07 - Operations on Paths

Opposite Path

Let $\gamma :[a,b]\to \mathbb{C}$ then

Opposite Path ${\gamma }^{-1}$ is defined as

$${\gamma }^{-}:[0,b-a]\to \mathbb{C}\text{ where }{\gamma }^{-1}(t)=\gamma (b-t)$$

Concatenated Paths

Let ${\gamma }_1[a,b]\to \mathbb{C}$ and ${\gamma }_2[c,d]\to \mathbb{C}$ be two paths such that

$${\gamma }_1(b)={\gamma }_2(c)$$

Concatenated Paths of ${\gamma }_1$ and ${\gamma }_2$ is defined as
${\gamma }_1\star {\gamma }_2:[a,b+d-c]$ where

$${\gamma }_1\star {\gamma }_2(t)=\{\begin{array}{ll}{\gamma }_1(t)& t\le b\\ {\gamma }_2(t-b+c)& t\ge b\end{array}$$

Note that a piecewise $C^1$ path is a finite concatenation of $C^1$ paths

Reparameterization of Paths

Let $\varphi :[a,b]\to [c,d]$ be continuously differentiable with

$$\varphi (a)=c\text{ and }\varphi (b)=d$$

Let $\gamma :[c,d]\to \mathbb{C}$ be a $C^1$-path

Setting $\tilde{\gamma }=\gamma \circ \varphi$ by Composite functions on paths then

$$\tilde{\gamma }:[a,b]\to \mathbb{C}\text{ is also a }{C}^1\text{-path}$$

which has the same image as $\gamma$

Hence $\tilde{\gamma }$ is a reparameterization of $\gamma$

Composite Functions on Paths lemma

Let $\gamma :[c,d]\to \mathbb{C}$
Let $s:[a,b]\to [c,d]$

Suppose that $s$ is differentiable at $t_0$ and $\gamma$ is differentiable at $s_0=s(t_0)$ then

$\gamma \circ s$ is differentiable at $t_0$ with derivative

$$(\gamma \circ s{)}^{\prime }(t_0)=s^{\prime }(t_0)\cdot {\gamma }^{\prime }(s(t_0))$$

Proof

Let $ϵ:[c,d]\to \mathbb{C}$ be given by $ϵ(s_0)=0$ and

$$\gamma (x)=\gamma (s_0)+{\gamma }^{\prime }(s_0)(x-s_0)+(x-s_0)ϵ(x)$$

so that this equation holds for all $x\in [c,d]$

By the definition of ${\gamma }^{\prime }(s_0)$ then $ϵ(x)\to 0$ as $x\to s_0$ so $ϵ$ is continuous at $t_0$

Substituting $x=s(t)$ then for all $t\ne t_0$

$$\frac{\gamma (s(t))-\gamma (s_0)}{t-t_0}=\frac{s(t)-s(t_0)}{t-t_0}\cdot ({\gamma }^{\prime }(s(t))+ϵ(s(t)))$$

As $s(t)$ is differentiable at $t_0$ then $s(t)$ is continuous at $t_0$ hence

$$ϵ(s(t))\to 0\text{ as }t\to t_0$$

Hence taking the limit as $t\to t_0$ then

$$(\gamma \circ s{)}^{\prime }(t_0)=s^{\prime }(t_0)({\gamma }^{\prime }(s_0)+0)=s^{\prime }(t_0){\gamma }^{\prime }(s(t_0))$$

Constant Path

Let $z_0\in \mathbb{C}$ then

Define $\gamma :[a,b]\to \mathbb{C}$ by

$$\gamma (t)=z_0\text{ for all }t\in [a,b]$$

Hence ${\gamma }^{\prime }(t)=0$ for all $t\in [a,b]$

Line Segment path

Let $a,b\in \mathbb{C}$ then

Line Segment Path ${\gamma }_{[a,b]}:[0,1]\to \mathbb{C}$ between $a$ and $b$ is defined by

$${\gamma }_{[a,b]}(t)=a+t(b-a)$$

Homotopic

Let $U$ be an open set in $\mathbb{C}$

Let $a,b\in U$
Let ${\gamma }_0:[0,1]\to U,{\gamma }_1:[0,1]\to U$ be two paths in $U$ such that

$${\gamma }_0(0)={\gamma }_1(0)=a\text{ and }{\gamma }_0(1)={\gamma }_1(1)=b$$

${\gamma }_1$ and ${\gamma }_2$ are homotopic in $U$ if there exists continuous function $h:[0,1]\times [0,1]\to U$ with

$$\begin{array}{rlrlrl}h(0,s)& =a& & h(1,s)& & =b\\ h(t,0)& ={\gamma }_0(t)& & h(t,1)& & ={\gamma }_1(t)\end{array}$$

Note that $h$ acts like a interpolation between the two curves

Null Homotopic

Let $U$ be an open set in $\mathbb{C}$

Closed curve $\gamma$ is null homotopic in $U$ if

$$\gamma \text{ is homotopic to a constant path}$$