K-THEORY AND THE ADAMS OPERATIONS

Steve Siu

Supervisor: Daniel Chan

School of Mathematics,The University of New South Wales.

November 2013

Submitted in partial fulfillment of the requirements of the degree ofBachelor of Science with Honours

Acknowledgements

First and foremost I want to thank my advisor Daniel for his patience and guidencethroughout the year.

Next I want to thank my friends in the honours room for their jokes, supportand demotivational rants on life in general (you know who you are). In particularAlec, Georgia, Jason, Kirsten, Lisa, Matt, Nick, Peter and Tim. You guys havemade the honours room a great place to learn mathematics throughout this year.

Last but not least I want to thank my parents for always being there for me.

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Introduction

Algebraic invariants are often constructed by defining some objects on a topologicalspace and turning the set of these objects into an algebraic object. Often these alge-braic invariants give topological information of the space. For example, recall thatthe fundamental group is constructed from the homotopy classes of loops definedon a topological space X. The topological information that the fundamental groupcarries are, intuitively, ’holes’ in X.

The most important use of algebraic invariants is that by defining these algebraicinvariants, alot of topological problems can be reduced to algebraic problems, whichare easier most of the time. For example a classical problem in algebraic topologyis whether the disk D2 can be continuously deformed onto its boundary S1 can besolved using such techniques.

A division algebra, intuitively, is an algebra over a field in which division ispossible. If we take R as the field, then the well-known division algebras over Rare R, C, the quaternions H and the octonions O, with corresponding dimensions1, 2, 4, 8. A natural question to ask is whether there are division algebras of higherdimension. It turn’s out that the only finite dimensional division algebras only Rmust be isomorphic to either R, C, H, O. So up to isomorohism, there are only fourdivision algebras only the real numbers. In 1960 Frank Adams and Michael Atiyahgave a very shorter proof of the above theorem using K-theory and the Adamsoperations. This thesis aims to give an exposition of the proof.

In essence, if a multiplication map exists on Rn, then by restricting to a spherewe get a continuous multiplication map on the sphere. If we want to show Rn is nota division algebra for n 6=1, 2, 4, 8, it suffices to show that the restricted continuousmap does not exist on sphere besides S0, S1, S3, S7. So to show Rn is in a divisionalgebra it is enough to show that a certain continuous map does not exist on Sn−1.

The idea is similar to showing that the disk cannot be continuously deformedonto the circle. For if that was true that we had a homotopy from the disk tothe circle, which is a certain kind continuous map. This continuous map inducesan isomorphism between π1(D) and π1(S

1) which is a contradiction as π1(D) =0 6= Z = π1(S

1). The idea is to assoiate a continous map to a homomorphismbetween some algebraic invariant of a topological space. It is easier to show thenonexistence of certain homomorphism then continuous maps since homomorphismpreserves algebraic structures so in general showing nonexistence of homomorphismis easier then showing nonexistence of continuous maps.

In a similar vein, for spheres Sn−1 we would like to associate the multiplicationmap with a homomorphism. To do this we need to define an algebraic invariant ofSn−1. It turns out that the invariant we need is the ring K(Sn−1), and this is whereK-theory comes into play in the proof of the above theorem.

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Author’s note

At the start of the year my goal for this thesis was to give an exposition of the proofof the non-existence of division algebras Rn over the reals besides n = 1, 2, 4, 8.During the first semester Daniel found an old paper of Atiyah in which Atiyahdefines the Adams operations in terms of the symmetric group. Atiyah proceeds toshow that ψk is a ring homomorphism using his new definition. My goal made achange in which then I focused most of my time on showing the remaining propertiesof the Adam’s operations using Atiyah’s definition.

In the end, after spending many hours in Daniel’s office, we were able to recovermost properties using Atiyah’s new definition. However we were only able to proveψkψl = ψlψk for any positive integers k, l which is a weaker property of ψkψl = ψkl.This weaker condition is enough to show the non-existence of division algebras sounfortunately we were not able to show everything.

One thing we got around is the splitting principle. It was not needed to provethe properties of the Adams operations.

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Contents

Chapter 1 Topological constructions 11.1 Wedge sum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Smash product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Suspension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.5 Reduced suspension . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.6 Additional properties . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Chapter 2 Vector bundles 32.1 Vector bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Operations on vector bundles . . . . . . . . . . . . . . . . . . . . . 52.3 Vector bundles on compact Hausdorff spaces . . . . . . . . . . . . . 62.4 Pullback bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Chapter 3 K-Theory 83.1 The group completion of N . . . . . . . . . . . . . . . . . . . . . . . 83.2 The ring K(X) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.3 The ring K(X) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.4 Ring structure on K(X) and K(X) . . . . . . . . . . . . . . . . . . 10

3.5 Functoriality of K(X) and K(X) . . . . . . . . . . . . . . . . . . . 10

Chapter 4 Adam’s operations 134.1 G-Bundles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.2 Adam’s operations . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.3 ψk is natural . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.4 ψk(L) = Lk for any line bundle L . . . . . . . . . . . . . . . . . . . 194.5 ψk(E) ∼= Ek mod p . . . . . . . . . . . . . . . . . . . . . . . . . . 194.6 ψkψl = ψlψk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Chapter 5 Division algebras 255.1 Divison algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

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Chapter 1

Topological constructions

In this prelimilary chapter we will introduce some topological constructions whichwill be used in the upcoming chapters. Throughout this chapter we let X and Ybe topological spaces with chosen base points {x0} and {y0}. For more of theseconstructions we refer the reader to [5].

1.1 Wedge sum

The wedge sum of X and Y , denoted by X ∨ Y , is defined as

X ∨ Y =X⊔Y

x0 ∼ y0,

equipped with the quotient topology. Intuitively the wedge sum are two spacesjoined at a point. For example, the wedge sum of two circles is an eight shapedfigure.

As a set, X ∨ Y = X × {y0} ∪ Y × {x0} which can be thought of as a subspacein X × Y .

1.2 Smash product

The smash product X ∧ Y is defined as the quotient

X ∧ Y =X × YX ∨ Y

1.3 Cone

The cone is defined as

CX =X × IX × {0}

One remark is that even if X is not contractible, the cone CX is contractibleby sending all elements to the point [X × {0}]

1.4 Suspension

The suspension of X, denoted by SX is defined as

SX =X × I

X × {0} ∪X × {1}

Basically we take the cylinder X × I and collpase the top and bottom ends totwo points. Intuitively taking the suspension is the same as attaching two cones ontop and bottom of X.

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One remark is that

1.5 Reduced suspension

For X with a given base point x0, we define the reduced suspension ΣX as

ΣX =X × I

X × {0} ∪X × {1} ∪ {x0} × I

This is equivalent to taking SX and collapsing the subspace {x0}×I to a point.

1.6 Additional properties

Some properties that will be needed are the following

(i) Sn ∧ X = ΣnX, that is the smash product of X with Sn is the reducedsuspension on X.

(ii) SmSn = Sm+n.

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Chapter 2

Vector bundles

We will first give definition of vector bundle. Intuitively a vector bundle is a topo-logical space which locally looks like a cartesian product. In some sense it is ageneralisation of vector spaces, in the senses that whatever we can do to vectorspaces, we can do to vector bundles.

2.1 Vector bundles

Definition 2.1.1. Given a compact Hausdorff space X, a vector bundle of rankn over X is a topological space E with a continuous map π : E → X such thateach π−1(x) is a complex n-dimensional vector space. We also demand that forevery x ∈ X there is an open cover {Uα} such that there is a homeomorphismhα : p−1(U) → U × Cn such that hα|x : p−1(x) → {x} × Cn is a vector spaceisomorphism for all x ∈ U .

A vector bundle of rank n is sometimes also called an n-dimensional vectorbundle. To properly define a homeomorphism hα : p−1(U) → U × Cn we needto specify a topology on U × Cn. We take the topology on Cn to be the normalEuclidean topology and we take the product topology on U × Cn. Each hα is thencalled a local trivilisation of E. For each x ∈ X the vector space π−1(x) is calledthe fiber over x, normally denoted by E|x. With this definition, we call X the basespace and E the total space. The notation we write is (E,X, π). For simplicity wenormally write E when the context is clear.

Definition 2.1.2. If all the fibers of a vector bundle E are one-dimensional, wecall E a line bundle.

Example 2.1.3. The product bundle X × Cn is a vector bundle. We call such abundle a trivial bundle. We will denote a trivial bundle of dimension n to beT n. Example 2.1.4. The tangent bundle of a manifold is a vector bundle. Since

manifolds are locally contractible, the tangent bundle is locally trivial. In generala vector bundle over a contractible space is trivial, a fact we will prove later.

