Gottlieb and Whitehead center groups of projective spaces

0.33 MB·English

by  Marek Golasinski

0 Gottlieb and Whitehead center groups of projective spaces
1
0
MAREKGOLASIN´SKI
2
JUNOMUKAI
n
a
J

6 ByuseofSiegel’smethodandtheclassicalresultsofhomotopygroupsofspheres
2 andLie groups,we determinesomeGottlieb groupsofprojectivespacesor give

the lower boundsof their orders. Furthermore, making use of the properties of
]
T Whitehead products, we determine some Whitehead center groupsof projective
A spaces.
.
h
55P05,55Q15,55Q40,55Q50,55R10;19L20
t
a
m
[
Introduction
1
v
3
Let X beaconnectedandpointedspace. The k-thGottliebgroup G (X) of X hasbeen
6 k
6 definedin[11]. Thisisthesubgroupofthe k-thhomotopygroup π (X) containing all
k
4 elementswhichcanberepresentedbyamapf :Sk → X suchthat f∨id : Sk∨X → X
. X
1 extends (up to homotopy) to a map F : Sk × X → X, where Sk is the k-sphere.
0
0 Throughout thepaperwedonotdistinguish betweenamapanditshomotopy class.
1
: For k ≥ 1, define the k-th Whitehead center group or P-group Pk(X) ⊆ πk(X) of
v
i elements α ∈πk(X) suchthattheWhiteheadproduct [α,β] = 0 forall β ∈ πl(X) and
X
l ≥ 1. Then, G (X) ⊆ P (X) for all k ≥ 1 [11]. Furthermore, if X is a Hopf space,
k k
r
a Gk(X) = Pk(X)= πk(X) forall k ≥ 1.
Let ι be the generator of π (Sn) represented by the identity map of Sn. Given
n n
α ∈π (Sn), α ∈G (Sn) ifandonlyif [α,ι ]= 0. Wenotethat G (Sn)= P (Sn) [9],
k k n k k
[47].
Let R and C bethefieldsofrealsandcomplexnumbers,respectivelyand H theskew
R-algebra of quaternions. Denote by FPn for the n-projective space over F. Put
d = dim F,write i : FPk ֒→ FPn for k ≤ n theinclusion mapandwritesimply
R k,n,FP
i = i :Sd ֒→ FPn.
F 1,n,FP
ThepurposeofthisnoteistodeterminesomeGottlieband P-groupsof FPn. Ouridea
is based on Lang’s result n!π2n+1(CPn) ⊆ G2n+1(CPn) [22], Barratt-James-Stein’s
resultabouttheWhitehead products [−, i ] [3].
F
2 MarekGolasin´skiandJunoMukai
ThemethodistouseSiegel’sresult[43],fibrationswith Sd(n+1)−1,FPn asbasespaces
and the results on the homotopy groups of spheres [28], [32], [31], [45], [46], the
classical groups[21],[25],[26],[27],[29],[30],[33],[34],[35],[36].
Forexample, Gk+n(RPn) ispartlydetermined and Pk+n(RPn) completely determined
for n ≥ 2 and k ≤ 7.
Weset
P′k(FPn) = Pk(FPn)∩(γn∗πk(Sd(n+1)−1))
and
P′k′(FPn) = Pk(FPn)∩iF∗(Eπk−1(Sd−1)).
Notice that P′′(FPn) = 0 if d = 1,2 and k ≥ d + 1. We determine the fact that
k
P4n+3+k(HPn) = P′4n+3+k(HPn) ⊕ P′4′n+3+k(HPn) and the group P′4n+3+k(HPn) for
0 ≤k ≤ 10.
Some particular cases of our results about the Gottlieb groups of FPn overlap with
thoseof[23]and[41].
1 Whitehead center groups of projective spaces
Let EX be the suspension of a space X and denote by E : πk(X) → πk+1(EX) the
suspension homomorphism. Write η ∈ π (S2), ν ∈ π (S4) and σ ∈ π (S8) for
2 3 4 7 8 15
theappropriate Hopfmaps, respectively. Weset ηn = En−2η2 ∈ πn+1(Sn) for n ≥ 2,
νn = En−4ν4 ∈ πn+3(Sn) for n ≥ 4 and σn = En−8σ8 ∈ πn+7(Sn) for n ≥ 8. Note
that νn for n ≥ 5 and σn for n ≥ 9 stand for generators of πn+3(Sn) ∼= Z24 and
πn+7(Sn) ∼= Z240, respectively. By abuse of notaion, these are also used to represent
generators ofthe 2-primarycomponents πn ∼= Z and πn ∼= Z ,respectively.
n+3 8 n+7 16
Denoteby γ = γ : S(n+1)d−1 → FPn thequotient mapand by q = q : FPn →
n n,F n n,F
Sdn themappinching FPn−1 tothebasepoint. Then,asitiswellknown,
(1−(−1)n)ι , if F = R,
n
(1.1) q γ = nη , if F = C,
n n 2n

