# 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 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. I find your writing very intuitive and kindly written.

Thanks!
June