Spherical classes and the Dickson algebra
Nguy^e"n H. V. Hu'ng Franklin P. Peterson
November 4, 1996
Abstract
We attack the conjecture that the only spherical classes in the homolog*
*y of
Q0S0 are Hopf invariant one and Kervaire invariant one elements. We do th*
*is
by computing products in the E2-term of the unstable Adams spectral sequen*
*ce
converging to ss*(Q0S0) using results about the Dickson algebra and by stu*
*dying
the Lannes-Zarati homomorphism.
1 Introduction
In this paper, we investigate the spherical classes in Q0S0, i.e. the elements *
*belonging to
the image of the Hurewicz homomorphism
H : sss*(S0) ~=ss*(Q0S0) ! H*(Q0S0; F2) :
We are interested in the following classical conjecture.
Conjecture 1.1 (conjecture on spherical classes). There are no spherical clas*
*ses in
Q0S0, except the Hopf invariant one and the Kervaire invariant one elements.
(See Curtis [7] and Wellington [21] for a discussion.)
Let Pk = F2[x1; : :;:xk] be the polynomial algebra on k generators x1; : :;:*
*xk, each of
dimension 1. The general linear group GLk = GL(k; F2) acts on Pk in the usual w*
*ay. Let
Dk be the Dickson algebra of k variables, i.e. the algebra of invariants
Dk := F2[x1; : :;:xk]GLk :
As the action of the (mod 2) Steenrod algebra, A, and that of GLk on Pk commute*
* with
each other, Dk is an algebra over A.
________________________________
11991 Mathematics Subject Classification. Primary 55P47, 55Q45, 55S10, 55T15.
2Key words and phrases. Spherical classes, Loop spaces, Adams spectral seque*
*nces, Steenrod algebra,
Invariant theory, Dickson algebra.
3The first-named author was supported in part by the DGU through the CRM (Ba*
*rcelona).
4The second-named author was supported in part by the NSF and the Alexander *
*von Humboldt
Foundation.
1
One way to attack Conjecture 1.1 is to study the Lannes-Zarati homomorphism
'k : Extk;k+iA(F2; F2) ! (F2 Dk)*i;
A
which is compatible with the Hurewicz homomorphism (see [10], [11, p. 46]). The*
* domain
of 'k is the E2-term of the stable Adams spectral sequence converging to sss*(S*
*0). Further-
more, by a theorem of Madsen [13], the range of 'k is E2k, the submodule of H*(*
*Q0S0; F2)
consisting of all A-annihilated primitives which correspond to Dyer-Lashof oper*
*ations of
length k. The direct sum of E2krunning over all k makes the filtration 0 line o*
*f the E2-
term of the unstable Adams spectral sequence converging to ss*(Q0S0). By compat*
*ibility
of 'k and the Hurewicz homomorphism we mean 'k is a "lifting" of the latter fro*
*m the
"E1 -level" to the "E2-level". r
Let hr denote the Adams element in Ext1;2A(F2; F2). Lannes and Zarati proved*
* in [11]
that '1 is an isomorphism with {'1(hr) | r 0} forming a basis of the dual of F*
*2 D1 and
*
* A
'2 is surjective with {'2(h2r)| r 0} forming a basis of the dual of F2 D2. Re*
*call that,
A
from Adams [1], the only Hopf invariant one elements are represented by h1; h2;*
* h3 of the
stems i = 2r - 1 = 1; 3; 7, respectively. Moreover, by Browder [5], the only d*
*imensions
where a Kervaire invariant one element would occur are 2(2r- 1), for r > 0, and*
* it really
occurs at this dimension if and only if h2ris a permanent cycle in the Adams sp*
*ectral
sequence for the spheres.
Therefore, Conjecture 1.1 is a consequence of the following conjecture.
Conjecture 1.2 'k = 0 in any positive stem i for k > 2.
One of our main results is that '3 = 0 in positive stems, '4 = 0 in positive*
* stems
< 89 and that 'k vanishes on decomposable elements for 2 < k < 10. The ingredie*
*nts to
prove this are:
(1) show that ' = 'k is a homomorphism of algebras (this is done in Section *
*3),
(2) show that E2khas a trivial algebra structure except for E21 E21! E22(thi*
*s is done
in Theorem 4.1), and
(3) show that '4 vanishes on known indecomposable elements (this is done in *
*Sec-
tion 5).
