# Proof about sum of indices of singular points

In Lee's Topological Manifolds, problem 8-10 discusses vector fields in the plane, and defines them as continuous maps $v:\mathbb{R}^2\rightarrow\mathbb{R}^2$. An isolated singular point is a point $p$ where $v(p)=0$, but there is an open set $p\in U$ so that $v^{-1}(0)\cap U=\{p\}$.

The first two parts of the problem show we can define the index of $p$ as follows: find a small circle $C\in U$ around $p$, and consider the map $v: C\rightarrow\mathbb{R}^2-\{0\}$. The index of $p$ is then the winding number of this map.

The third part of the problem is as follows (and where my question lies): Suppose $p_1,\ldots,p_n$ are all the isolated singular points of $v$, all contained in the open unit disk of $\mathbb{R}^2$. Let $S^1$ be the boundary of the unit disk. Show the sum of the indices of the $p_i$ is equal to the winding number of $v(S^1)$.

I have this idea that I could "shrink-wrap" $S^1$ down to loops around the $p_i$, and then show the parts connecting the loops cancel each other out, winding number-wise. But this is pretty hand-wavy. Is there a more formal way to prove this? I'm happy even to get a hint in the right direction.

(Sorry for all the setup, but there are lots of ways to define winding number, and I didn't want to use something not discussed in the book)

As Lee defines, if $v$ has an isolated zero at $p$, and if $C$ is a circle of radius $\varepsilon$ around $p$ in $\Bbb R^2$, then $v|_C : C \to \Bbb R^2 \setminus 0$ is a loop around the origin, and the index of $v$ at $p$ is precisely the winding number of $v|_C$ given by the formula (where the division is defined after identifying $\Bbb R^2$ with $\Bbb C$)

$$\iota_p(v) = \frac{1}{2\pi i} \int_C \frac{v'(s)}{v(s)} ds$$

It's worth noting that this formula works because, roughly, "$v'(s)/v(s) ds = d\log v(s)$" where $\log$ is a multivalued function defined on $\Bbb C \setminus 0$ which changes it's argument by $\pm 2\pi i$ if one goes a circle counterclockwise or clockwise around the origin. Therefore $\iota_p(v)$ is a sum over $\pm 1$'s corresponding to counterclockwise or clockwise rotations of $v|_C$ around the origin, which sums to the total number of counterclockwise rotations of $v|_C$ around $0$. Notice also that by the change of variables formula we could write $$\frac{1}{2\pi i} \int_C \frac{v'(s)}{v(s)} ds = \frac{1}{2\pi i}\int_{\gamma} \frac{dz}{z}$$ where $\gamma$ is the loop $v(C)$ parameterized by the map $v|_C$.

Suppose now that $p_1, \cdots, p_n$ are isolated zeroes of $v$ contained in the disk $D^2$. Let $C_1, \cdots, C_n$ be circles of radius $\varepsilon$ around these zeroes contained in $D^2$, which we can manage to do by choosing $\varepsilon > 0$ appropriately. Further, let's go by your suggestion in the question and delete arcs $\mathcal{A}_i$ and $\mathcal{A}'_i$ on $\partial D^2$ and $C_i$ respectively and add in tubes $T_i$ connecting $\partial D^2$ and $C_i$ by inserting it between the deleted arcs, roughly like the following picture (I haven't deleted the arcs while drawing, oops):

