New York Journal of Mathematics
New York J. Math. 6 (2000) 21{54.
The Chromatic E1-term H 0M12 for p > 3
Hirofumi Nakai
Abstract. In this paper, study the module structure of
Ext0
(
( 11 21 ))
which is regarded as the chromatic 1 -term converging to the second line of
the Adams-Novikov 2 -term for the Moore spectrum. The main diculty here
is to construct elements (
) from the Miller-Ravenel-Wilson elements
( 3 2 ) 2 0 21 . We achieve this by developing some inductive methods
of constructing (
) on .
BP BP
BP BP = p v
v
E
E
r
x sp =j k
s
x
r
=v
j pk
H M
r
x sp =j k
k
Contents
1. Introduction
2. BP -homology and Bockstein spectral sequence
3. Denitions of some elements
4. Preliminary calculations
5. Proof of the main theorem
References
1.
21
24
28
34
46
54
Introduction
Let BP be the Brown-Peterson spectrum for a xed prime p. As is well known,
the pair of homotopy groups BP and the BP -homology BP BP forms a Hopfalgebroid
( BP BP BP ) = ( (p) v1 v2 : : : ] BP t1 t2 : : : ] ):
The Adams-Novikov spectral sequence (ANSS) is one of the fundamental tools to
compute the p-component of the stable homotopy groups S X(p) for a spectrum
X:
E2 = ExtBP BP (BP BP X ) =) S X(p) :
Here, for any BP BP -comodule M , ExtBP BP (BP M ) is regarded as the right
derived functor of HomBP BP (BP M ). We abbreviate ExtsBP BP (BP M ) to
H s M as usual.
Z
Received May 5, 1999.
Mathematics Subject Classication. 55Q45 secondary 55Q51, 55T15, 55N22.
Key words and phrases. Stable homotopy of spheres, Adams-Novikov spectral sequence, chromatic spectral sequence.
21
ISSN 1076-9803/00
Hirofumi Nakai
22
As an important example of nite spectra, we have the Smith-Toda spectrum
V (n) for each prime p with BP V (n) = BP =(p v1 vn ), although its existence
is veried only for 0 n 3 and 2n + 1 p so far. (Recently, Lee S. Nave
has shown the non-existence of V ((p + 3)=2) for p > 5.) Then the E2 -term of the
ANSS for V (n) is H BP V (n). Miller, Ravenel and Wilson 2] have constructed
an algebraic spectral sequence converging to H BP V (n) as follows:
Denote the BP BP -comodules BP =(p v1 vn;1 ) by Nn0 . Then, dene Nnm
(m 1) inductively on m by short exact sequences
0 ;! Nnm ;! vm;1+n Nnm ;! Nnm+1 ;! 0:
We also dene Mnm by Mnm = vm;1+n Nnm . Indeed, they can be described directly as
Nnm = BP =(p vn;1 vn1 vn1+m;1 )
Mnm = vn;+1 m BP =(p vn;1 vn1 vn1+m;1 ):
Splicing the above short exact sequences, we get a long exact sequence:
0 ;! Nn0 ;! Mn0 ;! Mn1 ;! Mn2 ;! called the chromatic resolution of Nn0 . Applying H (;) to the above long exact
sequence, we obtain a spectral sequence converging to H Nn0 with E1st = H t Mns
and dr : Erst ! Ers+rt;r+1 , called the chromatic spectral sequence.
The simplest example in these E1 -terms is the 0-th cohomology of the n-th
Morava stabilizer group H 0 Mn0, which is isomorphic to =p vn1 ]. Moreover, H t Mn0
(1 t 2) has been computed by Ravenel 5]. In general the calculation of H t Mns
becomes terribly dicult as s + t increases, except for the following case:
Z
Theorem (Morava's vanishing theorem). If (p ; 1) n and t > n , then
2
-
H t Mn0 = 0:
Many chromatic E1 -terms are computed so far (cf. 12]), most of them are due
to Miller-Ravenel-Wilson 2], Ravenel 5] and Shimomura's works. In particular,
H 0 Mn1 is computed in 2, Theorem 5.10] and H 0 Mn2 (n 2 p > 2) is also done
in 10, Theorem 1.2]. The purpose of this paper is to prove the following theorem
about the k(1) -module structure of unknown H 0 M12 for p > 3.
Theorem. For each Miller-Ravenel-Wilson element xsr k =vjp
and p - s) and 1=v2mp
2H M
3
2
+
k
2HM
0
1
2
(p j
-
(p m), there exists an element
2 v3;1 BP =(p v21)
which is congruent to (xs3r =v2j )p mod (v1 ), and
1=X2mr 2 BP =(p v21 )
congruent to 1=v2mp mod (v1 ), so that as a k(1) -module
H 0 M12 =
n
o
k(1) x(spr =j k)=v1N (srjk) k 0 r 0 p s 2 and p j a3r
k(1) 1=v1a2 X2mr r 0 and p m 1 r
0
1
2
-
x(spr =j k)
k
r
-
r
-
Z
-
The Chromatic E1 -term H 0M12 for p > 3
23
where the integers a2r and a3r are dened in 2.3.1 and 3.3.1, and the integers
N (s r j k) are given as follows:
(1)
(p3 + p2 ; p ; 1)pk;3 ] + pk;3 ] for r = 0 and p - (s ; 1),
(2)
(pk+1 ; 1)=(p ; 1) for r = 0, p j (s ; 1) and s 62 N 1 ,
(3)
(2p2 ; 1)pk;2 ] + pk;2 ] for r = 0 and s 2 N 1 ,
(4)
(pk+1 ; 1)=(p ; 1) ; pk;1 ] for odd r 1, s 62 N 0 and j = a3r ; 1,
(5)
pk + pk;1 ; 1 ( = a2k ) for odd r 1, s 62 N 0 and j a3r ; 2
or even r 2 and s 62 N 0 ,
k
+1
k
(6)
p + p ; 1 ( = a2k+1 ) for r = 1, s 2 N 0 and j = p ; 1,
3
(7) (2p + p2 ; p ; 1)pk;3 ] + pk;3 ] for r = 1, s 2 N 0 and j p ; 2,
(8)
(p2 + p ; 1)pk;2 ] + pk;2 ] for even r 2 and s 2 N 0 ,
(9)
(p4 + p2 + p ; 1)pk;4 ] + pk;4 ] for odd r 3 and j = a3r ,
and for odd r 3, s 2 N 0 and j a3r ; 1, we have
for j = a3r ; 1,
for a3r ; p j a3r ; 2,
for j = a3r ; p ; 1,
for j = a3r ; p + 2,
or p - (j + 1) and j a3r ; p ; 2,
k
k
+1
(14)
p + (p ; 1)=(p ; 1) for p j (j + 1)
and a3r ; p2 j a3r ; 2p,
(15)
(p2 + p ; 1)pk;1 ] + pk;1 ] for j = a3r ; p2 ; p,
or p j (j + 1), p3 - (j + 1)(j + p + 1)
and j a3r ; p2 ; 3p,
(16) (p4 + p3 ; p2 + 1)pk;3 ] + 2pk;1 ] for j a3r ; 2p2 ; p and p2 j (j + 1),
(17)
pk+1 + pk for j a3r ; p2 ; 2p
and p2 j (j + p + 1),
Here x] is the greatest integer which does not exceed x, and
N0
= f ap ; 1 j p - a g N1
= (ap2 ; p ; 1)pr + 1 j p - a r 1 : odd :
2
(10)
(11)
(12)
(13)
(p3 + p2 ; 1)pk;3 ] + pk;3 ]
same as the case (1)
2pk
same as the case (7)
In Section 2 we shall review BP -theory and the Bockstein spectral sequence
and recall the structure of the chromatic E1 -term H 0 M21 computed in 2]. Then we
change the p -module basis of H 0 M21 and state the method of getting the structure
of H 0 M12 (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
unk , wnk and X3r , construct the new element 1=X2r , and give the dierentials
on these elements (some of them are introduced in 1] and 8]). In Section 4 we
F
24
Hirofumi Nakai
set up the elements X0 (v2j X3sr ) and X (v2j X3sr ), each of which is congruent to
X3sr =v2j mod (v1 ), and compute the dierentials. 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(spr =j k) and complete
the proof of the main theorem in Section 5.
We can deduce some applications to H 2 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 eld, 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 nished
these hard calculations without stimulation from them.
2.
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 M12 .
2.1. Summary of BP -homology and related maps. For a xed prime p, there
is a spectrum BP called the Brown-Peterson spectrum, which is characterized by
BP = (p) v1 v2 vn ]
BP BP = BP t1 t2 tn ]
where jvn j = jtn j = 2(pn ; 1). The pair (A ;) = (BP BP BP ) form a Hopf
algebroid with the following structure maps:
L (resp.R ) : A ! ; (left (resp. right) unit) c : ; ! ; (conjugation)
: ; ! A (augmentation) : ; ! ; A ; (coproduct):
Given a BP BP -comodule M and its BP BP -comodule structure : M ! M A;,
we use the following notation as usual:
H M = Ext; (A M ) = H ( M d)
where the cobar complex ( M d) is the double graded (p) -module with
n M = M A ; A A ; (n factors of ;)
dn (m 1 n ) = (m) 1 n
X
+ (;1)k m 1 (
k ) n
+ (;1)n+1 m 1 n 1:
Z
Z
In this paper we compute only the 0-th dierential d0 = R ; L : M12 ! M12 A ;.
By denition, d0 satises
d0 (xy) = d0 (x) y + R (x) d0 (y)
for any x y 2 A. This formula is frequently used for computations in this paper.
When d0 (x) z mod (p J ) for an element z 2 ; and an ideal J A, we also have
d0 (xp ) z p mod (p J p ):
The Chromatic E1 -term H 0M12 for p > 3
25
See also 2, Observation 5.8].
To compute d0 , we summarize some known results about R . Ravenel 4] has
shown the following congruence of formal group laws:
XF
XF p
ti R (vjp ) vi tj
mod p:
i
ij 0
i
ij 0
vnp ;1 t1 mod In;1 for the invariant prime ideal
In general R (vn ) vn + vn;1 tp1 ;
In;1 = (p v1 vn;2 ). Then, for instance, direct calculation shows that
(2.1.1) R (v2i ) (v2 + v1 tp1 )i ; i v1p v2i;1 t1 ; i(i ; 1) v1p+1 v2i;2 tp1+1 mod p v1p+2 :
;1
n
More precisely, R (v3 ) and R (v4 ) satisfy the following congruences (cf. 6](4.3.21)):
(2.1.2) R (v3 ) v3 + v2 tp1 ; v2p t1 + v1 tp2 ; v12 v2p;1 tp1 mod p v13 3
2
3
(2.1.3) R (v4 ) v4 + v3 tp1 ; v3p t1 + v2 tp2 ; v2p tp1 +1 mod p v1 v2p+1 :
2.2. Bockstein spectral sequence. In this paper will deduce the k(1) -module
structure of H 0 M12 from H 0 M21 , which has already been computed in 2]. By
denitions of comodules M21 and M12 , we obtain the short exact sequence
2
1=v1
v1 M 2 ;! 0:
0 ;! M21 ;!
M12 ;!
1
Applying H (;) to this sequence, we get the long exact sequence
1=v1
1=v1
v1 H 0 M 2 ;!
H 1 M 1 ;!
0 ;! H 0 M21 ;!
H 0 M12 ;!
:
1
2
Regarding this long sequence as an exact couple, we get a Bockstein type spectral
sequence in the usual way, leading from H M21 to H M12 . But as in 2] we compute
H 0 M12 directly by making use of the following lemma.
Lemma 2.2.1 (cf. 2], Remark 3.11). Assume that there exists a k(1) -submodule
B t of H t M12 for each t < N , such that the following sequence is exact:
=v1 0 v1 0 =v1
=v1 N ;1 v1 N ;1 0 ! H 0 M21 1;!
B ;! B ;! H 1 M21 1;!
1;!
B ;! B ;! H N M21 where : B t ! H t+1 M21 is the restriction of the coboundary map : H t M12 !
H t+1 M21. Then the inclusion map it : B t ! H t M12 is an isomorphism between
k(1) -modules for each t < N .
Sketch of the proof. Because H tM12 is a v1-torsion module, we can lter Bt and
H t M12 as
Pt (m) = fx 2 B t j v1m x = 0g and Qt (m) = fx 2 H t M12 j v1m x = 0g:
Assume that the inclusion ik is an isomorphism for k t ; 1 (the t = 0 case is
obvious), and consider the following commutative ladder diagram:
1=v1
v1 P (m ; 1) ;!
H t M21 ;!
Pt (m) ;!
H t+1 M21
t
k 1=v # it
# it
k
v
1
1
t
1
t
+1
H M2 ;! Qt (m) ;! Qt (m ; 1) ;! H M21 :
Using the Five Lemma, we can show that Pt (m) = Qt (m) (m 1) by induction
on m.
t;1
=B# it;1
H t;1 M12
;!
;!
Hirofumi Nakai
26
We shall construct B 0 satisfying the above lemma to determine the k(1) -module
structure of H 0 M12 . In order to construct a k(1) -module basis of B 0 , it is natural
to push each element of H 0 M21 to H 0 M12 and to divide it by v1 as many times as
possible. So we need to review the module structure of H 0 M21 .