 ±nν4n, if F = H.
Let ∆ = ∆FP :πk(FPn)→ πk−1(Sd−1) betheconnecting map. By[3],
∆(iF∗E)= idπk−1(Sd−1)
GottliebandWhiteheadcentergroupsofprojectivespaces 3
and
πk(FPn) = γn∗πk(Sd(n+1)−1)⊕iF∗Eπk−1(Sd−1).
By[14,Theorem(2.1)],
(1.2) [ξ◦α,η◦β] = 0if[ξ,η] = 0(ξ ∈ π (X),η ∈ π (X),α ∈π (Sm),β ∈ π(Sn)).
m n k l
According to [47], a map f : X → Y is cyclic if f ∨ id : X ∨ Y → Y extends to
Y
F : X×Y → Y. Theextension F iscalled theassociated mapwith f [22]. Werecall
from[43,Lemma2.1]and[47,Lemma1.3]:
Lemma 1.1 (Siegel-Varadarajan) Letf : X → Y becyclic. Then,thecomposite
f ◦g :W → Y iscyclicforanymapg :W → X.
Thefollowingisobtained from(1.2)andLemma1.1:
Lemma1.2 P (X)◦π (Sn)⊆ P (X) and G (X)◦π (Sn)⊆ G (X).
n k k n k k
By(1.2),[2,Corollary(7.4)] and[3,(4.1-3)],weobtainakeyformuladetermining the
Whiteheadcentergroupsof FPn.
Lemma 1.3 Let h α ∈ π (S2d(n+1)−3) bethe 0-thHopf-Hiltoninvariantfor α ∈
0 k
π (Sd(n+1)−1).Then:
k
0, if nisodd,
(1) [γ α,i ]=
n R (−1)kγ (−2α+[ι ,ι ]◦h α), if niseven;
n n n 0
(cid:26)
0, if nisodd,
(2) [γ α,i ]=
n C γn(η2n+1◦Eα+[ι2n+1,η2n+1]◦Eh0α), if niseven;
(cid:26)
(3) [γnα,iH]= ±(n+1)γn(ν4n+3◦E3α+[ι4n+3,ν4n+3]◦E3h0α).
′
Hereafter, we use the results and notations of [45] freely. Let ν and α (3) be the
1
generatorsofthe 2-primary π3 and 3-primary π (S3;3) componentsofπ (S3)∼= Z ,
6 6 6 12
respectively. Set ω = ν′ + α (3) and α (n) = En−3α (3) for n ≥ 3. We use the
1 1 1
relations
(1.3) [ι ,ι ]= 2ν −Eω
4 4 4
and En−3ω = 2ν for n ≥ 5.
n
4 MarekGolasin´skiandJunoMukai
Wewrite
O(n), for F =R;
O (n) = U(n), for F =C;
F

 Sp(n), for F =H,
where O(n), U(n) and Sp(n) denote the Lie groups of the orthogonal, unitary and

symplectic n×n matrices, respectively. Furthermore, weset
SO(n), for F = R;
SO (n) = SU(n), for F = C;
F