Our other main result is the definition and properties of some elements !k(n*
*) 2 Dk, of
dimension n, which are defined if 2k-ff(n)| n. These elements seem to be import*
*ant in the
structure of Dk as an A-module (see Arnon [3]). These elements are defined in S*
*ection 2
and used to prove the triviality of the algebra structure in E2kin Section 4. F*
*inally, in
Section 6 we give another example where Dk for k 2 differs very much from D1.
Acknowledgment. The authors would like to thank Manuel Castellet and all their *
*col-
leagues at the CRM, Barcelona, for the hospitality and for an excellent working*
* condition.
2 Some new elements in the Dickson algebra
Let Sk = Pkk = H*(BO(k); F2) be the subalgebra of Pk of elements invariant unde*
*r the
symmetric group. It is well-known that Sk = F2[oe1; : :;:oek], where oei denot*
*es the i-th
symmetric function on x1; : :;:xk.
2
Dickson [8] proved that Dk = F2[Qk-1; : :;:Q0], where |Qs| = 2k - 2s. Note t*
*hat Qs
depends on k and when necessary will be denoted Qk;s.
The following elements seem to be important in studying the Dickson algebra.
P i1 ik
Definition 2.1 Let !(n) = !k(n) = x1 : :x:k 2 Sk, wherePthe sum is over all s*
*e-
quences i1; : :;:ik with ij either 0 or a power of 2 and ij = n. This sum is*
* invariant
under k and hence lies in Sk.
Let ff(n) be the number of 1's occuring in the dyadic expansion of n.
We note that !k(n) = 0 if k < ff(n). The following theorem shows when !k(n) *
*2 Dk
if k ff(n).
Theorem 2.2 !k(n) 2 Dk if and only if 2k-ff(n)| n.
We need the following easy lemma from elementary linear algebra.
Lemma 2.3 If oe 2 Pk is invariant under the action of k and that of T , wher*
*e T (x1) =
x1 + x2 and T (xi) = xi, i 2, then oe is invariant under GLk, in other words, *
*oe 2 Dk.
Proof of Theorem 2.2.
We want to find conditions on !(n) so that (T - id)(!(n)) = 0. Here are four*
* easy
properties.
(a)(T - id)(xi22. .x.ikk) = 0
(b)If i1 6= i2, both > 0, then (T - id)(xi11xi22. .x.ikk+ xi21xi12. .x.ikk) =*
* 2xi1+i22xi33. .=.0.
(c)If i1 > 0 and i2 = 0, then (T - id)(xi11xi33.).=.xi12xi33... .
(d)If i2 > 0, i2 0(mod 2), Then (T - id)(xi2=21xi2=22xi33.).=.xi22xi33... .
Let ff(n) = k, then no ij = 0 and all are different powers of 2. Hence i1 6=*
* i2 and (b)
shows !(n) 2 Dk. Write the dyadic expansion of n = 2s-1+ 2t1+ 2t2+ . .+.2tff(n)*
*-1. Note
that T - id annihilates most monomials or pairs by (a) and (b), and by (c) and *
*(d), the
only partitions of n which do not cancel are those of the form (1; 0; 2r1; 2r2;*
* : :;:2rq) where
1 + 2r1+ . .+.2rq= 2s-1+ 2t1+ . .+.2tff(n)-1. However, q + 2 k, and to add up *
*to n we
must have q s - 1 + ff(n) - 1, or s k - ff(n). This happens only when 2k-ff(n*
*)6 | n. If
2k-ff(n)6 | n, then use the partition (1; 0; 1; 2; : :;:2s-2; 2t1; : :;:2tff(n)*
*-1).
The theorem is proved.
Corollary 2.4 If 2k-ff(n)+1| n, then !k(n) = (!k(n=2))2 2 Dk.
Corollary 2.5 If ff(n) = k, then Q0 | !(n) in Dk. If ff(n) = k and 2s | n, th*
*en
s
(Q0)2 | !(n) in Dk.
3
Proof. If ff(n) = k, then the above proof shows that no ij = 0 and hence x1 | !*
*(n). This
together with the fact that !(n) is invariant under GLk implies that x` | !(n) *
*and then
(x`+ `+1x`+1+ . .+.kxk) | !(n) for every 1 ` k, i2 F2. So, !(n) is divisible *
*by
the product of all the linear forms. This product is nothing but Q0 (see [8]).
