# A Precursor to Characteristic Classes

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 $$\mathbb{CP}^n$$.

## One thought on “A Precursor to Characteristic Classes”

1. June says:

I find your writing very intuitive and kindly written.

Thanks!
June