# Motivating Characteristic Classes Using $S^2$

Trying to understand characteristic classes, hoping someone can explain/fit my example below into the wider scheme of things:

Chern's book says

Characteristic classes are the simplest global invariants which measure the deviation of a local product structure from a product structure. They are intimately related to the notion of curvature in differential geometry. In fact, a real characteristic class is a "total curvature," according to a well-defined relationship.

Lets take a small examples - it seems to me that the example of going from

$$S^2 = \{(x,y,z) \in \mathbb{R}^3 | x^2 + y^2 + z^2 = 1 \}$$

to the functions

$$z_{\pm} = \pm \sqrt{1 - x^2 - y^2}$$

can be formulated in terms of fiber bundles, where

$$S^2 = \{(x,y,z) \in \mathbb{R}^3 | x^2 + y^2 + z^2 = 1 \}$$

is the total space $E$,

$$D^1 = \{(x,y) \in \mathbb{R}^2 | x^2 + y^2 \leq 1 \}$$

is the base $B$ of the bundle, the maps

$$z_{\pm} = \pm \sqrt{1 - x^2 - y^2}$$

are sections of the bundle, with fibers of the form $$F_x = \{ ((x,y),z_+(x,y)),((x,y),z_-(x,y))\}$$

(Hope I've set that up about right).

So we have two local product structures $D^1 \times z_+$ and $D^1 \times z_-$ representing the top and bottom of the sphere. My guess is that if we want to measure the deviation of local product structure from global product structure we want to measure how the transition functions between $z_+$ and $z_-$ interact right?

Is all of this about right so far? Can it be cleaned up if so/not?

Bolstering some of this intuition - in the first 3 minutes of this video it seems like they say that characteristic classes arise (in this example) in the form of some function $f(n)$ when you analyze the transition function gluing $z_+$ to $z_-$ on the equator, $$z_- = f(n)z_+$$

I can clarify this with the accompanying notes (notation a bit different):

That's a bit confusing, but trying to use some of it in my example, it seems like they are saying that the $-1$ in

$$z_-(x,y) = -1 z_+(x,y)$$

which can be found as a function of $n$ through something like

$$f(n) = e^{in \pi }$$

(for $n$ odd), in general this function lets us 'characterize' the fibers into 2 'classes', one representing the top of the sphere $S^2$ when $n$ is even giving us $z_+$, another representing the bottom half of the sphere when $n$ is odd giving us $z_-$.

Am I right in saying that this function tells us the bundle splits into two classes so that we've measured the deviation of local product structure from global product structure by noting you need to piece together two classes to form the global object?

How does this baby picture relate to grand statements like

Generally speaking, for a vector bundle on a manifold M, a characteristic class associates a cohomology class of M.

Characteristic classes are constructed as polynomials of the curvature $F = dA + A \wedge A$

In fact, a real characteristic class is a "total curvature," according to a well-defined relationship.

Characteristic classes are subsets of the cohomology classes of the base space and measure the non-triviality or twisting of a bundle. In this sense, they are obstructions which prevent a bundle from being a trivial bundle. Most of the characteristic classes are given by the de Rham cohomology classes.

In other words, if you put vector bundles on the sphere and start constructing things like Chern classes and set up curvatures and stuff, can those things be interpreted using my simple example here?

• Your mappings from the sphere to the disk do not constitute a fibre bundle: Interior points of the disk sit below two points of $S^{2}$, while boundary points sit below only one point. Instead, you want to look at trivial circle bundles over the disk, and ask how the fibres over the boundary in one copy of the disk are glued to fibres over the boundary of the other disk. This amounts to specifying a mapping $S^{1} \to S^{1}$, whose degree is (modulo sign) the first Chern class of the resulting circle bundle over $S^{2}$. (Haven't read your question any further than "Is this right so far?") – Andrew D. Hwang May 18 '16 at 23:05

While $S^2$ is perhaps the most convenient space for thinking about actual topology, some special features of $S^2$ seem to have thrown you for a loop. In particular, you're confusing the so-called clutching construction - which only works over spheres - with a fiber bundle. Andrew Hwang has illustrated why your construction is not a fiber bundle (the fibers at the equator are different), but I'd like to talk about the clutching construction a bit.

