 for positive integers e and e in two ways: d 0 : v3 /(p) v3 /(p) A Γ mod v e,ve, d 0 :v3 /(p, v ) v3 /(p, v ) A Γ mod v e. 4.. Some lemmas. Lemma 4... Assume that an element X satisfies d 0 (X) v e Xs 3,rt p /ve+ mod v e+, where r, e and e a 3,r. Then Y = X p v pe v 3 X sp 3,r /v(e+)p+ satisfies d 0 (Y ) v pe X sp 3,r (t v w, )/v pe+ mod v pe+.

15 The Chromatic E -term H 0 M for p>3 35 Proof. We observe that d 0 (v 3 X sp 3,r /v(e+)p+ )=d 0 (/v (e+)p+ ) v 3 X sp 3,r + η R(/v (e+)p+ ) d 0 (v 3 X sp 3,r ) v v 3 X sp 3,r tp /v(e+)p+ +(/v (e+)p+ v t p /v(e+)p+ ) d 0 (v 3 X sp 3,r ) mod p, v. Since (e +)p+<pa 3,r, it is sufficient to compute d 0 (v 3 X sp 3,r )mod v,v pa3,r. Using (..) and Proposition 3..3, wehave d 0 (v 3 X sp 3,r )=d 0(v 3 )X sp 3,r + η R(v 3 ) d 0 (X sp 3,r ) d 0(v 3 ) X sp 3,r (v t p vp t + v t p )Xsp 3,r. Observing the congruence w, v 3 t p /vp+ +(t p tp +p )/v p + tp+ /v (Remark 3..5), we obtain d 0 (v 3 X sp 3,r /v(e+)p+ ) X sp 3,r tp /v(e+)p X sp 3,r t /v pe+ + v X sp 3,r w,/v pe+. Now the result follows easy calculation. Given a positive integer j = cp b a 3,r (0 b<p), it is convenient to work with the element vx b sp 3,r /Xc, instead of X3,r/v s j. Define the integer a by a c (p) with 0 a<p. Lemma 4... Assume that c p for r =, that c a 3,r for odd r 3, or that c a 3,r p for even r 4. Then d 0 (v b X sp 3,r /Xc,) is expressed as X sp 3,r v j+b b E(b, i)} i= vp Xsp 3,r v j+ (a + b) t v [aw, +ab v 3t p ) v m n v p+ b(a + b ) tp+ ]} v mod v p+, where E(m, n) = ( m n (v t p )n. In particular, d 0 (X sp 3,r /Xc,) av p Xsp 3,r (t v w, )/v j+ mod (v p+ ). Proof. Notice that d 0 (vx b sp 3,r /Xc,) =d 0 (v)x b sp 3,r /Xc, + η R (v) b d 0 (X sp 3,r /Xc,), and that η R (v) b is given in (..). By the assumption on c, wehave d 0 (X3,r /v s ) c av X3,r t s p /vc+ mod (v). Then, Lemma 4.. gives d 0 (X sp 3,r /Xc,) mod(v p+ ). Similarly to the r case, we may consider the element X3,/X s, instead of for r =. X s 3,/v p Lemma d 0 ( X s 3, /X, ) (s ) v p Xs 3,t /v p+ + ( ) s v p vsp 3 t /v mod (v p+ ).

