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 ${\mathbb{RP}^n}$. In particular, we want to compute the Stiefel-Whitney classes of the tangent bundle ${T(\mathbb{RP}^n)}$. The cohomology ring of ${\mathbb{RP}^n}$ with ${\mathbb{Z}/2}$-coefficients is very nice: it’s ${\mathbb{Z}/2[t]/(t^{n+1})}$. We’d like to find what ${w(T(\mathbb{RP}^n)) \in \mathbb{Z}/2[t]/(t^{n+1})}$ is.

On ${\mathbb{RP}^n}$, we have a tautological line bundle ${\mathcal{L}}$ such that the fiber over ${x \in \mathbb{RP}^n}$ is the set of vectors that lie in the line represented by ${x}$. Let’s start by figuring out the Stiefel-Whitney classes of this. I claim that

$\displaystyle w(\mathcal{L}) = 1+t \in H^*(\mathbb{RP}^n, \mathbb{Z}/2).$

The reason is that, if ${\mathbb{RP}^1 \hookrightarrow \mathbb{RP}^n}$ is a linear embedding, then ${\mathcal{L}}$ pulls back to the tautological line bundle ${\mathcal{L}_1}$ on ${\mathbb{RP}^1}$. In particular, by the axioms, we know that ${w(\mathcal{L}_1) \neq 1}$, and in particular has nonzero ${w_1}$. This means that ${w_1(\mathcal{L}) \neq 0}$ by the naturality. As a result, ${w_1(\mathcal{L})}$ is forced to be ${t}$, 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 ${w(\mathcal{L})}$. Still, this is a long way of what we want—${w}$ of the tangent bundle. Nonetheless, we are going to see that if ${\mathcal{L}^{\perp}}$ is the bundle over ${\mathbb{RP}^n}$ whose fiber over a line ${x}$ is the set of vectors perpendicular to that line, then

$\displaystyle T(\mathbb{RP}^n) \simeq \hom(\mathcal{L}, \mathcal{L}^{\perp}).$

To see this, let us argue as follows. ${\mathbb{RP}^n}$ is the quotient of ${S^n}$ by the map ${x \rightarrow -x}$. Consequently the tangent space to ${\mathbb{RP}^n}$ can be identified with a quotient of ${T(S^n)}$ by this flip map. Let ${v \in T_x(S^n) \subset \mathbb{R}^{n+1}}$; then the pair ${(x,v ) \in T(S^n)}$ gets sent to ${(-x, -v)}$ under the flip map. In short, to give a tangent vector to ${\mathbb{RP}^n}$ is the same thing as giving an unordered pair

$\displaystyle \{(x,v) , (-x, -v) \}$

where ${x \in S^n}$ and ${v \in \mathbb{R}^{n+1}}$ is perpendicular to ${x}$. But to give an element of the fiber of ${\hom(\mathcal{L}, \mathcal{L}^{\perp})}$ is the same thing. Indeed, the line through ${x}$ gets mapped to the line through ${\pm v}$—this is thus clear.

So we have gotten the equality ${T(\mathbb{RP}^n) \simeq \hom(\mathcal{L}, \mathcal{L}^{\perp}).}$ This is good because we know what ${w(\mathcal{L} )}$ is. However, we don’t know what ${w}$ does when ${\hom}$‘s of vector bundles are applied. Nonetheless, we can argue as follows. Add ${\hom(\mathcal{L}, \mathcal{L})}$ to both sides:

$\displaystyle T(\mathbb{RP}^n) \oplus \hom(\mathcal{L}, \mathcal{L}) = \hom(\mathcal{L}, \mathcal{L}^{\perp} \oplus \mathcal{L})$

In general, ${\hom(\mathcal{L}, \mathcal{L})}$ is always the trivial line bundle whenever you have a line bundle, ${\hom}$ sends pairs of line bundles to line bundles, and ${1_{\mathcal{L}}}$ gives a global nonvanishing section of ${\hom(\mathcal{L}, \mathcal{L})}$.

