Theorem 1.47. There exist ring homomorphism Ψk
F : KF(X) → KF(X) defined for
all compact Hausdorff spaces X and integers k, which satisfy 1. naturality: Ψk
Ff
∗ =f∗Ψk
F,
2. ΨkF(L) =Lk, for L a line bundle†, 3. ΨkF◦ΨlF = ΨklF,
4. Ψp
F(α)≡α
p modp, for p prime, and
5. commutativity of the square
KR(X) KR(X) KC(X) KC(X). Ψk R c c Ψk C
The ring homomorphisms Ψk
Fare called Adams operations. Their existence is a well
known result, so we will omit the proof. See for example [10, Theorem 2.20] for the complex Adams operations and [1, Theorem 5.1] for the general case.
Corollary 1.48. The Adams operations on even dimension spheres
ΨkC:KeC(S2q)→KeC(S2q), (for arbitrary q)
ΨRk :KeR(S2q)→KeR(S2q), (for q even)
are given by
ΨkF(κ) =kqκ.
Proof. We first prove the complex case by induction on q. The real case will then follow. In the case q = 1, we know that
e
KC(S2)∼=Z{η−1}
†
1.11. ADAMS OPERATIONS 23 where η is the canonical line bundle on CP1 =S2, and moreover (η−1)2 = 0. Write 1 for 1, the multiplicative identity inKF(X). Then
ΨkC(η−1) = ΨkC(η)−ΨkC(1) =ηk−1 by property 2 of Theorem 1.47. But
ηk−1 = (η−1 + 1)k−1 = 1 +k(η−1)−1 =k(η−1). since (η−1)2= 0.
In the step case, we use Bott periodicityKeC(S2q+2)∼=KeC(S2)⊗KeC(S2q). Suppose
x∈KeC(S2q)∼=Z is a generator. Then (η−1)⊗x generates KeC(S2q+2)∼=Z and
ΨkC (η−1)⊗x
=k(η−1)⊗kqx
by the induction hypothesis. Sincek(η−1)⊗kqx=kq+1(η−1)⊗x, we have the result. Recall the complexification map c and the injection r from subsection 1.3.2. To prove the real case, we first observe thatrcsends a vector bundle E toE⊕E, so
Z∼=KR(S 2q)−→c K C(S 2q)−→r K R(S 2q)∼=Z
is multiplication by 2. Then the complexification mapc is non-zero and therefore it is multiplication by some non-zero integer. Sinceccommutes with the Adam’s operations (property 5 of Theorem 1.47), the real case follows.
Chapter 2
Constructing Vector Fields
In this chapter, we will determine a lower bound on the number of (non-vanishing, linearly independent tangent) vector fields on the (n−1)-sphereSn−1.
Definition 2.1. A k-field is an ordered set ofk point-wise orthonormal vector fields. A k-frame is an orthonormal set ofk vectors.
Using the Gram-Schmidt process, we can construct a k-field from an ordered set of k point-wise linearly independent vector fields. Therefore, our problem of finding k linearly independent vector fields is equivalent to finding a k-field.
Let K(n) be the maximal k such that there exists a k-field on Sn−1. We will construct a lower bound on K(n).
2.1
Clifford Algebras
Definition 2.2. DefineCk+to be the free, associativeR-algebra with generatorse1, ..., ek
and relations
e2i =−1 for alli, eiej+ejei = 0 for alli6=j.
Define Ck− to be the free, associative R-algebra with generatorse1, ..., ek and rela-
tions
e2i = 1 for all i, eiej +ejei= 0 for all i6=j. Proposition 2.3. We have the following R-algebra isomorphisms:
1. C1+∼=C,
2. C2+∼=H,
3. C1−∼=R⊕R,†
†
The direct sum of algebras X and Y over a field F is their direct sum as vector spaces, with multiplication defined by
(x1, y1)(x2, y2) = (x1x2, y1y2). Thus,R⊕Ris an algebra overR.
4. C2−∼=R(2),
where F(n) is the algebra ofn×n matrices over F.
Proof. 1. C1+ is 2-dimensional with e21 = −1. An arbitrary element of C1+ can be written in the forma+be1 fora, b∈R. Thus, we have an isomorphism
C1+ C, a+be1 a+bi.
∼ =
2. C2+ is 4-dimensional with basis {1, e1, e2, e1e2}. Define φ:C2+→Hon this basis φ(1) = 1, φ(e1) =i, φ(e2) =j, φ(e1e2) =k,
and extend linearly. To check that this is an algebra homomorphism, we compute:
φ(e21) =φ(−1) =−1 =i2=φ(e1)φ(e1), φ(e22) =φ(−1) =−1 =j2 =φ(e2)φ(e2), φ(e1e2) =k=ij =φ(e1)φ(e2), φ(e2e1) =φ(−e1e2) =−k=−ij =ji=φ(e2)φ(e1). Note that all of the relations inH are satisfied inC2+ and φis bijective.
