Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The presentation of the homology version of Cauchy's theorem in Ahlfors is slick, but sweeps a lot of the topology under the rug using clever arguments. This question is an attempt to reconcile Ahlfors' analytic notion of a curve being homologous to zero (presented in his book Complex Analysis and originally due to E. Artin, I believe) with the standard definition in homology as found in Hatcher.

We work in the complex plane and fix $a\in \mathbb C$. Let $\gamma$ be a continuous map $[0,1]\rightarrow C\backslash \{ a\}$. Following Munkres in his book Topology, we define the winding number of $\gamma$ with respect to the point $a$ by considering


This is clearly a loop in $S^1$ and corresponds to some multiple of of the generator of the fundamental group of $S^1$. If the generator is $\tau$ and $g(t)$ corresponds to $m\tau$, $m\in\mathbb Z$, we define the winding number $n(\gamma, a)$ to be $m$. This is the definition of winding number I will use in this question, but in case $\gamma$ is piecewise differentiable, it corresponds to the analytic definition by integration given in Ahlfors.

Ahlfors calls a curve contained in an open region $\Omega$ homologous to zero if $n(\gamma,a)=0$ for all $a\in \Omega^c$.

In homology theory, as I understand it, we would call a curve homologous to zero if it represents the zero element in $H_1(\Omega, \mathbb Z)$. That is, $\gamma$ is the boundary of some singular $2$-chain.

My question: Why do all of these notions agree? Why is Ahlfors' definition of being homologous to zero (using Munkres' definition of winding number) agree with the usual homological one?

I would like to use Munkres' definition because it works for continuous curves, not just piecewise differentiable ones, and it seems to me the equivalence should hold in this generality.

Edit: This result appears as proposition 1.9.13 in Berenstein and Gay's book on complex analysis.

share|cite|improve this question
I haven't looked at Ahlfors in awhile. Does the superscript $c$ mean closure (why doesn't he use a bar)? Also, do you mean $H_1(\Omega, \mathbb{Z})$ rather than the cohomology group? – Matt May 22 '13 at 3:08
@Matt It means complement. (Is there a less confusing notation I could use?) And yes, it should be a subscript. Let me fix that. – Potato May 22 '13 at 3:10
Sorry. That's what I originally thought it meant, but for some reason became convinced it had to mean something else. I was being dumb and what I thought was a counterexample really wasn't. – Matt May 22 '13 at 3:30

Imagine the curve as made of segments parallel to the coordinate axes. Create a grid on a large rectangle containing the curve in its boundary and interior so that every segment of our curve is an edge of some subrectangle. Now define a singular $2$-chain by using each subrectangle with the coefficient given by the winding number of our curve around the center of the subrectangle. Check that the boundary of this $2$-chain is our original curve.

share|cite|improve this answer
It tells you whether each subrectangle of your grid is "inside" the curve or not (and with multiplicities, too). We need a blackboard to draw pictures :) Draw a few sample curves and try it. Note that Ahlfors's definition says that the region can't have any holes inside the curve. – Ted Shifrin May 22 '13 at 3:27
P.S. And to get the other direction in the equivalence, use the integral formula for winding number and Stokes's Theorem. – Ted Shifrin May 22 '13 at 3:43
The integral formula for winding number won't work if the curve is merely continuous and not piecewise differentiable. – Potato May 22 '13 at 3:44
(Although I suppose we could attempt an approximation by a piecewise differentiable curve? I will need to think more.) – Potato May 22 '13 at 3:45
Well, actually, it'll work for rectifiable curves, but, personally, I'm happy to stay in the piecewise-$C^1$ setting. Yes, you should be able to make various approximation arguments (e.g., with mollifiers). – Ted Shifrin May 22 '13 at 4:04
up vote 2 down vote accepted

This result appears as proposition 1.9.13 in Berenstein and Gay's book on complex analysis.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.