16 36 Hirofumi Nakai Proof. Notice that d 0 (X3,/X s, )=d 0 (/X, ) X3, s + η R (/X, ) d 0 (X3,), s and that d 0 (/X, ) X3, s v p Xs 3,t /v p+ mod (v p+ ) by Proposition On the other hand, we have η R (/X, ) /v p + vp (v 3 v p t )/v p+ } mod v p+, d 0 (X3,) s sv p Xs 3, t p tp η R(v3 v 4) mod v p+,v p }, sv p Xs 3, t p3 d 0(v3 v 4) + ( ) s v p vsp 3 t mod v,v p+. Using these congruences, (..) and (..3), we obtain ( ) η R (/X, ) d 0 X s 3, sv p Xs 3,t /v p+ + ( ) s v p vsp 3 t /v mod v p+. Collecting terms gives the result. For an integer n with p n +, we denote the integer (n +)/p by n, and the set of integers s Z s = s p with p s } by N 0. Then we introduce two elements X 0 (v j,xs 3,r) and X(v j,xs 3,r) for s N 0, each of which is congruent to X s 3,r/v j mod (v ). Definition When s = s p N 0, we define X 0 (v j,xs 3,r) by X 0 (v j,xs 3,r) = X3,r/v s j + jv X3,r+/s s v j++a3,r+. Then we have: Lemma For odd r and p j a 3,r, d 0 (X 0 (v j,xs 3,r)) is expressed as ( j) v v a3,r j v (sp )pr 3 ζ mod v for r 3 and a 3,r p j a 3,r, and is expressed as ( ) j(j +)vx 3,r+t s p j + /s v j++a3,r+ v X3,rt s p /vj+ + ε(r, j) mod v 3 for r =and j p ; or for r 3 and either j = a 3,r p + or j a 3,r p, where ε(,j)=0and ε(r, j) =vv a3,r j X sp 3,r a(r)j w,3+b(r, j) t p ζ } for r 3 with integers a(3)=,a(r)=for r 5 and b(r, j) =j +( a(r)) j. Proof. Here we prove only for r 3 case (the r = case is easier). We compute the differential on each term of X 0 (v j,xs 3,r). For the first term, notice that d 0 (X3,r/v s j )=d 0(/v j )Xs 3,r + η R (/v j )d 0(X3,r). s Easy calculation shows that d 0 (X3,r) s v a3,r X sp 3,r (t v w, + vv tp w,) mod v,v 3 +a3,r. Using this and Proposition 3.3.4, we see that d 0 (X3,r/v s j ) is congruent to j v X3,rt s p v (j +)v t p }/vj+ + v v a3,r j X sp 3,r ζ v (j +)v t p } mod (v). 3 On the other hand, using Propositions 3..4 and 3.3.4, we observe that d 0 (jv X3,r+/s s v j++a3,r+ ) is congruent to j(j +)vx 3,r+t s p /s v j++a3,r+ + jv X3,rt s p v (j +)v t p }/vj+ + v v a3,r j X sp 3,r j v ζ +a(r)jv w,3 +j(j+3 a(r))v t p ζ } mod (v). 3 Collecting two terms gives the result for r 3.

17 The Chromatic E -term H 0 M for p>3 37 Definition For s = s p N 0 and j a 3,r, we define X(v j,xs 3,r) by X(v j,xs 3,r) =vx b sp 3,r /Xc, bv v b X s p 3,r /s X c+a3,r, + bv p+ X sp 3,r u,/v j+ (a(r) )b v p+ v a3,r j p X (sp )p 3,r u,3, where j = cp b and 0 b<p. Moreover, we define DX [m,n] (v e,ve+,x3,r) s v3 BP /(p, v ) A Γ for 0 m<p,0 n<p,s=s p N 0, and positive integers e and e by DX [m,n] (v e,ve+,x3,r) s = (m+n)v e X3,rt s /v e+ + n(m + n )v e 3,r+t Xs /s v e++a3,r+ + v e Xs 3,r/v e+ nζ +(m n)w, } Then we obtain the following lemma. Lemma Assume that r 3 is odd and c a 3,r. Then d 0 (X(v j,xs 3,r)) is expressed as DX [a,b] (v p+,v j+,x3,r)+ s A(v j,xs 3,r) + a(r)b v v a3,r j p v (sp )pr 3 (v p+ b ζ p η R(v b )+v p θ (v j )) mod v p+. Here b and c are the integers in 4..6, a c (p) with 0 a<p, and the element A(v j,xs 3,r) is defined to be 0 for 0 b and to be X sp 3,r v j+b