Let ${E}$ be a multiplicative cohomology theory. We say that ${E}$ is complex-oriented if one is given the data of an element ${t \in \widetilde{E}^2(\mathbb{CP}^\infty)}$ which restricts to the canonical generator of ${\widetilde{E}^2(\mathbb{CP}^1) \simeq \widetilde{E}^0(S^0)}$. It turns out that one has a bit more: a complex orientation gives on a functorial, multiplicative choice of Thom classes for complex vector bundles. In fact, this is a perhaps more natural definition of such a theory.

What does this mean? Given a vector bundle ${\zeta \rightarrow X}$, one can form the Thom space ${T(\zeta) = B(\zeta)/S(\zeta)}$: in other words, the quotient of the unit ball bundle ${B(\zeta)}$ in ${\zeta}$ (with respect to a choice of metric) by the unit sphere bundle ${S(\zeta)}$. When ${X}$ is compact, this is just the one-point compactification of ${\zeta}$.

Definition 1 The vector bundle ${\zeta}$ is orientable for a multiplicative cohomology theory ${E}$ if there exists an element ${\theta \in \widetilde{E}^*( T(\zeta)) = E^*(B(\zeta), S(\zeta))}$ which restricts to a generator on each fiberwise ${E^*(B^n, S^{n-1})}$, where ${\dim \zeta = n}$. Such a ${\theta}$ is called a Thom class.

Observe that for each point ${x \in X}$, there is a restriction map ${\widetilde{E}^*(T(\zeta)) \rightarrow E^*(B_x^n, S_x^{n-1})}$ if the dimension of ${\zeta}$ is ${n}$.

The existence of a Thom class implies a Thom isomorphism, as for ordinary homology.

Theorem 2 (Thom isomorphism) A Thom class ${\theta \in \widetilde{E}^*(T(\zeta))}$ induces an isomorphism

$\displaystyle E^*(X) \simeq \widetilde{E}^*(T(\zeta))$

given by cup-product with ${\theta}$.

In the case of ordinary homology, a Thom class is unique (up to sign) if it exists; in general, though, a Thom class is highly non-unique, and an orientation is additional data than simple orientability.

Here are a few basic cases:

1. Any vector bundle is orientable for ${\mathbb{Z}/2}$-cohomology.
2. An oriented (in the usual sense: i.e., the top wedge power is trivial) vector bundle is one oriented for ${\mathbb{Z}}$-cohomology.
3. Complex vector bundles are oriented for ${K}$-theory. We will see this below.
4. Spin bundles are oriented for ${KO}$-theory. An explicit construction of Thom classes can be made, as virtual bundles arising from Clifford modules: this is in Atiyah-Bott-Shapiro’s paper.
5. A trivial bundle is orientable for any cohomology theory (this is rather uninteresting: the Thom space is just a suspension). (more…)

Next, I would like to describe an alternative description of relative K-theory which is sometimes convenient (e.g. when describing the Thom isomorphism in K-theory). Let ${G}$ be a compact Lie group, as always. Let ${(X, A)}$ be a pair of compact ${G}$-spaces (with ${ A \subset X}$); then we have defined therelative ${K_G}$-theory via

$\displaystyle K_G(X, A) \equiv \widetilde{K}_G(X/A).$

Here ${X/A}$ is equipped with a distinguished basepoint (corresponding to ${A/A}$), and as a result this makes sense. As usual, we can use this definition to make ${K_G}$ into a cohomology theory on compact ${G}$-spaces.

To describe ${K_G(X, A)}$ without use of the space ${X/A}$, we can proceed as follows.

Definition 1 We let ${C_G(X, A)}$ be the category of complexes of ${G}$-vector bundles

$\displaystyle 0 \rightarrow E_0 \rightarrow \dots \rightarrow E_n \rightarrow 0$

on ${X}$, which when restricted to ${A}$ are exact. A morphism in ${C_G(X, A)}$ is a map of chain complexes.

The idea is that we are going to assign to every element ${E_\bullet}$ of ${C_G(X,A)}$ an element of the relative K-theory ${K_G(X, A) = \widetilde{K}_G(X/A)}$, by effectively taking the alternating sum ${\sum (-1)^i [E_i]}$. In order to do this, we will start by modifying the complex ${E_\bullet}$ by adding acyclic complexes. Namely, we start by adding complexes of the form

$\displaystyle 0 \rightarrow 0 \rightarrow \dots \rightarrow 0 \rightarrow F \rightarrow F \rightarrow 0 \rightarrow 0 \rightarrow \dots$

where ${F}$ is a ${G}$-vector bundle on ${A}$, to make all but the first term of ${E_\bullet}$ trivial (i.e. coming from an ${R(G)}$-representation). With this change made, we can assume that all but the first term of ${E_\bullet}$ is stably trivial. Then the first term of ${E_\bullet}$ is stably trivial when restricted to ${A}$ by exactness of ${E_\bullet|_A}$. Consequently, we can quotient all the terms by ${A}$ and get a complex of ${G}$-vector bundles ${E_\bullet|_{X/A}}$; this is exact at the basepoint of ${X/A}$. Now, taking the alternating sum as desired, we get a map

$\displaystyle C_G(X, A) \rightarrow K_G(X, A).$

This map does not see stable equivalence; that is, if we add to a complex ${E_\bullet}$ a complex of the form ${0 \rightarrow F \rightarrow F \rightarrow 0}$, the image in ${K_G(X, A)}$ does not change. Moreover, it is homotopy invariant.

