12 - Cycles

Cycle

Cycle $\Gamma$ is a finite formal sum of closed paths so

$$\Gamma =\sum _{i=1}^k{m}_i{\gamma }_i$$

where $m_i\in \mathbb{Z}\setminus \{0\}$ and ${\gamma }_i$ is a closed path for each $i\in \{1,\cdots ,k\}$

Support of a Cycle

Let $\Gamma$ be a cycle

Suppose of $\Gamma$ is defined as

$${\Gamma }^{\ast }:=\bigcup _{i=1}^k{\gamma }_i^{\ast }$$

Integral of a Cycle

Let $\Gamma$ be a cycle then

For any piecewise $C^1$ function $f$ defined on ${\Gamma }^{\ast }$ the integral over $\Gamma$ is defined as

$${\int }_{\Gamma }f(z)dz=\sum _{i=1}^k{m}_i{\int }_{{\gamma }_i}f(z)dz$$

Winding Number of a Cycle

Let $\Gamma$ be a cycle then

$$I(\Gamma ,z_0)=\sum _{i=1}^N{m}_{i}I({\gamma }_i,z_0)$$

Inside of a Cycle

Let $\Gamma$ be a cycle then

$$\mathrm{ins}(\Gamma ):=\{z\in C:z\notin {\Gamma }^{\ast }\text{ and }I(\Gamma ,z)\ne 0\}$$

Length of a Cycle

Let $\Gamma$ be a cycle then

$$\ell (\Gamma ):=\sum _{i=1}^k|m_i|\ell ({\gamma }_i)$$

Removal of a Point in a Cycle lemma

Let $\Gamma$ be a cycle in a domain $U$

Suppose $z_0\in {\Gamma }^{\ast }$ then
There exists cycle $\Sigma$ such that $z_0\notin {\Sigma }^{\ast }$ and

$${\int }_{\Sigma }f={\int }_{\Gamma }f$$

for every holomorphic function $f:U\to C$

Proof

Use induction on $k=k(\Gamma )$ where $k$ is the smallest integer such that $\Gamma$ can be in form

$$\sum _{i=1}^k{m}_i{\gamma }_i$$

Assuming that for all cycles ${\Gamma }^{\prime }$ with $k({\Gamma }^{\prime })<k(\Gamma )$ then

If $\Gamma ={\sum }_{i=1}^k{m}_i{\gamma }_i$ then let ${\Gamma }^{\prime }={\sum }_{i=1}^{k-1}{m}_i{\gamma }_i$
Let ${\Sigma }_1$ be a cycle such that

$$z_0\notin {\Sigma }_1^{\ast }\text{ and }{\int }_{{\Sigma }_1}fdz={\int }_{{\Gamma }_1}fdz$$

for all holomorphic functions $f$ defined on $U$

Let $\gamma ={\gamma }_k$
If $\gamma (t)=z_0$ for all $t$ then let $\Sigma ={\Sigma }^{\prime }$
Otherwise there exists $w\in {\gamma }^{\ast }$ with $w\ne z_0$ so shift parametrisation to get $\gamma (0)=w$

Let $r>0$ such that $B(z_0,r)\subseteq U$ and $|w-z_0|>r$
Since $[0,1]$ is compact then $\gamma$ is uniformly continuous

Hence there exists $N>0$ such that if $|s-t|<\frac{1}{N}$ then

$$\gamma (s)-\gamma (t)<r$$

Hence if image $\gamma \left(\left[\frac{k}{N},\frac{k+1}{N}\right]\right)$ of interval $\left[\frac{k}{N},\frac{k+1}{N}\right]$ contains $z_0$ then

$$\gamma \left(\left[\frac{k}{N},\frac{k+1}{N}\right]\right)\subseteq B(z_0,r)$$

Let $0=x_0<x_1<\cdots <x_m=1$ be the partition of $[0,1]$
By taking $x_{i}s$ to be $\left\{\frac{k}{N}:\gamma \left(\frac{k}{N}\right)\ne z_0,0\le k\le N\right\}$

Suppose $\gamma (t)=z_0$ for some $t\in (x_i,x_{i+1})$ then

$$\gamma {\mid }_{[x_i,x_{x+1}]}\text{ lies in }B(x_0,r)$$

So by Path Deformation Away from a Point
with $a=x_i$ and $b=x_{i+1}$ then replace it with path in $B(x_0,r)\setminus \{x_0\}$

As alteration to the path on each interval gives a path $\gamma$ which doesn’t pass through $z_0$
Then by Cauchy’s Theorem for Simply Connected Domains for the disk then

$${\int }_{\gamma }fdz={\int }_{\sigma }fdz\text{ for all holomorphic functions }f$$

Hence define

$$\Sigma :={\Sigma }_1+m_k\sigma$$

gives the required cycle

Path Deformation Away from a Point lemma

If $\gamma :[a,b]\to \mathbb{C}$ be a piece-wise $C^1$ path
where image ${\gamma }^{\ast }$ is contained in open ball $B(z_0,r)$ and $\gamma (a)\ne z_0\ne \gamma (b)$

Then

There exists path ${\gamma }_1:[a,b]\to B(z_0,r)$ with

$${\gamma }_1(a)=\gamma (a)\text{ and }{\gamma }_1(b)=\gamma (b)\text{ and }{\gamma }_1(t)\ne z_0\text{ for all}t\in (a,b)$$

Proof

By translation and scaling assume that $z_0=0$ and $r=1$

If necessary replace $\gamma$ with ${\gamma }^{-}$ (Opposite path) so that $0<|\gamma (a)|\le |\gamma (b)|$

If $|\gamma (t)|\ge |\gamma (a)|$ for all $t\in [0,1]$ then let ${\gamma }_1=\gamma$

Otherwise there exists $s_0\in [a,b]$ with $|\gamma (s_0)|<|\gamma (a)|$ so that for

$$L=\{t\in [a,b]:|\gamma (s)|\ge |\gamma (a)|\forall s\in [a,t]\}\text{ and }U=\{t\in [a,b]:|\gamma (s)|\ge \gamma (a)\forall s\in [t,b]\}$$

Then

$$a\in L\subseteq [a,s_0)\text{ and }b\in U\subseteq (s_0,b]$$

Hence

$$a\le t_0:=\sup (L)<t_1:=\inf (U)\le b$$

As $t\mapsto |\gamma (t)|$ is continuous then

$$|\gamma (t_0)|=|\gamma (a)|=|\gamma (t_1)|$$

If $\alpha ,\beta \in [-\pi ,\pi )$ given by

$$\gamma (t_0)=|\gamma (a)|\cdot e^{i\alpha }\text{ and }\gamma (t_1)=|\gamma (a)|\cdot e^{i\beta }$$

Let $g:[t_0,t_1]\to \mathbb{R}$ and ${\gamma }_1:[0,1]\to \mathbb{C}$ then

$$\gamma (t)=\frac{t_1-t}{t_1-t_0}\alpha +\frac{t-t_0}{t_1-t_0}\beta ,{\gamma }_1(t)=\{\begin{array}{ll}\gamma (t)& t\in [a,t_0]\cup [t_1,b]\\ |\gamma (a)|\exp (ig(t))& t_0\le t\le t_1\end{array}$$

where ${\gamma }_1$ is continuous and $|\gamma (1)|\ge d$ for all $t\in [a,b]$