2.3. H 0 M21 and changing its basis. We rst recall some notations dened in
2] to write down a k(2) -module basis of H 0 M21 . Hereafter we assume that p > 2.
Denition 2.3.1 (2], (5.11) and (5.13)). Dene integers a3k by
a30 = 1 a31 = p a32t = p a32t;1 a32t+1 = p a32t + p ; 1 for t 1
and elements x3k 2 v3;1 BP by
x30 = v3 x31 = v3p ; v2p v3;1 v4 x32t = xp32t;1 2
x32t+1 = xp32t ; v2a3 2 +1 ;p v3(p;1)p +1
for t 1.
Using these notations, Miller-Ravenel-Wilson (2, Theorem 5.10]) have shown
the following:
Theorem 2.3.2. As a k(2) -module,
=va3 j k 0 p m 2 1=vi j i 1 :
H 0 M21 = k(2) xm
p
3k
2
2
In this paper we will consider an analogue to Miller-Ravenel-Wilson construction
;1
of the elements xm
3k 2 v3 BP (2, Section 5]): The elements x3k have been dened
inductively on k with x30 = v3 , and each of them has the relation x3k v3p
mod (p v2N ) for a small enough integer N . Motivated by this, we shall construct
elements x(k) 2 v3;1 BP =(p v21 ) inductively on k with x(0) = xs3r =v2j so that
x(k) (xs3r =v2j )p mod (p v1N ) for a small enough N . Keeping it in mind, we now
change the p -basis of H 0 M21 to \a pk -power basis".
Lemma 2.3.3. Asna k(2) -module,
o
H 0 M21 = p (xs3r =v2j )p j k 0 r 0 p s 2 and p j a3r
p 1=v2i j i 1 :
Proof. It is sucient to prove that any p -module base xm3u =v2l (1 l a3u)
displayed in Theorem 2.3.2 can be written as a linear sum
X
(2.3.4)
xm3u =v2l = a x t
t
k
-
Z
F
k
k
F
k
F
-
Z
-
F
F
s
where a F p and each x has a form (x3r =v2j )p with p - s Z and p - j a3r .
We shall show it by induction on mpu . Notice that
xm3u =v2l = v3mp =v2l + (elements with v3 -exponents less than mpu ):
When mpu = 1, the only such a base is v3 =v2 , so it is clear. Suppose that
each base xa3b =v2c (1 apb < mpu ) can be expressed as (2.3.4) and l = dpe with
l a3u < pu+1 , we may assume that e u. Dene y = xm3u =v2l
;px-md. Because
p
d
H 0 M21 . Since the maximal v3 -exponent in y is less than mpu , y
3u;e =v2
l
has a form (2.3.4) by inductive assumption, hence so does xm
3u =v2 .
2
2
k
u
e
2
;
The Chromatic E1 -term H 0M12 for p > 3
27
2.4. The construction of the module B 0. Let x(spr =j k) and y(mpr ) be elements of v3;1 BP =(p v21 ) satisfying
x(spr =j k)=v1 = (xs3r =v2j )p =v1 and y(mpr )=v1 = 1=v1 v2mp
in H 0 M12, and N (s r j k) (resp. N (m r)) be the maximal integer such that each
x(spr =j k)=v1i (resp. y(mpr )=v1i ) with 1 i N (s r j k) (resp. 1 i N (m r))
is a cycle of H 0 M12 .
Proposition 2.4.1. As a k(1) -module,
n
o
B 0 = k(1) x(spr =j k)=v1N (srjk) k 0 r 0 p s 2 and p j a3r
n
o
k(1) y(mpr )=v1N (mr) r 0 and p m 1
r
k
-
Z
-
-
is isomorphic to H 0 M12 if it satises the following condition for the coboundary
map : B 0 ! H 1 M21 in Lemma 2.2.1:
n
x(spr =j k)=v1N (srjk)
o n y(mpr )=v1N (mr)
Proof. All exactness of the sequence 0
o
is F p -linearly independent.
! H M !=v
1 1
2
0
1
v1 B 0 !
H 1 M 1 is
B0 !
2
obvious, but Ker Im v1 . So we need to show only this inclusion. Separate the
p -basis of B 0 into two parts
n
F
o
o n
A = x(spr =j k)=v1N (srjk) y(mpr )=v1N (mr)
B = x(spr =j k)=v1l j 1 l < N (s r j k) y(mpr )=v1l j 1 l < N (m r) :
Then it is obvious that (x ) 6= 0 2 H 1P
M21 for x P
2 A, and that0(y ) = 0 2 H 1M21
for y 2 B . Thus
P for any element z = ax + by of B (a b 2 =p), we
have (z ) P
= a (x ). The condition implies that all a are zero when (z ) = 0,
and so v1 b y =v1 = z . This completes the proof.
Z
We will construct the element x(spr =j k) from (xs3r =v2j )p in Section 5, and the
element 1=X2mr from 1=v2mp as a candidate for y(mpr ) in Section 3.
k
r
2.5. p -linear independence and Coker . When we compute the coboundaries
: B 0 ! H 1 M21 of x(spr =j k)=v1N (srjk) and y(mpr )=v1N (mr) , we expect each of
these images to have an appropriate form so that we can judge whether the set
of -images is -linearly independent or not in H 1 M21 for use in Proposition 2.4.1.
Though the structure of H 1 M21 (p > 3) has already been computed by Shimomura
(9] and 11]), we don't need the whole structure of H 1 M21 for our purpose. We
follow the same method as in the proof of 2, Theorem 6.1 (p. 500)].
Consider the long exact sequence
F
F
=v
v H M ;!
H M ;!
:
;! H M ;!
H M ;!
0
1
2
1
0 1
3
2
1
1
2
2
1
1
2
Since the coboundaries on elements of H 0 M21 have; already been
computed in 2,
Proposition 5.17], we can obtain generators of Ker v2 j H 1 M21 . An easy calculation shows that Coker( : H 0 M21 ! H 1 M30 ) is isomorphic to the following direct
Hirofumi Nakai
28
sum as a p -vector space:
n sp
(2.5.1)
p v3 h0 j either s 62
o
o
F
r
n
F
N0
and even r 0, or p s and odd r 1
-
p vsp h j either s 62 and odd r 1, or p s and even r 0
n
o
pf h g p vtp; h j t 2 K (3) f g
= s 2 j p j s + 1 and p s + 1 , and hi is the cohomology class of
F
r
F
N0
1
3
F
1
3
1
-
Z
2
3
2
where 0
tp1 .
Let S be the set of elements
fAk = v3b ha =v2c j 0 k n 0 ak 2 c1 cn v3b ha 2 Coker g
and consider the following condition on S :
(2.5.2)
(ai bi ci ) 6= (aj bj cj ) for any two elements with i 6= j .
P
For a linear sum A = nk=1 k Ak , assume that A = 0 in H 1 M21 and c1 = c2 =
= cm for some m cn1;. 1Because H 1cM1;211 is a vP2-torsion
module, we can consider
b
the multiplication by v2 , so that v2 A = m
(
v
3 ha =v2 ) = 0 in Ker v2
k
=1 k
(
= Coker ). By the condition (2.5.2), the set fv3b ha g is linearly independent in
Coker and thus all k (1 k m) should be zero. Iterating this, we conclude
that all coecients k are zero and S is linearly independent in H 1 M21 .
N
i
Z
k
k
-
k
k
k
k
3.
k
k
k
Denitions of some elements
In this section we introduce the elements unk , wnk , X3r and 1=X2r . We will
use these to dene many elements in Section 4.
3.1. Moreira's element unk and Shimomura's element wnk .
Denition 3.1.1. As is done in 1, x6(4)] or 8, 2.8], we dene the element unk 2
vn;1 BP by the following recursive formula:
X
un0 = vn;1 and
vn+i upnj = 0 for k 1:
i
i+j =k
Remark 3.1.2. In 1] unk is dened only for k n and denoted as un k . By
denition
un1 = ; vn;p;1 vn+1 ;p2 ;1
un2 = vn;p2 ;p;1 vnp+1
vn+2 +1 ; vn
2
;p3 ;p2 ;p;1 p2 +p+1
vn+1 + vn;p3 ;p2 ;1 (vnp +1 vn+2 + vn+1 vnp +2 )
un3 ; vn
; vn;p3 ;1vn+3 mod (p)
+
and so on.
Computing the right unit R on unn, Moreira has shown that the K (n) -module
base n of H 1 Mn0 is homologous to np :
Proposition 3.1.3 (1], Theorem 6.2.1.1). R (unn);L(unn) n ;np mod In .
The Chromatic E1 -term H 0M12 for p > 3
29
We now recall the elements wnk introduced by Shimomura 8]. Dene elements
P ;1 t (vp ) Pj v T p ;1 mod (p) for j 1, and an
Tj (j 0) by T0 = 1 and ji=0
i R j ;i
i=1 i j ;i
element en (x) 2 vn;1 BP BP=In for x 2 vn;1 BP by the congruence R (x) en(x)
mod In .
Denition 3.1.4 (cf. 8], 2.10). Dene elements wnk 2 vn;1 BP BP (n 2) inductively on k by
i
wn0 = 0 and wnk =
Remark 3.1.5. By denition,
wn1 = vn;1 tp1 wn2 ; vn;p;1 vn+1 tp1
n
i
k
X
j =1
en(upnk;j )Tjp
j
;1
for k 1:
;2
n
;1
;1
n
+ vn;p (tp2
n
;1
; tp
1
n
+
p
n
;1
) + vn;1 tp1
mod p v1p
;1 +1
n
;2
n
:
In general, as an element of M21 A ;
;
w2k = ( 1)k;1
v3(p ;1)=(p;1) tp1 + (elements killed by v(p
2
v2(p ;1)=(p;1)
k
;1
k
k
;1)=(p;1);1
):
We note that w22 is similar to 2 = ; v2;p;1 v3 tp1 + v2;p (tp2 ; tp1 +p ) + v2;1 t2 . In
fact wnn (n 2) is related to n by the following congruence (cf. 8, 4.8]):
2
n ; wnn X
i<j n
1
0
1p ; ;
X
upnn;j @
tk c(tpj;k )A
n
j
;i;1
i
k
1
mod In :
ikj
v2;1 (t2 tp1+1 ) mod I2 for n = 2.
In particular we have 2 ; w22 ;
Using the notation wnk , Shimomura 8] has proved the following proposition,
which is a generalization of Moreira's result:
Proposition 3.1.6 (8], Proposition 2.2). For n 2,
X p p
R (unk ) uni tj ; wnk ; vn;1 wnk+1 R (vn;1 ) mod In;1 + (vnp ;1 ):
i
i+j =k
2
3.2. Shimomura's element X3r . The elements xnr 2 vn;1 BP (n 1 and
r 0) have been dened in 2, (5.11)] to express a basis of H 0 Mn1;1 (x3r was
listed in 2.3.1). As is shown in 2, Proposition 5.17], the dierentials d0 : vn;1 BP !
vn;1 BP A ; mod (p v1 v21+a3 ) are given by
r
d0 (x30 ) v2 tp1
and d0 (x3r ) v2a3 xp3r;1;1 tp1
for r 1
where (r) = 0 for odd r and 1 for even r. However, we need to calculate d0 (x3r )
mod (p v1 v2t ) with t > 1 + a3r for our computation. So we use the following
elements X3r dened by Shimomura 8] instead of x3r .
2
r
(r)
Hirofumi Nakai
30
Denition 3.2.1. cf. 8, (3.3)]] Dene elements X r 2 v; BP by
3
3
1
X30 = v3 X31 = X3p0 + v2p v3p u31 X32 = X3p1 ; v2p +p X3p;1 1 u32
2
X33 = X3p2 + v21+a3 3 X3p;2 1 u21 + v2b(3) X3p1;p;1 u33
2
X34 = X3p3 ; v2b(4) X3p2;p;1 (u22 ; u33 )
2
X3r = X3pr;1 + v21+a3 X3pr;;1 1u21 ; v2b(r) X3pr;;p2;1 (u22 ; u33 ) for odd r 5
2
X3r = X3pr;1 ; v2b(r) X3pr;;p2;1 (2 u22 ; u33 ) for even r 6:
with b(0) = 1, b(1) = p +1 and b(r) = a3r + a3r;1 +1 = (p2 + p +1)pr;2 for r 2.
Remark 3.2.2. By denition, it is obvious that X3r x3r mod (v21+a3 ). In
2
r
r
addition, they satisfy the congruences
X31 v3p mod v2p X32 X3p1 mod v2p +p 2
X32i+1 X3p2i mod v2a3 2 +1 ;p X32i+2 X3p2i+1 mod v2b(2i+2);p ;p;1
for i 1.
Then d0 (X3r ) is computed as follows. Let s be an integer with p s.
Proposition 3.2.3. Mod (p v1 v2b(r)+1), d0 (X3sr ) may be expressed as follows. For
small values of r,
8 s
2
>
sv2 v3 w31 ( sv2 v3s;1 tp1 )
for r = 0
<
d0 (X3sr ) = >sv2p v3sp;1 t1 ; s v2p+1 v3sp;1 w32 for r = 1
:sva3 2 X sp;1tp ; svb(2) vsp2 ;p;1(u20t2 ; w33) for r = 2.
