The first basic example of characteristic classes are the Stiefel-Whitney classes. Given a (real) ${n}$-dimensional vector bundle ${p: E \rightarrow B}$, the Stiefel-Whitney classes take values in the cohomology ring ${H^*(B, \mathbb{Z}/2)}$. They can be used to show that most projective spaces are not parallelizable.

So how do we get them? One way, as discussed last time, is to compute the ${\mathbb{Z}/2}$ cohomology of the infinite Grassmannian. This is possible by using an explicit cell decomposition into Schubert varieties. On the other hand, it seems more elegant to give the axiomatic formulation. That is, following Milnor-Stasheff, we’re just going to list a bunch of properties that we want the Stiefel-Whitney classes to have.

Let ${p: E \rightarrow B }$ be a bundle. The Stiefel-Whitney classes are characteristic classes ${w_i(E) \in H^i(B, \mathbb{Z}/2)}$ that satisfy the following properties.

First, ${w_i(E) = 0}$ when ${i > \dim E}$. When you compute the cohomology of ${\mathrm{Gr}_n(\mathbb{R}^{\infty})}$, the result is in fact a polynomial ring with ${n}$ generators. Consequently, we should only have ${n}$ characteristic classes of an ${n}$-dimensional vector bundle. In addition, we require that ${w_0 \equiv 1}$ always.

Second, like any characteristic class, the ${w_i}$ are natural: they commute with pulling back. If ${E \rightarrow B}$ is a bundle, ${f: B' \rightarrow B}$ is a map, then ${w_i(f^*E) = f^* w_i(E)}$. Without this, they would not be very interesting.

Third, they satisfy a product formula. Namely, if ${E, E'}$ are two bundles over ${B}$, then we have

$\displaystyle w_i(E \oplus E') = \sum_{j+k = i} w_j(E) \cup w_k(E'),$

which can be interpreted in the following way. Given any bundle ${E}$, we can form its total Stiefel-Whitney class ${w(E) = \sum w_i(E) \in H^*(B,\mathbb{Z}/2)}$. Then the formula becomes simply

$\displaystyle w(E \oplus E') = w(E) w(E').$

It follows that ${w}$ is a homomorphism from the semigroup of real vector bundles on ${E}$ into the units ${H^*(B, \mathbb{Z}/2)}$. The latter is a commutative group since we are using ${\mathbb{Z}/2}$-coefficients. Since ${w(E)}$ is always invertible, as ${w_0(E) \equiv 1}$, this in particular induces a map from the real ${K}$-group ${KO(B) \rightarrow H^*(B, \mathbb{Z}/2)}$.

Fourth, characteristic classes are supposed to measure the extent to which vector bundles aren’t trivial. So we require that ${w(E) = 1}$ for a trivial bundle.

Finally, we need one axiom to ensure that these classes are actually interesting in some way (i.e. not the ones with ${w(E) \equiv 1}$ for all ${E}$). For this, let us consider the tautological line bundle ${\mathcal{L}}$ over ${\mathbb{RP}^1}$. By definition, the fiber over a point ${x \in \mathbb{RP}^1}$ is the set of points lying in the line represented by ${x}$. We require that ${w_1(\mathcal{L}) \neq 0}$, so it is the unique nonzero element of ${H^1(\mathbb{RP}^1, \mathbb{Z}/2)}$.

Alright. So we have said that we want characteristic classes of this form. Unfortunately there doesn’t seem to be a very easy way to get these Stiefel-Whitney classes. Milnor-Stasheff get them via the Steenrod squaring operations combined with the Thom isomorphism.

So for now, let’s just quote:

Theorem 1 Stiefel-Whitney classes satisfying the above properties exist.

In fact, the above properties actually determine the Stiefel-Whitney classes, as one can check by doing some computations with the infinite Grassmannian.

What can we do with this? Well, the first thing I want to do in the next post is talk about when ${\mathbb{RP}^n}$ is parallelizable. But let us start with a simple example computation.

Namely, let us consider the tangent bundle ${TS^n \rightarrow S^n}$ and compute the Stiefel-Whitney classes. Note that ${S^n \subset \mathbb{R}^{n+1}}$ and consequently there is a normal bundle ${N \rightarrow S^n}$ with fiber over ${x \in S^n}$ consisting of vectors in ${\mathbb{R}^n}$ which are perpendicular to ${T_x(S^n)}$. By definition, ${TS^n \oplus N}$ is just the trivial bundle ${\mathbb{R}^{n+1} \times S^n \rightarrow S^n}$.

In particular, we find

$\displaystyle w(N) w(TS^n) = 1 \in H^*(S^n, \mathbb{Z}/2).$

But in fact ${N}$ is a trivial line bundle. The reason is that we have a globally defined nonzero section, because for each ${x \in S^n}$, the vector ${x}$ itself is in the fiber of ${N}$ over ${x}$. So the map ${x \rightarrow x}$ makes ${N}$ a trivial line bundle. Thus ${w(N) = 1}$.

In particular, we find

$\displaystyle w(TS^n) = 1,$

which means that Stiefel-Whitney classes are insufficient, at least by themselves, for determining when the sphere is parallelizable.