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?

  • $\begingroup$ I'd call it "shell" instead of "annulus". $\endgroup$ – Christian Blatter Jul 28 '13 at 9:38
  • $\begingroup$ 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? $\endgroup$ – 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.

  • $\begingroup$ Thanks Robert. Yes I get what you said, but why annulus..? $\endgroup$ – WishingFish Jul 28 '13 at 7:57
  • $\begingroup$ @WishingFish: not too sure I get the jist of your question, "why annulus?" That's the name of such regions between two concentric spheres as I know it. $\endgroup$ – Robert Lewis Jul 28 '13 at 8:07
  • $\begingroup$ 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..." $\endgroup$ – WishingFish Jul 28 '13 at 8:37
  • $\begingroup$ @WishingFish: same difference though, no? You can say it whichever way you like! $\endgroup$ – Robert Lewis Jul 28 '13 at 8:42
  • $\begingroup$ NO. I thought annulus is a ring without center circle. /_\ Or.. it actually means the sphere with greater radius, and the sphere with smaller radius, right? Now it all makes sense :D $\endgroup$ – WishingFish Jul 28 '13 at 18:53

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