3. Again, we can write an arbitrary element ofC1−of the forma+be1 wherea, b∈R and e21= 1. Define
C1+ R⊕R
a+be1 (a+b, a−b) ∼
=
It is straightforward to check that this is an algebra isomorphism.
4. C2− is 4-dimensional with basis {1, e1, e2, e1e2}. Define φ : C2− → R(2) on this basis φ(1) = " 1 0 0 1 # φ(e1) = " 1 0 0 −1 # φ(e2) = " 0 1 1 0 # φ(e1e2) = " 0 1 −1 0 #
and extend linearly. To check that this is an algebra homomorphism, we compute:
φ(e21) =φ(1) = " 1 0 0 1 # = " 1 0 0 −1 # " 1 0 0 −1 # =φ(e1)φ(e1), φ(e22) =φ(1) = " 1 0 0 1 # = " 0 1 1 0 # " 0 1 1 0 # =φ(e2)φ(e2), φ(e1e2) = " 0 1 −1 0 # = " 1 0 0 −1 # " 0 1 1 0 # =φ(e1)φ(e2), φ(e2e1) =φ(−e1e2) = " 0 −1 1 0 # = " 0 1 1 0 # " 1 0 0 −1 # =φ(e2)φ(e1).
2.1. CLIFFORD ALGEBRAS 27 The homomorphismφhas trivial kernel. Also, φ is surjective since
" a b c d # =φ a+ (d−a)e1+be2+ (c−b)e1e2 .
Theorem 2.4. For all k≥0, 1. Ck++2∼=Ck−⊗C2+,†
2. Ck−+2∼=Ck+⊗C2−.
Proof. Let{ei}ki=1+2be a basis forCk++2. Let{e 0 i}ki=1be the generators ofC − k and{e 00 1, e002} the generators ofC2+. Define anR-module homomorphismu:Ck++2 →C
− k ⊗C + 2 by u(ei) = 1⊗e00i ifi= 1,2, e0i−2⊗e001e002 otherwise,
and extend linearly. It is straightforward to check that u obeys the relations ofCk++2: u(ei)2 =−1 andu(ei)u0(ej) =−u(ej)u(ei) for i6=j.
So u is well defined. This also shows thatuis an algebra homomorphism.
We will show that uis an isomorphism. Notice thatusends distinct basis elements to distinct basis elements. It follows that u is injective. The dimensions of Ck++2 and Ck−⊗C2+ are equal. Thus,u is an isomorphism.
The second half of the theorem follows similarly: define v:Rk+2→C+
k ⊗C − 2 by v(ei) = 1⊗e00i ifi= 1,2, e0i−2⊗e001e002 otherwise,
where {ei}ki=1+2 is a basis for Rk+2, {e0i}ki=1 are the generators of C +
k and {e
00
1, e002} the generators ofC2−. By analogous working to the first half, we can show vextends to an isomorphism fromCk−+2.
Corollary 2.5. For all k≥0, 1. Ck++4∼=Ck+⊗C4+,
2. Ck−+4∼=Ck−⊗C4−.
Proof. Apply Theorem 2.4 twice:
Ck++4∼=Ck−+2⊗C2+∼=Ck+⊗C2−⊗C2+ ∼=Ck+⊗C4+. The second case follows analogously.
†
If X and Y are F-algebras thenX⊗Y is the tensor product of X andY as vector spaces with multiplication defined by
(x1⊗y1)(x2⊗y2) =x1x2⊗y1y2, and then extended linearly to all elements ofX⊗Y.
By inspection, we can determine 1. C1+∼=Cand C1− ∼=R⊕R, 2. C2+∼=Hand C2− ∼=R(2).
Then the following corollary is immediate.
Corollary 2.6. C4±=H⊗R(2)∼=H(2).
Corollary 2.7. For all k≥0, 1. Ck++8∼=Ck+⊗R(16),
2. Ck−+8∼=Ck−⊗R(16).
Proof. Apply corollary 2.5 twice:
Ck++8∼=Ck++4⊗C4+∼=Ck+⊗C4+⊗C4+. Now use corollary 2.6:
Ck+⊗C4+⊗C4+∼=Ck+⊗H⊗R(2)⊗H⊗R(2)∼=Ck+⊗R(16). The second case follows analogously.
From these results we obtain the following table of Clifford algebras:
k Ck+ Ck− 1 C R⊕R 2 H R(2) 3 H⊗(R⊕R)∼=H⊕H C⊗R(2)∼=C(2) 4 H⊗R(2)∼=H(2) H⊗R(2)∼=H(2) 5 H(2)⊗C∼=C(4) H(2)⊗(R⊕R)∼=H(2)⊕H(2) 6 H(2)⊗H∼=R(8) H(2)⊗R(2)∼=H(4) 7 H(2)⊗(H⊕H)∼=R(8)⊕R(8) H(2)⊗C(2)∼=C(8) 8 R(16) R(16)