# $\vec{v}(x)/|\vec{v}(x)|$ extends to the annulus bounded by the two spheres.

Guillemin and Pollack's Differential Topology Page 133:

$\vec{v}$ is a smooth map $\vec{v}: X \to \mathbb{R}^n$ such that $\forall x, \vec{v}(x) \in T_x(X)$. Assume that we are in $\mathbb{R}^k$ and that $\vec{v}$ has an isolated zero at the origin. The directional variation of $\vec{v}$ around $0$ is measured by the map $x \to \vec{v}(x)/|\vec{v}(x)|$ carrying any small sphere $S_\epsilon$ around $0$ into $S^{k-1}$. Choosing the radius $\epsilon$ so small that $\vec{v}$ has no zeros inside $S_\epsilon$ except at the origin, we define the index of $\vec{v}$ at 0, $\operatorname{ind}_0(\vec{v})$, to be the degree of this directional map $S_\epsilon \to S^{k-1}$.

As usual, the radius itself does not matter, for if $\epsilon^\prime$ is also suitable, then $\vec{v}(x)/|\vec{v}(x)|$ extends to the annulus bounded by the two spheres.

So I don't really get it here - why $\vec{v}(x)/|\vec{v}(x)|$ extends to the annulus bounded by the two spheres?

• I'd call it "shell" instead of "annulus". – Christian Blatter Jul 28 '13 at 9:38
• Oh I got it! So it actually means the sphere with greater radius, the sphere with smaller radius, and any sphere bounded between. Therefore it is called annulus, right @ChristianBlatter? – WishingFish Jul 28 '13 at 18:54

It's not so much that $\frac{v(x)}{\vert v(x) \vert}$ extends into the annulus bounded by $S_\epsilon$ and $S_{\epsilon'}$ as is that it's already defined there by virtue of the fact that $v(x)$ has no zero within the larger, hence also the smaller, of these two spheres.
• Sorry Robert, I wasn't clear. I thought it simply should be "$\vec{v}(x)/|\vec{v}(x)|$ extends to the sphere with greater radius..." – WishingFish Jul 28 '13 at 8:37