 Sp(n), for F = H.
Denote by in = in,F : SOF(n−1)֒→ SOF(n) and pn = pn,F : SOF(n) → Sdn−1 the
inclusion and projection maps, respectively. Then, we consider the exact sequence
inducedfromthefibration SO (n+1) S−O→F(n)Sd(n+1)−1:
F
(SFnk) ···−→πk+1(Sd(n+1)−1)−∆→F πk(SOF(n))−i→∗ πk(SOF(n+1))−p→∗ ··· ,
where i = i and p= p .
n+1,F n+1,F
Denote by r : U(n) → SO(2n), c : Sp(n) → SU(2n) the canonical map. Let J =
JR : πk(SO(n)) → πk+n(Sn) be the J-homomorphism, JF : πk(SOF(n)) → πk+dn(Sdn)
the complex or symplectic J-homomorphism defined as follows: JC = J ◦ r∗ :
πk(SU(n)) → πk+2n(S2n) and JH = J ◦r∗ ◦c∗ : πk(Sp(n)) → πk+4n(S4n). Then, we
seethat
(1.4) Ed−1◦J ◦∆ = J◦∆ for ∆ = ∆ .
F F R
Letωn,R ∈ πn−1(O(n)), ωn,C ∈ π2n(U(n))and ωn,H ∈π4n+2(Sp(n)) bethecharacteris-
ticelementsforappropriate bundles. Wenotethat ω = ω istheBlakeres-Massey’s
1,H
element, ωn,R =∆(ιn), ωn,C = ∆C(ι2n+1) and ωn,H =∆H(ι4n+3).
Asitiswell-known,
(1.5) i2n+1,Rrωn,C = ω2n+1,R and i2n+1,Ccωn,H = ω2n+1,C.
If ∆α = 0 for α ∈ π (Sn),thenthere exists alift [α] ∈ π (SO(n+1)) of α. Forthe
k k
inclusion
im,n = im,n,R = in,R◦···◦im+1,R : SO(m)֒→ SO(n)(m ≤ n−1),
weset [α] = i [α] ∈ π (SO(n)),where[α] ∈ π (SO(m))isaliftofα ∈ π (Sm−1).
n m,n∗ k k k
Noticethatπ (SO(4)) ={[η ] ,[ι ]} ∼= Z2andanyelementofπ (SO(4)) ∼= π (SO(3))⊕
3 2 4 3 k k
π (S3) isuniquelyrepresented bytwoelementsof π (S3):
k k
[η ] α+[ι ]β for α,β ∈ π (S3).
2 4 3 k
GottliebandWhiteheadcentergroupsofprojectivespaces 5
Noticethatwecantake J[ι ] =ν and J[η ] = ω.
3 4 2
Definethesubgroups of πk(Sd(n+1)−1) and πk−1(Sd−1) asfollows:
M (Sd(n+1)−1) =M (Sd(n+1)−1)= {α ∈ π (Sd(n+1)−1) |[γ α,i ] = 0},
k k,F k n F
L′k−1(Sd−1) = L′k−1,n(Sd−1)= {β ∈ πk−1(Sd−1)| [iFEβ,iF] = 0},
L′k′−1(Sd−1) = L′k′−1,n(Sd−1) = {β ∈πk−1(Sd−1) |[iFEβ,γn] = 0}
and
Lk−1(Sd−1) = Lk−1,n(Sd−1) = L′k−1,n(Sd−1)∩L′k′−1,n(Sd−1).
Wealsodefinethesubgroup Q (S3) of π (S3) by
k k
Q (S3) = {β ∈ π (S3) |hι ,βi = 0},
k k 3
where h−, −i standsfortheSamelsonproduct.
ForanabeliangroupG,wedenoteby(G;p)itsp-component. Forexample,(π (Sn);p) =
k
π (Sn;p).
k
Wewrite
′Lkd−−11 =L′k−1(Sd−1;2), ′′Ldk−−11 = L′k′−1(Sd−1;2), Ldk−−11 = Lk−1(Sd−1;2),
Pdk(n+1)−1 = Pk(Sd(n+1)−1;2), Qdk−−11 = Qk−1(Sd−1;2), Mkd(n+1)−1 = Mk(Sd(n+1)−1;2).
