Last time we gave the axiomatic description of the Stiefel-Whitney classes. Today, following Milnor-Stasheff, we want to look at what happens in the particular case of real projective space . In particular, we want to compute the Stiefel-Whitney classes of the tangent bundle . The cohomology ring of with -coefficients is very nice: it’s . We’d like to find what is.

On , we have a tautological line bundle such that the fiber over is the set of vectors that lie in the line represented by . Let’s start by figuring out the Stiefel-Whitney classes of this. I claim that

The reason is that, if is a linear embedding, then pulls back to the tautological line bundle on . In particular, by the axioms, we know that , and in particular has nonzero . This means that by the naturality. As a result, is forced to be , and there can be nothing in other dimensions since we are working with a 1-dimensional bundle. The claim is thus proved.

OK. So we have computed . Still, this is a long way of what we want— of the tangent bundle. Nonetheless, we are going to see that if is the bundle over whose fiber over a line is the set of vectors perpendicular to that line, then

To see this, let us argue as follows. is the quotient of by the map . Consequently the tangent space to can be identified with a quotient of by this flip map. Let ; then the pair gets sent to under the flip map. In short, to give a tangent vector to is the same thing as giving an unordered pair

where and is perpendicular to . But to give an element of the fiber of is the same thing. Indeed, the line through gets mapped to the line through —this is thus clear.

So we have gotten the equality This is good because we know what is. However, we don’t know what does when ‘s of vector bundles are applied. Nonetheless, we can argue as follows. Add to both sides:

In general, is always the trivial line bundle whenever you have a line bundle, sends pairs of line bundles to line bundles, and gives a global nonvanishing section of .

So if we denote by the trivial bundle, we get

But there’s more. Here was taken naturally as a subbundle of the trivial bundle over , and was the orthogonal complement. It follows that . So in total, we find that

Now we can apply to both sides. Since is just 1 for a trivial bundle, we find by the multiplicative property that

So we’re almost there. The problem is, we know , but we need . The good news is that is isomorphic to its dual —the reason is that we can choose a metric on (by paracompactness), and then the natural symmetric form

given by this metric, leads to the isomorphism .

Finally, with this we are done. We find that

Theorem 1We have

So we have computed the Stiefel-Whitney classes of the tangent bundle to . Now, we need to figure out when the projective spaces are parallelizable. Well, if were parallelizable, then the Stiefel-Whitney class to is clearly the unit. In particular, we have

which is the same thing as

The claim is that this is a very unusual event, and that in fact it can’t happen unless is a power of 2. Well, suppose in general that we had the equation

The claim is that then . To see this, note first of all that

where the dots denote higher terms; since squaring is a homomorphism modulo 2, we find

The only way this can be equal to is if .

So we have proved:

Theorem 2If is parallelizable, then is a power of .

A standard application of this is to real division algebras. Let be a finite-dimensional -vector space. Suppose is a **division algebra**. That is, there is a -linear associative multiplication law

such that any element has an inverse with .

Theorem 3Any finite-dimensional real division algebra has dimension a power of two.

The idea of the proof is to note that the multiplication map is necessarily continuous, since it is equivalent to a linear (and thus continuous) map . Moreover, it is even smooth. We can make , the projective space over , into a Lie group using the division algebra structure: the product of two lines through the origin is a line through the origin. But any Lie group is parallelizable, so is parallelizable.

This can only happen if is a power of 2, as we have seen.

As is known, the only associative examples are in dimensions 1, 2, 4 (the reals, the complexes, and the quaternions).

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