Obstruction cocycle of Stiefel manifold

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.

• 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?! – Manuel Jan 23 '12 at 4:53
• 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. – Ryan Budney Jan 23 '12 at 5:16
• 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". – Manuel Jan 23 '12 at 21:11
• Great. It sounds like you're on track. I think you can write out the homotopy from the HLP explicitly with enough effort. – Ryan Budney Jan 24 '12 at 2:04