The key result allowing the calculation of K(X) in nontrivial cases is a certain for-mula that computes K(X×S2) in terms of K(X) . For example, when X is a point this will yield a calculation of K(S2) . In the next section we will deduce other calculations, in particular K(Sn) for all n .
Let H be the canonical line bundle over S2= CP1. We showed in Example ?? that (H⊗H)⊕1 ≈ H⊕H . In K(S2) this is the formula H2+ 1 = 2H , so H2 = 2H − 1.
We can also write this as (H− 1)2 = 0, so we have a natural ring homomorphism Z[H]/(H − 1)2
→
K(S2) whose domain is the quotient ring of the polynomial ring Z[H] by the ideal generated by (H − 1)2. In particular, note that an additive basis for Z[H]/(H − 1)2 is{1, H}.We define a homomorphism µ as the composition
µ : K(X)⊗Z[H]/(H − 1)2
-→
K(X)⊗K(S2)-→
K(X×S2)where the second map is the external product.
T
heorem 2.2. The homomorphism µ : K(X)⊗Z[H]/(H − 1)2→
K(X×S2) is an iso-morphism of rings for all compact Hausdorff spaces X .Taking X to be a point we obtain:
C
orollary 2.3. The map Z[H]/(H − 1)2→
K(S2) is an isomorphism of rings.Thus if we regard eK(S2) as the kernel of K(S2)
→
K(x0) , then it is generated as an abelian group by H− 1. Since we have the relation (H − 1)2= 0, this means thatthe multiplication in eK(S2) is completely trivial: The product of any two elements is zero. Readers familiar with cup product in ordinary cohomology will recognize that the situation is exactly the same as in H∗(S2;Z) and eH∗(S2;Z), with H − 1 behaving exactly like the generator of H2(S2;Z). In the case of ordinary cohomology the cup product of a generator of H2(S2;Z) with itself is automatically zero since H4(S2;Z) = 0, whereas with K–theory a calculation is required.
The present section will be devoted entirely to the proof of the preceding theorem.
Nothing in the proof will be used elsewhere in the book except in the proof of Bott periodicity for real K–theory in §2.4, so the reader who wishes to defer a careful reading of the proof may skip ahead to§2.2 without any loss of continuity.
The main work in proving the theorem will be to prove the surjectivity of µ . Injectivity will then be proved by a closer examination of the surjectivity argument.
Clutching Functions
From the classification of vector bundles over spheres in§1.2 we know that vec-tor bundles over S2 correspond exactly to homotopy classes of maps S1
→
GLn(C),which we called clutching functions. To prove the theorem we will generalize this con-struction, creating vector bundles over X×S2 by gluing together two vector bundles over X×D2 by means of a generalized clutching function.
We begin by describing this more general clutching construction. Given a vector bundle p : E
→
X , let f : E×S1→
E×S1 be an automorphism of the product vector bundle p×11 : E×S1→
X×S1. Thus for each x ∈ X and z ∈ S1, f specifies an isomorphism f (x, z) : p−1(x)→
p−1(x) . From E and f we construct a vector bundle over X×S2 by taking two copies of E×D2 and identifying the subspaces E×S1 via f . We write this bundle as [E, f ] , and call f a clutching function for [E, f ] . If ft: E×S1→
E×S1 is a homotopy of clutching functions, then [E, f0]≈ [E, f1] since from the homotopy ft we can construct a vector bundle over X×S2×I restricting to [E, f0] and [E, f1] over X×S2×{0} and X×S2×{1}. From the definitions it is evident that [E1, f1]⊕[E2, f2]≈ [E1⊕E2, f1⊕ f2] .Here are some examples of bundles built using clutching functions:
1. [E,11] is the external product E∗ 1 = µ(E, 1), or equivalently the pullback of E via the projection X×S2
→
X .2. Taking X to be a point, then we showed in Example 1.10 that [1, z]≈ H where ‘1’
is the trivial line bundle over X , ‘ z ’ means scalar multiplication by z∈ S1⊂ C, and H is the canonical line bundle over S2= CP1. More generally we have [1, zn]≈ Hn, the n fold tensor product of H with itself. The formula [1, zn]≈ Hn holds also for negative n if we define H−1= [1, z−1] , which is justified by the fact that H⊗H−1≈ 1.
