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Many textbooks contain statements and proofs of the Jordan curve version, the homology version or the homotopy version of Cauchy's integral formula. In the first two cases, corresponding versions of the residue theorem are often also prsented, but apparently, no textbook states the residue theorem in terms of homotopy. If my impression is correct, why are homology or Jordan curve relevant to both theorems but homotopy is relevant only to one of them? What is the decisive difference between homotopy and the other two notions?


Background. Earlier this year, I was trying to re-learn complex analysis at undergraduate level from Cartan's Elementary Theory of Analytic Functions of One or Several Complex Variables. In his book, two versions of Cauchy's integral formula are presented. The first one, a homotopic form, is based on a notion of "primitive of a differential $1$-form along a homotopy". Its development and proof are very elegant. The assumptions in the theorem statement are amazingly weak: the closed curves are only required to be continuous; they are not required to be rectifiable or piecewise $C^1$.

Just as I was exclaiming in delight about the elegance of this homotopic form, Cartan presented an "alternative form of Cauchy's integral formula" in §2.II.8, theorem 5 (p.74):

Let $\Gamma$ be the oriented boundary of a compact subset $K$ of an open set $D$ and let $f(z)$ be a holomorphic function in $D$. Then, $$ \int_\Gamma f(z)dz=0; $$ if, moreover, $a$ is an interior point of $K$, then $$ \int_\Gamma \frac{f(z)dz}{z-a}=2\pi i f(a). $$

To my dismay, the proof of this alternative form relied on completely different machineries. In particular, the proof ultimately relied on the implicit function theorem as well as Green's theorem. The application of the former is so weird (at least to a non-expert like me) that I needed to ask a question on this site to understand what was going on, and the latter was stated without proof. Thus it seemed to me that the proof of this alternative form (or the book) wasn't self contained. Worse still, not only had Cartan not explained any relationship between the two versions of Cauchy's integral formula, but his proof of the residue theorem also relied on this alternative form rather than the homotopical one. At this point, I realized that I had never seen any textbook on complex analysis that presented a homotopical version of the residue theorem, and I wondered why.

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  • $\begingroup$ "apparently none of them states the residue theorem in terms of homotopy" and " but homotopy is relevant only to the residue theorem"? $\endgroup$
    – Paul Frost
    Nov 7, 2021 at 23:15
  • $\begingroup$ @PaulFrost Thanks. That of course is a typo. $\endgroup$ Nov 7, 2021 at 23:18
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    $\begingroup$ One issue that muddles the whole thing is that contour integrals are (traditionally) only defined on piecewise $C^1$ paths, while homotopy of paths only requires continuity. Various formulations of "homology" that do not attempt direct connections to homotopy can approach this in a different way... $\endgroup$ Nov 7, 2021 at 23:23

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While in Cauchy’s integral theorem, the two yields the same result, the truth is homotopy is just keeping too much informations for more complicated structures, which will nonetheless be lost when we integrate.

The simplest case to see the difference between homotopy and homology can be found in a plane with two holes. Its homology group is $\mathbb{Z}\times\mathbb{Z}$ abelian group of free rank 2. It keeps track of two numbers, meaning how many times the curve loops around hole A and how many times hole B. In contrast, its fundamental group is the free group of rank 2, non-abelian. It remembers in what order the curve loops around the holes.

Suppose we integrate over a curve $\gamma=\alpha+\beta$, where $+$ denotes the joining of curves. The integration can always be decomposed into two integrations, over $\alpha$ and over $\beta$. The operation that adds these two integrations, namely, the addition, is commutative. It is exactly this commutativity that filters out the information of the ordering that homotopy remembers.

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In the homotopical version we consider closed curves which are homotopic to zero, in the homological version cycles which are homologous to zero.

Each closed curve is a cycle, and if a closed curve is homotopic to zero, then it is also homologous to zero. This shows that the homotopical version of the residue theorem is just a special case of the homological version.

Update:

The OP asks why the homotopical version of Cauchy's integral theorem (= CIT) appears in textbooks, but not the homotopical version of the residue theorem.

I cannot say anything about the motivation of the various authors, but I think it has purely technical reasons. In my opinion the standard aproach goes essentially step by step as follows:

  1. Prove the CIT for the boundary curve of a rectangle (or triangle) contained in the domain of $f$.

  2. Generalize to arbitrary curves in nice domains like open disks or star-shaped domains.

  3. Generalize to arbitrary curves which are homotopic to zero (in arbitrary domains). This is the homotopical CIT which covers in particular the CIT for simply connected domains.

  4. Generalize to cycles which are homologous to zero. This is the homological CIT.

Perhaps one could shorten this sequence of steps, but for technical and didactical reasons it seems to be appropriate as it is.

The residue theorem is treated after the CIT. At that point the machinery of homology has been introduced, and there is no reason to start with curves which are homotopic to zero and then to generalize to cycles which are homologous to zero. Anyway, most basic applications of the residue theorem do not deal with arbitrary zero-homologous cycles, but with closed curves which are homotopic to zero.

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    $\begingroup$ Isn't the homotopy form of Cauchy's integral theorem also a special case of the homological form? Why then is the homotopical version of Cauchy's integral theorem presented in some textbooks but the homotopical version of the residue theorem is not? $\endgroup$ Nov 8, 2021 at 6:29
  • $\begingroup$ @RamenNii-chan See my update. $\endgroup$
    – Paul Frost
    Nov 8, 2021 at 11:05

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