Write (−,−) forthegreatest commondivisor.
Example1.4 M4n+3,H(S4n+3)= (24,2n4+1)π4n+3(S4n+3); L′3′,n(S3) = (24,2n4+1)π3(S3) for
n ≥1; Q (S3) = 12π (S3).
3 3
Weshowthefollowing,inwhich(6)wassuggested byK.Morisugi:
Lemma1.5 (1) M (Sd(n+1)−1)◦π (Sk) ⊆ M (Sd(n+1)−1),
k m m
L′k−1(Sd−1)◦πm−1(Sk−1) ⊆ L′m−1(Sd−1),
L′k′−1(Sd−1)◦πm−1(Sk−1) ⊆ L′m′−1(Sd−1),
Lk−1(Sd−1)◦πm−1(Sk−1) ⊆ Lm−1(Sd−1) andQk(S3)◦πm(Sk)⊆ Qm(S3);
(2) L′k′−1,n(S3)= E−4n−3(Ker(n+1)ν4n+3∗),where
ν4n+3∗ :πk+4n+2(S4n+6) → πk+4n+2(S4n+3) istheinducedhomomorphism.Inpartic-
ular,L′ (S3) ⊆L′′ (S3) for n odd;
k−1,n k−1,n
(3) Qk−1(S3)= L′k−1,n(S3) forn≥ 2;
6 MarekGolasin´skiandJunoMukai
(4)Letn ≥ 2.Then,L′ (S3;3) ⊆ L′′ (S3;3) and
k−1 k−1
L′k−1(S3;p) = L′k′−1(S3;p) = πk−1(S3;p) foranoddprimep ≥5;
(5) iH∗Pk(S4) ⊆ iH∗EQk−1(S3);Moreover,
iH∗Pk(S4)= iH∗EQk−1(S3) providedthat2E4Qk−1(S3) =0;
(6) L′k′−1,n(S3)= πk−1(S3) forn ≥ 1 if 5 ≤ k ≤ 4n+2.
Proof. Since hι ,β◦δi = hι ,βi◦E3δ [18,(6.3)],thelastof(1)follows. Therestof
3 3
(1)isadirectconsequence of(1.2).
Bytheproperties ofWhitehead products [48]andLemma1.3(3),
(−1)k[iHEβ,γn]= [γn,iHEβ] = [γn,iH]◦E4n+3β = ±(n+1)γnν4n+3E4n+3β.
Thisleadstothefirstof(2). If [Eβ,ι ] = 0,then [Eβ,ν ]= 0 (1.2). Thisleadstothe
4 4
secondof(2)for n =1. Since i ◦ν = 0for n ≥ 2 and [Eβ,ι ] = (2ν −Eω)◦E4β,
H 4 4 4
therelation i [Eβ,ι ]= 0 implies i E(ωE3β) = 0,andso ωE3β = 0. Hence,
H 4 H
(△) 2ν4n+3E4n+3β = 0
and (n+1)ν4n+3E4n+3β = 0 forodd n ≥ 3. Thisleadstothesecondof(2)for n ≥ 2.
Since hι ,βi = hι ,ι i◦E3β = ωE3β,weobtain(3).
3 3 3
(△) for β ∈ πk−1(S3;3) implies β ∈ L′k′−1(S3;3) andthefirsthalfof(4). Thesecond
halfisadirectconsequence ofthedefinitionof L′ (Sd−1) and L′′ (Sd−1).
k−1 k−1
Werecall
πk(S4)= Eπk−1(S3)⊕ν4∗πk(S7).
Suppose that β = Eβ1 +ν4β2 ∈ Pk(S4) for β1 ∈ πk−1(S3) and β2 ∈ πk(S7). Then,
bytherelation i β = i Eβ ,weobtain
H H 1
0 = i [ι ,β] = i [ι ,Eβ ] = (i E)(ωE3β ).
H 4 H 4 1 H 1
Thisimpliestheinclusion Pk(S4) ⊆ EQk−1(S3)⊕ν4∗πk(S7) andleadstothefirsthalf
of(5). Ontheotherhand,for β ∈ Qk−1(S3),weobtain [ι4,Eβ] = 2ν4E4β = 0 bythe
assumption. Thismeans EQk−1(S3)⊆ Pk(S4) andleadstothesecondhalfof(5).
We have ν4n+3E4n+3β = J([ι3]4n+3β). By the fact that πk−1(S3) is finite for k 6= 4
and
Z, if k ≡ 0(mod 4);
πk−1(SO(4n+3)) ∼= Z2, if k ≡ 1,2(mod 8);