So if we denote ${\mathbb{R}}$ by the trivial bundle, we get

$\displaystyle T(\mathbb{RP}^n) \oplus\mathbb{R} = \hom(\mathcal{L}, \mathcal{L}^{\perp} \oplus \mathcal{L}) .$

But there’s more. Here ${\mathcal{L}}$ was taken naturally as a subbundle of the trivial bundle ${\mathbb{R}^{n+1}}$ over ${\mathbb{RP}^n}$, and ${\mathcal{L}^{\perp}}$ was the orthogonal complement. It follows that ${\mathcal{L} \oplus \mathcal{L}^{\perp} = \mathbb{R}^{n+1}}$. So in total, we find that

$\displaystyle T(\mathbb{RP}^n) \oplus \mathbb{R} = \hom(\mathcal{L}, \mathbb{R}^{n+1}) = \bigoplus_{n+1} \hom(\mathcal{L}, \mathbb{R}).$

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

$\displaystyle w(T(\mathbb{RP}^n)) = w(\hom(\mathcal{L}, \mathbb{R}))^{n+1}.$

So we’re almost there. The problem is, we know ${w(\mathcal{L})}$, but we need ${w(\hom(\mathcal{L}, \mathbb{R}))}$. The good news is that ${\mathcal{L}}$ is isomorphic to its dual ${\hom(\mathcal{L}, \mathbb{R})}$—the reason is that we can choose a metric on ${\mathcal{L}}$ (by paracompactness), and then the natural symmetric form

$\displaystyle \mathcal{L} \otimes \mathcal{L} \rightarrow \mathbb{R}$

given by this metric, leads to the isomorphism ${\mathcal{L} \simeq \mathcal{L}^{\vee}}$.

Finally, with this we are done. We find that

Theorem 1 We have$\displaystyle w(T(\mathbb{RP}^n)) = w(\mathcal{L})^{n+1} = (1+t)^{n+1}.$

So we have computed the Stiefel-Whitney classes of the tangent bundle to ${\mathbb{RP}^n}$. Now, we need to figure out when the projective spaces are parallelizable. Well, if ${\mathbb{RP}^n}$ were parallelizable, then the Stiefel-Whitney class to ${T(\mathbb{RP}^n)}$ is clearly the unit. In particular, we have

$\displaystyle (1+t)^{n+1} = 1 \in \mathbb{Z}/2[t]/t^{n+1}$

which is the same thing as

$\displaystyle (1+t)^{n+1} = 1+t^{n+1} \in \mathbb{Z}/2[t].$

The claim is that this is a very unusual event, and that in fact it can’t happen unless ${n+1}$ is a power of 2. Well, suppose in general that we had the equation

$\displaystyle (1+t)^{2^r p} = 1 + t^{2^r p} \in \mathbb{Z}/2[t].$

The claim is that then ${p=1}$. To see this, note first of all that

$\displaystyle (1+t)^{p} = 1 + pt + \dots$

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

$\displaystyle (1+t)^{p 2^r} = 1 + p^{2^r} t^{2^r} + \dots.$

The only way this can be equal to ${1+t^{2^r p}}$ is if ${p=1}$.

So we have proved:

Theorem 2 If ${\mathbb{RP}^n}$ is parallelizable, then ${n+1}$ is a power of ${2}$.

A standard application of this is to real division algebras. Let ${V}$ be a finite-dimensional ${\mathbb{R}}$-vector space. Suppose ${V}$ is a division algebra. That is, there is a ${\mathbb{R}}$-linear associative multiplication law

$\displaystyle V \times V \rightarrow V$

such that any element ${v \in V}$ has an inverse ${v^{-1}}$ with ${vv^{-1} = v^{-1}v = 1}$.

Theorem 3 Any 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 ${V \otimes_{\mathbb{R}} V \rightarrow V}$. Moreover, it is even smooth. We can make ${\mathbb{RP}(V)}$, the projective space over ${V}$, 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 ${\mathbb{RP}(V)}$ is parallelizable.

This can only happen if ${\dim V}$ 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).