Example 2.1.5. We will now introduce an important line bundle called the tau-tological line bundle, denoted by H. We first introduce the complex projectiveline CP 1, defined as

CP 1 ={

(z0, z1) ∈ C2|(z0, z1) ∼ (λz0, λz1)}

Intuitively, CP 1 is the set of all lines in C\{0} through the origin, and is home-omorphic to the sphere. We will briefly show that there is a bijection from CP 1

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to the sphere. Let [z0, z1] be the equivalence class of (z0, z1) in CP 1. Notice thatby rescaling λ we can assume [z0, z1] = [z, 1] for some z ∈ C. Then all the pointsin CP 1 are of the form [z, 1] with an extra point [1, 0]. Intuitively this is C with apoint at infinity. Using sterographic projection we see that points on CP 1 have aone to one correspondence to points on the sphere. To define H, notice that everypoint on CP 1 is a line through the origin in C2 collapsed into a point. On everypoint we define the fiber to be the line that is quotiented out. If we view CP 1 asS2 then H is a line bundle on S2.

Similar to vector subspaces, we will now define a subbundle of a vector bundle.

Definition 2.1.6. A subbundle (F,X, π′) of a vector bundle (E,X, π) is a vectorbundle such that F ⊆ E and each π′−1(x) is a vector subspace of π−1(x).

We will now give an example of a subbundle inside a vector bundle. We firstdefine the Mobius bundle M to be the quotient

M =I × R

(0, t) ∼ (1,−t).

Roughly speaking this definition is an ’infinite’ Mobius band, an infinite cylinderwith a twist. Locally M is homeomoprhic to U × R where U is an open subset ofS1.

Example 2.1.7. Here we have a trivial bundle S1 × R2, with the Mobius bundlesitting inside as a subbundle.

The above example are vector bundles where the fibers are real vector spaces.Throughout this thesis we will generally be interested in complex vector bundlesonly.

We will now define a homomorphism between two vector bundles.

Definition 2.1.8. Given two vector bundles p : E → X, q : F → X over acompact Hausdorff space X, a homomorphism is a continuous map φ : E → Fsuch that

(i) q ◦ φ = p

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(ii) for each x ∈ X, φ|p−1(x) : p−1(x) → q−1(x) is a linear map between the twovector spaces p−1(x) and q−1(x).

Definition 2.1.9. Given vector bundles E, F over a compact Hausdorff spaceX, a homomorphism φ : E → F is an isomorphism if φ is bijective and φ−1 iscontinuous. If an isomorphism exists between E, F we say E and F are isomorphicand we write E ≈ F .

Intuitively an isomorphism between two vector bundles is a homeomorphismwhich also preserves the vector space structure on each fiber. Sometimes we willalso use the symbol ≈ to denote a homeomorphism between two topological spaces.We leave it to the reader to determine from context which is meant.

For vector spaces we know that an isomorphism is a bijective homomorphism.We have a similar result for vector bundles. In particular, a bijective homomorphismbetween vector bundles is indeed a vector bundle isomorphism. This is not at allobvious, as this implies that the inverse of the homomorphism is continuous. Thisresult will be useful for what is to come.

Lemma 2.1.10. A continuous map φ : E → F between two vector bundles is anisomorphism if φ|x : Ex → Fx is a linear isomorphism for all x ∈ X.

We now define the restriction of a vector bundle. Suppose that Y ⊆ X andp : E → X a vector bundle over X. We define the restriction of E at Y , denotedby E|Y , as the set p−1(Y ). This is essentially restricted a bundle over X to a’smaller’ bundle over a subspace Y ⊂ X.

We will conclude this section which is useful result.

Proposition 2.1.11. If E → X×I is a vector bundle and X is compact Hausdorff,then E|X×{0} ≈ F |X×{1}.

2.2 Operations on vector bundles

Given two vector spaces V , W , we can construct new vector spaces V ⊕W andV ⊗W . A natural question to ask here is whether we can extend these operationson vector spaces to operations on vector bundles. Given two vector bundles E, Fover the same space, suppose that E =

⊔x∈X Ex, F =

⊔x∈X Fx. We can naively

define E ⊕ F , E ⊗ F by taking the direct sums and tensor products of each fibersEx, Fx over each x ∈ X. One problem we need to do is to put a topology on thesesets and to ensure local triviality if we want to realise these sets as vector bundles.Here we will define these operations. For the explicit description of the topologieson these sets we refer the reader to [4],[2].

Let E and F be vector bundles over X. We will now give an example of threeoperations on vector bundles which will be important for the later chapters.

Example 2.2.1. The direct sum E ⊕ F is defined as the set

E ⊕ F =⊔x∈X

Ex ⊕ Fx

We have a similar definition for the tensor product

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Example 2.2.2. The tensor product E ⊗ F is defined as

E ⊗ F =⊔x∈X

Ex ⊗ Fx

Example 2.2.3. For each x ∈ X, the set of all vector space homomorphisms fromEx to Fx, Hom(Ex, Fx) has a vector space structure. We define

Hom(E,F ) =⊔x∈X

Hom(Ex, Fx)

One thing to note here is that the topologies on these vector bundles E ⊕ F ,E⊗F , Hom(E,F ) are natural in the sense that a vector space isomorphisms betweenthe fibers for each x ∈ X extends to a vector bundle isomorphism. For example,given vector bundles E and F , the fibers of E ⊕ F are defined as E|x ⊕ F |x. Forvector spaces we know that for each x, E|x⊕F |x is naturally isomorphic to E|x⊕F |x.These isomorphisms extend to an vector bundle isomorphism E ⊕ F ≈ F ⊕E. Wewill here state the following isomorphisms between vector bundles E, F , F ′:

(i) E ⊕ F ≈ F ⊕ E.(ii) E ⊗ F ∼= F ⊗ E.

(iii) E ⊗ (F ⊕ F ′) ≈ E ⊗ F ⊕ E ⊗ F ′.

2.3 Vector bundles on compact Hausdorff spaces

Given a vector space V , we can put the standard inner product of V by identifyingV with Cn. With this inner product and a given subspace W ⊂ V we can define theorthogonal complement W⊥. There is a natural isomorphism V = W ⊕W⊥. Onequestion is whether we can do the same thing for vector bundles. That is, given avector bundle E and a subbundle F ⊂ E, we ask whether there exists a subbundleF ′ such that E ≈ F ⊕F ′. For vector bundles over an arbitrary topological space Xthe answer is generally no. However, this is true if X is compact Hausdorff. Moreprecisely, we have

Proposition 2.3.1. If E → X is a vector bundle over a compact Hausdorff spaceX and F → X is a subbundle of E, then there exists a subbundle F ′ ⊂ E suchthat E ≈ F ⊕ F ′.

Proposition 2.3.2. For each vector bundle E → X with X compact Hausdorffthere exists a vector bundle F → X such that E ⊕ F is trivial.

To see why this can be true, consider example 2.1.7. We actually have S1×R2 ≈M⊕M . To see this, imagine putting another Mobius bundle in S1×R2 but rotatedby 90◦ about the base space S1. Then over each fiber we get the direct sum R⊕Rwhich is just R2. Hence we just get S1 × R2.

We wil now introduce some a lemma

Lemma 2.3.3. Let Y be a closed subspace of a compact Hausdorff space X, andlet E, F be two vector bundles over X. If f : E|Y → F |Y is an isomorphism, then

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there exists an open set U containing Y and an extension f : E|U → F |U which isan isomorphism.

2.4 Pullback bundles

We will now introduce the idea of a pullback. Given a continuous map f : X → Yand a vector bundle p : E → Y , we define a vector bundle on X, denoted by f ∗(E)as

f ∗(E) = (x, v) ∈ X × E|p(v) = f(x)

Intuitively on every point x ∈ X, we look at the point f(x) ∈ Y and we ’pullback’the fiber over Y onto X.

Example 2.4.1. Given Y ⊂ X, let i : Y → X be the inclusion and E → X avector bundle over X. Then we have i∗(E) = E|Y , which is just the vector bundlerestricted to Y .

We will now state some properties of pullback

(i) (fg)∗(E) ≈ g∗(f ∗(E))(ii) 1∗(E) ≈ E

(iii) f ∗(E1 ⊕ E2) ≈ f ∗(E1)⊕ f ∗(E2)(iv) f ∗(E1 ⊗ E2) ≈ f ∗(E1)⊗ f ∗(E2)(v) 1∗(E) = E

(vi) If f : X → Y and A ⊂ X, then (f |A)∗(E) = f ∗(E)|A.(vii) If f0, f1 are homotopic, then f ∗0 (E) ≈ f ∗1 (E).