2
31
1
2
3
For r = 3,
2
2
d0 (X3sr ) = sv2a3 3 X3sp2;1 t1 ; s v2b(3) X3sp1 ;p;1 fw22 + u20t2 ; w33 + u30 (tp1 t2 ; t3 )g:
For r 4 and even,
2
d0 (X3sr ) = sv2a3 X3spr;;11tp1 ; sv2b(r) X3spr;;2p;1 ( 2 + u20 t2 ; 3 ) :
For r 5 and odd,
;
2
d0 (X3sr ) = sv2a3 X3spr;;11t1 ; sv2b(r) X3spr;;2p;1 2 + w22 ; 3 :
2
i
-
r
r
Proof. Originally, this is proved in 8, Proposition 3.1]. Notice that it is sucient
to prove the s = 1 case because d0 (X3sr ) sX3sr;1d0 (X3r ).
2
For r = 0, d0 (X30 ) v2 tp1 mod p v1 v22 by (2.1.2). For r = 1, Proposition 3.1.6
shows that
2
d0 (v2p v3p u31 ) v2p v3p;1 t1 ; v2p tp1 ; v2p+1 v3p;1 w32
3
mod (p v1 v2p+2 ). Summing this with d0 (X3p0 ) v2p tp1 gives d0 (X31 ). For r = 2,
we have
2
2
2
2
d0 (;v2p +p v3p ;p u32) ;v2p +p v3p ;p (u31 tp1 + u30t2 ; w3p2 ; v2 v3;1 w33 )
The Chromatic E1 -term H 0M12 for p > 3
31
mod (p v1 v2p +p+2 ). On the other hand, using the congruence v3p ;p X3p;1 1 +
2
v2p v3p ;p u31 we have
2
2
2
d0 (X3p1 ) v2p X3p;1 1 tp1 + v2p +p v3p ;p (u31 tp1 ; w3p2 ):
Again, summing the terms gives d0 (X32 ).
For r = 3, we have
3
d0 (v2p +p X3p;2 1 u21) v21+a3 3 X3p;2 1 (u20 t1 ; w2p1 )
; v2b(3) X3p21;p;1 (e2(u21)tp1 + u20t2 + u30tp12 t2 ; w33 )
2
2
X
d0 (v2b(3) X3p1;p;1 u33) v2b(3) X3p1;p;1 (
2
2
2
u3i tp3;i ; w3p3 ):
i
i=0
2
b
(3)+1
mod (p v1 v2 ). On the other hand, by the congruences X3p1;p
2
2
3
2
2
2
v2p +p X3p1;p;1 u32 and v3p ;p ;p X3p1;p;1 + v2p X3p1;p;1 u31 , we have
X p; ;
32
1
X
w3p3 + u3i tp3;i ):
i=1
Using a congruence for w22 of Remark 3.1.5, we obtain d0 (X33 ).
d0 (X3p2 )
3
2
v2p X3p;2 1 tp1
;
2
v2b(3) X3p1;p;1 (up20 tp2
For r = 4, we have
2
d0 (;v2b(4) X3p2;p;1 u22 ) ; v2b(4)X3p22;p;1(;2p ; up20tp2 + 22 ; w33 ;
;
2
X
i
up33;j (ti c(tpj;i ))p2; )
j
;i;1
i
i
1i<j 3
2 ;p;1
b
(4)
p
p
v2 X32 (3 3 ) mod (p v1 v2b(4)+1 ). On the
2
3
2
3
2
other hand, by the congruences X3p2;p X3p;3 1 +v2p +p X3p2;p;1 u21 and X3p1;p ;p
2
2
3
2
X3p2;p;1 v2p +p X3p1;p ;p;1 u32 , d0 (X3p3 ) is congruent to
3
2
2
v2a3 4 X3p;3 1tp1 v2b(4) X3p2;p;1 u21tp1 + u32tp1 + w2p2 + up20tp2 w3p3 + up30(tp1 tp2 tp3 )
mod (p v vb(4)+1 ). Moreover, noticing the congruences
2
and d0 (v2b(4) X3p2;p;1 u33 )
;
1
;
;
f;
; g
;
2
;wp ; p ; X u itp;i X ip
p
p
p
p
;(u t + u t ) + u t t ;
u ;j ti c(tpj;i; )]p
i<j ;up tp ; X up ;j ti c(tpj;i; )]p
2
33
3
=0
3
31 2
30 3
30 1
i
3
3
j
2
i+1
+1
2;i
2;i
1
3
j
30 3
;i;1
33
1
3
i<j 3
1
;i;1
33
i+2
+2
2
we have
;
2
d0 (X3p3 ) v2a3 4 X3p;3 1 tp1 ; v2b(4) X3p2;p;1 ;u21tp1 + w2p2 + up20 tp2
; v2b(4)X3p22;p;1 (3 ; 3p ; X up33;;;j 1 X tk c(tpj;k )]p2; ):
j
i<j 3
1
k
i
i+1kj
i
Hirofumi Nakai
32
We obtain d0 (X34 ) in the desired expression by summing the terms and using the
congruences
;u21tp1 + w2p2 ; 2p ;2 + u20t2 X p ; ;1 X
u33;j (
tk c(tpj;k )p2; ) 3 ; w33 :
j
i
i<j 3
k
i
ikj
1
The r 5 cases are proved by induction on r. By Proposition 3.1.3, we have
congruences
2
2
d0 (v2b(r) X3pr;;p2;1 u22 ) v2b(r) X3pr;;p2;1 (2 ; 2p )
2
2
d0 (v2b(r) X3pr;;p2;1 u33 ) v2b(r) X3pr;;p2;1 (3 ; 3p ):
We also have the congruences X3(pr;;1)2 p X3pr;;1 1 and
2
d0 (v21+a3 X3pr;;1 1u21 ) v21+a3 X3pr;;1 1 (u20t1 ; w2p1 ) ; v2b(r) X3pr;;p2;1 t1 e2 (u21 )
2
for odd r, and X3(pr;;1)2 p X3pr;;1 1 +v21+a3 ;1 X3pr;;p2;1 u21 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 3.2.3.
Proposition 3.2.4. Mod (p v12 v2b(r);1), d0(X3sr ) is expressed as follows. For
small values of r, we get
sv2a3 2 X3sp1;1 tp1 r = 2
2
sv2a3 3 X3sp2;1 (t1 ; v1 w22 ) ; sv2b(3);1 X3sp1 ;p;1 v2 2 ; v1 tp1 (w22 + v2;1 t2 ) r = 3
2
sv2a3 4 X3sp3;1 tp1 ; sv2b(4);1 X3sp2 ;p;1 fv2 2 ; v1 (w23 + tp1 2 )g r = 4:
For r 5 and odd, we get
2
sv2a3 X3spr;;11 (t1 ; v1 w22 ) ; sv2b(r);1 X3spr;;2p;1 f2 v2 2 ; v1 (w23 + tp1 w22 )g :
For r 6 and even, we get
2
s v2a3 X3spr;;11 tp1 ; s v2b(r);1 X3spr;;2p;1 fv2 2 ; 2 v1 w23 g :
3.3. The element 1=X2mr . Here we modify x2r into X2r in the analogous way
of modifying x3r into X3r , and introduce new elements 1=X2mr .
Denition 3.3.1. Dene integers a2r by a20 = 1 and a2r = pr + pr;1 ; 1 for
r 1, and elements X2r 2 v2;1 BP by
X20 = v2 X21 = X2p0 + v1p v2p u21 X22 = X2p1 + v11+a2 2 X2p;1 1(u11 ; u22)
X2r = X2pr;1 + v11+a2 X2pr;;1 1 (2 u11 ; u22) for r 3:
Then we have:
Proposition 3.3.2. Mod (p v12+a2 ),
8mv vm;1 tp + ;mv2 vm;2t2p for r = 0
>
< 12 1 2 12 1
m
for r = 1
d0 (X2r ) >mv1p v2mp;1 (t1 ; v1 w22 )
:v1a2 v2(mp;1)p ;1 (2 t1 ; v12 ) for r 2.
r
r
r
r
r
r
r
r
r
The Chromatic E1 -term H 0M12 for p > 3
33
Proof. It is sucient to prove the s = 1 case because d (X mr ) mv m; p d (X r ).
0
2
The r = 0 case is given by (2.1.1). For r = 1, we have
d0 (v1p v2p u21) v1p f d0(u21 )v2p + R (u21 )d0 (v2p ) g
v1p v2pd0 (u21)
v1p (v2p;1t1 ; tp12 ) ; v1p+1v2p;1 w22
mod p v1p+2 . Summing this with d0 (X2p0 ) gives d0 (X21 )
(See also 2, Proposition 5.4].) For r = 2, we have
(
2
1)
r
0
2
vp vp; (t ; v w ).
1 2
1
1
1
22
d0 (v11+a2 X2p;1 1 (u11 ; u22 )) ;v1p X2p;1 1f (2v1p;1 t1 ; tp1 ) + v1p (w2p2 ; 2 ) g
mod p v12+a2 2 . Again, summing this with d0 (X2p1 ) gives
2
r
d0 (X22 ) v1a2 2 v2p ;p (2t1 ; v1 2 ):
2
For r 3, we have
d0 (v11+a2 X2pr;;1 1 (2u11 ; u22 )) v1a2 ;p+1 X2pr;;1 1f 2(v1p;1 t1 ; tp1 ) + v1p (2p ; 2 ) g
r
r
mod p v12+a2 . This gives d0 (X2r ) inductively on r.
r
0
2
Shimomura 7] has replaced 1=v1mp with 1=xm
11 in computing H M0 for p = 2.
mp
Analogously, we construct elements 1=X2mr as a substitute for 1=v2 . Notice that
our 1=X2mr should be dierent from X2;rm because there is no inverse element of
X2r (r 1) in v2;1 BP .
r
Denition 3.3.3. Dene elements 1=X mr 2 v; BP (r 0 and p m 1) by
2
2
1
-
2mp
m
1=X2m0 = 1=v2m and 1=X2mr = 2 (1=X2mp
r;1 ) ; X2r (1=X2r;1 )
For example, easy calculation shows the congruence
for r 1:
1=X2m1 1=v2mp + m v1p v3 =v2(m+1)p+1 mod (v12p ):
P
m k
Although our 1=X2mr is not equal to (X2mr );1 = 1
k=0 (1 ; X2r ) , the above
denition is justied by the congruences (X2m1 ; v2mp )2 0 mod (p v12p ) and (X2mr ;
2(a2 ;p)
2
X2mp
) for r 2. Hereafter we denote Y (1=X2mr ) by
r;1 ) 0 mod (p v1
m
m
l
l
Y=X2r and (1=X2r )=v1 by 1=v1X2mr for simplicity.
r
a ),
; m v1 tp1 =v2m+1 + m2+1 v12 t21p =v2m+2 for r = 0
m v1p (t1 v1 w22 )=v2mp+1
for r = 1
a
(mp+1)p ;1
2
for r 2.
m v1 (2 t1 v1 2 )=v2
Proposition 3.3.4. Mod (p v
8;
>
<
d (1=X mr ) >;
:;
0
2
2+ 2r
1
;
r
;
r
Proof. We prove that the all above dierentials have the form ; d (X mr )=v mp .
0
2
2
2
r
Hirofumi Nakai
34
The case r = 0 is easy. For r 1, it is shown by induction on r. In fact, we have
2mp
2mp
m
m
d0 (1=X2mr ) = 2 d0 (1=X2mp
r;1) ; d0 (X2r )=X2r;1 ; R (X2r )d0 (1=X2r;1 )
2 d0 (1=X2mr;1)p ; d0(X2mr )=v22mp ; v2mp d0(1=X22rm;1)p
;2 d0(X2mr;1)p =v22mp ; d0(X2mr )=v22mp + d0 (X22rm;1)p =v23mp
;d0(X2mr )=v22mp + f d0(X22rm;1) ; 2 v2mp ;1 d0(X2mr;1) gp=v23mp
r
r
r
r
r
r
r
r
mod p v12+a2 . By the congruence d0 (X2mr ) mv2(m;1)p d0 (X2r ), the second term
is trivial.
r
r
Then we directly obtain the v1 -divisibilities of 1=v1X2mr (= 1=v1v2mp ) in H 0 M12 .
r
Corollary
3.3.5.
1=v X mr can be divided by va2
a
1=v1
2r
1
2
1
2
1
r
;1
!
in H 0 M12 , and the image of
X2mr under the coboundary map : H 0 M H 1 M21 is
8
for r = 0
> m h1=v2m+1
<
a
mp
+1
2
m
for r = 1
(1=v1 X2r ) = > m h0 =v2
: 2m h =v(mp+1)p ;1 for r 2.
r
;
;
;
r
0
2
This corollary asserts that N (m r) = a2r when we choose 1=X2mr to be y(mpr )
in Proposition 2.4.1.
Using 2.5, we easily see that f (1=v1a2 X2mr ) g is linearly independent.
r
4.