 0, if k ≡ 3,5,6,7 (mod 8)
if k ≤ 4n+2 [6],wehave(6)for k 6≡ 1,2(mod 8).

GottliebandWhiteheadcentergroupsofprojectivespaces 7
Assume that ν4n+3E4n+3β 6= 0 for k ≡ 1 (mod 8). Then, in view of [21, Lemma
2], wehave ν4n+3E4n+3β = J([ι3]4n+3β) = J(δ′ηk−2), where δ′ ∈ πk−2(SO(4n+3))
is a generator. Since J : πk−1(SO(4n+3)) → πk+4n+2(S4n+3) is a monomorphism,
weobtain δ′ηk−2 = [ι3]4n+3β. By[7,Theorem 1.1], δ′ηk−2 has the SO(6)-of-origin.
Thisisacontradiction andleadsto(6)for k ≡ 1(mod 8).
By the parallel argument, we have the assertion for k ≡ 2 (mod 8) This leads to (6)
andcompletestheproof.
Another proof of Lemma 1.5(6). Toconsider k ≡ 1,2 (mod 8), write (according to
Adams’notations[1]) j fortheimageofthegeneratorin π (SO) under J : π (SO) →
r r r
πrS. First, assume that ν4n+3E4n+3β 6= 0 for k ≡ 1 (mod 8). Then, in view of [21,
Lemma 2], we have ν4n+3E4n+3β = νE∞β = jk−2η and jk−2η2 generates the J-
imageof π (SO) ∼=π (SO(4n+3)) ∼= Z . Ontheotherhand,by[45,Proposition3.1],
k k 2
β ∧η2 = E2(βηk−1) = η5E3β. This and the relation νη = 0 imply a contradictory
relation jk−2η2 = νE∞(βη) =νηE∞β = 0.
Now, assume that ν4n+3E4n+3β 6= 0 for 5 ≤ k ≤ 4n with k ≡ 2 (mod 8). Then,
by [21, Lemma 2], ν4n+3E4n+3β = νE∞β = νE∞β = jk−1. By the relation ([45,
Proposition 3.1])
(1.6) ν4∧β = ν7E7β = (E4β)νk+3,
wegetforthestableTodabracket
< jk−1,η,2ι >=<νE∞β,η,2ι >=<(E∞β)ν,η,2ι >
⊇ (E∞β) < ν,η,2ι >≡ 0(mod 2πS ).
k−1
Ontheotherhand,by[1,Corollary11.7]and[24,TheoremB.v)],wededucethat πS
k−1
containsthedirectsummandZ8 whichisgeneratedby< jk−1,j3,24ι > .(Wepointout
that[24,TheoremB.v)]hasamisprint. Itshouldbe e(ρ)= 2−p(j) (mod 2−p(j)−1).)
j
Because, j1 = η, 12j3 = η3 and jk−1 = jk−3◦η2,wederive
< jk−3,j3,24ι >⊆<jk−3,12j3,2ι >=< jk−3,η3,2ι >⊇< jk−1,η,2ι > .
Thus,
< jk−3,j3,24ι >≡< jk−1,η,2ι > (mod2π8Ss+3).
Thisisacontradiction withtheabovewhichleadstotheassertion.
Now,weshow:
8 MarekGolasin´skiandJunoMukai
Proposition1.6 P′k(FPn)= γn∗(Pk(Sd(n+1)−1)∩Mk(Sd(n+1)−1)) and
P′k′(FPn)= iF∗(ELk−1(Sd−1)).
Proof. For an element α ∈ π (Sd(n+1)−1), it holds that [γ α,γ ] = 0 if and only if
k n n
[α,ιd(n+1)−1]= 0. Thisleadstothefirsthalf.
Next, for an element β ∈ πk−1(Sd−1), it holds that iFEβ ∈ P′′(FPn) if and only if
β ∈ Lk−1(Sd−1). Thisleadstothesecondhalfandcompletestheproof.
We recall the order of [ι ,α] for α = ι ,η ,η2,Eω (n = 4),ν ,ν2,Eσ′ (n = 8) and
n n n n n n
σ [9].
n
1, if n= 1,3,7,
(1.7) ♯[ι ,ι ]= 2, if n isodd and n6= 1,3,7,
n n

