Definition of the sum of paths $\gamma_1:[a_1,b_1]\rightarrow \Omega$ and $\gamma_2:[a_2,b_2]\rightarrow \Omega$

complex-analysisgeneral-topology

Throught our lecture notes and textbox in complex analysis the sum of paths is given by

Given paths $\gamma_1:[a_1,b_1]\rightarrow \Omega$ and $\gamma_2:[a_2,b_2]\rightarrow \Omega$ such that $\gamma_1(b_1)=\gamma_2(a_2)$ we defined the path $\gamma_1+\gamma_2:[a_1,b_1,+b_2-a_2]\rightarrow \Omega$ by
$$
(\gamma_1+\gamma_2)(t)=
\begin{cases}
\gamma_1(t)&\text{if}\, t\in[a_1,b_1]\\
\gamma_2(t-b_1+a_2)&\text{if}\, t\in [b_1,b_1+b_2-a_2] \\
\end{cases}
$$

What is the reason for this? Why not just define the sums of path $\gamma_1+\gamma_2:[a_1,b_2]\rightarrow \Omega$ by

$$
(\gamma_1+\gamma_2)(t)=
\begin{cases}
\gamma_1(t)&\text{if}\, t\in[a_1,b_1]\\
\gamma_2(t)&\text{if}\, t\in [a_2,b_2] \\
\end{cases}
$$

Best Answer

The suggested solution has two issues:

  1. If $[a_1,b_1]$ and $[a_2,b_2]$ are disjoint, then the resulting function has a disconnected domain. This would not be a path since the domain for a path must be connected (an interval).

  2. If $[a_1,b_1]$ and $[a_2,b_2]$ overlap, then the definition is not well-defined because it is unclear if $(\gamma_1+\gamma_2)(t)$ would be $\gamma_1(t)$ or $\gamma_2(t)$.

While the original formula seems complicated, it is doing a very simple geometric operation. It is moving the interval $[a_2,b_2]$ so that it starts at $b_1$ and then gluing them together. Let's see why:

  • The length of $[a_1,b_1]$ is $b_1-a_1$, and, similarly, the length of $[a_2,b_2]$ is $b_2-a_2$. On the other hand, the length of the given interval for the sum is $(b_1+b_2-a_2)-a_1=(b_1-a_1)+(b_2-a_2)$. In other words, the length of the given interval is the sum of the original intervals.

  • What the piecewise formula is doing is traveling along $\gamma_1$ for $[a_1,b_1]$ and then traveling along $\gamma_2$ for the rest of the time. What looks messy here is that the interval $[a_2,b_2]$ has been shifted so that it comes right after $b_1$, by adding $b_1-a_2$ to the interval to get $[a_2+(b_1-a_2),b_2+(b_1-a_2)]=[b_1,b_1+b_2-a_2]$.

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