Let's remember the most important fact about vector bundles (and more general classes of bundles as well, in fact):

Fact 1: Every vector bundle over a contractible (nice) space is trivial.

You should know how to prove this, and what this result means intuitively. If you don't, please figure it out and then come back. This fact is pretty terrific - it tells us that what really matters in vector bundles are the transition maps!

So let's take some of the simplest examples of non-contractible spaces - the spheres! The spheres are super-convenient because they can be covered by two contractible sets. Write the 2-sphere as $S^2 = N \cup S$, where $N$ and $S$ are the north and south hemispheres. If $E \to S^2$ is a vector bundle (with structure group $G$), then we know $E|_N \to N$ and $E|_S \to S$ are trivial bundles. So all of the "topological information" must be contained in the transition functions $N \cap S \to G$. It's not hard to see that we only care about the space $N \cap S$ and the map $N \cap S \to G$ up to homotopy. But $N \cap S$ is homotopy equivalent to $S^1$ - our favorite topological space of all time!

(A note on structure groups: when I say structure group, you should think "automorphism group of a fiber". So for real vector bundles of rank $k$, this is just $GL_k\mathbb R$, and for complex vector bundles of complex rank $m$, this is just $GL_m\mathbb C$. It turns out that most of the discussion here works for principal $G$-bundles for nice Lie groups $G$.)

All that discussion, once made rigorous, and with a little more work for uniqueness and existence, gives us the following fact. (We assume all vector bundles come with an orientation; otherwise we've got a bit more work to do in the real-vector-bundle case.)

Fact 2, version 1: Vector bundles over $S^2$ are in one-to-one correspondence with homotopy classes of maps $S^1 \to G$.

Hold up - "homotopy classes of maps from $S^1$ into a space sounds familiar: isn't that just the fundamental group? So let's upgrade fact 2 again.

Fact 2, version 2: Vector bundles over $S^2$ are in one-to-one correspondence with elements of $\pi_1(G)$.

But what if we did this construction for $S^n$ instead of $S^2$? (Assume $n \ge 2$.) The same argument would go through, but our equatorial space would be $S^{n-1}$ - so we'd just get higher homotopy groups!

Fact 2, final version: Vector bundles over $S^n$ for $n \ge 2$ are in one-to-one correspondence with elements of $\pi_{n-1}(G)$.

How do we use Fact 2? Well, let's start with complex vector bundles. Any complex vector bundle of rank $k$ has structure group $GL_k\mathbb C$. So we need to find the homotopy groups of this space. It turns out that $U(k)$ is a deformation retract of $GL_k\mathbb C$, and since it's compact it's easier to work with.

Let's start with the easiest case: $k = 1$, so line bundles. $U(1)$ is another way to think about our favorite topological space - it's just $S^1$. So line bundles over $S^n$ for $n \ge 3$ are in one-to-one correspondence with elements of $\pi_{n-1}(S^1)$ - which is trivial (lift to the universal cover). So we've proved something nontrivial:

Fact 3: Every complex line bundle over $S^n$ for $n \ge 3$ is trivial.

But the real magic happens at $n = 2$. Here we get a classification by elements of $\pi_1(S^1) = \mathbb Z$. You've constructed some of these bundles by hand; all you need to know is that the Chern class is just the degree (or winding number) of the transition map $S^1 \to S^1$.

As an aside, Fact 2 is the main reason why physicists care about homotopy groups of $SO(k)$, $U(k)$, and $SU(k)$. We can reduce the (noncompact) $GL_k^+\mathbb R$ case of oriented real vector bundles to the (compact) $SO(k)$ calculation; there are several questions on this site that detail that equivalence. In the complex case, we can reduce (noncompact) $GL_k\mathbb C$ to (compact) $U(k)$. Then we can use a little algebraic topology (the long exact sequence in homotopy of a fibration) to reduce one step further to $SU(k)$.