Preliminary calculations
In Section 5 we shall construct elements x(spr =j k) 2 v3;1 BP =(p v21 ) and compute the dierentials on them. For the sake of this, we dene some elements and
compute dierentials in Subsection 4.1. Based on these results, we describe some
inductive methods of constructing the elements x(spr =j k) in Subsection 4.2. This
is the hardest part in this paper.
Hereafter we assume that p > 3, and that all elements are either in v3;1 BP =(p)
or in v3;1 BP =(p v21 ). We shall compute the dierential d0 for positive integers e1
and e2 in two ways:
d0 : v3;1 BP =(p) ;! v3;1 BP =(p) A ;
mod v1e1 v2e2 ;1
;1
1
1
d0 : v3 BP =(p v2 ) ;! v3 BP =(p v2 ) A ; mod v1e1 :
4.1. Some lemmas.
Lemma 4.1.1. Assume that an element X satises
d0 (X ) v1e1 X3sr tp1 =v2e2 +1 mod v1e1 +1 where r 1, e1 1 and 1 e2 a3r ; 2. Then Y = X p ; v1pe1 v3 X3spr =v2(e2 +1)p+1
satises
d0 (Y ) v1pe1 X3spr (t1 ; v1 w22 )=v2pe2 +1 mod v1pe1 +2 :
The Chromatic E1 -term H 0M12 for p > 3
35
Proof. We observe that
d0 (v3 X3spr =v2(e2 +1)p+1 ) = d0 (1=v2(e2 +1)p+1 ) v3 X3spr + R (1=v2(e2 +1)p+1 ) d0 (v3 X3spr )
; v1v3 X3spr tp1 =v2(e2 +1)p+2
+ (1=v2(e2 +1)p+1 ; v1 tp1 =v2(e2 +1)p+2 ) d0 (v3 X3spr )
mod p v12 . Since (e2 + 1)p + 2 < pa3r , it is sucient to compute d0 (v3 X3spr ) mod
v12 v2pa3 . Using (2.1.2) and Proposition 3.2.3, we have
d0 (v3 X3spr ) = d0 (v3 ) X3spr + R (v3 ) d0 (X3spr ) d0 (v3 ) X3spr
(v2tp12 ; v2pt1 + v1tp2 )X3spr :
r
Observing the congruence w22 ; v3 tp1 =v2p+1 + (tp2 ; tp1 +p )=v2p + tp1+1 =v2 (Remark 3.1.5), we obtain
2
d0 (v3 X3spr =v2(e2 +1)p+1 ) X3spr tp1 =v2(e2 +1)p ; X3spr t1 =v2pe2 +1 + v1 X3spr w22 =v2pe2 +1 :
Now the result follows easy calculation.
Given a positive integer j = cp ; b a3r (0 b < p), it is convenient to work
with the element v2b X3spr;1=X2c1 instead of X3sr =v2j . Dene the integer a by a c
(p) with 0 a < p.
Lemma 4.1.2. Assume that c p for r = 2, that c a3r;1 ; 2 for odd r 3, or
that c a3r;1 ; p ; 1 for even r 4. Then d0 (v2b X3spr;1=X2c1 ) is expressed as
2
b
X3spr;1 X
f
E (b i)g
v2j+b i=1
p
p+1
v1p X3spr;1
; vj+1 f(a + b) t1 ; v1 a w22 + ab vvp3+1t1 ; b(a + b ; 1) t1v2 ]g
2
2
;
p
+2
p
m
m
;n
mod v1 , where E (m n) = n v2 (v1 t1 )n . In particular,
d0 (X3spr;1 =X2c1) ; a v1p X3spr;1(t1 ; v1 w22 )=v2j+1 mod (v1p+2 ):
Proof. Notice that
d0 (v2b X3spr;1=X2c1 ) = d0 (v2b ) X3spr;1 =X2c1 + R (v2b ) d0 (X3spr;1=X2c1 )
and that R (v2b ) is given in (2.1.1). By the assumption on c, we have
d0 (X3sr;1 =v2c ) ; a v1 X3sr;1tp1 =v2c+1 mod (v12 ):
Then, Lemma 4.1.1 gives d0 (X3spr;1 =X2c1) mod (v1p+2 ).
Similarly to the r 2 case, we may consider the element X3s1 =X21 instead of
X3s1=v2p for r = 1.
Lemma 4.1.3.
;
d X s =X (s ; 1) vp X s t =vp
0
31
21
1
31 1
2
+1
s
+ 2 v1p v3sp;1 t21 =v2 mod (v1p+1 ):
Hirofumi Nakai
36
Proof. Notice that d (X s =X ) = d (1=X ) X s + R (1=X ) d (X s ), and
that d0 (1=X21) X3s1 ; v1p X3s1 t1 =v2p+1 mod (v1p+1 ) by Proposition 3.3.4. On the
other hand, we have
R (1=X21) 1=v2p + v1p (nv3 ; v2p t1 )=v22p+1 o
mod v1p+1 2
2
d0 (X3s1 ) s v1p X3s;1 1 ntp2 ; tp1 R (v3;1ov4 )
mod v1p+1 v2p ;
3
s v2pX3s;11 tp1 ; d0 (v3;1 v4) + 2s v22p v3sp;2 t21 mod v1 v22p+1 :
Using these congruences, (2.1.2) and (2.1.3), we obtain
; ;
R (1=X21) d0 X3s1 s v1p X3s1 t1 =v2p+1 + 2s v1p v3sp;1 t21 =v2 mod v1p+1 :
Collecting terms gives the result.
For an integer n with p j n + 1, we denote the integer (n + 1)=p by n0 , and the
set of integers fs 2 j s = s0 p ; 1 with p s0 g by 0 . Then we introduce two
elements X0 (v2j X3sr ) and X (v2j X3sr ) for s 2 0 , each of which is congruent to
X3sr =v2j mod (v1 ).
Denition 4.1.4. When s = s0 p ; 1 2 0, we dene X0(v2j X3sr ) by
X0 (v2j X3sr ) = X3sr =v2j + j v1 X3sr0 +1=s0 v2j+1+a3 +1 :
Then we have:
Lemma 4.1.5. For odd r and p j a3r , d0(X0(v2j X3sr )) is expressed as
0
31
21
0
Z
21
-
21
31
0
31
N
N
N
r
-
(1 ; j ) v1 v2a3 ;j v3(sp;1)p 2 mod v12
for r 3 and a3r ; p j a3r , and is expressed as
j + 1
0
j
+2+a3 +1
p
2
s
0
; j (j + 1) v1 X3r+1t1 =s v2
; 2 v12X3sr t21p=v2j+2 + "(r j ) mod v13
for r = 1 and j p ; 2 or for r 3 and either j = a3r ; p + 2 or j a3r ; p ; 1,
where "(1 j ) = 0 and "(r j ) = v12 v2a3 ;j;1 X3spr;;11 f a(r)j w23 + b(r j ) tp1 2 g for
r 3 with integers a(3) = 1, a(r) = 2 for r 5 and b(r j ) = j 2 + (2 ; a(r)) j ; 1.
Proof. Here we prove only for r 3 case (the r = 1 case is easier). We compute the dierential on each term of X0 (v2j X3sr ). For the rst term, notice that
d0 (X3sr =v2j ) = d0 (1=v2j )X3sr + R (1=v2j )d0 (X3sr ). Easy calculation shows that
d0 (X3sr ) ;v2a3 X3spr;;11 (t1 ; v1 w22 + v12 v2;1 tp1 w22 ) mod v13 v22+a3 :
Using this and Proposition 3.3.4, we see that d0 (X3sr =v2j ) is congruent to
;j v1 X3sr tp1 f2 v2 ; (j + 1) v1tp1 g =2v2j+2 + v1v2a3 ;j;1 X3spr;;112 fv2 ; (j + 1) v1tp1 g
mod (v13 ).0 On the other hand, using Propositions 3.2.4 and 3.3.4, we observe that
d0 (j v1 X3sr+1=s0 v2j+1+a3 +1 ) is congruent to
; j (j + 1)v12X3sr0 +1tp1 =s0v2j+2+a3 +1 + j v1 X3sr tp1 f v2 ; (j + 1) v1 tp1 g =v2j+2
+ v1 v2a3 ;j;1 X3spr;;11 f;j v2 2 + a(r)j v1 w23 + j ( j + 3 ; a(r) ) v1 tp1 2 g
mod (v13 ). Collecting two terms gives the result for r 3.
r
r
;1
r
r
r
r
r
r
r
r
The Chromatic E1 -term H 0M12 for p > 3
Denition 4.1.6. For s = s0 p ; 1 2
37
and j a3r , we dene X (v2j X3sr ) by
0
X (v2j X3sr ) = v2b X3spr;1=X2c1 ; b v1 v2b;1 X3srp =s0 X2c+1a3 + b v1p+1 X3spr;1u22 =v2j+1
; (a(r) ; 1)b v1p+1 v2a3 ;j;pX3(spr;;21)p u23
where j = cp ; b and 0 b < p. Moreover, we dene DXmn](v1e1 v2e2 +1 X3sr ) 2
v3;1 BP =(p v21 ) A ; for 0 m < p, 0 n < p, s = s0 p ; 1 2 0 , and positive
integers e1 and e2 by
DXmn] (v1e1 v2e2 +1 X3sr ) = ;(m + n) v1e1 ;1 X3sr t1 =v2e2 +1
0
+ n(m + n ; 2)v1e1 X3sr+1 t1 =s0 v2e2 +2+a3 +1
+ v1e1 X3sr =v2e2 +1 fn2 + (m ; n)w22 g
Then we obtain the following lemma.
Lemma 4.1.7. Assume that r 3 is odd and c a3r;1 ; 2. Then d0(X (v2j X3sr ))
is expressed as
DXab] (v1p+1 v2j+1 X3sr ) + A(v2j X3sr )
;1
+ a(r)b v1 v2a3 ;j;p;1 v3(sp;1)p (v2p+2;b 2p R (v2b;1 ) + v1p 1 (v2j ))
mod v1p+2 . Here b and c are the integers in 4.1.6, a c (p) with 0 a < p, and the
element A(v2j X3sr ) is dened to be 0 for 0 b 1 and to be
N0
r
r
N
r
r
r
b
b;1
X3spr;1 X
v1 X3srp X
f
(1
;
i
)
E
(
b
i
)
g
;
b
f
E (b ; 1 i)g
v2j+b i=2
s0 v2j+b+a3 +1 i=1
for b 2. Here E (m n) is as in Lemma 4.1.2 and
2
1 (v2j ) = (a ; 1)(v3 ; v2p t1 )2p + f(a + b ; 2) v2p t1 ; v2 tp1 gu22:
2
2
In particular, v2p ;1 1 (v2a3 ;p ;p ) = ; v3p+1 (t1 + v2 w32 )=v22 .
Proof. The dierential on the rst term of X (v2j X3sr ) has already0 been computed in Lemma 4.1.2. For the second term, notice that ;bv1 v2b;1 X3srp =s0 X2c+1a3
0
r
r
r
is congruent to
;bv X srp =s0vj
1
0
3
a
+1+ 3r+1
2
; b(a ; 1)vp
1
+1
v3 X3srp =s0v2j+p+2+a3
0
r+1
mod (v1p+2 ). Using Propositions 3.2.3, 3.2.4 and 3.3.4, we obtain
0
d0 (;bv1 v2b;1 X3srp =s0 X2c+1a3 )
;bv1X3sr0p =s0v2j+b+a3 +1 f(v2 + v1 tp1 )b;1 ; v2b;1 g
; bv1X3spr;1tp1 (v2 + v1 tp1 )b;1 =v2j+b
;1)p p
b;1
+ a(r)bv1 v2a3 ;j;b+1 X3(sp
r;2 2 R (v2 )
0
+ (a + b ; 2)bv1p+1 X3srp t1 =s0 v2j+2+a3 +1
+ bv1p+1 X3spr;1=v2j+1
fw2p2 + (b ; 1)tp1+1=v2 + (a ; 1)(tp12 +p ; tp2 )=v2p + (a ; 1)w22 g
r
r
r
r
Hirofumi Nakai
38
;1)p
+ bv1p+1 v2a3 ;j;p;1 X3(sp
r;2
f;v2tp12 2p + a(r)(a ; 1)2p (v3 ; v2p t1) ; ( a(r) ; 1 )v2w2p3g
mod (v1p+2 ). For the third term, notice that
d0 (bv1p+1 X3spr;1u22 =v2j+1 ) bv1p+1 d0 (X3spr;1 u22)=v2j+1
mod (v1p+2 ). Propositions 3.1.3 and 3.2.3 give d0 (X3spr;1 u22 ), and we obtain
d0 (bv1p+1 X3spr;1u22 =v2j+1 )
bv1p+1 X3spr;1(2 ; 2p )=v2j+1 ; bv1p+1v2a3 ;j;p X3(spr;;21)ptp12 (u22 ; 2p )
mod (v1p+2 ). For the last term, we may assume that r 5 because a(3) 2; 1 = 0.
a3 ;j ;p X (sp;1)p (u tp ; wp )
;1)p
Then it follows that d0 (v2a3 ;j;p X3(sp
22 1
23
r;2 u23 ) v2
3r ;2
mod v1 by Propositions 3.1.6 and 3.2.3, and hence
;1)p
d0 (; (a(r) ; 1)bv1p+1 v2a3 ;j;p X3(sp
r;2 u23 )
; (a(r) ; 1)v1p+1v2a3 ;j;p X3(spr;;21)p(u22tp12 ; w2p3 ):
Collecting four terms gives
d0 (X (v2j X3sr ))
X3spr;1 (v2 + v1 tp1 )b ; v2b ; bv1tp1 (v2 + v1tp1 )b;1 =v2j+b
; bv1X3sr0p (v2 + v1 tp1 )b;1 ; v2b;1 =s0v2j+b+a3 +1
;1)p p
b;1
+ a(r)bv1 v2a3 ;j;b+1 X3(sp
r;2 2 R (v2 )
0
+ b(a + b ; 2)v1p+1 X3srp t1 =s0 v2j+2+a3 +1
; (a + b)v1pX3spr;1t1 =v2j+1 + v1p+1nX3spr;1=v2j+1 fb2 + (a ; b)w22 go
;1)p
p
p
p2
+ a(r)bv1p+1 v2a3 ;j;p;1 X3(sp
r;2 (a ; 1)2 (v3 ; v2 t1 ) ; v2 t1 u22 :
Then, apply the equation
r
r
r
r
r
r
r
r
r
r
(X + Y )b
; ;
Xb
b
X
bY (X + Y )b;1 = (1
i=2
; i)
b
b;i i
i X Y
to the rst term, and the congruence
0
X3srp X3sr0 +1 + a(r)s0 v2b(r+1) X3spr;;11u22
mod v2b(r+1) for odd r 3
to the fourth term. Modifying these terms completes the proof.