 ∞, if n iseven;

1, if n≡ 3(mod4), n = 2orn = 6,
(1.8) ♯[ι ,η ]=
n n
2, ifotherwise;
(cid:26)
1, if n ≡2,3(mod4),
(1.9) ♯[ι ,η2]=
n n 2, ifotherwise;
(cid:26)
(1.10) ♯[ι ,Eω] = 6;
4
1, if n ≡ 7(mod8), n = 2i−3 ≥ 5,
2, if n ≡ 1,3,5(mod8) ≥ 9, n6= 2i−3,
(1.11) ♯[ιn,νn]=  12, if n ≡ 2(mod4) ≥ 6,n = 4,12,


 24, if n ≡ 0(mod4) ≥ 8, n 6=12;



(1.12) ♯[ι ,ν2]= 1ifandonlyifn ≡ 4,5,7(mod8)orn = 2i−5fori≥ 4;
n n
(1.13) ♯[ι ,Eσ′]= 60;
8
GottliebandWhiteheadcentergroupsofprojectivespaces 9
1, ifn = 11, n≡ 15(mod16),
2, ifn ≡ 1(mod2) ≥ 9, n6= 11, n 6≡ 15(mod16),
(1.14) ♯[ιn,σn]=  120, ifn = 8,



240, ifn ≡ 0(mod2) ≥ 10.