(viii) The pullback of a trivial bundle is trivial.

We will now denote the set of isomorphism classes of n-dimensional complexvector bundles on a space X by Vectn(X).

Recall that two spaces X, Y are homotopy equivalent if there exist twocontinuous maps f : X → Y , g : Y → X such that gf is homotopic to theidentity on X and fg is homotopic to the identity on Y . Intuitively two spaces arehomotopic equivalent if one can be continuously deformed into another. A spacewhich is homotopic equivalent to a point is called contractible.

Lemma 2.4.2.

(i) If f : X → Y is a homotopy equivalence, then f ∗ : Vect(Y ) → Vect(X) isbijective.

(ii) If X is contractible, every vector bundle over X is trivial.

Proof. To see that (i) implies (ii), by definition a contractible space is one whichis homotopy equivalenct to a point. Since all vector bundles over a point is trivial,it follows that all vector bundles over X are trivial.

To prove (i), we first check that the map f ∗ is well defined. It means if E andF are isomorphic vector bundles over Y , then f ∗(E) ∼= f ∗(F ). Now suppose thatf : X → Y and g : Y → X are continuous maps such that g ◦ f = idX andf ◦ g = idY . Then f ∗g∗ = (gf)∗ = (idX)∗ = idVect(X) and similarly g∗f ∗ = idVect(Y ).So we see that f ∗, g∗ are inverse of each other which implies that f ∗ is bijective.

To conclude the chapter, we will state a two results that will be used later on.

Proposition 2.4.3. All vector bundles on S1 is trivial.

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Chapter 3

K-Theory

Given a compact Hausdorff space X, we will now define the ring K(X). Recall thatthe set VectC(X) has two natural operations: the direct sum and tensor product.With respect to these two operations we see that VectC(X) has an addition and amultiplication structure. The additive identity is the vector bundle X ×{0}, wherethe fibers are just the zero vector. The multiplicative identity on the other hand isthe trivial bundle X×C, since for any complex vector space V , we have V ⊗C ∼= Vand this isomorphsim extends to vector bundles.

For a general vector bundle E of rank n in VectC(X), E has neither an additiveinverse nor a multiplicative inverse. In some sense the set VectC(X) is similar tothe natural numbers N. In fact if Y is a compact Hausdorff space in which all thevector bundles on Y are trivial e.g. when Y is just a point, then we can make theidentification φ : VectC(Y )→ N by [εn]→ n, where [εn] is the isomorphism class oftrivial bundles of rank n. Under this identification the additive and multiplicationstructure of VectC(Y ) and N are exactly the same.

We will first motivate the construction of K(X) by first describing a procedurethat turns N into a ring, namely Z.

3.1 The group completion of NThe canonical way of turning N into a group is to introduce the negative integers.This can be done by introducing an equivalence relation on the set N× N by

(m,n) ∼ (m′, n′)⇐⇒ m+ n′ = m′ + n.

The idea is to introduce each integer as a difference between two natural num-bers, since m+n′ = m′+n =⇒ m−n = m′−n′. Under the set N×N/ ∼ elementscan be represented by [m− n], which intuitively is Z. Given two elements [m− n],[m′, n′] ∈ N×N/ ∼, we define addition by [m−n] + [m′−n′] = [m+m′− (n+n′)]and multiplication by [m− n]× [m′ − n′] = [mm′ + nn′ − (mn′ +m′n)]. Then as aring N× N/ ∼ is isomorphic to Z under the isomorphism

[n− 0] 7→ n

[0− n] 7→ −n.

The idea of turning Vect(X) into a ring is similar to the previous construction.There is however more details involved.

First we write p : εn → X to be the trivial bundle of rank n. Given two vectorbundles E, F over X, we say E and F are stably isomorphic, written as E ≈s Fif E ⊕ εn = F ⊕ εn for some n. Similarly we define E ∼ F if E ⊕ εn = F ⊕ εm for

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some m and n. Checking ≈s is an equivalence classes is easy. To see that the directsum is well-defined, if E ′ ∈ [E] and F ′ ∈ [F ] then suppose that E ⊕ εm = E ′ ⊕ εm,F ⊕ εn = F ′ ⊕ εn. We see that

E ⊕ F ⊕ εm+n = E ⊕ εm ⊕ F ⊕ εn = E ′ ⊕ εm ⊕ F ′ ⊕ εn = ⊕E ′ ⊕ F ′ ⊕ εm+n

since the direct sum is commutative. It also easy to see that direct sum is associative.These equivalnce relations defines us the K rings.

3.2 The ring K(X)

For ≈s, only the trivial bundle has an inverse since if [E⊕F ] = [ε0] then E⊕F⊕εn =εn for some n which can happen only when E = F = εn. However we have acancellation property that E ⊕ E ′ ≈s F ⊕ E ′ implies E ≈s F since we can add abundle E ′′ such that E ′ ⊕ E ′′ is trivial.

Now, similar to the group completion construction of N, we consider the setVectC(X)×VectC(X) modulo an equivalence relation ≈ defined as (E,F ) ≈ (E ′, F ′)iff E ⊕ F ′ ≈s E ′ ⊕ F . Checking reflecxive and symmetric is easy. TO see that≈ is transitive, suppose that (E,F ) ≈ (E ′, F ′) and (E ′, F ′) ≈ (E ′′, F ′′). Thenby definition E ⊕ F ′ ≈s E ′ ⊕ F and E ′ ⊕ F ′′ ≈s E ′′ ⊕ F ′. Then we see thatE ⊕ F ′ ≈s E ′′ ⊕ F ′. By adding both sides a bundle F0 such that F ′ ⊕ F0 = T n,we see that E ⊕ T n ≈s E ′′T n which implies that E ≈s E ′′ and similarly F ≈s F ′′.Thus we see that E ⊕ F ′′ ≈s E ′′ ⊕ F which implies that (E,F ) ≈0 (E ′′, F ′′).

We now define addition by (E − F ) + (E ′ − F ′) = (E ⊕ E ′)− (F ⊕ F ′). Thenthe set VectC(X) × VectC(X)/ ≈ is then a group with respect to the additiondefined above. We denote this group by K(X). The zero element is the equivalenceclass of (ε0 − ε0) (or in general, (E − E) for any vector bundle) and the inverseof (E − F ) is (F − E). Notice that for any element (E − F ) ∈ K(X), let F ′

be the bundle such that F ⊕ F ′ = εn. Then since (F ′ − F ′) = 0, we see that(E −F ) = (E ⊕F ′−F ⊕F ′) = (E ⊕F ′− εn). Hence any element in K(X) can berepresentated as (E − εn) for some vector bundle E and some trivial bundle εn.

3.3 The ring K(X)

Proposition 3.3.1. If X is compact Hausdorff, then the set VectC(X) modulo ∼is an abelien group with respect to ⊕. We note this group by K(X).

Proof. The identity is the trivial vector bundle ε0. Clearly taking direct sum of anyvector bundle is still a vector bundle so it remains to check that every element hasan inverse. If all the fibers of X have the same dimension, then prop1.4(hatcher).Otherwise let Xi = {x ∈ X : dim π−1x = i}. To see that Xi is open, every x ∈ Xi

has a neighbourhood Ux such that π−1(Ux) = Ux × Ci. It is obvious that Ux ∈ X.Then Xi =

⋃x∈X Ux which is open. We see that {Xi}i∈Z is an open cover for X

which by the compactness of X must be finite. Hence over each Xi we can producea suitable vector bundle which E ⊕ E ′ is trivial, and that all the fibers have thesame dimension.

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We can actually look at K as a subgroup of K(X), to see this observe that thereis a natural map

φ : K(X)→ K(X)

(E − εn) 7→ [E]

Here [E] denotes the ∼-class of E as an element in K(X). To see that this mapis well-defined, suppose that (E − εn) = (F − εm). Then by definition we see thatE ⊕ εm ≈s F ⊕ εn =⇒ E ⊕ εm ⊕ εk = F ⊕ εn ⊕ εk =⇒ E ∼ F =⇒ [E] = [F ].Clearly φ is surjective, and if (E − εn) ∈ kerφ then [E] = [ε0] which implies thatE ⊕ εm = ε0 ⊕ εn. Now, the rank of E is non-negative, this implies m ≤ n. Writem = n+ n′ and we see that E ⊕ εm ≈ εn ⊕ εn′ which implies that E ≈s εn

′. Hence

the kernel of φ consists of the subset {εm − εn} ⊆ K(X).