The second part of the corollary is similarly proved.
The elements !k(n) have been used by Arnon [3] to study Dk. One of his resul*
*ts is
the following theorem.
i j
Theorem 2.6 (Arnon [3]). If !(n) 2 Dk, then Sqi(!(n)) = n+i2i!(n + i).
We note that this formula is not true in general in Sk.
Corollary 2.7 {!k(n)} Dk form a cyclic A-module generated by !k(2k-1) = Qk-1.
Arnon also gives a formula expressing !k(n) in terms of the Q's. This is giv*
*en in the
following theorem.
P Q ` i k (r -s )j2si
Theorem 2.8 (Arnon [3]). !k(n) = i=1 !k(2 - 2 i i) where the sum is o*
*ver
all `-tuples (r1; : :;:r`) with k ri- si 0 for every i and (s1; : :;:s`) is a*
* fixed `-tuple
with 2s1+ . .+.2s`= 2`-1 for ` large.
In Section 4, we will need the following corollary of this result. (A direct*
* proof can be
given of this corollary, but it is long.)
*
* P I
Corollary 2.9 Denote QI = Qik-1k-1.Q.i.00for I = (ik-1; :::; i0). Write !k(n)*
* = Q .
Then each I has at most two odd numbers.
Proof. This comes from the fact that if 2`-1 is written as a sum of ` powers of*
* 2, at most
two of them are one.
3 The Lannes-Zarati map is an algebra homomor-
phism
We would like to recall how to define the Lannes-Zarati map
'k : ExtkA(-kF2; F2) ! (F2 Dk)*;
A
which is compatible with the Hurewicz map (see [10], [11]). To this end, we fo*
*llow the
explanation by the first-named author of the present paper [15].
First we sketch Lannes-Zarati's work [11] on the derived functors of the des*
*tabilization.
Let D be the destabilization functor, which sends an A-module M to the unstable*
* A-
module D(M) = M=B(M), where B(M) is the submodule of M generated by all Sqiu
with u 2 M, i > |u|.
D is a right exact functor. Let Dk be its k-th derived functor for k 0.
4
Suppose M1; M2 are A-modules. Lannes and Zarati defined in [11, x2] a homom*
*or-
phism
\ : ExtrA(M1; M2) Ds(M1) ! Ds-r(M2)
(f; z) 7! f \ z
as follows. Let F*(Mi) be a free resolution of Mi, i = 1; 2. A class f 2 ExtrA(*
*M1; M2) can
be represented by a chain map F : F*(M1) ! F*-r(M2) of homological degree -r. We
write f = [F ]. If z = [Z] is represented by Z 2 F*(M1), then by definition f \*
* z = [F (Z)].
Let M be an A-module. We set r = s = k, M1 = -kM, M2 = Pk M, where as
before Pk = F2[x1; : :;:xk], and get the homomorphism
\ : ExtkA(-kM; Pk M) Dk(-kM) ! Pk M :
Now we need to define the Singer element ek(M) 2 ExtkA(-kM; Pk M) (see Sing*
*er
[18, p. 498]). Let bP1be the submodule of F2[x; x-1] spanned by all powers xiwi*
*th i -1,
where |x| = 1. The A-module structure on F2[x; x-1] is extending that of P1 = F*
*2[x] (see
Adams [2], Wilkerson [22]). The inclusion P1 bP1gives rise to a short exact se*
*quence of
A-modules:
0 ! P1 ! bP1! -1F2 ! 0 :
Denote by e1 the corresponding element in Ext1A(-1F2; P1).
Definition 3.1 (Singer [18]).
(i)ek = e1__._._.e1-z_____"2 ExtkA(-kF2; Pk).
k times
(ii)ek(M) = ek M 2 ExtkA(-kM; Pk M) ; for M an A-module.
Here we also denote by M the identity map of M.
The cap-product with ek(M) gives rise to the homomorphism
ek(M) : Dk(-kM) ! D0(Pk M) Pk M
ek(M)(z) = ek(M) \ z :
As n-1F2 is an unstable A-module for n 1, the following theorem is a specia*
*l case
of the main result in [11].