b ( i)e(b, i)} b i= v X s p 3,r s v j+b+a3,r+ for b. Here E(m, n) is as in Lemma 4.. and E(b,i)} b θ (v j )=(a )(v 3 v p t )ζ p + (a + b ) vp t v t p }u,. In particular, v p θ (v a3,r p p )= v p+ 3 (t + v w 3, )/v. Proof. The differential on the first term of X(v j,xs 3,r) has already been computed in Lemma 4... For the second term, notice that bv v b X s p 3,r /s X c+a3,r, is congruent to bv X s p 3,r /s v j++a3,r+ b(a )v p+ v 3 X s p 3,r /s v j+p++a3,r+ mod (v p+ ). Using Propositions 3..3, 3..4 and 3.3.4, we obtain d 0 ( bv v b X s p 3,r /s X c+a3,r, ) bv X s p 3,r /s v j+b+a3,r+ (v + v t p )b v b } bv X sp 3,r tp (v + v t p )b /v j+b + a(r)bv v a3,r j b+ ζ p η R(v b X (sp )p 3,r ) +(a+b )bv p+ X s p 3,r t /s v j++a3,r+ + bv p+ X sp 3,r /vj+ w p, +(b )tp+ /v +(a )(t p +p t p )/vp +(a )w,} i=

18 38 Hirofumi Nakai + bv p+ v a3,r j p X (sp )p 3,r v t p ζp +a(r)(a )ζp (v 3 v p t ) ( a(r) )v w p,3 } mod (v p+ ). For the third term, notice that d 0 (bv p+ X sp 3,r u,/v j+ ) bv p+ d 0 (X sp 3,r u,)/v j+ mod (v p+ ). Propositions 3..3 and 3..3 give d 0 (X sp 3,r u,), and we obtain d 0 (bv p+ X sp 3,r u,/v j+ ) bv p+ X sp 3,r (ζ ζ p )/vj+ bv p+ v a3,r j p X (sp )p 3,r t p (u, ζ p ) mod (v p+ ). For the last term, we may assume that r 5 because a(3) =0. Then it follows that d 0 (v a3,r j p X (sp )p 3,r u,3 ) v a3,r j p X (sp )p 3,r (u, t p wp,3 ) mod v by Propositions 3..6 and 3..3, and hence d 0 ( (a(r) )bv p+ v a3,r j p X (sp )p 3,r u,3 ) (a(r) )v p+ v a3,r j p X (sp )p 3,r (u, t p wp,3 ). Collecting four terms gives d 0 (X(v j,xs 3,r)) X sp 3,r bv X s p 3,r (v + v t p )b v b bv t p (v + v t p )b } /v j+b + a(r)bv v a3,r j b+ (v + v t p )b v b } /s v j+b+a3,r+ X (sp )p 3,r ζ p η R(v b + b(a + b )v p+ X s p 3,r t /s v j++a3,r+ (a + b)v p Xsp 3,r t /v j+ + v p+ +a(r)bv p+ v a3,r j p X (sp )p 3,r Then, apply the equation (X + Y ) b X b by (X + Y ) b = ) X sp 3,r /vj+ bζ +(a b)w, } (a )ζ p (v 3 v p t ) v t p u, b ( ) b ( i) X b i Y i i to the first term, and the congruence X s p 3,r Xs 3,r+ + a(r)s v b(r+) X sp 3,r u, mod v b(r+) for odd r 3 to the fourth term. Modifying these terms completes the proof. We also set X(v j,xs 3,)tovX b 3,/X s, bv v b X3,/s s X p+, for j = p b. Then we have: Lemma d 0 (X(v p,x3,)) s is expressed as 3v p Xs 3,t /v p mod vp+, while d 0 (X(v p,x3,)) s is expressed as vx 3,(t s p vp t )/s v p +p vx 3,/v s p t p /v +4v p t v p (ζ +t /v + C )} + v p+ Z/v p i= }.

19 The Chromatic E -term H 0 M for p>3 39 mod v p+ for some Z v3 BP BP, where C =(u 3,0 t p + w 3, )t p +(u 3, w p 3, )tp +up 3,0 tp 3. Proof. Notice that /X, m /v mp + mv p v 3/v (m+)p+ mod (v p ). So we may compute d 0 on X3,/v s p b + v p v 3X3,/v s p+ b bv X3,/s s v p +p+ b bv p+ v 3 X3,/s s v p +p+ b instead of d 0 (X(v p b,x3,)). s For b = case, we consider the differential mod (v p+ ), so that the last term can be omitted. Because X3, s v sp 3 + vp v(s )p 3 v 4 mod v p and X3, s X s p 3, mod +p vp, we may compute d 0 on v (v3/v s ) p + v p vsp+ 3 /v p + vp v(s )p 3 v 4 /v p v (X3,/s