3.4 Ring structure on K(X) and K(X)

As promised, the groups K(X) and K(X) has a well-definedd ring structure withrespect to the tensor product. For two elements (E − F ), (E ′ − F ′) ∈ K(X), wedefine

(E − F )(E ′ − F ′) = E ⊗ E ′ − E ⊗ F ′ − E ′ ⊗ F + F ⊗ F ′

The rings K(X), K(X) can also be thought of as functors from the categoryof compact Hausdorff spaces to the category of rings. Suppose that we have acontinuous map f : X → Y . Then this induces a ring homomorphism

f ∗ : K(Y )→ K(X)

E − F 7→ f ∗(E)− f ∗(F )

3.5 Functoriality of K(X) and K(X)

Given a continuous map f : X → Y , we can define an induced map

f ∗ : K(Y ) −→ K(X)

[E − F ] 7→ [f ∗(E)− f ∗(F )].

Since f ∗(E ⊕ F ) ≈ f ∗(E)⊕ f ∗(F ). Given two elements [E − F ], [E ′ − F ′], we seethat

f ∗([E − F ] + [E ′ − F ′]) = f ∗([E ⊕ E ′]− [F ⊕ F ′])= [f ∗(E ⊕ E ′)]− [f ∗(F ⊕ F ′)]= [f ∗(E)⊕ f ∗(E ′)]− [f ∗(F )⊕ f ∗(F ′)]= [f ∗(E)− f ∗(F )] + [f ∗(E ′)− f ∗(F ′)]= f ∗([E − F ]) + f ∗([E ′ − F ′])

The fact that f ∗ is multiplicative, (fg)∗ = g∗f ∗ and 1∗ = 1 easily follows from theother properties of pullback.

Now choose a base point x0 ∈ X and consider the projection map X → {x0}.This induces a ring homomorphism i∗ : K(x0) → K(X) where the image is the

10

subgroup {εm − εn}, since the pullback of vector bundles over a point is trivial.This map is clearly injective and so we have a short exact sequence

0 −→ K(x0)p∗−→ K(X)

φ−→ K(X) −→ 0.

Now consider the map induced by i : {x0} → X

i∗ : K(X) −→ K(x0)

(E − F ) −→ (E|x − F |x)

This is clearly a ring homomorphism and moreover ψi∗ = id and so the shortexact sequence splits. In other words, K(X) ∼= K(X) ⊕ K(x0) = K(X) ⊕ Zdepending on x0. In this way, K(X) can be thought of as the kernel of the mapi∗ : K(X)→ K(x0).

We can now define a map called the external product

µ : K(X)⊗K(Y )→ K(X × Y )

α⊗ β 7→ p∗X(α)⊗ p∗Y (β)

where pX : X×Y → X, pY : X×Y → Y are the projections. µ is a ring homomor-phism since µ(a⊗b)(c⊗d) = µ(ac⊗bd) = p∗X(ac)p∗Y (bd) = p∗X(a)p∗X(c)p∗Y (b)p∗Y (d) =p∗X(a)p∗Y (b)p∗X(c)p∗Y (d) = µ(a ⊗ b)µ(c ⊗ d) by properties of pullback. If we takeY = S2 we have the following

µ : K(X)⊗K(S2)→ K(X × S2).

The product theorem, which we will state without proof, asserts that the aboveµ is an isomorphism.

Theorem 3.5.1. The homomorphism µ : K(X) ⊗ Z[H](H−1)2 → K(X × S2) is an

isomorphism of rings for all compact Hausdorff spaces X.

If we take X to be a point we get

Corollary 3.5.2. The map Z[H](H−1)2 → K(S2) is an isomorphism of rings.

The above theorem is non-trivial, and without surprise this theorem has manyimportant applications. In particular, it allows us to deduce that the reduced K-theory of spheres are two periodic.

Theorem 3.5.3. The homomorphism

β : K(X) −→ K(S2X)

a 7→ (H − 1) ∗ a

is an isomorpism for all compact Hausdorff spaces X. We now have Now, recall

that K(S1) = 0 and K(S2) = Z. Since SmSn = Sm+1, We have the followingcorollary.

Corollary 3.5.4. K(S2n+1) = 0 and K(S2n) = Z, generated by the n-fold reducedexternal product (H − 1) ∗ · · · ∗ (H − 1).

11

In general, K(X) and K(X) are very hard to compute, since the isomorphismclasses of vector bundles on an arbitrary compact Hausdorff space is not very wellunderstood. The power of the Bott periodicity theorem is that to understand thereduced K-theory of a sphere it is enough to compute the reduced K-theory of thecircle and sphere, since the reduced K-theory of spheres are two periodic.

12

Chapter 4

Adam’s operations

In this chapter we will primarily deal with a class of operations on K(X) calledthe Adams operations. These operations turns out to be ring homomorphismssatisfying additional properties. We know that given any compact Hausdorff space,K(X) is a ring. One natural question is that whether K(X) can be any ring. Usingthe existence of these Adams operation we can show that there is no compactHausdorff space X such that K(X) = Z[i]. Hence the existence of these operationsin some sense shows that the ring K(X) has more ’structure’. The most importantapplication of the Adam’s operations is on the Hopf invariant problem, which is ourmain use of these ring homomorphisms.

Adam’s operation were first introduced by Frank Adams to prove the Hopfinvariant one problem. Adam’s defined this class of ring homomorphism throughthe use of Newton polynomials. In 1960 Atiyah defined the Adams operations interms of the symmetric group. We will follows Atiyah’s definition and proceed toshow the properties of these ring homomorphism.

One point we want to emphasize for the upcoming proofs involving isomorphismof vector bundles is that all the maps that are to be considered will be continuous,a fact which we will omit in the proofs. Hence by Lemma 2.1.10 we will only showthat each fibers are isomorphic.

4.1 G-Bundles

We will now proceed by defining the notion of a finite groupG acting on a topologicalspace Y .

Definition 4.1.1. Let Y be a topological space and G be a finite group. We callY a G-space if there is a continuous map

f : G× Y −→ Y

(g, y) 7→ g · y.

We will now define the notion of a finite group acting on a vector bundle overX.

Definition 4.1.2. Let p : E → X be a vector bundle over a compact Hausdorffspace X and let G be a finite group. Suppose that E, X are G-spaces. We say Eis a G-vector bundle if

(i) g · p(v) = p(g · v) for all g ∈ G and v ∈ E, that is the action of G commuteswith the projection map.

(ii) For each g ∈ G the map Ex → Eg·x is linear.

13

In the remaining chapter we will only look at G-vector bundles with a trivialaction on the base space X. That is g · x = x for all x ∈ X.

A G-vector bundle is therefore a vector bundle where each fiber is a represen-tation space of G. We will now define a homomorphism between two G-bundles.Recall that a homomorphism between two vector bundles is a continuous map thatis also a homomorphism when restricted to each fiber. For G-bundles, every fiber isa representation space of G, so we want the homomorphism to preserve the actionof G.

Definition 4.1.3. Given two G-bundles E,F , a homomorphism φ : E → F is G-linear if φ is a vector bundle homomorphism such that for all v ∈ E, φ(g·v) = g·φ(v).

We can now make precise when two G-bundles are isomorphic.

Definition 4.1.4. Two G-bundles E,F are isomorphic as G-bundles if thereexists a G-linear isomorphsim φ : E → F .

We can now consider the set of all isomorphisms classes of G-bundles whichforms an abelian monoid. Similar to the contruction of K(X) from VecC(X) wecan construct the Grothendieck group of the set of all isomorphisms classes of G-bundles, in which we will call KG(X).

In module theory we know that any representation of G over C can be decom-posed into a direct sum of irreducible representations of G. Now a G-bundle E is avector bundle where each fiber is a representation space of G. A natural questionto ask is whether we can write E as a direct sum of subbundles, where the fibersof the sub-bundles are irreducible representation space of G. If a vector bundle istrivial, then we can indeed do this. In general this is more complicated.

We now let {[Wπ]} be the set of isomorphism classes of irreducible representa-tions of G. We define R(G) =

∑π Z[Wπ], the free abelian group over {[Wπ]}. This

set R(G) is a ring under the tensor product. To see this, the tensor prodcut of anytwo irreducible representations Vπ⊗Vρ is also a representation of G, and so Vπ⊗Vρcan be decomposed into a direct sum of irreducible representations.

Now suppose that V,E are G-bundles. We define HomG(V,E) to be the vectorbundle where each fiber consists of HomG(Vx, Ex), the set of G-linear vector spacehomomorphisms from Vx to Ex. Then we have

Lemma 4.1.5. The map

φ : V ⊗ HomG(V,E)→ E

v ⊗ f 7→ f(v)

is a G-linear homomorphism.

Proof. First we want to show that the map ψ : V × HomG(V,E) → E sending(v, f) 7→ f(v) is bilinear. We have

ψ(λv + v′, f) = f(λ(v + v′)) = λf(v) + f(v′) = λψ(v, f) + ψ(v′, f)

and similarly ψ(v, λf + f ′) = λψ(v, f) + ψ(v, f ′).