Theorem 3.2 (Lannes-Zarati [11]). Let Dk Pk be the Dickson algebra of k vari*
*ables.
Then ek(nF2) : Dk(n-kF2) ! nDk is an isomorphism of internal degree 0.
Let N be an A-module. By definition of the functor D, we have a natural hom*
*o-
morphism: D(N) ! F2 N. Suppose F*(N) is a free resolution of N. Then the above
A
natural homomorphism induces a commutative diagram
. . ._____-DFk(N) _______- DFk-1(N) ______-. . .
| |
| |
|ik |ik-1
| |
|? |?
. . ._____-F2A Fk(N)_______- F2A Fk-1(N) ______-. .:.
5
Here the horizontal arrows are induced from the differential in F*(N), and
ik[Z] = [1 Z]
A
for Z 2 Fk(N). Passing to homology, we get a homomorphism
ik : F2 Dk(N) ! TorAk(F2; N)
A
1 [Z] 7! [1 Z] :
A A
Taking N = 1-kF2, we obtain a homomorphism
ik : F2 Dk(1-kF2) ! TorAk(F2; 1-kF2) :
A
Note that the suspension : F2 Dk ! F2 Dk and the desuspension
A A
~= A -k
-1 : TorAk(F2; 1-kF2) -! Tork(F2; F2)
are isomorphisms of internal degree 1 and (-1), respectively. This leads us to
Definition 3.3 (Lannes-Zarati [11]). The homomorphism 'k of internal degree 0 i*
*s the
dual of
'*k= -1ik 1 e-1k(F2) : F2 Dk ! TorAk(F2; -kF2) :
A A
Let k = p + q. The usual inclusion GLk GLp x GLq and the usual diagonal
: Pk ! Pp Pq induce the diagonal
__
: F2 Dk ! (F2 Dp) (F2 Dq) :
A A A
As consequence, (F2 Dk)* is equipped with an algebra structure. On the other*
* hand,
k A
ExtkA(-kF2; F2) has the usual algebra structure with respect to the cup produ*
*ct.
k
The following proposition was stated to the authors by J. Lannes.
Proposition 3.4 The Lannes-Zarati map
M M k M
' = 'k : ExtA(-kF2; F2) ! (F2 Dk)*
k k A
is an algebra homomorphism.
Proof. We prove the dual statement that
M M M A
'* = '*k: (F2 Dk) ! Tork(F2; -kF2)
k k A k
is a coalgebra homomorphism. The key point is to show that '*kis also defined b*
*y means
of 2 instead of .
Step 1. From Theorem 3.2, ek(2F2) : Dk(2-kF2) ! 2Dk is an isomorphism. By
definition, we have the commutative diagram
6
1 e-1(F ) ______ i (1-kF )
F2 Dk ____________-Ak2D (1)_________-k2TorA(F ; 1-kF )
A k k 2 2
| | |
| | |
| | |
| | |
|? |? |?
1Ae-1(2F2) ______ ik(2-kF2)
F2 2Dk ____________-kD (2)___________-TorA(F ; 2-kF ) ;
A k k 2 2
______ ______
where Dk(1) := F2 Dk(1-kF2) and Dk(2) := F2 Dk(2-kF2).
A A
Therefore
'*k= -2ik(2-kF2)(1 e-1k(2F2))2:
A
Step 2. We discuss the coproduct on the involving objects.
Suppose k = p + q.
(a)The usual diagonal : Dk ! Dp Dq induces a diagonal
: 2Dk ! Dp Dq:
(b)Let M; N be two left A-modules and F*(M), F*(N) their free resolutions. T*
*hen
F*(M) F*(N) is a free resolution of M N. We can choose F*(M N) =
F*(M) F*(N). Passing to the derived functors D* we get a diagonal
= p;q: Dk(M N) ! Dp(M) Dq(N) :
(c)Similarly, we have
: TorAk(F2; M N) ! TorAp(F2; M) TorAq(F2; N) :
Note that, for M = rF2, N = sF2, this is the usual coproduct on TorA*(F2;*
* F2).
