What is a positively oriented Jordan curve?

Question

I'm reading the book "Methods of Nonlinear Analysis" written by Pavel Drábek. In this book there's the following proposition:

Let $\gamma$ be a positively oriented Jordan curve, $\sigma(B)\subset \text{int}\, \gamma $ ($\sigma(B)$ is the set of all eigenvalues of $B$), and let $f$ be a holomorphic function on a neighborhood of $\overline{\text{int}\, \gamma}$. Then

$$f(B)x=\frac{1}{2\pi i}\int _\gamma f(w)\left(w I-B\right)^{-1}xdw,\, \, \, \, x\in X$$

However, the book doesn't say what a positively oriented Jordan curve is.

Question 1: What does it mean precisely to say that $\gamma$ is a positively oriented Jordan curve?

In the book "Noncommutative Functional Calculus", written by Fabrizio Colombo, there's the following proposition:

Let $U$ be an open bounded set in $C$ such set $\partial U$ is a finite union of continuously differentiable Jordan curves. Let $f:U\cup \partial U\to X$ be a holomorphic function. Then

$$\int_{\partial U}f(z)dz=0$$

Question 2: What does it mean precisely to say that a set $\partial U$ is the finite union of Jordan curves?

Thank you for your attention!

Answer 1

The wikipedia page on the Jordan Curve Theorem (which roughly speaking proves, after significant effort, that the "inside" and "outside" of a Jordan curve is a well-defined notion) also defines what a Jordan curve is.

Your answer is correct; another way of saying the same thing is to say that $\gamma$ is a closed, simple curve.

The answer of paw88789 can be made rigourous if e.g. $\gamma \in C^1$ so that $\gamma$ has a well defined tangent vector $T=T(x)$, continuous as a function in $x \in \Omega$, and therefore by rotating 90º anti-clockwise, a well-defined normal $N=N(x)$. Then, if for every $ x \in \Omega = \gamma ([a,b])$, there exists $\epsilon_0 = \epsilon_0(x) >0$ such that for every $0<\epsilon<\epsilon_0$, $ x + \epsilon N(x)$ belongs to the "inside“ of the curve, we say that $\gamma$ (or $\Omega$, which are identified by an abuse of notation) is positively oriented.

The definition you give based on the index is equivalent to the definition above for $C^1$ functions, but has the advantage that the integral can be defined even for less regular curves. So it provides an extension of the definition of an orientation even to situations where a tangent vector cannot be specified.

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response to comment,

Could you elaborate on how the index definition extends to non $C^1$ curves?

Well right off the bat its easy to extend the definition to piecewise $C^1$ curves. But also I believe you can do the following. (Essentially, you replace $\gamma$ by a good polygonal approximation and let Cauchy's theorem take the wheel.)

Let $\gamma \in C^0$ be contained in the domain $U$ of an analytic function $f$. By compactness, cover $\gamma$ with finitely many small enough balls $B_1,\dots ,B_N, B_{N+1}:=B_1$ such that $$ \bigcup_{i=1}^N B_i \subset U,\\ 1\le i \le N \implies B_i \cap B_{i+1} \neq \emptyset.$$ whose union is contained in the domain $U$ of the integrand $f$. On each of these sets we have a local antiderivative $F_i$ of $f$. Pick an arbitrary point $x_i$ in each $B_i\cap B_{i+1}$, $x_{N+1} := x_1$, and define $$ \int_{\gamma} f(z) dz:= \sum_{i=1}^N\Big( F_i(x_{i+1}) - F_i(x_i) \Big)$$ None of the choices made matter - the values $F_i(x_{i+1}) - F_i(x_i)$ are independent of the choice of the antiderivative, and therefore independent of $B_i$. You can change the points $x_i$, or the number of points, by Cauchy's Theorem for polygonal sets. So the integral is well-defined, and extends the integral for $C^1$ curves.

Answer 2

About the index of a continuous non-differentiable curve:

Say $\gamma:[a,b]\to\Bbb C$ is a closed curve and $z\in\Bbb C\setminus\gamma([a,b])$. A little mumbling about the exponential being a covering map shows that there exist continuous functions $r:[a,b]\to(0,\infty)$ and $\theta:[a,b]\to\Bbb R$ such that $$\gamma(t) - z = r(t)e^{i\theta(t)}.$$The index of $\gamma$ about $z$ is now just $(\theta(b)-\theta(a))/2\pi$. (The net increase in the argument of $\gamma(t)-z$ as $t$ increases from $a$ to $b$.)

Answer 3

Searching I was able to find the following definitions:

A set $\Omega\subset \mathbb{C}$ is said to be a Jordan curve if there is a continuous function $\gamma:[a,b]\to\mathbb{C}$ such that the following propositions are true:

1. $\gamma([a,b])=\Omega$;
2. $\gamma(a)=\gamma(b)$;
3. The restriction $\gamma \vert_{[a,b)}:[a,b)\to\mathbb{C}$ is injective.

Besides, if $\gamma$ is of class $C^1$, then $\gamma$ is positively oriented if $\text{Ind}_\gamma(z):=\frac{1}{2\pi i}\int_\gamma \frac{dw}{w-z}$ is positive for some $z\in \text{int}\, \gamma([a,b])$.

Answer 4

That's a Jordan curve. Also, positively-oriented means traversed in a counterclockwise direction.

Answer 5

Imagine walking around the curve. The orientation is positive if the inside of the curve is on your left.