Then we have a loop $\ell$ given by the union of all these little pieces (i.e., traversing $\partial D^2 - \bigcup_i \mathcal{A}_i$, the tubes $T_i$ and then the circles $C_i - \mathcal{A}_i'$, in counterclockwise order) which does not contain any zero of $v$ inside it by construction. Therefore the winding number of $v$ along $\ell$ is $$\frac{1}{2\pi i} \int_{\ell} \frac{v'(s)}{v(s)} ds = \frac{1}{2\pi i} \int_{v(\ell)} \frac{dz}{z} = 0$$ by Cauchy's theorem as $1/z$ is holomorphic inside $v(\ell)$, which is a loop that does not contain the origin. We can expand the integral out as follows (here the integral over the circles $C_i$ gets a sign because if $\ell$ is traversed counterclockwise then $C_i$ is oriented clockwise)

$$\frac{1}{2\pi i} \left ( \int_{\partial D^2 - \bigcup_i \mathcal{A}_i} \frac{v'(s)}{v(s)} ds + \sum_{i} \int_{T_i} \frac{v'(s)}{v(s)} ds + \sum_{i} -\int_{C_i - \mathcal{A}_i'} \frac{v'(s)}{v(s)} ds \right ) = 0$$

If we write $T_i = I^1_i \cup I^2_i$ where $I^1_i$ and $I_i^2$ are the parallel lines that constitute the tube (oppositely oriented), then as we decrease the width of the tube i.e., let $\delta \to 0$, we get

$$\int_{T_i} \frac{v'(s)}{v(s)} ds = \int_{I^1_i} \frac{v'(s)}{v(s)} ds - \int_{I^2_i} \frac{v'(s)}{v(s)} ds \to 0$$

Therefore we can take the middle term to vanish and as letting $\delta \to 0$ also shrinks the removed arcs $\mathcal{A}_i, \mathcal{A}_i'$, we can forget about them too. Therefore rearranging the rest gives $$\frac{1}{2\pi i} \int_{\partial D^2} \frac{v'(s)}{v(s)} ds = \frac1{2\pi i} \left (\int_{C_1} + \cdots + \int_{C_n} \right ) \frac{v'(s)}{v(s)} ds$$ which proves that the winding number of $v$ along $\partial D^2$ is $\iota_{p_1}(v) + \cdots + \iota_{p_n}(v)$, as required.

• @BalarkaSan: Very nice presentation. (+1) – Markus Scheuer Feb 8 '18 at 18:41
• Wow, thank you! I actually understand what is going on (finally). Just for my own clarity: I think we can do this without integrals. $v(\ell)$ is a loop that has an extension to a disk, so by a previous exercise, has winding number zero. And it's pretty easy to see that $\delta D^2$, union the $T_i$, is homotopic to $\delta D^2$, so gives the same winding number $v(\delta D^2)$. Finally, I think I can show that concatenating these loops like this is just adding up the respective winding numbers. Thanks again! – Hempelicious Feb 8 '18 at 21:48
• @Hempelicious Well, it depends on how you define the winding number. I checked Lee and problem 8-9 seems to define it as an integral. So to prove that the winding number of $v$ around a loop that extends to a disk without zeroes of $v$ in it is zero you still have to use something like Stokes' theorem (or Green's theorem, in particular) somewhere. I have secretly used that as Cauchy's theorem in concluding $\int dz/z = 0$ over $v(\ell)$. I can add a note on the differential forms interpretation of this proof if you're familiar with those (and if you want me to do so) – Balarka Sen Feb 8 '18 at 21:52
• Oh sorry, Lee defines the winding number of $f(S^1)$ as the degree of the composition $r\circ f$, where $r$ retracts the punctured plane to the unit circle. – Hempelicious Feb 8 '18 at 22:06
• @Hempelicious Ah... well, yes, then it's a lot easier, as you said. Basically boils down to a lemma that if $f : D^2 \to S^1$ is a map, $f|\partial D^2 : S^1 \to S^1$ has degree zero. Sorry for assuming it was defined otherwise. But there's actually an issue; I don't know how to say that the winding number of $v(\ell)$ is indeed the sum of the winding numbers of $v(\partial D^2)$ and sum of (signed) $v(C_i)$'s purely using fundamental groups. You can't introduce those tubes without uttering the words "homologous" in the process. So I think there's a reason Lee introduces the integral in 8-9 – Balarka Sen Feb 8 '18 at 23:03