In complex analysis what is meant by a simple closed path?

2.1k Views Asked by At

I'm a bit confused by what a simple closed path is in complex analysis.

I understand that "closed" means the parameterising function is equal at the endpoints. However, what does simple actually mean in terms of the path? Does it mean there are no singular points in the parameterising function? Also is path synonymous with contour?

3

There are 3 best solutions below

0
On BEST ANSWER

Let $\gamma\colon[a,b]\longrightarrow\mathbb C$ be a path. We say that it is a closed path (or a loop) if $\gamma(a)=\gamma(b)$. If it is a closed path, we say that it is a simple closed path if the restriction of $\gamma$ to $[a,b)$ (or to $(a,b]$) is injective.

0
On

A simple closed path in your space $X$ (I'd guess you're considering $X=\mathbb{C}$) can be described either as:

  1. A continuous function $\alpha\colon[0,1]\to X$ such that $\alpha$ is injective on $[0,1)$ and $\alpha(0)=\alpha(1)$

or

  1. A continuous injective function $\beta\colon\mathbb{S}^1\to X$, where $\mathbb{S}^1$ is the circle.

These two descriptions are equivalent by relating the functions $\alpha$ and $\beta$ above by the formula $\alpha(t)=\beta(e^{2\pi i t})$.

In the first description you can acually consider any other closed interval $[a,b]$ instead of $[0,1]$ (as they are all homeomorphic).

In the second description, if $X$ is Hausdorff (as is the case for $\mathbb{C}$), then $\beta$ is a homeomorphism of $\mathbb{S}^1$ onto $\beta(\mathbb{S}^1)$.

0
On

To also address the path/contour difference:

1)A path may not be closed in general.

2)A contour is not a path in the sense that a contour is always a closed curve simple or not and is in addition oriented. It is thus viewed as a directed closed curve and is not synonymous to a path in general.