I was reading about the interpretation of Stiefel Whitney classes as obstructions from Milnor-Stasheff's book and I got stuck at a step. The context is the following. Let $E \to B$ be a vector bundle of rank $n$ and let $V_k(\mathbb{R}^n)$ be the Stiefel manifold, i.e. the manifold whose points are $k$-tuples of linearly independent vectors in $\mathbb{R}^n$. The main theorem I want to understand is that the reduction mod 2 of the obstruction class $c_j(E)$ is equal to the Stiefel Whitney class $w_j(E) \in H^j(B; \mathbb{Z}/2)$.

Let $\gamma^n \to G_n(\mathbb{R}^{n+1}) \cong \mathbb{R}P^n$ be the canonical bundle over the Grassmannian of $n$-planes in $\mathbb{R}^{n+1}$ and consider the bundle $V_1 (\gamma^n) \to G_n(\mathbb{R}^{n+1})$ whose fibers are $V_1(\mathbb{R}^n)$. They use the following fact in their proof:

"The obstruction cocycle of the bundle $V_1 (\gamma^n) \to G_n(\mathbb{R}^{n+1})$ clearly assigns to the $n$-cell of $\mathbb{R}P^n \cong G_n(\mathbb{R}^{n+1})$ a generator of the cyclic group $\pi_{n-1}(V_1(\mathbb{R}^n))=\pi_{n-1}(\mathbb{R}^n - 0) = \mathbb{Z}$"

I understand why $\pi_{n-1}(V_1(\mathbb{R}^n))=\mathbb{Z}$. The n-th obstruction cocycle of the bundle $V_1 (\gamma^n) \to G_n(\mathbb{R}^{n+1}) \cong \mathbb{R}P^n $ assigns to the n-cell of $\mathbb{R}P^n$ an element of $\pi_{n-1}(V_1 (\gamma^n))$. So if we compose with the map induced by restriction to the fiber we get an element of $\pi_{n-1}(V_1(\mathbb{R}^n))$. Why is this element a generator of such group?

Thank you!


For your bundle $V_1(\gamma^n) \to G_n(\mathbb R^{n+1})$, $V_1(\gamma^n)$ is the space $\{(L,v) : L \subset \mathbb R^{n+1}, v \in L, |v|=1, dim(L)=n \}$, and the map is "forgetting $v$".

I'm not sure if Milnor and Stasheff do it this way, but you can make a direct computation to justify their claim, using something like Schubert calculus techniques. Since $G_n(\mathbb R^{n+1}) \equiv G_1(\mathbb R^{n+1})$ by orthogonal complements, you can think of the bundle as

$$\{(L,v) : L \subset \mathbb R^{n+1}, dim(L)=1, v \perp L, |v|=1\}$$

Given a 1-dimensional subspace $L$ of $\mathbb R^{n+1}$, as long as it's not the $x$-axis, in the unit sphere orthogonal complement there is a unique vector $v$ such that it's x-component is minimal (most negative). Choose that vector. With some patience, you can think of this as a map from $D^n$ to this bundle, the idea is to think of $D^n$ as the $x=1$ plane in $\mathbb R^{n+1}$ suitably compactified with a point at infinity, and "blown up" along the x-axis. For every line $L$, you can ask where it intersects this disc (provided the line isn't the x-axis), and take the $v$ with minimal $x$-coordinate.

It's a fairly straightforward geometry exercise to show that for $L$ intersecting $x=1$ near the x-axis, it's a degree 1 map of spheres.

Does that sort of make sense? I imagine Milnor has a slicker argument but it's totally possible to "manhandle" this problem into submission.

  • $\begingroup$ It sort of makes sense, but I am still confused. Think of $G_n(\mathbb{R}^{n+1})$ as $\mathbb{R}P^n$. Let $x_0$ be a point in the middle of the $n$-cell $e^n$. A section of $\gamma^n$ which is non zero except at $x_0$ is given by $\{x,-x\} \mapsto x_0-<x_0,x>x$. This gives a section $s_{k-1}$ of $V_1(\gamma^n)$ over the $n-1$ skeleton $\mathbb{R}P^{n-1}$. Then, the obstruction cocylce takes $e^n$ to $\phi: \partial e^n=S^{n-1} \to \mathbb{R}P^{n-1} \to V_1(\mathbb{R}^n)=S^{n-1}$, where the 1st map is given by attaching and the 2nd by $s_{k-1}$. Why is $[\phi] \in \mathbb{Z}$ a generator?! $\endgroup$ – Manuel Jan 23 '12 at 4:53
  • $\begingroup$ This map that I'm describing, it's giving you a degree 1 map $S^{n-1} \to S^{n-1}$. But for the extension to exist, this map would have to be degree 0. $\endgroup$ – Ryan Budney Jan 23 '12 at 5:16
  • $\begingroup$ I understood my confusion. I'll make it clear: Consider the section $s_{k-1}$ described above. We can consider the characteristic map of $e^n$ as a map $F: S^{n-1} \times I \to \mathbb{R}P^n$ where $F_0$ is the attaching map and $F_1$ is the constant map to $x_0$. The section $s_{k-1}$ gives us an initial lift of $F$ so by the HLP we can lift $F$ to $F': S^{n-1} \times I \to V_1(\gamma^n)$. Since $F_1$ is constant, $\text{Im}F'_1$ is contained in the fiber above $x_0$. This gives a map $F'_1: S^{n-1} \to S^{n-1}$ of deg 1. Intuitively, $F'$ "approaches $x_0$ in each direction". $\endgroup$ – Manuel Jan 23 '12 at 21:11
  • $\begingroup$ Great. It sounds like you're on track. I think you can write out the homotopy from the HLP explicitly with enough effort. $\endgroup$ – Ryan Budney Jan 24 '12 at 2:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.