Weshow:
Lemma1.7 (1) M (Sd(2n+2)−1)= π (Sd(2n+2)−1) forF = R,C;
k k
(2)(i) Eπk−1(S2n−1)∩Ker2ι2n∗ ⊆ Mk,R(S2n),where2ι2n∗ : πk(S2n)→ πk(S2n)isthe
inducedhomomorphism; Mk,R(S2n) =Ker2ι2n∗ ifk ≤ 4n−2;
(ii) Eπk−1(S4n)∩E−1(Kerη4n+1∗) ⊆Mk,C(S4n+1),whereη4n+1∗ :πk+1(S4n+2) →
πk+1(S4n+1) ; Mk,C(S4n+1) = E−1(Kerη4n+1∗) ifk ≤ 8n;
(iii) Eπk−1(S4n+2) ∩ E−3(Ker (n + 1)ν4n+3∗) ⊆ Mk,H(S4n+3), where ν4n+3∗ :
πk+3(S4n+6)→ πk+3(S4n+3); Mk,H(S4n+3) = E−3(Ker(n+1)ν4n+3∗) ifk ≤ 8n+4.
Inparticular,M (S4n+3) = π (S4n+3) ifn+1 ≡ 0(mod 24);
k,H k
(iv) Eπk−1(S8n+2)∩E−3(Ker(2n+1)ν8n+3∗) ⊆ Mk,H(S8n+3);
Mk,H(S8n+3) = E−3(Ker(2n+1)ν8n+3∗) if k ≤ 16n+4;
(v) Mk,H(S8n+7) = E−3(Ker2(n+1)ν8n+7∗);
(3) [ι2n,ι2n] ∈ M4n−1,R(S2n);
(4) M (S2) = π (S2) exceptk = 2 andM (S2)= 0;
k,R k 2,R
(5) Pk+2n(S2n) ⊆ Mk+2n,R(S2n) ifk ≤ 2n−2;
(6)(i) P (S8n+3;p) = M (S8n+3;p) = π (S8n+3;p) foranoddprimep≥ 5;
k k,H k
(ii) M (S8n+3;3) = π (S8n+3;3) ifn ≡ 1(mod 3).
k,H k
8π8n+3, if k = 0;
8n+3
(7) M8n+3 = π8n+3 , if k = 1,2,4,5,7,8,9,10;
k+8n+3  k+8n+3
 2πk8+n+8n3+3, if k = 3,6.
Proof. (1)isadirectconsequence ofLemma1.3(1);(2).
By Lemma1.3(1), Eα ∈ M (S2n) if and only if 2Eα = 2ι ◦Eα = 0. This leads
k,R 2n
tothefirsthalfof(2)-(i). ThesecondhalfisobtainedfromtheFreudenthalsuspension
theorem. By Lemma 1.3(2), Eα ∈ Mk,C(S4n+1) if and only if η4n+1E2α = 0 for
α ∈πk−1(S4n). Thisleadsto(2)-(ii).
10 MarekGolasin´skiandJunoMukai
ByLemma1.3(3), Eα∈ Mk,H(S4n+2) ifandonlyif (n+1)ν4n+3E4α = 0. Thisleads
to (2)-(iii). By the parallel argument, we have (2)-(iv). By Lemma 1.3(3) and the
relation [ι8n+7,ν8n+7] = 0 (1.11),weobtain(2)-(v).
(3)isadirectconsequence of[13,Proposition 2].
By Lemma 1.3(1), [γ ,i ] = −2γ and [γ η ,i ] = 0. This implies M (S2) = 0
2 R 2 2 2 R 2,R
and η ∈ M (S2). ByLemma1.5(1), π (S2) = η ◦π (S3) ⊆ M (S2). Thisleads
2 3,R k 2 k k,R
to(4).
Suppose that Eα ∈ P2n+k(S2n). Then, 0 = [ι2n,Eα] = [ι2n,ι2n] ◦ E2nα. By the
fact that H[ι2n,ι2n] = ±2ι4n−1 for the Hopf homomorphism H : π4n−1(S2n) →
π4n−1(S4n−1),wehave 0 = H[ι2n,ι2n]◦E2nα = 2E2nα. Since E2n−1 : πk+2n(S2n) →
πk+4n−1(S4n−1) isanisomorphism, weobtain 2Eα = 0. Hence,(2)-(i)leadsto(5).
Since s[ι8n+3,α] = 0 for s = 2,p and tα1(8n + 3)E3α = 0 for t = 3,p. This
leads to (6)-(i). The assumption on n implies 2n + 1 ≡ 0 (mod 3). We have
u[ι8n+3,ν8n+3]◦E3h0(α) = 0 for u= 2,3. Thisleadsto(6)-(ii).
Next, we show (7). By Lemma 1.3(3), we get that α ∈ M8n+3 if and only if
k+8n+3
ν8n+3E3α = 0. Hence, M88nn++33 = 8π88nn++33 and in view of the relation ν6η9 = 0 [45,
(5.9)],wegetthat M8n+3 = π8n+3 for k = 1,2.
k+8n+3 k+8n+3
Because π88nn++63 = {ν8n+3} and ♯ν82n+3 = 2,wederivethat M88nn++63 = 2π88nn++63.
Trivially, M8n+3 = M8n+3 = 0.
8n+7 8n+8
Since ♯ν3 = 2 and π8n+3 = {ν2 },wehave M8n+3 = 2π8n+3 = 0.
8n+3 8n+9 8n+3 8n+9 8n+9
Because π88nn++130 = {σ8n+3},inviewoftherelation ν11σ14 = 0[45,(7.20)],wededuce
M8n+3 = π8n+3 .
8n+10 8n+10
Invirtueoftherelation [45,(7.17)and(7.18)]
(1.15) ν ε = ν ν¯ = [ι ,ν2],
6 9 6 9 6 6
we obtain ν ε = ν ν¯ = 0. So we get M8n+3 = π8n+3 . Trivially, M8n+3 =
7 10 7 10 8n+11 8n+11 8n+12
π8n+3 .
8n+12
Finally,[45,(5.9)]leadsto M8n+3 = π8n+3 andtheproofiscompleted.
8n+13 8n+13
By Lemma 1.7 and the fact that η53 = 12ν5, η5ν6 = 0, πn+6(Sn) = {νn2} ∼= Z2 for
n ≥5,weobtain:
Example1.8 M7,C(S5) = 0,M12,C(S9)= π12(S9) and M4n+6,H(S4n+3) = (2,n2+1)π4n+6(S4n+3).