3. [E, zn]≈ E ∗ Hn= µ(E, Hn) for n∈ Z.
4. Generalizing this, [E, znf ]≈ [E, f ]⊗ bHn where bHn denotes the pullback of Hn via the projection X×S2
→
S2.Every vector bundle E0
→
X×S2 is isomorphic to [E, f ] for some E and f . To see this, let the unit circle S1⊂ C ∪ {∞} = S2 decompose S2 into the two disks D0 and D∞, and let Eα for α = 0, ∞ be the restriction of E0 over X×Dα, with E the restriction of E0 over X×{1}. The projection X×Dα→
X×{1} is homotopic to the identity map of X×Dα, so the bundle Eα is isomorphic to the pullback of E by the projection, and this pullback is E×Dα, so we have an isomorphism hα: Eα→
E×Dα.Then f = h0h−1∞ is a clutching function for E0.
We may assume a clutching function f is normalized to be the identity over X×{1} since we may normalize any isomorphism hα: Eα
→
E×Dα by composing it over each X×{z} with the inverse of its restriction over X×{1}. Any two choices of normalized hαare homotopic through normalized hα’s since they differ by a map gα from Dα to the automorphisms of E , with gα(1)=11 , and such a gα is homotopic to the constant map 11 by composing it with a deformation retraction of Dα to 1 . Thus any two choices f0 and f1 of normalized clutching functions are joined by a homotopy of normalized clutching functions ft.The strategy of the proof will be to reduce from arbitrary clutching functions to successively simpler clutching functions. The first step is to reduce to Laurent polynomial clutching functions, which have the form `(x, z)=P
|i|≤nai(x)zi where ai: E
→
E restricts to a linear transformation ai(x) in each fiber p−1(x) . We call such an ai an endomorphism of E since the linear transformations ai(x) need not be invertible, though their linear combinationPiai(x)zi is since clutching functions are automorphisms.
P
roposition 2.4. Every vector bundle [E, f ] is isomorphic to [E, `] for some Laurent polynomial clutching function ` . Laurent polynomial clutching functions `0 and`1 which are homotopic through clutching functions are homotopic by a Laurent polynomial clutching function homotopy `t(x, z)=P
iai(x, t)zi.
Before beginning the proof we need a lemma. For a compact space X we wish to approximate a continuous function f : X×S1
→
C by Laurent polynomial functions P|n|≤Nan(x)zn=P
|n|≤Nan(x)einθ, where each an is a continuous function X
→
C.Motivated by Fourier series, we set an(x)= 1
2π Z
S1f (x, θ)e−inθdθ For positive real r let u(x, r , θ)=P
n∈Zan(x)r|n|einθ. For fixed r < 1 , this series converges absolutely and uniformly as (x, θ) ranges over X×S1, by comparison with the geometric series P
nrn, since compactness of X×S1 implies that |f (x, θ)| is bounded and hence also|an(x)|. If we can show that u(x, r , θ) approaches f (x, θ) uniformly in x and θ as r goes to 1 , then sums of finitely many terms in the series for u(x, r , θ) with r near 1 will give the desired approximations to f by Laurent polynomial functions.
L
emma 2.5. As r→
1 , u(x, r , θ)→
f (x, θ) uniformly in x and θ .P
roof: For r < 1 we have u(x, r , θ)=X∞ n=−∞
1 2π
Z
S1r|n|ein(θ−t)f (x, t) dt
= Z
S1
1 2π
X∞ n=−∞
r|n|ein(θ−t)f (x, t) dt
where the order of summation and integration can be interchanged since the series in the latter formula converges uniformly, by comparison with the geometric series P
nrn. Define the Poisson kernel P (r , ϕ)= 1
2π X∞ n=−∞
r|n|einϕ for 0≤ r < 1 and ϕ ∈ R
Then u(x, r , θ)=
S1P (r , θ− t)f (x, t) dt . By summing the two geometric series for positive and negative n in the formula for P (r , ϕ) , one computes that
P (r , ϕ)= 1
2π · 1− r2 1− 2r cos ϕ + r2 Three basic facts about P (r , ϕ) which we shall need are:
(a) As a function of ϕ , P (r , ϕ) is even, of period 2π , and monotone decreasing on [0, π ] , since the same is true of cos ϕ which appears in the denominator of P (r , ϕ) with a minus sign. In particular we have P (r , ϕ)≥ P(r , π) > 0 for all r < 1 .