14

Since ψ is bilinear, the universal property of tensor product implies that thereis a unique homomorphism sending v⊗ f to f(v). We see that the map is preciselyφ. It remains to check that φ is G-linear. Notice that

φ(g · v ⊗ f) = φ(gv ⊗ f) = f(gv) = g · f(v) = g · φ(v ⊗ f)

and this completes the proof.We will need an easy lemma which we will state without proof

Lemma 4.1.6. Suppose A,B,C are representation space of G, and φ : B → C isa G-linear isomorphism. Then the map

Φ : HomG(A,B)→ HomG(A,C)

f 7→ φ ◦ f

is a vector space isomorphism.

With the above lemma we can now establish

Lemma 4.1.7. Suppose that E, F , H are G-vector bundles, and φ : F → H is aG-linear isomorphism. Then the map

Φ : HomG(E,F )→ HomG(E,H)

f 7→ φ ◦ f

is a vector bundle isomorphism.

Proof. By Lemma 4.1.6, for each x ∈ X the fibers HomG(Ex, Fx), HomG(Ex, Hx)are isomorphic. We assume Φ is continuous so by Lemma 2.1.10 we see that Φ isan vector bundle isomorphism.

Lemma 4.1.8. Suppose that {Wπ} is a set of irreducible representations of G.Let Vπ = X×Wπ be the trivial G-vector bundle. Then the following G-bundles areisomorphic: ∑

π

Vπ ⊗ HomG(Vπ, E) ∼= E

Proof. We first want to show the fibers are isomorphic. Since Ex is a representationof G, we can write Ex as a direct sum of irreducible representations of G, that isEx ∼=

∑πW

nππ . By Lemma 4.1.6 we can write∑

π

Wπ ⊗ HomG(Wπ, Ex) ∼=∑π

Wπ ⊗ HomG(Wπ,∑ρ

W nρρ )

∼=∑π

Wπ ⊗∑ρ

HomG(Wπ,Wρ)nρ universal property of Hom,

∼=∑π

Wπ ⊗ Cnπ Schur’s lemma,

∼=∑π

W nππ

∼= Ex

15

Recall that R(G) is the free abelien group generated by the irreducible repre-sentations Wπ of G. By letting Vπ = X ×Wπ we see that the free abelien groupgenerated by the isomorphism classes {[Vπ]} has the same ring structure as R(G).Hence we can look at elements in R(G) as trivial G-vector bundles Vπ = X ×Wπ.

We can now establish

Proposition 4.1.9. The following rings are isomorphic

φ : R(G)⊗Z K(X)→ KG(X)

Proof. Let V = X×W where W is a representation of G. Consider the two maps

Φ : R(G)⊗Z K(X)→ KG(X) Ψ : KG(X)→ R(G)⊗Z K(X)

[V ]⊗ [E] 7→ [V ⊗ E] [E] 7→∑π

[Vπ]⊗ [HomG(Vπ, E)]

We first need to define a G-action on [V ⊗E]. The obvious action is that over eachfiber, we define g · (v⊗ e) = g(v)⊗ e ∈ Vx ⊗Ex. With this set up we see that Φ, Ψare inverses of each other.

With all of these setup, we can proceed to define Adam’s operations. For anyvector bundle E, we will first define a natural action of Sk on E⊗k. Given σ ∈ Sk,for any element v ⊗ · · · ⊗ v ∈ E⊗k we define

σ · v1 ⊗ · · · ⊗ vk = vσ(1) ⊗ · · · ⊗ vσ(k)

Hence by eariler proposition we can look at E⊗k as an Sk bundle and we get thefollowing Sk-bundle isomorphism

E⊗k '∑π

Vπ ⊗ HomSk(Vπ, E⊗k)

Lemma 4.1.10. Let {Vπ} be the isomorphism classes of the set of irreduciblerepresentations of G and let χπ : G → C be the character of Vπ. Then for anyg ∈ G the map

tr(g) : R(G)→ ZVπ 7→ χπ(g)

is a ring homomorphism.

Proof. This follows from the fact that

χVπ⊕Vρ(g) = χVπ(g) + χVρ(g)

andχVπ⊗Vρ(g) = χVπ(g)χVρ(g).

16

For any k, let σ ∈ Sk be a k-cycle. Given a representation V of Sk with characterχ, we define ck = χ(σ). Notice that all the k-cycles in Sk are conjugate to eachother and thus gives the same trace so this map is well-defined.

Another lemma which we will need but we won’t prove is the following

Lemma 4.1.11. Suppose that φ : A → A′ and ψ : B → B′ are ring homomor-phisms. Then the map

φ⊗ ψ : A⊗B → A′ ⊗B′

a⊗ b 7→ φ(a)⊗ ψ(b)

is well-defined and is a ring homomophism.

Now suppose that E is an Sk bundle and σ ∈ Sk is a k-cycle. We know thatE ∼=

∑π Vπ ⊗ HomSk(Vπ, E). Now let ι : K(X) → K(X) be the identity ring

homomorphism. We now define the map

ck ⊗ ι : R(Sk)⊗K(X)→ Z⊗K(X) ∼= K(X)∑π

[Vπ]⊗ [HomSk(Vπ, E)] 7→∑π

ck(Vπ) · [HomSk(Vπ, E)]

For this map to be well-defined, we need ck(Vπ) ∈ Z since K(X) is an abelian group.Indeed, we have

Lemma 4.1.12. All characters of Sk are integer-valued.

To prove this lemma we need to introduce some Galois theory.

Lemma 4.1.13. Suppose that ζ is an n-root of unity. Then the set of ring auto-morphisms from Q[ζ]→ Q that fixes Q is of the form

ζm : Q[ζ] −→ Qζ 7→ ζm

for some 1 ≤ m < n where m,n are coprime.

We can now go back and prove Lemma 4.1.12Proof. [Lemma 4.1.12]. Suppose that g ∈ Sk, |g| = n. Let ρ be a representation ofSk with character χ. Since ρ(σ) is diagonalisable, we can write ρ(σ) = QDQ−1 forsome diagonal matrix D, where the eigenvalues are roots of unity satisfying ζn = 1.Now let m < n be such that m is coprime to the cycles lengths of σ, so that g and gm

are conjugate in Sk. We then have ρ(gm) = ρ(g)m = QDmQ−1. Hence χ(σm) is theimage of the element χσ under the ring automorphism ζm : Q[ζ] → Q[ζ], ζ 7→ ζm.That is ζm(χ(σ)) = χ(σm). But g and gm are conjugate in Sk so χ(σ) = χ(σm).This implies that χ(σ) is fixed by σm which implies that χ(σ) ∈ Q. Since χ(σ) isan algebraic integer, we conlude that χ(σ) ∈ Z.

Remark 4.1.14. We see that ck = tr(σ) ⊗ ι so by lemma 2, ck is a ring homo-morphism.

17

4.2 Adam’s operations

Before defining the Adam’s operations we need the following

Lemma 4.2.1. The power map

⊗k : K(X) −→ KSk(X)

[E] 7→ [E⊗k]

is well-defined.

Definition 4.2.2. (Adam’s operations) For each integer k and any [E] ∈ K(X),define

ψk : K(X)→ R(Sk)⊗K(X)→ K(X)

[E]→ [E⊗k]→ ck ⊗ ι(E⊗k)

For a general element [E −F ], we take its tensor power [(E −F )⊗k], where we get

an integer combination of E⊗i ⊗ F⊗k−i, decompose each of these vector bundles asan Sk vector bundle and apply ck ⊗ ι to each.

The Adam’s operations have a number of properties. We will now proceed toprove the following proposition.

Proposition 4.2.3. For each positive integer k, the map

ψk : K(X)→ K(X)

satsifies the following properties:

(i) ψk : K(X)→ K(X) is a ring homomorphism.(ii) ψk is natural, that is for any continuous map f : X → Y , ψkf ∗ = f ∗ψk

(iii) ψk([L]) = [L]k for any line bundle [L].(iv) ψk ◦ ψl = ψl ◦ ψk for any positive integers k, l.(v) ψp(α) ≡ αp for any prime p.

Foe the last property, it means ψp(α) = αp + pβ for some β ∈ K(X). Atiyahproved that ψk is a ring homoorphism, that is ψk() so we will omit his proof here.We will prove the remaining properties of the adams operations.