s v p+ ) p. Then, we obtain the desired result using (..) (..3), Proposition 3..3 and Proposition Similar but harder calculation shows the b = case. Next we define X + (v jp,xsp 3,r ) to modify the differential on X 0(v j,xs 3,r) p. Let A 0 = ( ) j+ v p v 3X sp 3,r /v(j+)p+ + j(j +)v p v 3X s p 3,r+ /s v (j++a3,r+)p+ }, A = v p v(a3,r j )p X (sp )p 3,r a(r)jv u,3 (j +)v 3 u,, A = v p v (sp )p+ 3 /v. Using these elements, we define X + (v jp,xsp 3,r ) as follows. For r =, X + (v jp,xsp 3,r )= A 0 for j p 3 A 0 + A for j = p. If r 3 is odd and either j = a 3,r p +orj a 3,r p, then X + (v jp,xsp 3,r )=A 0+A. Then we obtain the following lemma. Lemma Mod (v p+ ), d 0 (X 0 (v j,xs 3,r) p + X + (v jp,xsp 3,r )) ( ) j(j+) vp 3,r+t Xs j + v p Xsp 3,r t s v jp++a3,r+ v jp+ (j )v p v(a3,r j )p X (sp )p 3,r θ (v j ), where θ (v j )=0for r =and θ (v j )=(j+ )(v 3 v p t )ζ p + (j +)vp t v t p }u, for odd r 3. In particular, v p θ (v a3,r p )= v p+ 3 (t + v w 3, )/v. Proof. Here we prove only for r 3 case. Other cases are similarly proved. Notice that d 0 (A 0 ) is congruent to ( j + ) v p d 0(v 3X sp 3,r )/v(j+)p+ + j(j +)v p d 0(v 3 X s p 3,r+ )/s v (j++a3,r+)p+

20 40 Hirofumi Nakai mod (v p+ ), so that it is sufficient to compute d 0 (v3x sp 3,r )mod(v,v (j+)p+ ) and d 0 (v 3 X s p 3,r+ )mod(v,v (j++a3,r+)p+ ). Easy computations show that d 0 (v3x sp 3,r ) (v t p + v p t +v v 3 t p vp v 3t v p+ t p + )X sp 3,r, d 0 (v 3 X s p 3,r+ ) (v t p vp t )X s p 3,r+ + s v pa3,r+ X sp 3,r tp (v 3 + v t p vp t ) s v b(r+) X (sp )p 3,r ζ p (v 3 + v t p vp t ). Using these and the congruence X3,r+ s X s p 3,r+ s v a3,r+ p v 3 X sp 3,r s v b(r+) X sp 3,r u,, we can show that d 0 (A 0 ) is congruent to ( ) j + v p Xsp 3,r (tp v p t ) v (j+)p + j(j +) vp p (Xs 3,r+ tp vp X3,r+t s ) s v (j++a3,r+)p j(j +)v p v(a3,r j )p 3,r ζ p (v 3 + v t p vp t )+v p t u, }. On the other hand, d 0 (A ) is congruent to X (sp )p jv p v(a3,r j )p d 0 (a(r) X (sp )p 3,r u,3 ) (j +)v p v(a3,r j )p d 0 (v 3 X (sp )p 3,r u, ). Noticing that (a 3,r j )p (p 3)p, wehave d 0 (X (sp )p 3,r u,3 ) d 0 (u,3 )X (sp )p 3,r + η R (u,3 )d 0 (X (sp )p 3,r ) (u, t p wp,3 )X(sp )p 3,r +(u,3 +u, t p wp,3 )d 0(X (sp )p 3,r ) mod (v ). Because d 0 (X (sp )p 3,r ) v pa3,r X (sp p )p 3,r t p mod (v,v pa3,r + ), the second term remains only for r = 3. Observe that u,3 and w p,3 can be replaced with v p3 v p 3 u, and v p3 v p 3 ζp respectively. So we obtain d 0 (a(r) X (sp )p 3,r u,3 ) a(r) X (sp )p 3,r (u, t p wp,3 ) + a(r) v (a3,r p )p v p p )p 3 X(sp 3,r t p (u, ζ p ) a(r) X (sp )p 3,r (u, t p wp,3 )+( a(r)) X(sp )p 3,r t p (u, ζ p ) X (sp )p 3,r u, t p a(r) wp,3 ( a(r)) tp ζp }. Moreover, it is easy to see that d 0 (v 3 X (sp )p 3,r u, ) X (sp )p 3,r v t p u, v p t u, ζ p (v 3 + v t p vp t )}. Collecting terms gives d 0 (X + (v jp )) as ( ) j + v p Xsp 3,r (tp v p t ) v (j+)p,xsp 3,r + j(j +) vp p (Xs 3,r+ tp vp X3,r+t s ) s v (j++a3,r+)p ε(r, j) p (j )v p v(a3,r j )p X (sp )p 3,r θ (v j ). This completes the proof for r 3 case.