(b)
S1P (r , ϕ) dϕ = 1 for each r < 1, as one sees by integrating the series for P (r , ϕ) term by term.
(c) For fixed ϕ ∈ (0, π), P(r , ϕ)
→
0 as r→
1 since the numerator of P (r , ϕ) ap-proaches 0 and the denominator apap-proaches 2− 2 cos ϕ ≠ 0.Now to show uniform convergence of u(x, r , θ) to f (x, θ) we first observe that, using (b), we have
u(x, r , θ)− f (x, θ) = Z
S1P (r , θ− t)f (x, t) dt − Z
S1P (r , θ− t)f (x, θ) dt
≤ Z
S1P (r , θ− t) f (x, t)− f (x, θ) dt
Given ε > 0 , there exists a δ > 0 such that|f (x, t) − f (x, θ)| < ε for |t − θ| < δ and all x , since f is uniformly continuous on the compact space X×S1. Let Iδ denote the integral
P (r , θ− t) |f (x, t) − f (x, θ)| dt over the interval |t − θ| ≤ δ and let Iδ0
denote this integral over the rest of S1. Then we have Iδ≤
Z
|t−θ|≤δP (r , θ− t) ε dt ≤ ε Z
S1P (r , θ− t) dt = ε By (a) the maximum value of P (r , θ− t) on |t − θ| ≥ δ is P(r , δ). So
Iδ0 ≤ P(r , δ) Z
S1|f (x, t) − f (x, θ)| dt
The integral here has a uniform bound for all x and θ since f is bounded. Thus by (c) we can make Iδ0 ≤ ε by taking r close enough to 1. Therefore |u(x, r , θ) − f (x, θ)| ≤
Iδ+ Iδ0 ≤ 2ε . tu
P
roof of Proposition 2.4: Choosing a Hermitian inner product on E , the endomor-phisms of E×S1 form a vector space End(E×S1) with a normkαk = sup|v|=1|α(v)|.The triangle inequality holds for this norm, so balls in End(E×S1) are convex. The subspace Aut(E×S1) of automorphisms is open in the topology defined by this norm since it is the preimage of (0,∞) under the continuous map End(E×S1)
→
[0,∞),α
,
inf(x,z)∈X×S1| det(α(x, z))|. Thus to prove the first statement of the lemma it will suffice to show that Laurent polynomials are dense in End(E×S1) , since a suf-ficiently close Laurent polynomial approximation ` to f will then be homotopic to f via the linear homotopy t`+ (1 − t)f through clutching functions. The second statement follows similarly by approximating a homotopy from `0 to `1, viewed as an automorphism of E×S1×I , by a Laurent polynomial homotopy `0t, then combin-ing this with linear homotopies from `0 to `00 and `1 to `01 to obtain a homotopy `t from `0 to `1.To show that every f ∈ End(E×S1) can be approximated by Laurent polynomial endomorphisms, first choose open sets Ui covering X together with isomorphisms hi: p−1(Ui)
→
Ui×Cni. We may assume hi takes the chosen inner product in p−1(Ui) to the standard inner product inCni, by applying the Gram-Schmidt process to h−1i of the standard basis vectors. Let {ϕi} be a partition of unity subordinate to {Ui} and let Xi be the support of ϕi, a compact set in Ui. Via hi, the linear maps f (x, z) for x∈ Xi can be viewed as matrices. The entries of these matrices define functions Xi×S1→
C. By the lemma we can find Laurent polynomial matrices `i(x, z) whose entries uniformly approximate those of f (x, z) for x∈ Xi. It follows easily that `i approximates f in the k · k norm. From the Laurent polynomial approximations `iover Xi we form the convex linear combination ` =P
iϕi`i, a Laurent polynomial
approximating f over all of X×S1. tu
A Laurent polynomial clutching function can be written `= z−mq for a polyno-mial clutching function q , and then we have [E, `]≈ [E, q]⊗ bH−m. The next step is to reduce polynomial clutching functions to linear clutching functions.