4.3 ψk is natural

Proof. [Proposition 4.2.3(ii)]Let π(E) = HomSk(Vπ, E

⊗k) and let ck : R(G) −→ Z be the trace of a k-cycle.We know that E⊗k =

∑π Vπ ⊗ π(E). Now, since f ∗(E ⊕ F ) = f ∗(E) ⊕ f ∗(F ),

f ∗(E ⊗F ) = f ∗(E)⊗ f ∗(F ) and that the pull back of a trivial bundle is trivial, wesee that

f ∗(E⊗k) = f ∗

(∑π

Vπ ⊗ π(E)

)=∑π

f ∗(Vπ)⊗ f ∗(π(E)) =∑π

Vπ ⊗ f ∗(π(E))

and hence in K(X) we have

18

ψk(f ∗[E]) = ψk([f ∗(E)]) = ck([f∗(E)⊗k]) = ck([f

∗(E⊗k)])

= ck

(∑π

Vπ ⊗ f ∗([π(E)])

)=∑π

tr(Vπ)[f ∗(π(E)].

On the other hand,

f ∗ψk([E]) = f ∗ck([E⊗k]) = f ∗(

∑π

tr(Vπ)[π(E)]) =∑π

tr(Vπ)[f ∗(π(E))]

So we see that ψkf ∗([E]) = f ∗ψk([E]) for the element [E] ∈ K(X). In gen-eral any element in K(X) is of the form [E] − [F ] and since ψk, f ∗ are ringhomomorphisms we see that ψkf ∗([E] − [F ]) = f ∗ψk([E] − [F ]) for any element[E]− [F ] ∈ K(X).

4.4 ψk(L) = Lk for any line bundle L

Proof. (Proposition 1(ii)) Since L is one-dimensional, L⊗k is one-dimensional also.We see that Sk acts trivially on every fiber of L⊗k. Let V1 be the trivial represen-tation of Sk, and let V = X × V1 be the one-dimensional trivial bundle. We seethat L⊗k = V1 ⊗ HomSk(V, L

⊗k). But the trivial one dimensional vector bundle isthe identity with respect to the tensor product in VectC(X), so we see that L⊗k =V1 ⊗ HomSk(V, L

⊗k) = HomSk(V, L⊗k). Thus ψ(L) = [HomSk(V, L

⊗k)] = [L⊗k]

4.5 ψk(E) ∼= Ek mod p

For any prime p, denote σ to be a p-cycle in Sp. We want to show that p|χV (1)−χV (σ) for any irreducible representation of Sp.

Firstly we have

Lemma 4.5.1. Suppose that E =∑

π Vπ ⊗ HomSk(Vπ, E⊗k). Then

[E⊗] =∑π

dim(Vπ)[HomSk(Vπ, E⊗k] ∈ K(X)

Proof. We have

[E⊗k] =∑π

Vπ ⊗ HomSk(Vπ, E⊗k)

=∑π

[HomSk(Vπ, E⊗k)]dim(Vπ)

=∑π

dim(Vπ)[HomSk(Vπ, E⊗k)].

Hence the element [E⊗k] is the same if we took the trace of the identity elementof Vπ. To show that ψk(E) ≡ E⊗k mod p it suffices to show that p|χπ(1)− χπ(σ),where χπ is the character of Vπ.

19

Lemma 4.5.2. The map σk : Z[ζ]→ Z[ζ] sending ζ 7→ ζk is a ring automorphismfor any k ∈ {1, ..., p− 1}Proof. Since (k, p) = 1, σk is bijective. σk(ζmζn) = ζk(m+n) = ζkmζkn =σk(ζm)σk(ζn).

Lemma 4.5.3. Let ζ be a p-th root of unity . Then p =∏p−1

k=1(1− ζk).Proof. By factorising the expression xp − 1 we have

∏p−1k=1(x − ζk) =

∑p−1k=0 x

k.Substitute x = 1 and the result follows.

We now come to the lemma

Lemma 4.5.4. Suppose that ρ : Sk → GL(V ) is an irreducible representation ofSk with character χ. Then p|χV (1)− χV (σ).

Proof. Suppose dim(V ) = χV (1) = d. Let ζ 6= 1 be a p-th root of unity. Weknow that the eigenvalues of ρ(σ) are of the form ζk, k ∈ {1, .., p − 1}. So we canwrite χV (1)− χV (σ) =

∑dkd=1 1− ζkd . But 1− ζ|1− ζk for any k, so we can write

χV (1)−χV (σ) = (1−ζ)ν for some ν ∈ Z[ζ]. Since the characters of Sn takes integervalues, we can write (1− ζ)ν = χV (1)− χV (σ) = n for some n ∈ Z.

Now, apply the ring automorphism σk : Z[ζ]→ Z[ζ] to the equation (1−ζ)ν = nfor each k = {1, ..., p− 1}. We see that σk((1− ζ)ν) = (1− ζk)σk(ν) = n for eachk. Multiply all these together we get

∏p−1k=1(1− ζk)σk(ν) = np−1 which implies that

p∏p−1

k=1 σk(ν) = np−1. Since p is prime, we conclude that p|n = χV (1)− χV (σ).

So now we have shown that ψp([E]) ≡ [E⊗k] mod p for [E] ∈ K(X). We willnow proceed to show

Proposition 4.5.5. For a general element [E]− [F ] ∈ K(X), we have

ψp([E]− [F ]) ≡ ([E]− [F ])p mod p

Proof. We first consider the case when p = 2. We then have

ψ2([E − F ]) = ψ2([E])− ψ2([F ])

≡ [E⊗2]− [F⊗2]

≡ [E⊗2]− 2[E ⊗ F ]− [F⊗2] + 2[F⊗2]

= [(E − F )⊗2] = [E − F ]2 mod 2.

Now, for p ≥ 3, we have

ψp([E − F ]) = ψp([E])− ψp([F ]) mod p

≡ [E⊗p]− [F⊗p] mod p

≡ [E⊗p] +

p−1∑i=1

(−1)i(pi

)[E⊗i ⊗ F⊗p−i]− [F⊗p] mod p

since p∣∣∣ ( p

i

),

≡ [(E − F )⊗p] = [E − F ]p mod p

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and this completes the proof.

4.6 ψkψl = ψlψk

Lemma 4.6.1. For any finite groups G,H, we have

R(G×H) ∼= R(G)⊗Z R(H)

Proof. Firstly notice that for any g, g′ ∈ G and h, h′ ∈ H, we have

(g′, h′)(g, h)((g′)−1, (h′)−1) = (g′g(g′)−1, h′h(h′)−1)

so we see that the conjugacy class of (g, h), denoted by C(g,h) is precisely Cg × Ch.Suppose that there are m,n number of conjugacy classes of G,H respectively. Thenwe see that the number of conjugacy classes of G×H is mn.

Suppose that ρ : G → GL(V ), π : H → GL(V ) are irreducible representationsof G,H respectively. Then we can define a representation space, ρ× π, for G×Hby

ρ× π : G×H → GL(V ⊗W )

(g, h) 7→ ρ(g)⊗ π(h)

(check ρ(g)⊗ π(h) is indeed an automorphism of V ⊗W )Now, let χ be the character of ρ × π. Then by definition we have χ(g, h) =

χρ(g)χπ(h). Recall that 〈χ, χ〉 is the orthogonality relation. We then have

〈χ, χ〉 =1

|G×H|∑

(g,h)∈G×H

χ(g−1, h−1)χ(g, h)

=1

|G||H|∑g∈G

∑h∈H

χρ(g−1)χρ(g)χπ(h−1)χπ(h)

=1

|G|

(∑g∈G

χρ(g−1)χρ(g)

)1

|H|

(∑h∈H

χπ(h−1)χπ(h)

)= 〈χρ, χρ〉〈χπ, χπ〉.

Since ρ, π are irreducibles, we see that 〈χ, χ〉 = 1 =⇒ ρ × π is an irreduciblerepresentation of G×H.

Now, let {ρi}, {πj} be the set of irreducible representations of G,H respec-tively. From above we see that the set {ρi × πj} is a set of irreducible repre-sentations of G × H. Similar to the above formula 〈χ, χ〉 = 〈χρ, χρ〉〈χπ, χπ〉, ifρ1, ρ2 and π1, π2 are irreducible representations of G and H, then 〈χρ1×π1 , χρ2×π2〉 =〈χρ1 , χρ2〉〈χπ1 , χπ2〉 = 1 iff ρ1 = ρ2 and π1 = π2. We also know that there are mnconjugacy classes of G×H so {ρi×πj} is precisely the set of irreducible representa-tions of G×H. Hence we see that the map ρi⊗πj → ρi×πj is a bijection betweenthe generators of R(G)⊗R(H) and R(G×H).