21 The Chromatic E -term H 0 M for p> Some inductive methods of constructing elements x(sp r /j; k). Here we describe some inductive methods of constructing x(k) =x(sp r /j; k). Because (x s 3,r/v j =(X3,r/v s j )pk for j a )pk 3,r, we may start with X3,r/v s j (p j a 3,r). In general, each x(k) (k ) is constructed by adding some appropriate terms to x(k ) p so that we can find an element of Coker δ (.5.) from the numerator of the leading term of d 0 (x(k)). There are some patterns for adding terms and constructing x(k) for a large enough k. We first observe the case with small j and s N 0, which is the simplest example of such patterns. Proposition 4... Assume that an element W satisfies d 0 (W ) jv p Xsp 3,r t /v jp+ mod v p+ for j a 3,r, then W k (k ) defined inductively on k by satisfies W k = W p + jva, p v X sp 3,r /Xjp+, for k =, = W p k +jva,k p X spk 3,r for k 3 d 0 (W k ) jv a,k X spk 3,r t /v (jp+)pk mod v a,k+. Proof. Using Propositions 3..3 and 3.3.4, for k =, wehave ( ) d 0 jv a, p v X sp 3,r /Xjp+, jv a, p+ X sp 3,r (tp vp t )/v (jp+)p, while for k 3, we have ( ) d 0 jv a,k p X spk /v(jp+)pk 3,r jv a,k p+ X spk 3,r (tp vp t )/v (jp+)pk. The result now follows by an easy calculation. We will see that each x(sp r /j; ) with r, j a 3,r and s N 0 satisfies the condition for W in the above proposition. In general, we speculate on the existence of some rules in constructing elements x(k) for large k as the above case. Next we observe another pattern, which occurs in most cases. To state this, we first introduce some elements. Definition 4... For e, e, s = s p N 0 and odd r 3, we define Y [a,b] (v ep,v (e+)p,x3,r+) s as follows: for s N 0,itisgivenby (a+b) v(e )p v X sp 3,r X e+, + a vep X sp 3,r u, v (e+)p (a + b)(b ) v ep X s 3,r+ s v (e+)p+a3,r+ ; for s N 0,itisgivenby (above terms) + a v ep X3,r+3 s s v (e+)p+b(r+3).

22 4 Hirofumi Nakai Moreover, we define Z(v epi,v (e+)pi,x s 3,r+i )by v epi v 3 X s p 3,r+i s v (e+)pi +p++a 3,r+i+ v epi X sp 3,r+i u, v (e+)pi i vep X3,r+i s v (e+)pi i vep X sp 3,r+i u, v (e+)pi (3u, u 3,3 ) fors N 0, and DZ(v epi,v (e+)pi,x s 3,r+i ) v 3 BP /(p, v ) A Γby v epi X s 3,r+i+ t s v (e+)pi ++a 3,r+i+ v epi X3,r+i+ s tp s v (e+)pi ++a 3,r+i+ i vep X3,r+i s v (e+)pi i + vep X s 3,r+i v (e+)pi i + vep X3,r+i s v (e+)pi for s N 0 and odd r + i, for s N 0 and even r + i, (w, ζ ) for s N 0 and odd r + i, ( t ) ζ v (3ζ ζ 3 ) fors N 0. for s N 0 and even r + i, Notice that each of these leading terms includes an element of Coker δ. Proposition Assume that an element x(k) satisfies d 0 (x(k)) DX [a,b] (v e,ve+,x3,r)+v s e Z k/v e mod v e+ for some Z k v3 BP BP, with e b(p)and odd r 3. Let x(k+)=x(k) p +Y [a,b] (v ep,v (e+)p,x3,r+). s We compute d 0 (x(k + )). Fora 0,weget d 0 (x(k+ )) adz(v ep,v (e+)p,x3,r+)+v s ep Z k+ /v (e+)p mod v ep+ for some Z k+ v 3 BP BP. Fora=0,weget d 0 (x(k+ )) bv ep+ X3,r+t s p /v(e+)p+ + v ep Z k+/v (e+)p mod v ep+ for some Z k+ v 3 BP BP. Proof. Here we may ignore elements killed by v (e+)p. For the first term of Y [a,b] (v ep,v (e+)p,x3,r+), s notice that (a+b)v (e )p v X sp 3,r /Xe+, is congruent to (a + b)v (e )p v X sp 3,r /v(e+)p (b )v p v 3/v (e+)p+ } mod (p, v (e+)p ). Using (..), Propositions 3..4 and 3.3.4, wehave d 0 ((a + b)v (e )p v X sp 3,r /Xe+, ) (a + b)v (e )p X sp 3,r v p tp +(b )vp v p t +(b )v p +p (A) (tp t p + v v p (B) t p )}/v (e+)p (C)