P
roposition 2.6. If q is a polynomial clutching function of degree at most n , then [E, q]⊕[nE,11]≈ [(n + 1)E, Lnq] for a linear clutching function Lnq .P
roof: Let q(x, z)= an(x)zn+ ··· + a0(x) . Each of the matricesA=
1 −z 0 ··· 0 0
0 1 −z ··· 0 0
0 0 1 ··· 0 0
... ... ... ... ...
0 0 0 ··· 1 −z
an an−1 an−2 ··· a1 a0
B=
1 0 0 ··· 0 0 0 1 0 ··· 0 0 0 0 1 ··· 0 0 ... ... ... ... ... 0 0 0 ··· 1 0 0 0 0 ··· 0 q
defines an endomorphism of (n+ 1)E by interpreting the (i, j) entry of the matrix as a linear map from the jthsummand of (n+1)E to the ithsummand, with the entries 1 denoting the identity E
→
E and z denoting z times the identity, for z∈ S1. We can pass from the matrix A to the matrix B by a sequence of elementary row and column operations in the following way. In A , add z times the first column to the second column, then z times the second column to the third, and so on. This produces 0 ’s above the diagonal and the polynomial q in the lower right corner. Then for each i≤ n, subtract the appropriate multiple of the ith row from the last row to make all the entries in the last row 0 except for the final q . These row and column operations are not quite elementary row and column operations in the traditional sense since the entries of the matrices are not numbers but linear maps. However, by restricting to a fiber of E and choosing a basis in this fiber, each entry in A becomes a matrix itself, and then each of the preceding row and column operations can be realized by a sequence of traditional row and column operations on the expanded matrices.The matrix B is a clutching function for [nE,11]⊕[E, q] , hence in each fiber the expanded version of B has nonzero determinant. Since elementary row and column operations preserve determinant, the expanded version of A is also invertible in each fiber. This means that A is an automorphism of (n+ 1)E for each z ∈ S1, and therefore determines a clutching function which we denote by Lnq . Since Lnq has the form A(x)z+ B(x), it is a linear clutching function. The matrices A and B define homotopic clutching functions since the elementary row and column operations can be achieved by continuous one-parameter families of such operations. For example the first operation can be achieved by adding tz times the first column to the second, with t ranging from 0 to 1 . Since homotopic clutching functions produce isomorphic bundles, we obtain an isomorphism [E, q]⊕[nE,11]≈ [(n + 1)E, Lnq] . tu
Linear Clutching Functions
For linear clutching functions a(x)z+ b(x) we have the following key fact:
P
roposition 2.7. Given a bundle [E, a(x)z+ b(x)], there is a splitting E ≈ E+⊕E− with [E, a(x)z+ b(x)] ≈ [E+,11]⊕[E−, z] .P
roof: The first step is to reduce to the case that a(x) is the identity for all x . Consider the expression:(∗) (1+ tz)
a(x) z+ t
1+ tz+ b(x)
=
a(x)+ tb(x)
z+ ta(x) + b(x)
When t = 0 this equals a(x)z + b(x). For 0 ≤ t < 1, (∗) defines an invertible linear transformation since the left-hand side is obtained from a(x)z+ b(x) by first applying the substitution z
,
(z+ t)/(1 + tz) which takes S1 to itself (because if|z| = 1 then |(z+t)/(1+tz)| = |z(z+t)/(1+tz)| = |(1+tz)/(1+tz)| = |w/w| = 1), and then multiplying by the nonzero scalar 1+tz . Therefore (∗) defines a homotopy of clutching functions as t goes from 0 to t0 < 1 . In the right-hand side of (∗) the term a(x)+ tb(x) is invertible for t = 1 since it is the restriction of a(x)z + b(x) to z= 1. Therefore a(x) + tb(x) is invertible for t = t0 near 1 , as the continuous function t
,
infx∈X det[a(x)+ tb(x)] is nonzero for t= 1, hence also for t near 1 . Now we use the simple fact that [E, f g]≈ [E, f ] for any isomorphism g : E→
E .This allows us to replace the clutching function on the right-hand side of (∗) by the clutching function z+ [t0a(x)+ b(x)][a(x) + t0b(x)]−1, reducing to the case of clutching functions of the form z+ b(x).