21

Now consider the map

Φ : R(G)×R(H)→ R(G×H)

ρi × πj → ρi ⊗ πj

Here we will look at R(G), R(H), R(G×H) as a subring of the ring of all complexclass functions on G,H,G × H respectively. Thus, a basis of R(G) is {χρi} andsimilarly for R(H). Now, by definition as above we see that a basis for R(G×H)is {χρiχπj}. It is now easy to check that Φ is a bilinear homomorphism so by the

universal property of tensor product there is a ring isomorphism Φ : R(G)⊗R(H)→R(G×H).

We will now show that ψk ◦ ψl = ψl ◦ ψk for all k, l ∈ Z.Proof. Recall that given an element [E] ∈ K(X), ψk(E) is obtained by

ψk : K(X)→ R(Sk)⊗K(X)→ Z⊗K(X) ∼= K(X)

E 7→ E⊗k 7→ ck(E⊗k)

Now consider the element ψk ◦ ψl(E). This element is obtained through

ψk ◦ ψl : K(X)→ R(Sl)⊗K(X)→ K(X)→ R(Sk)⊗K(X)→ K(X)

E 7→ E⊗l 7→ cl(E⊗l) 7→ (cl(E

⊗l))⊗k 7→ ck((cl(E⊗l))⊗k)

Now since cl is a ring homomorphism, we have ck((cl(E⊗l))⊗k) = ck ◦cl((E⊗l)⊗k)

and so we can redefine ψk ◦ ψl as

ψk ◦ ψl : K(X)→ R(Sl)⊗K(X)→ R(Sk)⊗R(Sl)⊗K(X)→ R(Sk)⊗K(X)→ K(X)

E 7→ E⊗l 7→ (E⊗l)⊗k 7→ cl((E⊗l)⊗k) 7→ ck ◦ cl((E⊗l)⊗k)

Now, since R(Sk)⊗R(Sl) ∼= R(Sk × Sl), we can look at (E⊗l)⊗k as an Sk × Sl-bundle. Suppose that {Vi}, {Wj} are the set of irreducible representations of Sk, Slrespectively. We know that the set Vi × Wj are precisely the set of irreduciblerepresentations of Sk × Sl so we can decompose (E⊗l)⊗k as

E⊗kl ∼=∑i,j

Vi ×Wj ⊗HomSk×Sl(Vi ⊗Wj, (E⊗l)⊗k)

Pick the isomorphism φ : R(Sk × Sl) → R(Sk) ⊗ R(Sl), ρi × πj → ρi ⊗ πj. Thisinduces an isomorphism

φ : R(Sk × Sl)⊗K(X)→ R(Sk)⊗R(Sl)⊗K(X)

where∑i,j

Vi×Wj⊗HomSk×Sl(Vi⊗Wj, (E⊗l)⊗k) 7→

∑i,j

Vi⊗Wj⊗HomSk×Sl(Vi⊗Wj, (E⊗l)⊗k)

Here we assume Vi is the representation space of ρi and Wi be the representationspace of πj. If we now take the trace of Wj and Vi we get ψk ◦ ψl(E). The same

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is true for ψl ◦ ψk however there is a difference. To obtain ψk ◦ ψl(E) we first takeE⊗l and then (E⊗l)⊗k. On the other hand if we were to compute ψl ◦ ψk we firsttake E⊗k then (E⊗k)⊗l. As vector bundles both are isomorphic to E⊗kl. Howeverthe action of Sk × Sl on these vector bundles are different and hence (E⊗l)⊗k and(E⊗k)⊗l are not necessarily isomorphic as Sk × Sk-bundles.

We will now proceed to show (E⊗l)⊗k and (E⊗k)⊗l are indeed isomorphic asSk × Sl-bundles and thus by Lemma 4.1.7∑i,j

Vi⊗Wj⊗HomSk×Sl(Vi⊗Wj, (E⊗l)⊗k) ∼=

∑i,j

Vi⊗Wj⊗HomSk×Sl(Vi⊗Wj, (E⊗k)⊗l)

To do this it suffices to show that every fiber of (E⊗l)⊗k and (E⊗k)⊗l areisomorphic as Sk × Sl-modules and that the isomorphism is continuous. Let Vbe the fiber of E, so that V ⊗kl is the fiber of E⊗kl. For convenience we write(V ⊗l)⊗k =

⊗l,ki=1,j=1 Vij where each Vij = V and similarly (V ⊗k)⊗l =

⊗l,ki=1,j=1 Vji.

The action of Sk × Sl on (V ⊗l)⊗k and (V ⊗k)⊗l are illustrated below.

V11 · · · V1k...

. . ....

.... . .

...Vl1 · · · Vlk

V11 · · · · · · V1l

.... . . · · · ...

Vk1 · · · · · · Vkl

For the left one (V ⊗l)⊗k, Sl acts on the columns while Sk acts on the rows. On theright side we have (V ⊗k)⊗l, where Sl acts on the rows and Sk acts on the columns.Now let vij be vectors in Vij. We let

Φ :

l,k⊗i=1,j=1

Vij →l,k⊗

i=1,j=1

Vji

vij 7→ vji

Clearly this is a vector space homomophism, so it remains to show that Φ is Sk×Sl-linear. So suppose that σ ∈ Sk, τ ∈ Sl. Then we have

Φ((σ, τ) · Vij) = Φ(Vσ(i)τ(j)) = Vτ(j)σ(i) = (σ, τ)Vji = (σ, τ)Φ(Vij).

In some sense, ψk ◦ ψl and ψl ◦ ψk differs only in the order of taking traces. Wewill now conclude this chapter by showing the following remark.

Proposition 4.6.2. There is no compact Hausdorff space X such that K(X) =Z[i].

Proof. Suppose for a contradiction that X is compact Hausdorff and K(X) =Z[i]. We want to apply the ring homomorphism ψ2 to elements of Z[i] and get acontradiction using the extra properties of ψ2. So suppose that ψ2(i) = a + bi for

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some a, b ∈ Z. Since ψ2(i) ≡ i2 = −1 mod 2, we see that a is odd and b is even.Now

0 = ψ2(0) = ψ2(1 + i2) = ψ2(1) + (ψ2(i))2 = 1 + (a+ bi)2 = 1 + a2 + 2abi− b2

which implies that 2ab = 0 =⇒ a = 0 or b = 0. But a is odd, so we must haveb = 0. This implies that 1 + a2 = 0 which is a contradiction.

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Chapter 5

Division algebras

Some examples of division algebras are the real numbers R, the complex numbersC,x the Quaternions H and the Octonions O, corresponding to dimensions 1, 2, 4, 8.

In fact we have

Theorem 5.0.3. Rn is a division algebra only when n = 1, 2, 4, 8

5.1 Divison algebra

An Division algebra D over R is the vector space Rn with a continuous multi-plication map φ : Rn × Rn → Rn such that if we fix an a ∈ Rn then the mapsx 7→ φ(a, x) and x 7→ φ(x, a) are linear in x and invertible if a 6= 0. Since the mapsare linear, invertibility implies that the multiplication map has no zero divisors. Wecan assume that the multiplcation map has a two-sided identity element as follows.Choose a unit vector e ∈ Rn. Suppose that M ∈ GLn(R) takes e2 to e. Thenletting φ = M ◦ φ we can assume φ(e, e) = e. Now, let p, q : Rn → Rn be the mapsx 7→ φ(x, e), x 7→ φ(e, x) respectively. We see that p(e) = q(e) = φ(e2) = e. Thenthe map

ψ : Rn × Rn −→ Rn × Rn −→ Rn

(x, y) 7→ (p−1(x), q−1(y)) 7→ φ(p−1(x), q−1(y))

maps (x, e) to (p−1(x), e) to x and similarly it sends (e, y) to y. For any a in thedivisions algebra, the maps (a, x) 7→ e and (x, a) 7→ e are bijective, hence everyelement a has multiplicative inverse on both left and right.

The strategy here is that we can consider the sphere Sn−1 ⊂ Rn. By rescalingthe multiplication map from R into the sphere we get a continuous multiplicationmap on the sphere, defined by

µ : Sn−1 × Sn−1

(x, y) 7→ xy

|xy|

The bulk of this chapter is to show that such a map does not exists on the sphereif n 6= 1, 2, 4, 8. The strategy is that instead of showing directly that such a mapdoes not exist, we pass it to an an induced homomorphism on the K rings and showthat such a homomorphism does not exist. This is similar to showing that there isno deformation retract from the disk onto the circle.

We will first deduce some consequences of the Bott periodicity theorem

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(i) We know that K(S2n−1) = 0 for n odd and S2n. This follows from that fact

that K(S1) = 0, K(S2) = Z and the isomorphism K(X) = K(S2X). Theisomorphism is the map a 7→ a ∗ (H − 1). Hence we see that a generator of

K(S2n) is the k-fold external product (H − 1) ∗ · · · ∗ (H − 1).