23 The Chromatic E -term H 0 M for p>3 43 mod (v ep+ ). For the second term, we can easily obtain d 0 (av ep X sp 3,r u,/v (e+)p ) av ep X sp 3,r (ζ ζ p )/v(e+)p mod (v ep+ ) by Proposition For the third term, using the congruence X3,r+ s X s p 3,r+ s v a3,r+ p v 3 X sp 3,r mod v b(r+) p p, (..), Propositions 3..3 and 3.3.4, wehave d 0 ( (a+b)(b )v ep X3,r+/s s v (e+)p+a3,r+ ) (a+b)(b )v ep X s p 3,r+ tp /s v (e+)p++a3,r+ +(a+b)(b )v ep X sp 3,r vp t (A) v (t p +p t p ) (B) v v p t p (C) }/v (e+)p mod (v ep+ ). Collecting three terms gives (A) = 0, (B) = v ep X sp 3,r /v(e+)p (a + b)(t p +p t p )/vp + a(ζ ζ p )}, (C) = (a+b)v ep+ X sp 3,r tp /v(e+)p+. Thus d 0 (Y [a,b] (v ep,v (e+)p,x3,r+)) s is congruent to (a + b)(b )v ep X s p 3,r+ tp /s v (e+)p++a3,r+ +(a+b)v (e )p X sp 3,r tp /v(e+)p + v ep X sp 3,r /v(e+)p (a+b)(t p +p t p )/vp + a(ζ ζ p )} mod v ep+ for a 0, and d 0 (Y [0,b] (v ep,v (e+)p,x3,r+)) s is congruent to b(b )v ep X s p 3,r+ tp /s v (e+)p++a3,r+ + bv (e )p X sp 3,r vp tp + vp +p (tp t p )+vp+ v p t p }/v(e+)p mod v ep+ for a = 0. Now, recall definitions of ζ and w, and notice the congruences X s p 3,r+ X3,r+ s + s v a3,r+ p v 3 X3,r+ s mod v b(r+) p p, X sp 3,r X3,r+ s mod v b(r+) p p. Modifying the differentials using the above two congruences completes the proof of this proposition for the case s N 0. For s N 0 case, the result is proved using the congruence d 0 (av ep X3,r+3/s s v (e+)p+b(r+3) ) av ep Y, where Y = X3,r+t s p /v(e+)p++a3,r+ X3,r+(ζ s ζ 3 + t /v )/v (e+)p + Z/v (e+)p for some Z v3 BP BP. We define Y (v p3 +p p,v p p,x sp 3, )tobe v p3 +p p X sp 3, (4v + v p+ u, )/X p, for s N 0, (above terms) + 4v p3 +p p X3,4/s s v p3 +p +a 3,4 for s N 0. (B)

24 44 Hirofumi Nakai By routine calculation, we have: Lemma For s N 0, d 0 (Y (v p3 +p p,v p p,x sp 3, )) is expressed as (v p3 +p p X sp 3, /v p )4vp tp + vp +p (4tp 5t p )+vp vp (ζ ζ p + C w 3,3 )} + v p3 +p p Z/v p p mod v p3 +p p+.fors N 0, it is expressed as (above terms)+4v p3 +p p X s 3,3t p /vp3 +p 4v p3 +p p X sp 3, (ζ ζ 3 + v t )/v p p + v p3 +p p Z /v p p mod v p3 +p p+. Here, Z, Z v3 BP BP, and C =( u 3,i t pi j )tp + up 3,0 tp (tp +p tp3 ). i+j= Proof. By Proposition 3.3.4, d 0 (v X sp 3, /X p, )=d 0(v X sp 3, )/X p, + η R(v X sp 3, )d 0 (/X p, ) d 0 (v X sp 3, )(/v p p v p v 3/v p + ) + η R (v X sp 3, ) v p (t v w, )/v p p mod (v p+ ). It is sufficient to compute d 0 (v X sp 3, )mod(v,v p + ) and mod (v p+,v p p ). Notice that d 0 (v X sp 3, )=d 0 (v )X sp 3, + η R (v )d 0 (X sp 3, ), d 0 (v )v (sp )p 3 + η R (v )d 0 (v (sp )p 3 ) mod v p+,v p p. Using the results of 3., we see that d 0 (4 v p3 +p p v X sp, to (4 v p3 +p p X sp 3, /v p p )t p + vp +p (tp t p )/vp }. On the other hand, we have 3, /X p d 0 (v p3 +p p X sp 3, u, /v p p ) v p3 +p p d 0 (X sp 3, u, )/v p p mod (v p3 +p p+ ), and d 0 (X sp 3, u, )=d 0 (u, )X sp 3, + η R (u, )d 0 (X sp 3, ) (ζ ζ p )Xsp 3, +(u, +ζ ζ p )d 0(X sp 3, ) mod (v,v ). According to Proposition 3..3, d 0 (X sp 3, ) v p Xsp p 3, t p + vp +p+ v sp3 p p 3 (u,0 t w 3,3 ) ) is congruent