Taking M = 1-iF2, N = 1-jF2 and consider
p;q: Dk(2-(i+j)F2) ! Dp(1-iF2) Dq(1-jF2) ;
p-i;q-j: Dk-(i+j)(2Pi+j) ! Dp-i(Pi) Dq-j(Pj) :
We claim that
3.5
(f g) \ p;q(x) = p-i;q-j[(f g) \ x]
in Dp-i(Pi) Dq-j(Pj) for f 2 ExtiA(1-iF2; Pi), g 2 ExtjA(1-jF2; Pj) and x 2
Dk(2-(i+j)F2). (Compare with Singer [18, p. 501].) To prove it, represent x as *
*a cycle
in D(F*(1-iF2) F*(1-jF2)).
Now we apply this for i = p, j = q, f = ep(F2), g = eq(F2). By definition,
ek(2F2) = ep(F2) eq(F2). So, formula 3.5 leads us to the commutative diagram
7
-1(2F2)_
(F2 2Dk) ______________________-1AekDk(2)
| |
| |p;q
| |
| |
|? |?
___-1_ -1 ___*
*___
(F2 Dp) (F2 Dq) ______________________-1AepD(F2)1Aeq((F2)1) D (*
*1):
A A p q
Furthermore, by definition of ik(-), we easily check that the following diag*
*ram is
commutative.
______ ik(2-kF2) A
Dk(2) __________________- Tork (F2; 2-kF2)
| |
|p;q |
| |
| |
|? |?
____________ ip(1-pF2)iq(1-qF2) A 1-p A 1-q
Dp(1) Dq(1) __________________-Torp(F2; F2) Torq(F2; F2) :
Combining the two commutative diagrams with the formula at the end of Step 1*
* we
have proved that '* is a coalgebra homomorphism. The proposition is proved.
4 The diagonal map
In this section we study the_diagonal_map__= p;q:_Dk ! Dp Dq, for k = p + q, a*
*nd
the induced homomorphism : D k! D p Dq, where D k= F2 Dk. Our main result is
A
the following theorem.
__
Theorem 4.1 Let k > 2. Then = 0 in positive dimensions unless either p = *
*0 or
q = 0.
__ 2s-1 2s-1 2s-1
Remark 4.2 When p = q = 1, then it is easy to see that (Q2;1 ) = Q1;0 Q1;0*
* .
s-1 2s-1 __ __
Note that Q21;0 and Q2;1 are non-zero in D 1 and D 2 respectively, so the case*
*s k = 2
and k > 2 are different.
One of the motivations for defining !k(n) is the following easy theorem.
P
Theorem 4.3 If !k(n) 2 Dk, then (!k(n)) = n1+n2=n!p(n1) !q(n2).
Note: Since 2k-ff(n)| n, it is easy to check that both !p(n1) 2 Dp and !q(n2) 2*
* Dq or one
of them is 0 using the fact that ff(n1 + n2) ff(n1) + ff(n2).
We now study when p > 0 and q > 0.
Proposition 4.4 (!k(2k-1)) 2 (Dp)2 Dq + Dp (Qq;0) + 1 (Dq)2.
8
Proof. The term 1!q(2k-1) 2 1(Dq)2 because q < k. Let s = p-ff(n1) and assume 2*
*s |
n1 but 2s+1 6 | n1. Hence !p(n1) 62 (Dp)2, 2k-1 = n1+n2 = 2k-2+2k-3+. .+.2s+1+2*
*s+2s,
ff(n1) + ff(n2) = k - s = k - p + ff(n1). Thus ff(n2) = k - p = q and by Coroll*
*ary 2.5,
s
Q2q;0| !q(n2).
Since (Dp)2 and (Qq;0) are closed under the Steenrod algebra, we get the fol*
*lowing
corollary using Corollary 2.7.
Proposition 4.5 (!k(n)) 2 (Dp)2 Dq + Dp (Qq;0) + 1 (Dq)2.
P K L
Lemma 4.6 If (Qk;0) = Q Q ; then either k0 = 1 and L is even, or `0 = 1 *
*and
K is even.
Proof. By Theorem 2.2, X
Qk;0= xi11. .x.ikk
i1+...+ik=2k-1
is=0 or a 2-power
The unique (up to an order) decomposition of 2k - 1 into a sum of k 2-powers is*
* 2k - 1 =
1 + 2 + . .+.2k-1. Then each term in the expansion is divisible by (x1. .x.k) *
*and not
divisible by (x1. .x.k)2. So, for each term QK QL in Qk;0, we have (x1. .x.p) *
*| QK
and (x1. .x.q) | QL. Hence k0 > 0; `0 > 0. Furthermore, we get min(k0; `0) = 1,*
* otherwise
(x1. .x.k)2 | QK QL, a contradiction.