We also set X (v2j X3s1 ) to v2b X3s1 =X21 ; bv1v2b;1 X3s02 =s0 X2p+1
1 for j = p ; b. Then
we have:
Lemma 4.1.8. d0 (X (v2p;1 X3s1)) is expressed as ;3v1pX3s1t1=v2p mod v1p+1 , while
d0 (X (v2p;2 X3s1)) is expressed as
; 2v12X3s02(tp1 ; 2v1p;1 t1)=s0v2p2 +p
; v12 X3s1=v2p;1 ft21p=v2 + 4v1p;2t1 ; v1p;1 (2 + t2=v2 + C1)g
+ v1p+1 Z=v2p;2
The Chromatic E1 -term H 0M12 for p > 3
39
mod v1p+2 for some Z 2 v3;1 BP BP , where
2
C1 = (u30 tp1 +1 ; w32 )tp1 + (u31 ; w3p1 )tp2 + up30 tp3 :
Proof. Notice that 1=X m
21
1=vmp + mvp v =v m
1 3
2
(
2
p
+1) +1
mod (v12p ). So we may
compute d0 on
2
2
X3s1=v2p;b + v1p v3 X3s1 =v22p+1;b ; b v1 X3s02=s0 v2p +p+1;b ; b v1p+1 v3 X3s02 =s0 v2p +2p+2;b
instead of d0 (X (v2p;b X3s1 )). For b = 1 case, we consider the dierential mod
(v1p+1 ), so that the last term can be omitted. Because X3s1 v3sp + v2p v3(s;1)p;1 v4
0
2
mod v22p and X3s02 X3s1p mod v2p +p , we may compute d0 on
v2 (v3s =v2 )p + v1p v3sp+1 =v22p + v1p v3(s;1)p v4 =v2p ; v1 (X3s01 =s0 v2p+1 )p :
Then, we obtain the desired result using (2.1.1) { (2.1.3), Proposition 3.2.3 and
Proposition 3.3.4.
Similar but harder calculation shows the b = 2 case.
Next we dene X+ (v2jp X3spr ) to modify the dierential on X0 (v2j X3sr )p . Let
; v2p v2 X sp =v(j+2)p+2 + j (j + 1)v2p v X s0p =s0v(j+2+a3 +1 )p+1
A0 = ; j+1
3 3r +1
3 3r
1
2
1
2
2
n
o
(sp;1)p
2p (a3 ;j ;1)p;1
A1 = v1 v2
X3r;1 a(r)jv2 u23 ; (j + 1)v3 u22 A2 = v12p;1 v3(sp;1)p+2 =v2 :
Using these elements, we dene X+ (v2jp X3spr ) as follows. For r = 1,
(
for j p ; 3
jp
sp
X+ (v2 X3r ) = A0
A0 + A2 for j = p ; 2:
If r 3 is odd and either j = a3r ; p + 2 or j a3r ; p ; 1, then
X+ (v2jp X3spr ) = A0 + A1 :
Then we obtain the following lemma.
Lemma 4.1.9. Mod (v12p+1 ),
d0 (X0 (v2j X3sr )p + X+ (v2jp X3spr ))
j + 1 v2p X sp t21
2p
s0
1
3r
3r +2 t1
;
;j (j + 1) sv0v1jpX+2+
a3 +2
jp
+2
2
v2
2
(sp;1)p
2p (a3 ;j ;1)p;1
; (j ; 1)v1 v2
X3r;1 2 (v2j )
where 2 (v2j ) = 0 for r = 1 and
2
2 (v2j ) = (j + 1)(v3 ; v2p t1 )2p + f (j + 1)v2p t1 ; v2 tp1 gu22
2
for odd r 3. In particular, v2p ;1 2 (v2a3 ;p;1 ) = ;v3p+1 (t1 + v2 w32 )=v22 .
Proof. Here we prove only for r 3 case. Other cases are similarly proved.
Notice that d0 (A0 ) is congruent to
j + 1
; 2 v12p d0 (v32X3spr )=v2(j+2)p+2 + j (j + 1)v12p d0(v3 X3sr0p+1)=s0v2(j+2+a3 +1 )p+1
r
r
r
r
r
r
Hirofumi Nakai
40
mod (v12p+1 ), so that it is sucient to compute d0 (v32 X3spr ) mod (v1 v2(j+2)p+2 ) and
0
d0 (v3 X3srp+1) mod (v1 v2(j+2+a3 +1 )p+1 ). Easy computations show that
r
(v22 t21p + v22p t21 + 2 v2 v3 tp1 ; 2 v2p v3 t1 ; 2 v2p+1 tp1 +1 )X3spr 2
0
2
2
(v2 tp1 ; v2p t1 )X3srp+1 + s0 v2pa3 +1 X3spr tp1 (v3 + v2 tp1 ; v2p t1 )
;s0v2b(r+2)X3(spr;;11)p2p (v3 + v2 tp12 ; v2pt1 ):
Using these and the congruence
d0 (v32 X3spr )
0
d0 (v3 X3srp+1)
2
2
2
r
X3sr0 +2 X3srp+1 ; s0 v2a3 +2 ;p v3 X3spr ; s0 v2b(r+2) X3spr;1 u22 we can show that d0 (A0 ) is congruent to
0
2
j + 1 v2p X sp (t2p2 ; v2p;2 t21)
v12p (X3srp+1tp1 ; v2p;1 X3sr0 +2 t1 )
1
3r 1
2
+ j (j + 1)
2
v2(j+2)p
s0 v2(j+2+a3 +1 )p
; j (j + 1)v12p v2(a3 ;j;1)p;1 X3(spr;;11)pf2p(v3 + v2 tp12 ; v2p t1) + v2pt1 u22g:
On the other hand, d0 (A1 ) is congruent to
;1)p
;1)p
2p (a3 ;j ;1)p;1
d0 (v3 X3(sp
jv12p v2(a3 ;j;1)p d0 (a(r) X3(sp
r;1 u23 ) ; (j +1)v1 v2
r;1 u22 ):
Noticing that (a3r ; j ; 1)p (p ; 3)p, we have
;1)p
(sp;1)p
(sp;1)p
d0 (X3(sp
r;1 u23 ) d0 (u23 )X3r;1 + R (u23 )d0 (X3r;1 )
(u22tp12 ; w2p3 )X3(spr;;11)p + (u23 + u22tp12 ; w2p3)d0 (X3(spr;;11)p)
0
r
r
r
r
r
pa3 ;1 X (sp ;p;1)p tp mod (v v pa3 ;1 +1 ),
;1)p
mod (v1 ). Because d0 (X3(sp
1
r;1 ) ;v2
3r ;2
1
2
the second term remains only for r = 3. Observe that u23 and w2p3 can be replaced
3
2
3
2
with ;v2;p v3p u22 and ;v2;p v3p 2p respectively. So we obtain
;1)p
d0 (a(r)X3(sp
r;1 u23 )
a(r) X3(spr;;11)p(u22tp12 ; w2p3 )
2
2
2
2
+ a(r) v2(a3 ;1 ;p )p v3p X3(rsp;;2 p;1)p tp1 (u22 ; 2p )
a(r) X3(spr;;11)p(u22tp12 ; w2p3 ) + (2 ; a(r)) X3(spr;;11)ptp12 (u22 ; 2p)
X3(spr;;11)pf 2 u22tp12 ; a(r) w2p3 ; (2 ; a(r)) tp12 2p g:
Moreover, it is easy to see that
;1)p
(sp;1)p
p2
p
p
p2
p
d0 (v3 X3(sp
r;1 u22 ) X3r;1 fv2 t1 u22 ; v2 t1 u22 ; 2 (v3 + v2 t1 ; v2 t1 )g:
Collecting terms gives d0 (X+ (v2jp X3spr )) as
2
r
2
r
r
0
j + 1 v pX spr (t p ; v p; t )
v p (X srp tp ; vp; X sr0 t )
+ j (j + 1)
2
2
1
3
2 2
1
vj
p
2
2
2 2
1
2
1
3
( +2)
2
; "(r j )p ; (j ; 1)v pv a ;j; p; X spr;;
This completes the proof for r 3 case.
2
1
( 3r
2
1)
1
(
3
2
1
3
+1 1
2
0 ( +2+ 3r+1 )
2
svj
p (v j ):
2 2
1)
1
a
p
+2 1
The Chromatic E1 -term H 0M12 for p > 3
41
4.2. Some inductive methods of constructing elements x(spr =j k). Here
we describe some inductive methods of constructing x(k) = x(spr =j k). Because
(xs3r =v2j )p = (X3sr =v2j )p for j a3r , we may start with X3sr =v2j (p j a3r ).
In general, each x(k) (k 1) is constructed by adding some appropriate terms to
x(k ; 1)p so that we can nd an element of Coker (2.5.1) from the numerator
of the leading term of d0 (x(k)). There are some patterns for adding terms and
constructing x(k) for a large enough k. We rst observe the case with small j and
s 62 0 , which is the simplest example of such patterns.
Proposition 4.2.1. Assume that an element W1 satises
k
k
-
N
d0 (W1 ) ;jv1p X3spr t1 =v2jp+1 mod v1p+1
for 1 j a3r ; 2, then Wk (k 2) dened inductively on k by
Wk = W1p + jv1a2 2 ;p v2 X3spr =X2jp1+1
for k = 2
a
(jp+1)p ;1 ;1
p
2 ;p sp
= Wk;1 + 2jv1
X3r =v2
for k 3
2
k
k
k
satises
d0 (Wk ) ;2jv1a2 X3spr t1 =v2(jp+1)p
mod v1a2
Proof. Using Propositions 3.2.3 and 3.3.4, for k = 2, we have
d0 jv1a2 2 ;p v2 X3spr =X2jp1+1
2
k
k
k
jva
1
p
22 ; +1
;1
k +1
:
X3spr (tp1 ; 2v1p;1 t1 )=v2(jp+1)p 2
while for k 3, we have
;1
;1
d0 2jv1a2 ;p X3spr =v2(jp+1)p ;1 2jv1a2 ;p+1 X3spr (tp1 ; v1p;1 t1 )=v2(jp+1)p :
k
k
k
k
k
k
The result now follows by an easy calculation.
We will see that each x(spr =j 1) with r 1, 1 j a3r ; 2 and s 62 0 satises
the condition for W1 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 rst introduce some elements.
Denition 4.2.2. For e1 1, e2 1, s = s0p ; 1 2 0 and odd r 3, we dene
Yab](v1e1 p v2(e2 +1)p X3sr+1) as follows: for s0 62 0 , it is given by
N
N
N
v1(e1 ;1)p;1 v2 X3spr v1e1 p X3spr u22
v1e1 p;1 X3sr0 +2
(a + b)
+ a (e2 +1)p ; (a + b)(b ; 2) 0 (e2 +1)p+a3 +2 X2e21+1
v2
s v2
r
for s0 2
N0
, it is given by
v1e1 p X3sr00 +3
(above terms) + 2a 00 (e2 +1)p+b(r+3) :
s v2
Hirofumi Nakai
42
Moreover, we dene Z (v1e1 p v2(e2 +1)p X3sr+i ) by
i
i
;2 s0 v e
v1e1 p v3 X3srp+i
0
i
p +p+1+a3
( 2 +1)
2
i
r+i+1
e1 p sp
; v1 vX(e32r+1)+i;p 1u22 for s0 62
2
i
i
N0
and odd r + i
v1e1 p X3spr+i;1 u22
for s0 62 0 and even r + i
v2(e2 +1)p
e1 p s
fors0 2 0 ; v1v(e2X+1)3rp+i (3u22 ; 2u33)
2
and DZ (v1e1 p v2(e2 +1)p X3sr+i ) 2 v3;1 BP =(p v21 ) A ; by
ve1 p X sr0 +i+1t1
v1e1 p X3sr+i
2 0 (e12 +1)p3+1+
+
(2w22 ; 2 ) for s0 62 0 and odd r + i
a3 + +1
v2(e2 +1)p
s v2
ve1 p X sr0 +i+1 tp1
v1e1 p X3sr+i t2
2 0 (e12 +1)p3+1+
+
2
;
for s0 62 0 and even r + i
2
(e2 +1)p
a
3 + +1
v
2
v2
s v2
v1e1 p X3sr+i
fors0 2 0 :
; v(e2 +1)p (32 ; 23 )
2
Notice that each of these leading terms includes an element of Coker .
