I’ll assume that you know what a line bundle is and are comfortable with the following equivalences; if you aren’t familiar with the notation in these equivalences, John Baez might help. Note that integral cohomology := cohomology with coefficients in (\mathbb{Z}).

(U(1) \simeq S^1 \simeq K(\mathbb{Z}, 1))

(BU(1) \simeq CP^\infty \simeq K(\mathbb{Z}, 2))

The aim of this post is to give you a taste of the beautiful world of characteristic classes and their intimate relationship to line bundles via the concrete example of how the second integral cohomology group of a space is actually the isomorphism classes of line bundles over that space.

That’s right! (H^2(X; Z) \simeq) the isomorphism classes of (complex) line bundles over X. It is in fact, a group homomorphism — the group operations being tensor product of line bundles and the usual addition on cohomology. This isn’t something that I understood at first glance. I mean, hot damn, it’s unexpectedly rich.

Let’s talk about line bundles.

• (RP^1) consists of all lines that intersect the origin of (R^2).
• (CP^1) consists of all complex lines that intersect the origin of (C^2).

Let’s look at (C^2): Draw a line (r) parallel to one of the axes, each line (L) through the origin will intersect this line at a unique point (x). This point characterizes (L).

Only one line, the line parallel to our line (r) will not intersect (r) — we can say that this line is characterized by the point at (\infty).

(CP^1), the collection of all complex lines through the origin of (C^2), is then isomorphic to all of the points (x) (including the point at infinity). In other words, (CP^1 \simeq S^2).

Recall that the complex line has 2 real dimensions — this powerful isomorphism is simply due to the one-point compactification of (\mathbb{R}^2) that we know and love. Note that we can also get from (CP^1) to (S^2) via stereographic projection.

“Canonical” line bundles

The elements of (CP^1) are the points (x), thus we can describe a line bundle over (CP^1) as follows: its points are the pairs ((a,x)), where (a) is a point on the line (L) (characterized by x) (\in \mathbb{C}^2). The base space is (C^1) + (pt) at infinity, and each fiber is (L).

This line bundle is called the canonical line bundle of (CP^1).

This story holds for all (n). In general, each point (x) in (\mathbb{CP}^n) is line (L) through the origin in (\mathbb{C}^{n+1}). Let (\ell^n) := the canonical line bundle of (CP^n).

I hope we can agree that we can describe a line bundle over (X) as follows: to each element of (X) (a point), we associate an element of (\mathbb{CP}^\infty) (a line).

Saying that the line bundle over (X) we know and love is a way to associate a line to every point in (X) seems obvious and trivial — but asking “where do lines live?” has some beautiful consequences. I want you to feel this in your bones, so I’ll spell it out a bit more explicitly.

What are line bundles over a topological space (X)?

A line bundle (f) is:

• a map from each point in (X)
• to a line (i.e. an element of (\mathbb{CP}^\infty)).

I’ll repeat this again: a line bundle (f) is a map of type (X \to \mathbb{CP}^\infty).

todo: add a bit about (\ell^\infty) here

In other words, any complex line bundle (L) over (X) is a pullback of (\ell^\infty) by the map (f).

Cohomology is a Representable Functor

The homotopy classes of maps from a space (X) to the nth Eilenberg-MacLane space (B^n(G)) of a group (G) is isomorphic to the (n)th cohomology group of a space (X), with coefficients in the group (G). In other words:

([X, B^n(Z)] \simeq H^n(X; Z))

This is a special case of a theorem, the Brown Representability Theorem, which states that all cohomology theories are represented by spectra, and vice versa. But that’s a whole ‘nother story! Let’s see how this connects to line bundles:

• ([X, B^n(Z)] \simeq H^n(X; Z))
• ([X, B^2(Z)] \simeq H^2(X; Z))

As you’ll recall from the equivalences listed at the beginning of this post, (B^2(Z)),the 2nd Eilenberg-MacLane space of the integers as a group, is isomorphic to (CP^\infty), thus:

• ([X, CP^\infty] \simeq H^2(X; Z))

Warning: (B^2(\mathbb{Z})) is non-standard notation, and is usually written as (K(\mathbb{Z},2)) here’s a paper that explains how to compute Eilenberg-MacLane spaces.

Hey, you said that there would be characteristic classes! Where do those come in?

I did say that this post was a precursor to characteristic classes, but let’s look at a piece of the map. ((\to) := correspond to)

• complex line bundles (\to) elements of (H^2) over (Z) (“Chern classes”)
• real line bundles (\to) elements of (H^2) over (Z/2) (“Stiefel-whitney classes”)
• quarternionic line bundles (\to) elements of (H^4) over (Z) (“Pontryagin classes”)

This works for BU(n) when n=1. What about other n?

This section is also a teaser. I’d like to suggest that the story generalizes from complex line bundles to complex vector bundles. I don’t quite understand the details of this generalization, but I wish to share with you what I do understand.

Note that a complex n-dimensional vector bundle + a choice of hermitian metric = a (U(n))-principle bundle.

So, it makes sense that classifying complex n-dimensional vector bundles (which I’ll denote (E \to X)) is closely related to the story of classifying their associated principle (U(n))-bundles (which I’ll write as (\hat{E} \to X)). This motivates us considering that our previous picture…

… might just be a special case of a more general phenomena! But how?

Well we have a complex vector bundle — so what is our associated frame bundle?

Let the classifying map of (\hat{E} \to X) be (f: X \to BU(n)).

Note that (BU := \text{colim}_n BU(n)).

If you’d like to learn more about characteristic classes, I’ve found Milnor and Stasheff to be of great help.

Thank you to Peter Teichner for patiently explaining why (\mathbb{CP}^1 \simeq S^2) and consequently the cell decomposition of (\mathbb{CP}^n).