Suppose `0 = 1. Each term in QK is a part of some term in the above expansio*
*n of
Qk;0. So ff(m) = p, where m = dim QK . Also, by the expansion and the fact `0*
* = 1,
(x1. .x.p)2 | QK . Hence 2p-ff(m)+1= 21 | m. Then, by Corollary 2.4, QK = !p(m=*
*2)2. As
!p(m=2) satisfies 2p-ff(m=2)| m=2, it is a GLp-invariant. Thus K is even.
The case_of k0 = 1 is symmetrically considered. The lemma is proved. __ *
* __
Let A be the submodule of all elements of positive degree in A. Hence D p= D*
*p=A Dp.
Proposition 4.7 Let QJP2 Dp; J = (jp-1; :::; j0)_with_j0 > 0. If jm 0(mod *
*2) for
some m > 0 or h(J) := jm 0(mod 2), then QJ 2 A Dp.
Proof. We prove this by induction on the smallest m > 0 with jm 0(mod 2). If *
*m = 1,
then Sq1Q(jp-1;:::;j1+1;j0-1)= QJ. For the induction step,
m-1 (j ;:::;jm+1;j -1;:::;jJ)X L
Sq2 Q p-1 m-1 = Q + 0 Q ;
where `m-1 0(mod 2) and `0 > 0. The last step of the induction is slightly di*
*fferent. If
all jm 1(mod 2); m > 0, but h(J) 0(mod 2), then
p-1 (j -1;j ;:::;jJ) X L
Sq2 Q p-1 k-2 = Q 0+ Q ;
where `p-1 0(mod 2) and `0 > 0. The proposition is proved.
Theorem 4.8 If p q, then
__ __ q-1
Im A Dp Dq + Dp ADq + Dp (Qq;0) :
9
Q P
Proof. Let QI 2 Dk. QI = i!k(ni): Let (QI) =_ uv. We call v the "Dq-coefficie*
*nt"__
of u in (QI). We study the terms where u 62 A Dp. Then we want to show either v*
* 2 A Dq
or Qq-1q;0| v. __
Let u = QJ be such a term in the sum. We first show that either v 2 A Dq or *
*j0 > 0.
By Proposition 4.5, (!k(n)) 2 (Qp;0) Dq + Dp (Dq)2 + (Dp)2_ 1. In u, if there*
* is
no term from (Qp;0) Dq, then v is a square and hence v 2 A Dq. Thus, we have l*
*eft to
consider the case j0 > 0. By Proposition 4.7, we may assume j` 1(mod 2) for `*
* > 0.
>From Corollary 2.9, there are at most two odd exponents for Qp;`and Qp;m (`*
* 6= m)
appearing in a term of Qk;sfor any s. In case if there are exactly such two odd*
* exponents
with ` > 0 and m > 0, then we have s > 0, by Lemma 4.6. Therefore, by dimension*
*al
information and Proposition 4.5, the "Dq-coefficient" of the term containing th*
*e two odd
exponents should be divisible by Q2q;0.
Otherwise, in case if there is only one odd exponent for Qp;`(` > 0) appeari*
*ng in a
term of some Qk;r, then the "Dq-coefficient" of this term is divisible by Qq;0,*
* according
to Proposition 4.5.
Consequently, since QJ has (p - 1) odd exponents j` with ` > 0, the "Dq-coef*
*ficient"
v of QJ in QI is divisible by Qp-1q;0. Hence, Qq-1q;0divides v, as p q. The th*
*eorem is
proved.
Recently, Giambalvo and Peterson [9] have proved the following theorem, whic*
*h was
conjectured in Remark 4.8 of the earlier version [17] of the present paper.
__
Theorem 4.9 Let q > 1 and assume that Qq-1q;0| QJ in Dq. Then QJ 2 A Dq.
Proof of Theorem 4.1. Theorems 4.8 and 4.9 prove 4.1 in case both p and q are *
* 2.
