Suppose we have a vector field v on $\Bbb R^n$ with exactly two isolated zeros (call them $p,q$) which are connected by a flow-line of the vector field. Furthermore, assume one can modify the vector field in a compact nbhd of the flow line and merge $p,q$ to give a new vector field w with a single isolated zero $r$. Show that the index of $r$ must be the sum of the indices of $p$ and $q$.

My approach: my first attempt would be to show that there is a relation between the vector field v and the vector field w. If we where working on a compact manifold I believe this exercise is an easy corollary of the Poincaré-Hopf theorem. But since we are working with a non-compact manifold, I can't use it here.

Here was asked the same question: Show that the index of $r$ must be the sum of the indices of $p$ and $q$.

but with no answer, and actually it's pointed out that the solution of this is quite complicated. It's mentioned that it's a consequence of the Boundary theorem but the only boundary theorem I know is about the incidence number of two sub-manifolds one of which bounds a mfld.


Let $S^{n-1}$ be the unit $n-1$ sphere centered on the origin. Let $S_p,S_q$ be a small $n-1$ spheres centered on $p,q$ respectively, so small $S_p,S_q$ are disjoint and each is contained in the outside of the other. Let $S_{pq}$ be a large $n-1$ sphere whose inside contains $S_p,S_q$. Choose orientations on all three spheres so that the normal direction points outward from the sphere.

Since each of these spheres is disjoint from the set of zeroes $\{p,q\}$, it follows that the vector field $\bf{v}$ defines continuous functions $f_p : S_p \to S^{n-1}$ and $f_q : S_q \to S^{n-1}$ and $f_{pq} : S_{pq} \to S^{n-1}$. These functions are all just restrictions of the same function $x \mapsto \frac{{\bf{v}}(x)}{|{\bf{v}}(x)|}$ for $x \in \mathbb{R}^n - \{p,q\}$.

By definition, the index of $\bf{v}$ at $p$ equals the degree of the map $f_p$, and the index of $\bf{v}$ at $q$ equals the degree of the map $f_q$.

Next, there is a "boundary theorem" to apply here, although the one you know might not be the same as the one we need. The boundary theorem one needs says that if you have a compact, oriented $n$-manifold with boundary $M$ --- in this case $M$ is the portion of $\mathbb{R}^n$ outside of $S_p$ and $S_q$ and inside of $S_{pq}$ --- and if the boundary $\partial M$ has the induced orientation with normal vector pointing outward from the manifold, and if you have a continuous function from $M$ to $S^{n-1}$ --- in this case, the restriction of the function $x \mapsto \frac{{\bf v}(x)}{|{\bf v}(x)|}$ --- then the sum of the degrees of the maps on the boundary components equals zero. It follows that degree of $f_{pq}$ equals the sum of the degrees of $f_p$ and $f_q$, because we compute the degree on $f_{pq}$ using the orientation induced from $M$ but we compute the degrees of $f_p$ and $f_q$ using the opposite orientation from the one induced from $M$. Thus the degree of $f_{pq}$ equals the sum of the indices of $\bf{v}$ at $p$ and $q$.

Now modify $\bf{v}$ as described to obtain $\bf{w}$. This "modification" is a homotopy from $\bf v$ to $\bf w$, and $S_{pq}$ is disjoint from the support of this homotopy, so the function $f_{pq}$ is unchanged under this homotopy. When the homotopy is complete, the point $r$ is inside the sphere $S_{pq}$, and so, again by definition, the index of $\bf w$ at $r$ is equal to the degree of $f_{pq}$.

Putting it all together, the index of $\bf{w}$ at $r$ equals the sum of the indices of $\bf{v}$ at $p$ and at $q$.

By the way, this point of view on vector fields is well explained in the book of Guilleman and Pollack, which I recommend for more details.

  • $\begingroup$ I see, thanks for the careful description. I had in mind an approach similar to this one, but I was quite confused by the fact that they stressed the existence of this flow line connecting the two points. is there a reason for that? what if such line doesn't exists? $\endgroup$ – Luigi M Aug 22 '17 at 16:29
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    $\begingroup$ All you need to make this argument work is a sphere such that the only zeroes of the vector field on or inside the sphere are $p$ and $q$. The sphere does not even have to be an ordinary round sphere, it can be just a smoothly embedded sphere in $\mathbb{R}^n$. The only role being played by the flow line connecting the points is that any neighborhood of that flow line connecting $p$ and $q$ contains an appropriate sphere of this type. But, you could carry that argument out with any embedded path connecting $p$ and $q$. $\endgroup$ – Lee Mosher Aug 22 '17 at 18:37

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