(ii) K(S2k) ⊗ K(X) = K(S2kX) = K(S2k ∧X) by iterating the bott periodicitytheorem

(iii) Recall that

K(S2k)⊗K(X) = (K(S2k)⊗ K(X))⊕ K(S2k)⊕ K(X)⊕ ZK(S2k ×X) = K(S2k ∧X)⊕ K(S2k)⊕ K(X)⊕ Z.

From (ii) we see that K(S2k×X) = K(S2k), and the isomorphism is given bythe external product.

Proposition 5.1.1. Rn is not a division algebra for any odd n > 1.

We will outline the proof here. To show this proposition it suffices to show thatµ : Sn−1×Sn−1 → Sn−1 for any odd n > 1. In this case n−1 is even so we can replaceSn−1 with S2n. So suppose that µ : S2n×S2n → S2n is a continuous map with a twosided identity. We then get a induced homomorphism µ∗ : K(S2n)→ K(S2n×S2n)which by eariler proposition, a homomorphism between the following rings

µ∗ :Z[γ]

〈γ2〉−→ Z[α, β]

〈α2, β2〉.

It turns out that µ∗(γ) = α+ β + kαβ for some k ∈ Z, which is a contradiction, asµ∗(γ2) = (α + β + kαβ)2 = 2αβ since α2 = β2 = 0. But this is a contradiction asµ∗(γ2) = 0 6= 2αβ.

This is a good example of turning a topological problem into an algebraic prob-lem. Often it is easier to show the non-existence of homomorphisms since they mustpreserve structure, which is easier to show the non-existence of continuous maps.

The previous theorem takes care of Rn for odd n. We cannot use the above ideabecause the crux of the proof above was the isomorphism

K(S2n)⊗K(S2n) ∼= K(S2n × S2n)

which allowed us to compute the ring K(S2n × S2n). We do not have a similarisomorphism theorem for K(S2n−1× S2n−1) and K groups in general are very hardto compute. What we do now is to associate any continuous map g : Sn−1 × Sn−1to a map g : S2n−1 → Sn. In the end we will construct a map between spheres ofeven dimensions and get a contradiction.

We first regard S2n−1 as ∂(Dn×Dn). To see this, notice that Dn is homeomor-phic to In, where I is the closed unit interval. Then Dn ×Dn is homeomorphic toI2n which is D2n and so S2n−1 is precisely the boundary of D2n. SO now we writeS2n−1 = ∂(Dn×Dn) = ∂Dn×Dn ∪Dn× ∂Dn and Sn = Dn

+ ∪Dn−. We then define

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g+ : ∂Dn ×Dn → Dn+ g− : ∂Dn ×Dn → Dn

−

(x, y) 7→ |y|g(x,

y

|y|

)(x, y) 7→ |x|g

(x

|x|, y

)this gives us a map g : S2n−1 → Sn.We now define a Hopf invariant associated to g. We now replace n with 2n since

we are interested in spheres of odd dimensions. Given a map f : S4n−1 → S2n, letCf be S2n with a cell e4n attached by f . The quotient Cf/S

2n is then S4n, and

since K(S4n) = K(S2n) = 0, the exact sequence becomes a short exact sequence

0→ K(S4n)q∗→ K(Cf )

i∗→ K(S2n)→ 0

Suppose that q∗((H − 1)2n) = α ∈ K(Cf ) and i∗(β) = (H − 1)n. Then i∗(β2) =i∗(β)2 = ((H − 1)n)2 = 0 =⇒ β2 ∈ ker i∗ = Im q∗ by exactness. Thus we see thatβ2 = hα for some integer h. To see that h is well defined, we need to show that forany other β′ such that i∗(β′) = (H − 1)n. Then since i∗(β − β′) = 0 =⇒ β − β′ ∈ker i∗ = Im q∗ =⇒ β′ = β + mα. It follows that (β + mα)2 = β2 + 2mαβ sinceα2 = 0. Since α ∈ Im q∗ = ker i∗, we see that i∗(α) = 0 and so does αβ. Henceαβ = kα for some k. Then kαβ = αβ2 = hα2 = 0. Thus kαβ = 0 which impliesαβ = 0 since αβ belongs to the cyclic subgroup of K(Cf ) generated by the imageof α.

A key theorem that solves the problem is

Lemma 5.1.2. [4] If g : S2n−1 × S2n−1 → S2n−1 is a H-space multiplication, thenthe associated map g : S4n−1 → S2n has Hopf invariant ±1

Theorem 5.1.3. There exists a map f : S4n−1 → S2n of Hopf invariant ±1 onlywhen n = 1, 2 or 4.

To prove this theorem we will need certain homomorphisms on K(X). The oneswe need are precisely the Adam’s operations defined in Chapter 3.

We will now show that ψk restricts to a ring homomophism on the subring K(X).

To see that recall that K(X) is the kernel of the homomorphism i∗ : K(X)→ K(x0)induced by the inclusion map {x0} ↪→ X. By the naturality of ψk,

α ∈ K(X) =⇒ i∗(α) = 0 =⇒ ψki∗(α) = 0 =⇒ i∗ψk(α) = 0 =⇒ ψk(α) ∈ K(X)

Also, for the external product, we have

ψk(α ∗ β) = ψk(p∗1(α)p∗2(β)) = ψk(p∗1(α))ψk(p∗2(β))

= p∗1ψk(α)p∗2ψ

k(β) = ψk(α) ∗ ψk(β).

We can now determine the map ψk : K(X)→ K(X) explicitly.

Proposition 5.1.4. ψk : K(S2n)→ K(S2n) is multiplication by kn.

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Proof. We first consider the case when n = 1. Since ψk is additive and K(S2) isgenerated by α = H − 1, it suffices to show ψk(α) = kn(α). Now we have

ψk(α) = ψk(H)− 1 = Hk − 1 = (1 + α)k − 1

= 1 + α− 1 since ai = 0 for all i > 1

= kα

For n > 1 we use the external product K(S2) ⊗ K(S2n−2) → K(S2S2n−1) =

K(S2n) and prove by induction. Suppose that ψk : K(S2n−2) → K(S2n−2) ismultiplication by kn−1. Then φk(α ∗ β) = ψk(α) ∗ ψk(β) = kα ∗ kn−1 = kn(α ∗ β)and this completes the proof.Proof. (Adams theorem) Recall that we have the exact sequence

0 −→ K(S4n)q∗−→ K(Cf )

i∗−→ K(S2n) −→ 0.

Suppose again that q∗(H − 1)2n = α and i∗(β) = (H − 1)n. By naturality we have

ψk(α) = ψk(q∗(H − 1)2n) = q∗(ψk((H − 1)2n)) = q∗(k2n(H − 1)2n) = k2nα.

Also,i∗ψk(β) = ψk(i∗(β)) = kn(H − 1)n = i∗(knβ)

so we see that ψk(β) − knβ ∈ ker i∗ = Im q∗. Hence ψk(β) = knβ + µkα for someµk ∈ Z since Im q∗ is generated by α. Hence

ψkψl(β) = ψk(lnβ + µlα) = ln(knβ + µkα) + k2nµlα = lnknβ + (k2nµl + lnµk)α

But ψkψl(β) = ψlψk(β), so we see that

lnknβ + (k2nµl + lnµk)α = lnknβ + (l2nµk + knµl)α

which implies that kn(kn − 1)µl = ln(ln − 1)µk. Now, since ψ2(β) ≡ β2 = hαmod 2, from ψ2(β) = 2nβ+µ2α we see that µ2 ≡ h mod 2 so µ2 is odd as h = ±1.From above we see that 2n(2n − 1)µ3 = 3n(3n − 1)µ2 and since 3n and µ2 are odd,we see that 2n|3n − 1. The proof will be completed by the following lemma.

Lemma 5.1.5. [4] If 2n divides 3n − 1, then n = 1, 2, or 4.

In some sense the modulo property of the Adams operations turns these topo-logical questions into number theory problems, which is often easier.

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References

[1] Adams, J.F.; Atiyah, M.F., K-Theory and the Hopf Invariant, The QuarterlyJournal of Mathematics 17 (1): 3138, 1966

[2] Atiyah, Michael Francis, K-theory, Advanced Book Classics (2nd ed.), Addison-Wesley, 1989

[3] Atiyah, Michael Francis, Power Operations in K-Theory, Quant. J. of Math.(Oxford), 17 (1966), 165-193

[4] Allen Hatcher, Vector Bundles and K-Theory, 2003[5] Allen Hatcher, Algebraic topology Cambridge University Press, Cambridge, 2002

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