Proposition 4.2.3. Assume that an element x(k) satises
d0 (x(k)) DXab](v1e1 v2e2 +1 X3sr ) + v1e1 Zk =v2e2
mod v1e1 +1
for some Zk 2 v3;1 BP BP , with e2 ;b (p) and odd r 3. Let
x(k + 1) = x(k)p + Yab] (v1e1 p v2(e2 +1)p X3sr+1):
We compute d0 (x(k + 1)). For a 6= 0, we get
d0 (x(k + 1)) aDZ (v1e1 p v2(e2 +1)p X3sr+1 ) + v1e1 p Zk+1 =v2(e2 +1)p;1 mod v1e1 p+1
i
N
i
i
N
i
i
i
i
i
i
r
i
i
i
N
i
N
i
i
r
i
i
i
N
for some Zk+1 2 v3;1 BP BP . For a = 0, we get
d0 (x(k + 1)) bv1e1 p+1 X3sr+1tp1 =v2(e2 +1)p+1 + v1e1 p Zk0 +1 =v2(e2 +1)p;1 mod v1e1 p+2
for some Zk0 +1 2 v3;1 BP BP .
Proof. Here we may ignore elements killed by v2(e2 +1)p;1. For the rst term of
Yab](v1e1 p v2(e2 +1)p X3sr+1), notice that (a + b)v1(e1 ;1)p;1 v2 X3spr =X2e21+1 is congruent
to
(a + b)v1(e1 ;1)p;1 v2 X3spr f1=v2(e2 +1)p ; (b ; 1)v1p v3 =v2(e2 +2)p+1 g
mod (p v1(e1 +1)p;1 ). Using (2.1.2), Propositions 3.2.4 and 3.3.4, we have
d0 ((a + b)v1(e1 ;1)p;1 v2 X3spr =X2e21+1 )
(a + b)v1(e1 ;1)pX3spr
fv2ptp1 + (b ; 2)v1p;1 v2pt1(A) + (b ; 1)v1p (tp12 +p ; tp2 (B) + v1 v2p;1tp1 (C))g=v2(e2 +2)p
The Chromatic E1 -term H 0M12 for p > 3
43
mod (v1e1 p+2 ). For the second term, we can easily obtain
d0 (av1e1 p X3spr u22 =v2(e2 +1)p ) av1e1 p X3spr (2 ; 2p )=v2(e2 +1)p (B)
mod (v1e1 p+1 ) by Proposition 3.1.3. For the third term, using the congruence
2
0
X3sr0 +2 X3srp+1 ; s0 v2a3 +2 ;p v3 X3spr mod v2b(r+2);p ;p;1 (2.1.2), Propositions 3.2.3 and 3.3.4, we have
d0 (;(a + b)(b ; 2)v1e1 p;1 X3sr0 +2 =s0v2(e2 +1)p+a3 +2 )
;(a + b)(b ; 2)v1e1 pX3sr0p+1tp1 =s0v2(e2 +1)p+1+a3 +2
2
+ (a + b)(b ; 2)v1e1 p;1 X3spr f;v2pt1 (A) ;v1 (tp1 +p ; tp2 )(B) ;v12 v2p;1 tp1 (C ) g=v2(e2 +2)p
r
r
r
mod (v1e1 p+2 ). Collecting three terms gives
(A) = 0
2
(B ) = v1e1 p X3spr =v2(e2 +1)p f(a + b)(tp1 +p ; tp2 )=v2p + a(2 ; 2p )g
(C ) = (a + b)v1e1 p+1 X3spr tp1 =v2(e2 +1)p+1 :
Thus d0 (Yab] (v1e1 p v2(e2 +1)p X3sr+1)) is congruent to
; (a + b)(b ; 2)v1e1 pX3sr0p+1tp1 =s0v2(e2 +1)p+1+a3 +2 + (a + b)v1(e1 ;1)pX3spr tp1 =v2(e2+1)p
2
+ v1e1 p X3spr =v2(e2 +1)p f(a + b)(tp1 +p ; tp2 )=v2p + a(2 ; 2p )g
mod v1e1 p+1 for a 6= 0, and d0 (Y0b] (v1e1 p v2(e2 +1)p X3sr+1)) is congruent to
; b(b ; 2)v1e1 pX3sr0p+1tp1 =s0v2(e2 +1)p+1+a3 +2
2
+ bv1(e1 ;1)p X3spr fv2p tp1 + v1p (tp1 +p ; tp2 ) + v1p+1 v2p;1 tp1 g=v2(e2 +2)p
mod v1e1 p+2 for a = 0. Now, recall denitions of 2 and w22 and notice the congruences
2
0
X3srp+1 X3sr0 +2 + s0 v2a3 +2 ;p v3 X3sr+1 mod v2b(r+2);p2 ;p;1 X3spr X3sr+1
mod v2b(r+1);p ;p;1 :
Modifying the dierentials using the above two congruences completes the proof of
this proposition for the case s0 62 0 .
For s0 2 0 case, the result is proved using the congruence
d0 (2av1e1 p X3sr00 +3 =s00 v2(e2 +1)p+b(r+3) ) 2av1e1 p Y
where
Y = X3sr0 +2tp1 =v2(e2 +1)p+1+a3 +2 ; X3sr+1(2 ; 3 + t2 =v2 )=v2(e2 +1)p + Z=v2(e2 +1)p;1
for some Z 2 v3;1 BP BP .
r
r
r
N
N
r
We dene Y (v1p +p ;p v2p ;p X3sp2;1 ) to be
3
2
3+ 2
2
2
8 p p ; p; sp;
<v
X (4v + vp u )=X p;
:(above terms) + 4vp p ;pX s0 =s0vp p
1
1
32
1
2
1
3+ 2
1
+1
34
22
21
2
for s 62
1
a3 4 for s
3+ 2+
2
N0
,
N0
:
Hirofumi Nakai
44
By routine calculation, we have:
Lemma 4.2.4. For s 62 , d (Y (vp3
p p v p2 ;p X sp;1 )) is expressed as
2
32
3
2
2
2
(v1p +p ;2p X3sp2;1 =v2p ) 4v2p tp1 + v1p (4tp1 +p 5tp2 ) + v1p v2p (22 2p + C2 w33 )
3
2
2
+ vp +p ;p Z=vp ;p;1
N0
0
1
+ 2;
f
1
;
;
;
g
2
mod v1p +p ;p+1 . For s 2 N 0 , it is expressed as
3
2
(above terms) + 4v1p +p ;p X3s3 tp1 =v2p +p
; 4v1p3 +p2;p X3sp2;1(2 ; 3 + v2;1t2 )=v2p2;p + v1p3 +p2;p Z 0=v2p2;p;1
3
2
3
2
mod v1p +p ;p+1 . Here, Z Z 0 2 v3;1 BP BP , and
3
2
C2 = (
X
i+j =2
u3i tp2;j )tp1 + up30tp1 (tp2
i
2
2
; tp
1
p ):
3+ 2
Proof. By Proposition 3.3.4,
d0 (v2 X3sp2;1 =X2p;1 1) = d0 (v2 X3sp2;1 )=X2p;1 1 + R (v2 X3sp2;1 )d0 (1=X2p;1 1)
d0 (v2 X3sp2;1)(1=v2p2;p ; v1pv3 =v2p2+1)
2
+ R (v2 X3sp2;1 ) v1p (t1 ; v1 w22 )=v2p ;p;1
mod (v1p+2 ). It is sucient to compute d0 (v2 X3sp2;1 ) mod (v12 v2p +1 ) and mod
2
(v1p+2 v2p ;p ). Notice that
2
d0 (v2 X3sp2;1 ) = d0 (v2 )X3sp2;1 + R (v2 )d0 (X3sp2;1 )
d0(v2 )v3(sp;1)p2 + R (v2 )d0 (v3(sp;1)p2 ) mod v1p+2 v2p2 ;p:
Using the results of 3.2, we see that d0 (4 v1p +p ;2p;1 v2 X3sp2;1 =X2p;1 1) is congruent
to
3
2
2
2
(4 v1p +p ;2p X3sp2;1 =v2p ;p )ftp1 + v1p (tp1 +p ; tp2 )=v2p g:
On the other hand, we have
3
2
d0 (v1p +p ;p X3sp2;1 u22 =v2p ;p ) v1p +p ;p d0 (X3sp2;1 u22)=v2p ;p
3
2
2
3
2
2
mod (v1p +p ;p+1 ), and
3
2
d0 (X3sp2;1 u22) = d0 (u22 )X3sp2;1 + R (u22 )d0 (X3sp2;1 )
(2 ; 2p )X3sp2;1 + (u22 + 2 ; 2p )d0(X3sp2;1 )
mod (v1 v2 ). According to Proposition 3.2.3,
2
2
2
3
2
d0 (X3sp2;1 ) ;v2p X3sp1 ;p;1 tp1 + v2p +p+1 v3sp ;p ;p;1 (u20 t2 ; w33 )
The Chromatic E1 -term H 0M12 for p > 3
45
mod (v1 v2p +p+2 ). Using this congruence and Denition 3.2.1, we obtain
d0
2
v1p +p ;p X3sp2;1 u22
v2p2 ;p
3
2
!
v
p3 +p2 ;2p X sp;1 32
22
2 ;p
p
v
1
2
p
; p ; vt p
2
2
2
+ C2 ; w33
p p p
3+ 2;
+ v1 p2 ;p;1Z
v2
for some Z 2 v3 BP BP . Collecting terms gives the result for s 362 20 case.
3
2
For s 2 0 case, it is easy to see that d0 (4 v1p +p ;p X3s04 =s0 v2p +p +a3 4 ) is congruent to
;1
N
N
2
v1p +p ;p X3s3tp1 v1p +p ;p X3sp2;1 tp1
;
4
(2 ;3 +u20t2 )+(elements killed by v2p ;p;1 ):
3 +p2
2 ;p
p
p
v2
v2
3
4
3
2
2
Proposition 4.2.5. Assume that an element x(2) satises
v1p +p;2 X3sp1;1 t1
v1p +p;1 X3s2 t1
+
2(
s
;
1)
v2p;1
v2p2 +p
s vp2 +p;1X sp;1 t2 vp2 +p;1 Z
+ 2 1 p;1 31 2 + v + C1 + 1 p;2
v2
v2
2
2
mod (v1p +p ) for; some Z 2 v3;1 BP BP , where C1 is as in Lemma 4.1.8. Then
3
2
2
x(3) = x(2)p + 2s Y (v1p +p ;p v2p ;p X3sp2;1 ) satises
3
2
2
3
2
2
d0 (x(3)) s(s ; 1)DZ (v1p +p ;p v2p ;p X3sp2;1 ) + v1p +p ;p Z3=v2p ;p;1
3
2
mod (v1p +p ;p+1 ) for some Z3 2 v3;1 BP BP .
Proof. Collect the dierentials on x(2)p and ;2sY (v1p3 +p2 ;p v2p2 ;p X3sp2;1) using
d0 (x(2)) ;2s(s ; 1)
the congruences
2
2
C1p + C2 w33 mod I3 3
3
2
X3s3 X3sp2 ; sv2p ;1 v3 X3sp2;1 mod v2p +p :
As the next step of Proposition 4.2.3 or 4.2.5, we show the following proposition.
Here we state only the result.
Proposition 4.2.6. Assume that an element x(k + 1) satises
d0 (x(k + 1)) DZ (v1e1 p v2(e2 +1)p X3sr+1 ) + v1e1 p Zk+1 =v2(e2 +1)p;1 mod v1e1 p+1
for some Zk+1 2 v3;1 BP BP , where e1 and e2 are positive integers and r 1.
Then x(k + i) dened inductively on i 2 by
x(k + i) = x(k + i ; 1)p + Z (v1e1 p v2(e2 +1)p X3sr+i )
satises
d0 (x(k + i)) DZ (v1e1 p v2(e2 +1)p X3sr+i ) + v1e1 p Zk+i =v2(e2 +1)p ;1 mod v1e1 p +1
i
i
i
i
i
i
i
Hirofumi Nakai
46
for some Zk+i 2 v3;1 BP BP .