25 The Chromatic E -term H 0 M for p>3 45 mod (v,v p +p+ ). Using this congruence and Definition 3.., we obtain ( ) v p 3 +p p X sp 3, u 3, d 0 vp +p p X sp ( ) } 3, ζ v p p v p p ζ p tp v p + C w 3,3 3 + vp +p p Z v p p for some Z v3 BP BP. Collecting terms gives the result for s N 0 case. For s N 0 case, it is easy to see that d 0 (4 v p3 +p p X3,4/s s v p3 +p +a 3,4 ) is congruent to 3 4 vp +p p v p3 +p X s 3,3t p 3 4 vp +p p X sp 3, t p v p p Proposition Assume that an element x() satisfies d 0 (x()) s(s ) vp +p X sp 3, t + (ζ ζ 3 +u,0 t )+(elements killed by v p p ). v p ( ) s v p +p X sp 3, v p +(s ) vp +p X3,t s ( ζ + t v + C v p +p ) + vp +p Z v p mod (v p +p ) for some Z v3 BP BP, where C is as in Lemma Then x(3) = x() p + ( ) s Y (v p 3 +p p,v p p,x sp 3, ) satisfies d 0 (x(3)) s(s )DZ(v p3 +p p,v p p,x sp 3, )+v p3 +p p Z 3 /v p p mod (v p3 +p p+ ) for some Z 3 v3 BP BP. Proof. Collect the differentials on x() p and ( s the congruences C p + C w 3,3 mod I 3, ) Y (v p 3 +p p,v p p,x sp 3, ) using X3,3 s X sp 3, svp3 v 3 X sp 3, mod v p3 +p. As the next step of Proposition 4..3 or 4..5, we show the following proposition. Here we state only the result. Proposition Assume that an element x(k +) satisfies d 0 (x(k + )) DZ(v ep,v (e+)p,x3,r+)+v s ep Z k+ /v (e+)p mod v ep+ for some Z k+ v3 BP BP, where e and e are positive integers and r. Then x(k + i) defined inductively on i by satisfies x(k + i) =x(k+i ) p + Z(v epi,v (e+)pi,x s 3,r+i) d 0 (x(k + i)) DZ(v epi,v (e+)pi,x3,r+i)+v s epi Z k+i /v (e+)pi mod v epi +

26 46 Hirofumi Nakai for some Z k+i v3 BP BP. Proposition 4..6 is regarded as the last step in this pattern. We will see in Section 5 that x(k) is constructed in this way in many cases. Actually we have other two inductive ways of constructing x(k), introduced in the following two propositions. Their proofs are relatively easy, so we leave the proofs to the reader. Proposition Assume that an element x(k) satisfies d 0 (x(k)) v ep v p3 p v (sp )pr+i 3 θ i (v a3,r p3 i p i ) p mod v e+ (= v e ve3p 3 t p /vp + ve Z k/v p ) for θ i (v a3,r p3 i p i ) (i =,) as in Lemma 4..7 or 4..9, and some Z k v3 BP BP. Then x(k + i) defined inductively on i by x(k + i) = x(k +i ) p v epi v e3pi+ + 3 /v pi+ +p i+ pa 3,i++ for odd i, = x(k + i ) p for even i satisfies d 0 (x(k + i)) v epi v e3pi+ 3 t pδ(i) /v pi+ +p i+ a 3,i+ + v epi Z k+i /v pi+ +p i mod v epi + for some Z k+i v3 BP BP, where δ(i) =0for odd i and for even i. Proposition Assume that an element x(k) satisfies d 0 (x(k)) v e ve ve3pr 3 w,t mod v e+ with Maxp t p r, 0} + P (t )+ e. Then x(k + i) defined inductively on i by x(k + i) =x(k+i ) p +( ) i v epi +P (i,) v epi P (i,) v e3pr+i 3 u,t+i satisfies d 0 (x(k + i)) ( ) i v epi +P (i ) v epi P (i ) v e3pr+i 3 w,t+i mod v epi +P (i )+, where P (i, j) =p j (p i j+ )/(p ) = p i + +p j for i j. We will abbreviate P (i, 0) to P (i). Our guide to constructing x(k) is to add suitable elements to x(k ) p so that the differential has one of the forms in the assumptions of Propositions 4.., 4..6, 4..7 and In fact, we will observe that each case follows one of the above four patterns by k =5. 5. Proof of the main theorem In this section we prove our main theorem by defining x(k) (=x(sp r /j; k)) for all cases using the preparatory computations displayed in Section 4. Notice that the smallest integer N with d 0 (x(k)) 0modp, v N+ gives the v -divisibility N(k) (= N(s, r, j; k)) of x(k).