Since z+ b(x) is invertible for all x , b(x) has no eigenvalues on the unit circle S1.
L
emma 2.8. Let b : E→
E be an endomorphism having no eigenvalues on the unit circle S1. Then there are unique subbundles E+ and E− of E such that :(a) E= E+⊕E−. (b) b(E±)⊂ E±.
(c) The eigenvalues of b ||E+ all lie outside S1 and the eigenvalues of b ||E− all lie inside S1.
P
roof: Consider first the algebraic situation of a linear transformation T : V→
V withcharacteristic polynomial q(t) . Assuming q(t) has no roots on S1, we may factor q(t) in C[t] as q+(t)q−(t) where q+(t) has all its roots outside S1 and q−(t) has all its roots inside S1. Let V± be the kernel of q±(T ) : V
→
V . Since q+and q− are relatively prime inC[t], there are polynomials r and s with r q++sq−= 1. From q+(T )q−(T )= q(T ) = 0, we have Im q−(T ) ⊂ Ker q+(T ) , and the opposite inclusion follows from r (T )q+(T )+ q−(T )s(T )=11 . Thus Ker q+(T )= Im q−(T ) , and similarly Ker q−(T )= Im q+(T ) . From q+(T )r (T )+ q−(T )s(T )=11 we see that Im q+(T )+ Im q−(T )= V , and from r (T )q+(T )+ s(T )q−(T ) = 11 we deduce that Ker q+(T )∩ Ker q−(T ) = 0.Hence V = V+⊕V−. We have T (V±)⊂ V±since q±(T )(v)= 0 implies q±(T )(T (v))= T (q±(T )(v))= 0. All eigenvalues of T ||V± are roots of q± since q±(T )= 0 on V±. Thus conditions (a)–(c) hold for V+ and V−.
To see the uniqueness of V+ and V− satisfying (a)–(c), let q±0 be the characteristic polynomial of T ||V±, so q = q0+q0−. All the linear factors of q0± must be factors of q± by condition (c), so the factorizations q= q0+q0− and q= q+q− must coincide up to scalar factors. Since q0±(T ) is identically zero on V±, so must be q±(T ) , hence V± ⊂ Ker q±(T ) . Since V = V+⊕V− and V = Ker q+(T )⊕Ker q−(T ) , we must have V±= Ker q±(T ) . This establishes the uniqueness of V±.
As T varies continuously through linear transformations without eigenvalues on S1, its characteristic polynomial q(t) varies continuously through polynomials with-out roots in S1. In this situation we assert that the factors q± of q vary continuously with q , assuming that q , q+, and q− are normalized to be monic polynomials. To see this we shall use the fact that for any circle C in C disjoint from the roots of q, the number of roots of q inside C , counted with multiplicity, equals the degree of the map γ : C
→
S1, γ(z)= q(z)/|q(z)|. To prove this fact it suffices to consider the case of a small circle C about a root z= a of multiplicity m, so q(t) = p(t)(t − a)m with p(a)≠ 0. The homotopyγs(z)= p(sa+ (1 − s)z)(z − a)m
|p(sa + (1 − s)z)(z − a)m|
gives a reduction to the case (t− a)m, where it is clear that the degree is m .
Thus for a small circle C about a root z= a of q of multiplicity m, small per-turbations of q produce polynomials q0 which also have m roots a1,··· , am inside C , so the factor (z− a)m of q becomes a factor (z− a1)··· (z − am) of the nearby q0. Since the ai’s are near a , these factors of q and q0 are close, and so q0± is close to q±.
Next we observe that as T varies continuously through transformations without eigenvalues in S1, the splitting V = V+⊕V− also varies continuously. To see this, recall that V+= Im q−(T ) and V−= Im q+(T ) . Choose a basis v1,··· , vn for V such that q−(T )(v1),··· , q−(T )(vk) is a basis for V+ and q+(T )(vk+1),··· , q+(T )(vn) is a basis for V−. For nearby T these vectors vary continuously, hence remain indepen-dent. Thus the splitting V = Im q−(T )⊕Im q+(T ) continues to hold for nearby T , and so the splitting V = V+⊕V− varies continuously with T .