Proposition 4.2.6 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 4.2.7. Assume that an element x(k) satises
d0 (x(k)) v1e1 p v2p ;p v3(sp;1)p i (v2a3 ;p
(= v1e1 v3e3 p tp1 =v22p + v1e1 Zk =v2p )
;1
r+i
3
r
p
3;i ; 2;i
)p mod v1e1 +1
3;
2;
for i (v2a3 ;p ;p ) (i = 1 2) as in Lemma 4.1.7 or 4.1.9, and some Zk
v3;1 BP BP . Then x(k + i) dened inductively on i 1 by
r
i
i
x(k + i) = x(k + i ; 1)p ; v1e1 p v3e3 p
= x(k + i ; 1)p
i+1
i
+1
=v2p
i+2
p +1 ;pa3
+2
i
2
for odd i
for even i
i+1 +1
satises
d0 (x(k + i)) v1e1 p v3e3 p tp1 =v2p +2p ;a3 +2
+1
;1
+ v1e1 p Zk+i =v2p +p
mod v1e1 p +1
for some Zk+i 2 v3;1 BP BP , where (i) = 0 for odd i and 1 for even i.
Proposition 4.2.8. Assume that an element x(k) satises
d0 (x(k)) v1e1 v2e2 v3e3 p w2t mod v1e1 +1
with Maxfpt;1 ; pr 0g + P (t ; 2) + 1 e2 . Then x(k + i) dened inductively on
i 1 by
+
x(k + i) = x(k + i ; 1)p + (;1)i;1 v1e1 p +P (i;11) v2e2 p ;P (i;11) v3e3 p u2t+i;1
i
i+1
i+2
(i)
i
i
i+1
i
i
i
r
i
i
r
i
satises
d0 (x(k + i)) (;1)i v1e1 p +P (i;1) v2e2 p ;P (i;1) v3e3 p w2t+i mod v1e1 p +P (i;1)+1 where P (i j ) = pj (pi;j+1 ; 1)=(p ; 1) = pi + + pj 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 ; 1)p so that
i
i
r+i
i
the dierential has one of the forms in the assumptions of Propositions 4.2.1, 4.2.6,
4.2.7 and 4.2.8. 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 dening x(k) (= x(spr =j k)) for
all cases using the preparatory computations displayed in Section 4. Notice that
the smallest integer N with d0 (x(k)) 6 0 mod p v1N +1 gives the v1 -divisibility N (k)
(= N (s r j k)) of x(k).
The Chromatic E1 -term H 0M12 for p > 3
47
5.1. Denitions of x(0) and the dierentials. Now we start with X3sr =v2j
(p j a3r and p s 2 ). Using Propositions 3.2.4 and 3.3.4, we can easily
calculate d0 (X3sr =v2j ) mod v12 :
-
-
Z
d0 (X3s0 =v2j ) ;v1 v3s tp1 =v22 + sv1 v3s;1 (tp2 ; tp1 +p )=v2 and for r = 1 or for even r 2,
d0 (X3sr =v2j ) ;jv1 X3sr tp1 =v2j+1 :
Finally, for odd r 3,
2
d0 (X3sr =v2j ) ;jv1 X3sr tp1 =v2j+1 ; sv1 v2a3 ;j X3spr;;11 jtp1+1 =v2 + w22 :
r
Moreover, we have
d0 (X3s0 =v2 )
v1 v2p;1 v3s;1 w22
mod (v12 )
v1 v2p;2 v3s;1 2 (v2 ; v1 tp1 ) mod (v13 )
for p j (s ; 1). Then the coboundary : H 0 M12 ! H 1 M21 on X3sr =v1 v2j is
(X3sr =v1 v2j ) = ;jv3sp tp1 =v2j+1 + (elements killed by v2j ):
According to 2.5, its numerator belongs to Coker, when s 62 0 = fs0 p ; 1 j p s0 g
and r 1 is odd or when r 0 is even by (2.5.1). In this case, we can set x(0)
to X3sr =v2j with the v1 -divisibility N (0) = 1. In another case, we need to modify
X3sr =v2j so that its coboundary includes an element of Coker.
5.1.1. For odd r 3 and s 2 0 .
(i) For a3r ; p j a3r . We set x(0) to X0 (v2j X3sr ). By Lemma 4.1.5, we
r
N
-
N
have
;1
d0 (x(0)) (1 ; j )v1 v2a3 ;j v3(sp;1)p 2 mod v12 r
r
and so N (0) = 1. Notice that this dierential is trivial either for j = a3r ;
p + 2 1 (p) or for j a3r ; p ; 1.
(ii) Either for j = a3r ; p + 2 or for j a3r ; p ; 1. We also set x(0) to
X0 (v2j X3sr ). Then, Lemma 4.1.5 shows that
d0 (x(0)) ;j (j + 1)v12 X3sr0 +1 tp1 =s0 v2j+2+a3 +1 ; j +2 1 v12 X3sr t21p =v2j+2 + "(r j )
mod (v13 ), and so N (0) = 2. Note that all elements but "(r j ) are vanished
when p j (j + 1), and "(r j ) = 0 unless r 3 is odd and a3r ; p2 ; p j a3r ; 2p, in which case
;1
"(r j ) = ;a(r)v12 v2a3 ;j;1 v3(sp;1)p w23 :
(iii) For p j (j +1), p2 (j + p +1) and j a3r ; p2 ; p. Set x(0) = X (v2j X3sr ).
2
2
Notice that we replace X0 (v2a3 ;p ;p X3sr ) with X (v2a3 ;p ;p X3sr ) for j =
a3r ; p2 ; p although "(r a3r ; p2 ; p) 6= 0. Then Lemma 4.1.7 shows that
;1
d0 (x(0)) DXj0 1] (v1p+1 v2j+1 X3sr ) + a(r)v1p+1 v2a3 ;j;p;1 v3(sp;1)p 1 (v2j )
r
r
r
-
r
r
r
r
Hirofumi Nakai
48
mod v1p+2 . Note that d0 (x(0)) 6 0 mod (v1p+1 ), and so N (0) = p when
p2 (j + p + 1).
(iv) For p2 j (j + p + 1) and j a3r ; p2 ; 2p. We rst dene B (j ) by
-
B (j ) = X (v2j X3sr ) for j a3r ; p2 ; 3p
2
;1
B (a3r ; p2 ; 2p) = X (v2a3 ;p ;2p X3sr ) ; a(r)v1p+1 v3(sp;1)p +p+2 =2v22 :
r
r
Then, let
(
for s0 62 0
x(0) = B (j )
00
j
+1+
b
(
r
+2)
B (j ) ; 2 v1p+1 X3sr+2=s00 v2
for s0 2 0 :
N
N
Easy calculation shows that d0 (x(0)) ;DZ (v1p+1 v2j+1 X3sr ) mod (v1p+2 ),
and so N (0) = p + 1.
5.1.2. For r = 1 and s 2
Lemma 4.1.5 shows that
N 0.
For j p ; 2, we set x(0) to X0 (v2j X3s1 ). Then
d0 (x(0)) ;j (j + 1)v12 X3s02 tp1 =s0 v2j+2+a3 2 ; j +2 1 v12 X3s1 t21p =v2j+2
mod (v13 ), and so N (0) = 2. For j = p ; 1, we set x(0) to X (v2p;1 X3s1 ). By
Lemma 4.1.8, we have d0 (x(0)) ;3v1p X3s1 t1 =v2p mod (v1p+1 ), and so N (0) = p.
5.2. Denitions of x(k) (k 1) and the dierentials. We complete the
proof of our main theorem by dening x(k) and computing the dierentials for all
cases, based on the computations in k = 0 case.
5.2.1. For r = 0.
(i) For p (s ; 1).
-
We set x(1) to X3s1 =X21 . Then Lemma 4.1.3 shows that
d0 (x(1)) (s ; 1)v1p X3s1t1 =v2p+1 + 2s v1p v3sp;1 t21 =v2
mod v1p+1 , and so N (1) = p. For k = 2, we set
x(2) = x(1)p +
Then Lemma 4.1.8 implies
d0
s
s
p ;2
p;2 sp;1
2 v1 X (v2 X31 ):
2
p ;2
p;2 sp;1
2 v1 X (v2 X31 )
2
v1p X3s2 p
;(s ; 1) vp2 +p (t1 ; 2v1p;1t1)
2
s v1p2 X3sp1;1 ( t2p
t
) vp2 +p;1Z
2
p
;2
p
;1
1
; 2 vp;1 v2 + 4v1 t1 ; v1 2 + v2 + C1 + 1 vp;2 2
2
2
2
The Chromatic E1 -term H 0M12 for p > 3
49
mod v1p +p , and hence d0 (x(2)) is congruent to
2
p +p;2 sp;1
p +p;1 s
; 2s(s ; 1) v1 vpX;131 t1 + 2(s ; 1) v1 vp2 +Xp32t1
2
2
s vp2 +p;1X sp;1 t2 vp2 +p;1Z2
+ 2 1 p;1 31 2 + v + C1 + 1 p;2
v2
v2
2
2
2
mod v1p +p . Notice that d0 (x(2)) 6 0 mod
(v1p 3+p;21 ), and2 so N (2) = p2 +
;
p ; 2. For k = 3, we set x(3) to x(2)p + 2s Y (v1p +p ;p v2p ;p X3sp2;1 ). Then
Lemma 4.2.5 shows that
3
2
2
3
2
2
d0 (x(3)) s(s ; 1)DZ (v1p +p ;p v2p ;p X3sp2;1 ) + v1p +p ;p Z3 =v2p ;p;1
2
2
mod (v1p +p ;p+1 ) for some Z3 2 v3;1 BP BP , and so N (3) = p3 + p2 ; p.
Applying Proposition 4.2.6, we can dene x(k) inductively on k 4 and
obtain N (k) = pk + pk;1 ; pk;2 .
3
2
Recall that 1 = (ap2 ; p ; 1)pr + 1 j p a r 1 : odd .
(ii) For p j (s ; 1) and s 62 1 . We have already computed d0 (x(0)) v1 v2p;1 v3s;1 w22 mod v12 . Thus we can dene x(k) inductively on k 1 using
Proposition 4.2.8 and obtain N (k) = P (k).
N
-
N
(iii) For s 2 1 . In this case s is expressed as (s0 p ; 1)pr0 + 1 (s0 2
r0 1). We set x(1) to
N
N0
and odd
x(0)p ; 1=a(r0 + 2)v1p;1 X (v2(a3 0 +1 ;p+2)p X3sr0 0 +2 ) ; v12p v2p ;2p;1 v3(s;1)p+1 u22
with a(3) = 1 and a(r0 + 2) = 3 for r0 3. Then the dierential on the
2
r
second term is given by Lemma 4.1.7. For the third term, we have
d0 (;v12p v2p ;2p;1 v3(s;1)p+1 u22 )
v12p v2p2;2p;1 v3(s;1)p fv32p + (v2p t1 ; v2 tp12 )(u22 ; 2p )g
mod (v12p+1 ). Collecting three terms gives
2
d0 (x(1)) ;1=a(r0 + 2)DX21](v12p v2(a3 0 +1 ;p+2)p X3sr0 0 +2 )
mod v12p+1 , and so N (1) = 2p ; 1. Applying Propositions 4.2.3 and 4.2.6, we
can dene x(k) inductively on k 2 and obtain N (k) = 2pk .
r
5.2.2.
r 2. Dene x(1) by
x(1) = X3spr =X2a31 ;1 + sv1p v3(sp;1)p +2 =2v22 for j = a3r ; 1
= X3spr =X2j1
for j a3r ; 2:
For even
r
r
Then, in both cases we have
d0 (x(1)) ;jv1p X3sr+1(t1 ; v1 w22 )=v2jp+1
Hirofumi Nakai
50
mod (v1p+2 ). Notice that d0 (x(1)) 6 0 mod (v1p+1 ), and so N (1) = p when s 62 0 .
Then we can dene x(k) inductively on k 2 by Proposition 4.2.1 and obtain
N (k) = a2k . When s 2 0 , notice that the above dierential can be rewritten as
d0 (x(1)) DXj0](v1p+1 v2jp+1 X3sr+1)
and so N (1) = p. Applying Propositions 4.2.3 and 4.2.6, we can dene x(k) inductively on k 2 and obtain N (k) = pk + pk;1 .
N
N
r 1 and s 62 0 . Dene x(1) by
x(1) = X3sp1 =X2j1
for r = 1
(a3 ;j )p (sp;1)p
sp
j
p
= X3r =X21 ; sv1 v2
v3
u22 for odd r 3:
For odd r 3 and j a3r ; p ; 1, d0 (X3spr =X2j1) has already been computed in
Lemma 4.1.2. We also obtain d0 (X3spr =X2j1 ) for other cases by easy computations.
We rst give the case r = 1:
d0 (X3sp1 =X2p;1 1) = v1p X3sp1 t1 =v2(p;1)p+1 ; sv1p v3(sp;1)p+1 tp1 =v2 and for j p ; 2,
d0 (X3sp1 =X2j1) = ;jv1p X3sp1 t1 =v2jp+1 :
Now let r 3 be odd. Then
d0 (X3spr =X2a31 ) = v1p X3spr t1 =v2pa3 +1 ; sv1p v3(sp;1)p (2p ; w22 )
d0 (X3spr =X2a31 ;1 ) = 2v1p X3spr t1 =v2(a3 ;1)p+1 ; sv1p v3(sp;1)p (v2p 2p + 2v3 tp1 =v2 )
and for j a3r ; 2, we have
5.2.3.