We have left to consider the case p > q = 1. Note that !1(n) 6= 0 implies n is *
*a power
of 2. Hence, in Proposition 4.4 we get (!k(2k-1)) 2 (Dk-1)2 D1 + !k-1(2k-1 - 1)
(Q1;0) + 1 (D1)2, and !k-1(2k-1 - 1) 2 (Qk-1;0). Thus, in Proposition 4.5 we *
*get
(!k(n)) 2 (Dk-1)2 D1 + (Qk-1;0) (Q1;0) + 1 (D1)2. Now use the same argument
as that_in the proof of Theorem 4.8 to see that Qk-2k-1;0| u. But k 3, so k - *
*1 2 and
u 2 A Dk-1 by Theorem 4.9. The theorem is proved.
Remark 4.10 Recently Campbell, Hughes, Shank and Wehlau have shown that the i*
*mage
of the trace, tr : Pk ! Dk is the ideal (Qk-1k;0). Presumably there is some rel*
*ation between
this and Theorem 4.9.
5 On the vanishing of 'k
Theorem 5.1 '3 : Ext3;3+iA(F2; F2) ! (F2 D3)*iis zero in positive stems.
A
Proof. Let Vk be an elementary abelian 2-group of rank k.
>From Boardman [4], Ext3;3+iA(F2; F2) ~=F2 P Hi(BV3; F2) via the algebraic*
* transfer
GL3
defined by Singer [18], where P H*(BVk; F2) denotes the submodule of all A-anni*
*hilated
elements in H*(BVk; F2). Therefore, the following proof is a version of that of*
* Hu'ng [15,
Th. 3.2].
10
1;2r 3;*
Let hr 2 ExtA (F2; F2) be the Adams element. According to Wang[20], ExtA (F*
*2; F2)
has a basis consisting of some products of the form hrhsht, where r; s; t are n*
*on-negative
integers (but not all such appear), and some elements ci (i 0) with stem(ci) =*
* 2i+3+
2i+1+ 2i- 3.
By Theorem 4.1, any decomposable element in (F2 D3)*iis zero. Then, since '*
* = 'k
A
is an algebra homomorphism, '3 sends any element of the form hrhsht to zero.
On the other hand, we showed in [16] that F2 D3 is concentrated in dimensio*
*ns
A
2s+2 - 4 (s 0) and 2r+2 + 2s+1 - 3 (r > s > 0). A routine computation shows t*
*hat
these dimensions are different from stem(ci) for any i. Then '3 also sends cito*
* zero. The
theorem is proved.
So far as we know, Ext4;*A(F2; F2) is unknown. In the literature, its value *
*is available
only for stems < 89. (See Tangora [19], Bruner [6].)
Theorem 5.2 '4 : Ext4;4+iA(F2; F2) ! (F2 D4)*iis zero in positive stem i < 8*
*9.
A
Proof. Using Theorem 4.1 and the same argument as that given in the proof of th*
*e last
theorem, we know that '4 vanishes on the decomposable part.
According to Tangora [19], Ext4;*A(F2; F2) has the following indecomposable *
*generators
up to stem * - 4 < 71:
d0(14); e0(17); f0(18); g1(20); d1(32); p0(33); e1(38);
f1(40); g2(44); D3(61); d2(68); p0(69); p1(70) :
Here the number in parenthesis shows the stem of the corresponding element, for*
* example
stem(d0) = 14. (Accidentally, the element in stem 61 is also denoted by D3, bu*
*t it
is not the Dickson algebra of 3 variables.) Furthermore, the chart in Bruner [*
*6] shows
that Ext4;*A(F2; F2), for 71 * - 4 < 89, has at most indecomposable generators*
* in stems
71; 80; 82; 84. Personally, he informed us that the only indecomposable generat*
*ors in these
stems are:
e2(80); f2(84) :
On the other hand, we proved in [16] that F2 D4 is concentrated in dimensio*
*ns
A
2s+3- 8 (s 0)
2r+3+ 2s+2- 6 (r > s > 0)
2t+3+ 2r+2+ 2s+1- 4 (t > r > s > 1)
2r+3+ 2s+1- 4 (r > s + 1 > 2) :
One can check that these dimensions are completly different from the stems o*
*f the
indecomposable generators listed above. For instance, we will check it here for*
* stems 68,
84 and leave the remaining cases to the reader. (Our claim is true even for ste*
*ms 71 and
82, in which all generators are reported to be decomposable.)
