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).