For odd
N
r
r
r
r
r
r
r
r
d0 (X3spr =X2j1) = ;jv1p X3spr t1 =v2jp+1 ; sv1p v2(a3 ;j)p v3(sp;1)p 2p :
r
r
Additionally, we consider the element ;sv1p v2(a3 ;j)p v3(sp;1)p u22 for odd r 3, on
which the dierential is congruent to
;sv1pv2(a3 ;j)p v3(sp;1)p (2 ; 2p )
mod v1p+1 . Collecting terms shows that d0 (x(1)) is congruent to
v1p X3spr t1 =v2pa3 +1 + sv1p v3(sp;1)p (tp1+1 ; t2 )=v2 for odd r 3 and j = a3r ;jv1ppX3spspr t1=v2(jpa3+1;1)p+1 ; sv1p v3(sp;1)p +1tp1 =v2 for j = a3r ; 1
;jv1 X3r t1=v2
for j a3r ; 2
r
r
r
r
r
r
r
mod v1p+1 , and so N (1) = p.
(i) For k 2 and j a3r ; 2.
inductively on k so that
r
Using Proposition 4.2.1, we can dene x(k)
d0 (x(k)) ;2jv1a2 X3spr t1 =v2(jp+1)p
mod (v11+a2 ) and obtain N (k) = a2k .
k
k
k
k
;1
The Chromatic E1 -term H 0M12 for p > 3
(ii) For k 2 and j = a3r ; 1.
51
Let
x(2) = x(1)p + jv1a2 2 ;p X3spr =v2a3
2
r+2
;p2
+1
2
2
+ sv1p v2p v3(sp;1)p u22 :
r
Then easy calculation shows that
d0 (x(2)) ;sv1p +1 v2p ;1 v3(sp;1)p w23
2
mod (v1p +2 ), and so N (2) = p2 +1. Applying Proposition 4.2.8, we can dene
x(k) inductively on k 3 and obtain N (k) = P (k) ; pk;1 .2
(iii) For k 2, odd r 3 and j = a3r . Let x(2) = x(1)p ; v1p ;1 X3sr+2 =v2a3 +2 ,
+2
3
and x(3) = x(2)p + sv1p +p v3(sp;1)p +1 =v2p+1 . Then easy computations show
2
r+1
2
r
r
that
d0 (x(2)) ;sv1p +1 v3(sp;1)p tp1 =v2 mod v1p +2 +2
3
3
d0 (x(3)) ;sv1p +p v3(sp;1)p (t1 ; v1 w22 )=v2 mod v1p +p+2
3
( sDX10](v1p +p+1 v2 X3spr;+21 ))
and so N (2) = p2 + 1 and N (3) = p3 + p. Applying Propositions 4.2.3
and 4.2.6, we can dene x(k) inductively on k 4 and obtain N (k) = pk +
pk;2 + pk;3 .
r+1
2
2
r
5.2.4. For r = 1 and s 2 0 .
2
(i) For j = p ; 1. First we dene xe by xe = x(0)p + 3v1a2 2 ;p X3sp1 =v2p ;1 .
2
Observe that d0 (xe) ;3v1a2 2 X3sp1 t1 =v2p mod (v11+a2 2 ). Then, let x(1) =
2
xe+3v1a2 2 X3s03 =s0v2p +a3 3 , x(2) = xep +3v1a2 3 ;p X3sp2 =v2a3 3 ;p , and x(3) = x(2)p +
4
4
4
3v1a2 4 ;p v3sp =v2p ;1 + 3v1a2 4 X3s05 =s0v2p +a3 5 . Easy computations show that
N
d0 (x(1)) ;3v1a2 2 v2 v3(sp;3 1)p 2 d0 (x(2)) ;3v1a2 3 v3sp t1 =v2p d0 (x(3)) 3v1a2 4 v3(sp;1)p +p t1 =v2p
mod v11+a2 +1 , and so N (k) = a2k+1 (1 k 3). For k 4, we dene x(k)
;2
;2
inductively on k by x(k) = x(k ; 1)p ; 3v1a2 +1 ;p v3(sp;1)p +p =v2p ;1 .
3
3
k
k
k
k
k
Parallel calculation to the proof of Proposition 4.2.1 shows that
d0 (x(k)) 3v1a2
v3(sp;1)p +p t1 =v2p
mod (v11+a2 +1 ), and so we obtain N (k) = a2k+1 .
(ii) For j p ; 2. We set x(1) to X0 (v2j X3s1 )p + X+ (v2jp X3sp1 ). As is already
k
k+1
k
;2
k
;2
k
computed in Proposition 4.1.9,
d0 (x(1)) ;
j + 1
; 2 v p X s t =vjp
mod (v p ), and so N (1) = 2p. For k 2, we can construct x(k) as in
5.2.5 (ii) and obtain N (k) = 2pk + pk; ; pk; .
2 +1
1
j (j + 1)v p X3s03 t1 =s0 v2jp+2+a3 3
2
1
1
2
2
1
2
32 1
2
+2
Hirofumi Nakai
52
5.2.5. For odd r 3 and s 2 0 .
(i) For a3r ; p j a3r (j 6= a3r ; p + 2). We set
x(1) = x(0)p + (1 ; j )v1p v2(a3 ;j)p v3(sp;1)p u22:
Easy calculation shows that for a3r ; 1 j a3r ,
d0 (x(1)) (1 ; j )v1p v2(a3 ;j)p v3(sp;1)p 2 mod v1p+1 and for a3r ; p j a3r ; 2,
N
r
r
r
r
d0 (x(1)) (j ; 1)v1p+1 v2(a3 ;j)p;1 v3(sp;1)p w23 mod v1p+2 :
For j = a3r and k 2, we can dene x(k) as in 5.2.3 (iii). For j = a3r ; 1,
2
we set x(2) to x(1)p ; 3=2v1p ;1 X (v2(a3 +1 ;p+1)p;1 X3sr+2). Then d0 (x(2)) is
r
r
r
congruent to
+1 2
2
2
2
3v1p +p;1 X3sr+2(2t1 ; v1 2 )=2v2(a3 +1 ;p+1)p + 3v1p +p v2p ;p v3(sp;1)p tp1 u22
2
2
(= ;3=2DX11](v1p +p v2(a3 +1 ;p+1)p X3sr+2) + 3v1p +p Z2 =v22p+1 )
r
r
r
mod (v1p +p+1 ) for some Z2 2 v3;1 BP BP , and so N (2) = p2 + p ; 1. Applying
Propositions 4.2.3 and 4.2.6, we can dene x(k) inductively on k 3 and
obtain N (k) = pk + pk;1 . On the other hand, we can apply Proposition 4.2.8
to a3r ; p j a3r ; 2 case and obtain N (k) = P (k).
2
(ii) Either for j = a3r ; p + 2, or for j a3r ; p ; 1 and p (j + 1). For
j = a3r ; p ; 2, we set
x(1) = X0 (v2a3 ;p;2 X3sr )p + X+ (v2(a3 ;p;2)p X3spr ) + 2v12p v3(sp;1)p +p+2 =v22 :
Otherwise, we set
x(1) = X0 (v2j X3sr )p + X+ (v2jp X3spr ):
We now use Proposition 4.1.9, to compute d0 (x(1)) mod (v12p+1 ). For j =
a3r ; p ; 1,
-
r
r
r
2p
s0
v2p X s t21
2
v
a3 ;p;1 ):
;1)p
1 Xr +2 t1
d0 (x(1)) ;2 0 b(r+2);p2 ;p+1 ; a31 +1 ;3rp2+1;p+2
+ 3v12p v2p ;1 X3(sp
r;1 2 (v2
s v2
v2
r
r
Otherwise,
j + 1 v2pX3sr+1t21
2p
s0
v
1 Xr +2 t1
d0 (x(1)) ;j (j + 1) 0 jp+2+a3 +2 ; 2 1 jp+2 :
v2
s v2
So N (1) = 2p. For k = 2, we set
j + 1 2
p
x(2) = x(1) ;
v2p ;2 X (v(jp+2)p;2 X s
r
3r +2 ):
2
2 1
Using Proposition 4.1.7, we now compute d0 (x(2)), getting
+1
2
3
2
3v;12p v2p ;p v3(sp;1)p2 2 (v2a3 ;p;1 )p
mod v12p2 +1 for j = a3r ; p ; 1
; j+12 DX22](v12p +p;1 v2(jp+2)p;1 X3sr+2) mod v12p +p otherwise:
r
r
The Chromatic E1 -term H 0M12 for p > 3
53
Thus, N (2) = 2p2 for j = a3r ; p ; 1, and N (2) = 2p2 + p ; 2 for other cases.
In the former case, we can construct x(k) inductively on k 3 using Proposition 4.2.7 and obtain N (k) = 2pk . In the latter case, Propositions 4.2.3
and 4.2.6 work well, and so we obtain N (k) = 2pk + pk;1 ; pk;2 .
(iii) For j a3r ; 2p and p j (j + 1). For a3r ; p2 j a3r ; 2p, we have
already shown that
;1
d0 (x(0)) ;a(r)v12 v2a3 ;j;1 v3(sp;1)p w23
mod (v13 ) in 5.1.1 (ii). Then we can dene x(k) inductively on k 1 using
Proposition 4.2.8 and obtain N (k) = pk + P (k). For j a3r ; p2 ; p,
5.1.1 (iii) and (iv) also show how to compute the congruence class of d0 (x(0))
mod p v1p+2 . For p2 (j + p + 1), we get
;1
DXj0 1](v1p+1 v2j+1 X3sr ) + a(r)v1p+1 v2a3 ;j;p;1 v3(sp;1)p 1 (v2j ):
For p2 j (j + p + 1), we get
;DZ (v1p+1 v2j+1 X3sr ):
When p2 j(j + p + 1), we can construct x(k) inductively on k 1 using Proposition 4.2.6 and obtain
N (k) = pk+1 + pk . On the other hand, we set x(1) to
2
(
j
x(0)p + Yj0 1] (v1p +p v2 +1)p X3sr+1) when p2 (j + p + 1). Then d0 (x(1)) is
expressed as follows. For p2 (j + 1),
2
2
d0 (x(1)) j 0 DZ (v1p +p v2(j+1)p X3sr+1) mod v1p +p+1 :
r
r
-
r
r
-
-
For j = a3r ; p2 ; p,
2
+1
2
3
d0 (x(1)) a(r)v1p +p v2p ;p v3(sp;1)p 1 (v2a3 ;p ;p )p
For p2 j (j + 1) and j a3r ; 2p2 ; p,
r
r
mod v1p +p+1 :
2
d0 (x(1)) v1p +p+1 X3sr+1 tp1 =v2(j+1)p+1 mod v1p +p+2 :
When p2 (j + 1), we construct x(k) inductively on k 2 using Proposition 4.2.6 and obtain N (k) = pk+1 + pk . When j = a3r ; p2 ; p, we can
construct x(k) inductively on k 2 as in j = a3r ; p ; 1 case and k 3
(discussed in 5.2.5 (ii)), and obtain N (k) = pk+1 + pk . When p2 j (2j + 1)
3
2
and j a3r ; 2p2 ; p, we set x(2) to x(1)p ; v1p +p +p v3 X3spr+1 =v2(j+1)p +p+1 .
2
2
-
Since
d0 (;v1p +p +p v3 X3spr+1=v2(j+1)p +p+1 )
n
o
;v1p3 +p2+p X3sr+2=v2(j+1)p2 +p (tp12 ; v2p;1 t1) + v1v2p;1 w22
3
2
2
mod (v1p +p +p+2 ), we have
2
3
2
d0 (x(2)) ;DX10](v1p +p +p+1 v2(j+1)p +1 X3sr+2)
and so N (2) = p3 + p2 + p. Applying Propositions 4.2.3 and 4.2.6, we can
dene x(k) inductively on k 3 and obtain N (k) = pk+1 + pk + pk;1 + pk;2 .
Finally we have completed the computations for all cases.
3
2
Hirofumi Nakai
54
Now we can prove our main theorem.
Proof of Theorem. For each p-basic element (xs3r =v2j )p of Lemma 2.3.3, we
have already constructed the element x(k) (= x(spr =j k)), which satises the congruence x(spr =j k) (xs3r =v2j )p mod (v1 ), and determined the v1 -divisibility N (k)
(= N (s r j k)) of x(k) as the smallest integer with d0 (x(k)) 6 0 mod (v1N (k)+1 ).
We also dene y(mpr ) by y(mpr ) = 1=X2mr and obtain N (m r) = a2r (Corollary 3.3.5). Using these results, let B 0 be the direct sum of the cyclic modules as
in Proposition 2.4.1. The linear independence of the set
f(x(spr =j k)=v1N (srjk) )g f(1=v1a2 X2mr )g
is veried by checking with condition (2.5.2). This completes the proof.
k
F
k
r
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BP BP
E
p >
BP BP
E
H
M
E
H Mn
V
n
E
H M
V
E
Department of Mathematics, Faculty of Science, Osaka City University
nakai@sci.osaka-cu.ac.jp http://www.sci.osaka-cu.ac.jp/~nakai/
This paper is available via http://nyjm.albany.edu:8000/j/2000/6-2.html.