It follows that the union E± of the subspaces V± in all the fibers V of E is a subbundle, and so the proof of the lemma is complete. tu
To finish the proof of Proposition 2.7, note that the lemma gives a splitting [E, z+ b(x)] ≈ [E+, z+ b+(x)]⊕[E−, z+ b−(x)] where b+ and b− are the restrictions of b . Since b+(x) has all its eigenvalues outside S1, the formula tz+ b+(x) for 0 ≤ t ≤ 1 defines a homotopy of clutching functions from z + b+(x) to b+(x) . Hence [E+, z+ b+(x)]≈ [E+, b+(x)]≈ [E+,11] . Similarly, z+ tb−(x) defines a homotopy of clutching functions from z+ b−(x) to z , so [E−, z+ b−(x)]≈ [E−, z] . tu
For future reference we note that the splitting [E, az+ b] ≈ [E+,11]⊕[E−, z]
constructed in the proof of Proposition 2.7 preserves direct sums, in the sense that the splitting for a sum [E1⊕E2, (a1z+b1)⊕(a2z+b2)] has (E1⊕E2)±= (E1)±⊕(E2)±. This is because the first step of reducing to the case a=11 clearly respects sums, and the uniqueness of the± splitting in Lemma 2.8 guarantees that it preserves sums.
Conclusion of the Proof
Now we address the question of showing that the homomorphism µ : K(X)⊗Z[H]/(H − 1)2
→
K(X×S2is an isomorphism. The preceding propositions imply that in K(X×S2) we have [E, f ]= [E, z−mq]
= [E, q]⊗ bH−m
= [(n + 1)E, Lnq]⊗ bH−m− [nE,11]⊗ bH−m
= [((n + 1)E)+,11]⊗ bH−m+ [((n + 1)E)−, z]⊗ bH−m− [nE,11]⊗ bH−m
= ((n + 1)E)+⊗H−m+ ((n + 1)E)−⊗H1−m− nE⊗H−m
This last expression is in the image of µ : K(X)⊗K(S2)
→
K(X×S2) . Since every vector bundle over X×S2 has the form [E, f ] , it follows that µ is surjective.To show µ is injective we shall construct ν : K(X×S2)
→
K(X)⊗Z[H]/(H − 1)2 such that νµ=11 . The idea will be to define ν([E, f ]) as some linear combination of terms E⊗Hk and ((n+ 1)E)±⊗Hk which is independent of all choices.To investigate the dependence of the terms in the formula for [E, f ] displayed above on m and n we first derive the following two formulas, where deg q≤ n:
(1) [(n+ 2)E, Ln+1q]≈ [(n + 1)E, Lnq]⊕[E,11]
(2) [(n+ 2)E, Ln+1(zq)]≈ [(n + 1)E, Lnq]⊕[E, z]
The matrix representations of Ln+1q and Ln+1(zq) are:
1 −z 0 ··· 0
0 1 −z ··· 0
... ... ... ...
0 0 0 1 −z
0 an an−1 ··· a0
and
1 −z 0 ··· 0 0
0 1 −z ··· 0 0
... ... ... ... ...
0 0 0 ··· 1 −z
an an−1 an−2 ··· a0 0
In the first matrix we can add z times the first column to the second column to eliminate the−z in the first row, and then the first row and column give the summand [E,11] while the rest of the matrix gives [(n+ 1)E, Lnq] . This proves (1). Similarly, in the second matrix we add z−1 times the last column to the next-to-last column to make the−z in the last column have all zeros in its row and column, which gives the splitting in (2) since [E,−z] ≈ [E, z], the clutching function −z being the composition of the clutching function z with the automorphism −1 of E .
In view of the appearance of the correction terms [E,11] and [E, z] in (1) and (2), it will be useful to know the ‘±’ splittings for these two bundles:
(3) For [E,11] the summand E− is 0 and E+= E . (4) For [E, z] the summand E+ is 0 and E−= E .