Let's step back and think about the general theory of characteristic classes for a bit. Here's the most general recipe I know:

Fix $G$ a Lie group, and consider $G$-bundles over a space $X$. Construct $BG$, the classifying space of $G$, and compute its cohomology $H^{\bullet}(BG)$. The classifying space has the property that $G$-bundles over $X$ give rise to maps $X \to BG$, and two such maps are homotopic if and only if the $G$-bundles they arise from are isomorphic. Hence given any $G$-bundle $P \to X$ we get a map $X \stackrel{P}{\rightarrow} BG$; if $P$ is the trivial bundle then the induced map $X \stackrel{P}{\rightarrow} BG$ is null-homotopic.

Now take cohomology; this gives us a map $P^{\bullet}: H^{\bullet}(BG) \to H^{\bullet}(X)$. Then characteristic classes of $P$ are just the images in $H^{\bullet}(X)$ of generators of $H^*(BG)$. And if $P$ is trivial then the map is zero on cohomology so all characteristic classes must vanish - this is how characteristic classes provide an obstruction to a bundle being trivial.

For a concrete example, take complex line bundles, so $G = GL_1\mathbb C \simeq U(1)$. It turns out that $BU(1) = \mathbb{CP}^{\infty}$, and $H^{\bullet}(\mathbb{CP}^{\infty}) = \mathbb Z[x]$ as a ring, where $x$ has degree 2. So when we map into $H^{\bullet}(X)$, the image of $x$ is something in $H^2(X)$ - it's just $c_1(P)$, the first Chern class of your bundle!

In addition, all the higher Chern classes vanish in $H^{\bullet}(BU(1))$, just as the axioms would predict. And if you look at the cohomology of $BU(n)$ for larger $n$, you get the first $n$ Chern classes $c_1, \cdots, c_n$ as generators.

For (oriented) real vector bundles the story is similar. At a first pass, it's easier to work over $\mathbb{Z}/2$, and then $H^{\bullet}(BSO(n);\mathbb{Z}/2) = \mathbb{Z}/2[w_2, \cdots, w_n]$. (Here $w_1$ must vanish because we assume our bundles are oriented, hence orientable.) And if we do the hard work of lifting to integer coefficients, we get all those goofy Pontrjagin and Euler classes.

The last story that you're asking about - getting characteristic classes from curvature of connections - is called Chern-Weil theory. I don't know this story well enough to write a good post on it, but there are quite a few references that you can find via your favorite search engine. (There are even some posts on M.SE about it that should give you good starting places.)

• Any favourite texts to learn everything you just mentioned? I would very much like to have this much knowledge! – snulty May 25 '16 at 17:01
• Some combination of Milnor-Stasheff Characteristic Classes, Hatcher's book project Vector Bundles and K-Theory, the nLab, M.SE/MO, and other sources I can't remember. But the first two are really excellent. – Thurmond May 26 '16 at 20:07
• @Thurmond This is probably a stupid question, but, regarding Fact 2, Version 2, which element of $\pi_1(SO(2)) \cong \pi_1(S^1) \cong \mathbb{Z}$ corresponds to the the tangent bundle of $S^2$? (Here I'm using the fact that $G = GL_2(\mathbb{R})$ strong deformation retracts onto $SO(2)$ via the Gram-Schmidt process.) – Jeffrey Rolland Dec 24 '16 at 0:08
• (Reading through $\textit{Vector Bundles and K-Theory}$ by Hatcher, on page 22 it appears to say that 2∈ℤ≅[S1→G] is the characteristic class of $T(S^2)$; please feel free to correct me if this is wrong.) (Additional dumb question: Why does the vector field Hatcher chose on page 22 produce the correct clutching function? Why doesn't a vector field where all vector point parallel, forming parallels of lattitude for integral curves, work to produce the clutching function of the tangent bundle as well, for instance?) – Jeffrey Rolland Dec 24 '16 at 1:01