In fact, one can give a presentation of the group ${K_G(X, A)}$ in this way. We start with the set of all such complexes in ${C_G(X, A)}$ as above. We identify complexes which are chain homotopic to each other. Then, we mod out by the relation of (geometric) homotopy: if one has complexes ${E_\bullet, F_\bullet \in C_G(X, A)}$ which can be obtained by restriction to the end faces of a complex in ${C_G(X \times [0, 1], A \times [0, 1])}$, then they should both be identified. Given these identifications, one gets precisely the group ${K_G(X, A)}$.

I don’t really want to prove these things in detail, partially because I don’t want to get too bogged down with this project. (more…)

I’ve been trying to fix the (many) gaps in my knowledge of classical algebraic topology as of late, and will probably do a few posts in the near future on vector bundles, K-theory, and characteristic classes.

Let ${B}$ be a base space, and let ${p: E \rightarrow B}$ be a real vector bundle. There are numerous constructions for the characteristic classes of ${B}$. Recall that these are elements in the cohomology ring ${H^*(B; R)}$ (for ${R}$ some ring) that measure, in some sense, the twisting or nontriviality of the bundle ${B}$.

Over a smooth manifold ${B}$, with ${E}$ a smooth vector bundle, a construction can be made in de Rham cohomology. Namely, one chooses a connection ${\nabla}$ on ${E}$, computes the curvature tensor of ${E}$ (which is an ${\hom(E,E)}$-valued 2-form ${\Theta}$ on ${B}$), and then applies a suitable polynomial from matrices to polynomials to the curvature ${\Theta}$. One can show that this gives closed forms, whose de Rham cohomology class does not depend on the choice of connection. This is the subject of Chern-Weil theory, and it applies more generally to principal ${G}$-bundles on a manifold for ${G}$ a Lie group.

But there is something that this approach misses: torsion. By working with de Rham cohomology (or equivalently, cohomology with ${\mathbb{R}}$-coefficients), the very interesting torsion phenomena that algebraic topologists care about is lost. For the purposes of this post, we’re interested in cohomology classes where the ground ring is ${R = \mathbb{Z}/2}$, and so de Rham cohomology is out. However, in return, we have cohomology operations. We can use them instead. (more…)

Suppose ${B}$ is a topological space, and ${p: E \rightarrow B}$ is a vector bundle over ${B}$. Here ${B}$ is a “reasonable” space, for instance a manifold or a CW complex. There are many familiar examples of how such bundles arise “in nature,” for instance the tangent bundle to a manifold. It is of interest to understand how the apparatus of algebraic topology applies to the bundle.

Now, clearly there is a deformation retraction of ${E}$ given by ${(t,v) \rightarrow tv}$ that deformation retracts ${E}$ to the image of the zero section. In particular, ${E}$ is homotopy equivalent to ${B}$. So ${E}$ by itself might not all be that interesting.

Nonetheless, let us suppose that ${E}$ is a normed bundle, e.g. the tangent bundle to a manifold with a Riemannian metric. In this case, we can consider the set of elements of norm one; it is no longer a vector bundle over ${B}$, but it is a fiber bundle (with fibers various spheres). This is homotopically very different from ${B}$ in general.

For a general bundle ${E}$, we can still consider the subset ${\dot{E}}$ consisting of nonzero elements. Then there is a map ${p: \dot{E} \rightarrow B}$ whose fibers are of the form ${\mathbb{R}^n - \left\{0\right\}}$. It is easy to see that if ${E}$ is a normed bundle, then ${\dot{E}}$ deformation retracts onto the associated sphere bundle. So we might as well see what ${\dot{E}}$ is like. By using the long exact sequence, we might try studying the pair ${(E, \dot{E})}$.

Today, I want to talk about the cohomology of the pair ${(E, \dot{E})}$. Namely, the main result is:

Theorem 1 (Thom) For all ${k}$, there is an isomorphism (to be described)$\displaystyle H^k(B; \mathbb{Z}/2) \simeq H^{k+n}(E, \dot{E}; \mathbb{Z}/2)$

if ${E}$ has dimension ${n}$.

(To ease notation, all cohomology will henceforth be with ${\mathbb{Z}/2}$-coefficients.) (more…)