27 The Chromatic E -term H 0 M for p> Definitions of x(0) and the differentials. Now we start with X s 3,r/v j (p j a 3,r and p s Z). Using Propositions 3..4 and 3.3.4, we can easily calculate d 0 (X s 3,r/v j )modv : d 0 (X3,0/v s j ) v v3t s p /v + sv v3 s (t p +p tp )/v, and for r = or for even r, d 0 (X3,r/v s j ) jv X3,rt s p /vj+. Finally, for odd r 3, d 0 (X3,r/v s j ) jv X3,rt s p /vj+ sv v a3,r j X sp 3,r ( jt p+ /v + w, ). Moreover, we have d 0 (X3,0/v s ) v v p v3 s w, mod (v), v v p v3 s ζ (v v t p ) mod (v3 ) for p (s ). Then the coboundary δ : H 0 M H M on X3,r/v s v j is δ(x3,r/v s v j )= jvspr 3 t p /vj+ + (elements killed by v j ). According to.5, its numerator belongs to Cokerδ, when s N 0 = s p p s } and r is odd or when r 0isevenby(.5.). In this case, we can set x(0) to X3,r/v s j with the v -divisibility N(0) =. In another case, we need to modify X3,r/v s j so that its coboundary includes an element of Cokerδ For odd r 3 and s N 0. (i) For a 3,r p j a 3,r. We set x(0) to X 0 (v j,xs 3,r). By Lemma 4..5, we have d 0 (x(0)) ( j)v v a3,r j v (sp )pr 3 ζ mod v, and so N(0) =. Notice that this differential is trivial either for j = a 3,r p + (p)orforj a 3,r p. (ii) Either for j = a 3,r p + or for j a 3,r p. We also set x(0) to X 0 (v j,xs 3,r). Then, Lemma 4..5 shows that ( ) d 0 (x(0)) j(j+)vx 3,r+t s p j + /s v j++a3,r+ v X3,rt s p /vj+ + ε(r, j) mod (v), 3 and so N(0) =. Note that all elements but ε(r, j) are vanished when p (j + ), and ε(r, j) = 0 unless r 3 is odd and a 3,r p p j a 3,r p, in which case ε(r, j) = a(r)vv a3,r j v (sp )pr 3 w,3. (iii) For p (j +), p (j +p+) and j a 3,r p p. Set x(0) = X(v j,xs 3,r). Notice that we replace X 0 (v a3,r p p,x3,r) s with X(v a3,r p p,x3,r) s for j = a 3,r p p although ε(r, a 3,r p p) 0. Then Lemma 4..7 shows that d 0 (x(0)) DX [j,](v p+,v j+,x s 3,r)+a(r)v p+ v a3,r j p v (sp )pr 3 θ (v j )

28 48 Hirofumi Nakai mod v p+. Note that d 0 (x(0)) 0mod(v p+ ), and so N(0) = p when p (j + p + ). (iv) For p (j + p + ) and j a 3,r p p. We first define B(j) by B(j)=X(v j,xs 3,r) for j a 3,r p 3p; B(a 3,r p p) =X(v a3,r p p,x3,r) s a(r)v p+ v (sp )pr +p+ 3 /v. Then, let B(j) for s N 0 x(0) = B(j) v p+ X3,r+/s s v j++b(r+) for s N 0. Easy calculation shows that d 0 (x(0)) DZ(v p+,v j+,x s 3,r) mod(v p+ ), and so N(0) = p For r =and s N 0. For j p, we set x(0) to X 0 (v j,xs 3,). Then Lemma 4..5 shows that ( ) d 0 (x(0)) j(j+)vx 3,t s p j + /s v j++a3, v X3,t s p /vj+ mod (v), 3 and so N(0) =. For j = p, we set x(0) to X(v p,x3,). s By Lemma 4..8, wehaved 0 (x(0)) 3v p Xs 3,t /v p mod (vp+ ), and so N(0) = p. 5.. Definitions of x(k) (k ) and the differentials. We complete the proof of our main theorem by defining x(k) and computing the differentials for all cases, based on the computations in k = 0 case For r =0. (i) For p (s ). We set x() to X3,/X s,. Then Lemma 4..3 shows that ( ) d 0 (x()) (s )v p Xs 3,t /v p+ s + v p vsp 3 t /v mod v p+, and so N() = p. Fork=,weset ( ) s x() = x() p + v p X(v p,x sp 3, ). Then Lemma 4..8 implies (( ) ) s d 0 v p X(v p,x sp 3, ) (s ) vp Xs 3, v p +p ( ) s v p Xsp 3, v p (t p vp t ) t p v +4v p t v p ( ζ + t v +C ) } + vp +p Z v p