11
For stem 68:
2s+3 = 8 + 68 = 76; no solution.
2r+3+ 2s+2 = 6 + 68 = 64 + 8 + 2; no solution.
2t+3+ 2r+2+ 2s+1 = 4 + 68 = 64 + 8; no solution.
2r+3+ 2s+1 = 4 + 68 = 64 + 8; r = 3; s = 2;
it does not satisfyr > s + 1:
For stem 84:
2s+3 = 8 + 84 = 92; no solution.
2r+3+ 2s+2 = 6 + 84 = 64 + 16 + 8 + 2; no solution.
2t+3+ 2r+2+ 2s+1 = 4 + 84 = 64 + 16 + 8; t = 3; r = 2; s = 2;
it does not satisfyr > s:
2r+3+ 2s+1 = 4 + 84 = 64 + 16 + 8; no solution.
So, '4 vanishes on the above indecomposable generators. The theorem is proved.
Remark 5.3 R. Bruner informed us an unpublished result by himself which claim*
*s that
Ext 4;*A(F2; F2), for 89 * - 4 < 120, has exactly one indecomposable generator*
* in stem
92. It is easy to check by the same argument as given above that D4 has no A-ge*
*nerator
in dimension 92. Therefore, if Bruner's result is correct, Theorem 5.2 is exten*
*dingly valid
up to positive stems < 120.
The following theorem is an immediate consequence of Theorem 4.1.
Theorem 5.4 'k : Extk;k+iA(F2; F2) ! (F2 Dk)*ivanishes on decomposable eleme*
*nts for
A
k > 2.
6 A-generators for a localization of the Dickson alge-
bra
In this section we prove a curious result which gives another instance where Dk*
* is quite
different when k = 1 from k > 1 (4.1 and 4.8 already show such.)
Let D^k= Dk[Q-10], that is the Dickson algebra localized by inverting Q0. >*
*From
Hu'ng [14], D^kis an A-module by defining
Sq(Q-10) = (SqQ0)-1 = [Q0(1 + Qk-1 + Qk-2 + . .+.Q0)]-1
= Q-10(1 + Qk-1 + . .+.Q0)-1
= Q-10+ Q-10(Qk-1 + . .+.Q0) + Q-10(Qk-1 + . .+.Q0)2 + . .:.
W. H. Lin [12] proved that F2 D^1has one generator, Q-10.
A
12
Theorem 6.1 If k > 1, then F2 D^k= 0.
A
Proof. We follow the notation of [16]. Let QI, I = (ik-1; : :;:i0), ij 0 if j*
* 1 and
i0 2 Z, be a monomial in D^k. Set fj = s(ij) + j - 1 for j 1 and f(I) = min{fj*
*}, j 1.
Given j 2 with fj = f(I), then
f(I)(i ;:::;i +2s(ij);i -2s(ij);:::;iI)L
Sq2 Q k-1 j j-1= Q + Q : 0
Now s(ij-1)+j -2 = fj-1 fj = s(ij)+j -1, so s(ij-1) s(ij)+1. Hence s(`j-1) s*
*(ij)
and thus f(`j-1) = s(`j-1) + j - 2 < s(ij) + j - 1 = f(ij) and we have f(L) < f*
*(I). To
finish the induction step, we must consider the case f1 = f(I) and fj > f(I) if*
* 1 < j < k.
In this case f s(i ) s(i )
Sq2 1Q(ik-1;:::;i1+2 1;i0-2= Q1)I+ QL:
f1
If no f(`j), 1 j < k, is < f1, then Sq2 acts only to increase the power of Q0*
*. However,
2k - 1 6 | 2f1if k > 1.
To start the induction, assume f(I) = 0. Then i1 0(mod 2), and
Sq1Q(ik-1;:::;i1+1;i0-1)= QI:
The theorem is proved.
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Department of Mathematics, National University of Hanoi, 90 Nguy^en Trai Street*
*, Hanoi,
Vietnam.
E-mail address: nhvhung@it-hu.ac.vn
Department of Mathematics, Massachusetts Institute of Technology, Cambridge,
MA 01239.
E-mail address: fpp@math.mit.edu
14