Statement (4) is obvious from the definitions since the clutching function z is already in the form z+ b(x) with b(x) = 0, so 0 is the only eigenvalue of b(x) and hence
E+ = 0. To obtain (3) we first apply the procedure at the beginning of the proof of Proposition 2.7 which replaces a clutching function a(x)z+ b(x) by the clutching function z+ [t0a(x)+ b(x)][a(x) + t0b(x)]−1 with 0 < t0< 1 . Specializing to the case a(x)z+ b(x) =11 this yields z+ t0−111 . Since t0−111 has only the one eigenvalue t0−1> 1 , we have E−= 0.
Formulas (1) and (3) give ((n+ 2)E)− ≈ ((n + 1)E)−, using the fact that the
± splitting preserves direct sums. So the ‘minus’ summand is independent of n.
Suppose we define
ν([E, z−mq])= ((n + 1)E)−⊗(H − 1) + E ⊗H−m∈ K(X)⊗Z[H]/(H − 1)2 for n ≥ deg q. We claim that this is well-defined. We have just noted that ‘minus’
summands are independent of n , so ν([E, z−mq]) does not depend on n . To see that it is independent of m we must see that it is unchanged when z−mq is replaced by z−m−1(zq) . By (2) and (4) we have the first of the following equalities:
ν([E, z−m−1(zq)])= ((n + 1)E)−⊗(H − 1) + E⊗(H − 1) + E⊗H−m−1
= ((n + 1)E)−⊗(H − 1) + E⊗(H−m− H−m−1)+ E⊗H−m−1
= ((n + 1)E)−⊗(H − 1) + E⊗H−m
= ν([E, z−mq])
To obtain the second equality we use the relation (H− 1)2= 0 which implies H(H − 1)= H −1 and hence H −1 = H−m−H−m−1 for all m . The third and fourth equalities are evident.
Another choice which might perhaps affect the value of ν([E, z−mq]) is the con-stant t0 < 1 in the proof of Proposition 2.7. This could be any number sufficiently close to 1 , so varying t0gives a homotopy of the endomorphism b in Lemma 2.8. This has no effect on the ± splitting since we can apply Lemma 2.8 to the endomorphism of E×I given by the homotopy. Hence the choice of t0 does not affect ν([E, z−mq]) . It remains to see that ν([E, z−mq]) depends only on the bundle [E, z−mq] , not on the clutching function z−mq for this bundle. We showed that every bundle over X×S2 has the form [E, f ] for a normalized clutching function f which was unique up to homotopy, and in Proposition 2.5 we showed that Laurent polynomial approximations to homotopic f ’s are Laurent-polynomial-homotopic. If we apply Propositions 2.6 and 2.7 over X×I with a Laurent polynomial homotopy as clutching function, we conclude that the two bundles ((n+ 1)E)− over X×{0} and X×{1} are isomorphic.
This finishes the verification that ν([E, z−mq]) is well-defined.
It is easy to check through the definitions to see that ν takes sums to sums since Ln(q1⊕q2)= Lnq1⊕Lnq2and, as previously noted, the± splitting in Proposition 2.7 preserves sums. So ν extends to a homomorphism K(X×S2)
→
K(X)⊗Z[H]/(H −1)2.
The last thing to verify is that νµ=11 . The groupZ[H]/(H − 1)2 is generated by 1 and H , so in view of the relation H+ H−1 = 2, which follows from (H − 1)2= 0,
we see that K(S2) is also generated by 1 and H−1. Thus it suffices to show νµ=11 on elements E⊗H−m for m ≥ 0. We have νµ(E ⊗H−m)= ν([E, z−m])= E−⊗(H − 1)+ E ⊗H−m= E ⊗H−m since E−= 0, the polynomial q being 11 so that (3) applies.
This completes the proof of Theorem 2.2. tu
2.2 Bott Periodicity
In this section we develop a few basic tools that make it possible to compute K(X) for more complicated spaces X , and in particular for all spheres, yielding the Bott Periodicity Theorem. These tools are formally very similar to the basic machinery of algebraic topology involving cohomology, and in fact they make K–theory into what was classically called a generalized cohomology theory, but now is called simply a cohomology theory.