# Why are Stiefel-Whitney classes not represented in de Rham cohomology?

Stiefel-Whitney classes of a real vector bundle $(E, \pi, M)$ are defined axiomatically in singular cohomology $H^*(M, \mathbb{Z}/2\mathbb{Z})$ (compare wiki).

Now various books state, that Stiefel-Whitney classes are not represented in de Rham cohomology (e.g. "The Stiefel Whitney classes are not de Rham cohomology classes" compare click).

Why is this the case? Naivly I would think, that the composition of the coefficient homomorphism with the de Rham iso $H^*(M, \mathbb{Z}/2\mathbb{Z}) \rightarrow H^*(M, \mathbb{R}) \rightarrow H_{dR}(M)$ should map Stiefel-Whitney classes on classes of de Rham cohomology satisfying the axioms of Stiefel-Whitney classes. Why does this fail? Can you give a reference?

• Do you know what $H^*(\Bbb RP^2, \Bbb R)$ is? – PVAL-inactive Jan 13 '17 at 15:13

## 2 Answers

There is no (nonzero) map $f : H^*(M; \mathbb{Z}/2\mathbb{Z}) \to H^*(M; \mathbb{R})$ as you claimed. The field $\mathbb{Z}/2\mathbb{Z}$ has characteristic $2$, but the field $\mathbb{R}$ has characteristic $0$. In $H^*(M; \mathbb{Z}/2\mathbb{Z})$, all the elements $\alpha$ satisfy $2 \alpha = 0$. So under your map $f$ you get $2 f(\alpha) = 0$, but over $\mathbb{R}$ you can divide by $2$ and you get $f(\alpha) = 0$...

More abstractly, the universal Stiefel-Whitney class $w_n$ lives in $H^n(\operatorname{Gr}_n(\mathbb{R}^\infty); \mathbb{Z}/2\mathbb{Z})$, the $n$th cohomology group of the Grassmanian of $n$-subspaces in $\mathbb{R}^\infty$. All the other Stiefel-Whitney classes $w_n(\xi)$ are pullbacks of this one. So if you wanted "de Rham" Stiefel-Whitney classes, you'd be looking at $H^n(\operatorname{Gr}_n(\mathbb{R}^\infty); \mathbb{R})$. But for example $\operatorname{Gr}_1(\mathbb{R}^\infty)$ is the infinite-dimensional real projective space $\mathbb{RP}^\infty$, and $H^1(\mathbb{RP}^\infty; \mathbb{R}) = 0$.

• Sorry, I have not a very good algebraic topological background. I understood why $H^n(\mathbb{R}P^\infty; \mathbb{Z})= 0$. But why is $H^n(\mathbb{R}P^\infty; \mathbb{R} ) = 0$? – warpfel Jan 27 '17 at 11:46
• One should be able to see it with the universal coefficient theorem in algebraic topology. – Herman Chu Nov 6 '18 at 6:04

Let's look at $w_1$: it takes 1-cycles to numbers $0$ or $1$. What, in $H^1(M, \Bbb R)$ would be the cocycle $W$ corresponding to that? In particular, for surfaces, $w_1$ measures non-orientability: if $c$ is a curve that reverses orientation, then $w_1(c) = 1$. So for $RP^2$, $w_1$ is nonzero, but what will $W$ be? Since $H^1(RP^2, \Bbb R) = 0$, $W$ pretty much can't be anything interesting.

@Najib's answer makes this simple example more general.