Cauchy's Theorem: Understanding Complex Analysis

[Silviu]

Hello! I am reading a book on complex analysis and I came across this: If $G \in \mathbb{C}$ is a region, a function f is holomorphic in G and $\gamma$ is a piecewise smooth path with $\gamma \sim_G 0$ then $\int_\gamma f = 0$. I want to make sure I understand. First of all, $\gamma \sim_G 0$ means that if G doesn't have "holes", any closed loop is homotopic to a point? And this also means that if G doesn't have holes, the integral of any holomorphic function over a closed loop is 0? Which means that in G any holomorphic function has an antiderivative? Thank you!

 

[FactChecker]

Hello! I am reading a book on complex analysis and I came across this: If $G \in \mathbb{C}$ is a region, a function f is holomorphic in G and $\gamma$ is a piecewise smooth path with $\gamma \sim_G 0$ then $\int_\gamma f = 0$. I want to make sure I understand. First of all, $\gamma \sim_G 0$ means that if G doesn't have "holes", any closed loop is homotopic to a point?

What it specifically says is that the particular path $\gamma$ can be shrunk to a point within G. It doesn't say that all paths can be shrunk to a point within G.
(But in the special case of G being simply connected, any closed path can be shrunk to a point within G.)

And this also means that if G doesn't have holes, the integral of any holomorphic function over a closed loop is 0?

Right.

Which means that in G any holomorphic function has an antiderivative?

Right.

 

[Silviu]

What it specifically says is that the particular path $\gamma$ can be shrunk to a point within G. It doesn't say that all paths can be shrunk to a point within G.
(But in the special case of G being simply connected, any closed path can be shrunk to a point within G.)
Right.Right.

Thank you! One more question, as far as I remember a circle and a point are not topologically equivalent as they have a different number of holes. So topologically a circle can't be shrunk to a point. But homotopically it can. What is the difference between the two? In the case of a homotopy you ignore the holes? (I understand mathematically the definition of homotopy and homeomorphism I am just not sure I can visualize the difference from a geometrical point of view).

 

[fresh_42]

Thank you! One more question, as far as I remember a circle and a point are not topologically equivalent as they have a different number of holes. So topologically a circle can't be shrunk to a point. But homotopically it can. What is the difference between the two? In the case of a homotopy you ignore the holes? (I understand mathematically the definition of homotopy and homeomorphism I am just not sure I can visualize the difference from a geometrical point of view).

A point and a circle are not homeomorph since they are of different dimensions. You will lose information. Or formally, you cannot establish a bijection. A circle and a point, if regarded as the graphs of two functions are homotopic, since there is a continuous function (the shrinking) that transforms one into the other. However, if you cut out a point in the inner area of the circle, they won't be homotopic anymore, since the "shrinking" would have to "jump" across that whole, i.e. it cannot be done continuously anymore.

 

[FactChecker]

Topology is not my strength, but here is a thought about terminology.
We should distinguish between "shrink to a point" and "shrink to become a point". The first would mean that, given any open set containing the point, the path becomes completely contained in that open set. The second would mean that there is a homotopy between the line and a point, which is not possible(?).
Perhaps there is already better terminology than "shrinks to a point".

 

[fresh_42]

The second would mean that there is a homotopy between the line and a point, which is not possible(?).

It is possible: $H : \mathbb{S}^1 \times [0,1] \longrightarrow \{0\}\, , \,H(\varphi ,t)=(1-t)\cdot \begin{bmatrix}\cos \varphi \\ \sin \varphi \end{bmatrix}$ is continuous, $\{H(\varphi,0)\,\vert \,\varphi \in [0,2\pi)\} = \mathbb{S}^1\; \textrm{ and } \; \{H(\varphi ,1)\,\vert \,\varphi \in [0,2\pi)\} = \{(0,0)\}$ is a homotopy that shrinks the circle.

 

[WWGD]

Thank you! One more question, as far as I remember a circle and a point are not topologically equivalent as they have a different number of holes. So topologically a circle can't be shrunk to a point. But homotopically it can. What is the difference between the two? In the case of a homotopy you ignore the holes? (I understand mathematically the definition of homotopy and homeomorphism I am just not sure I can visualize the difference from a geometrical point of view).

By cardinality reasons alone, a circle and a point are not topologically equivalent/ homeomorphic.

 

[lavinia]

A space that can be shrunk to a point is called "contractible". This means that it can be shrunk to a point inside itself. That is: there is a homotopy i.e. a continuous map $H:M×[0,1]→M$ such that at time $0$ $H$ is the identity map and at time $1$ it projects the space $M$ onto a point $p$ in $M$.

An example of a contractible space is Euclidean space. The continuous map $H(x,t) = (1-t)x$ is the identity at time $0$ and maps everything to the origin at time $1$. Another example is a set of line segments that share a common end point.

If the topological space $M$ is a subset of another space $N$ then one says that $M$ is null homotopic in $N$ if it can be shrunk to a point in $N$. In this case the homotopy maps $M×[0,1]$ into $N$. That is: $H: M×[0,1]→N$ is the inclusion of $M$ in $N$ at time $0$ and maps $M$ to a point in $N$ at time $1$.

A circle is not contractible. There is no homotopy $H:S^1×[0,1]→S^1$ that is the identity at time $0$ and projects the circle onto one of its points at time $1$.

But a circle inside Euclidean space is homotopic to a point since the same homotopy that shrinks all of Euclidean space to a point shrinks any subset to a point. That is: every subset of Euclidean space is null homotopic.

In complex analysis one considers regions of the plane that are not contractible. Typical examples are contractible open sets - such as a disk - minus a finite number of points or minus a finite number of disks, for instance an annulus. Within such regions, there are closed curves that are not null homotopic - for instance in a disk minus a point, a closed curve that winds around the missing point a finite number of times. Also there are closed curves that are null homotopic - for instance any closed curve that does not wind around the missing point. None of these closed curves can be shrunk to a point in themselves but some of them can be shrunk to a point within the region.

Observations:

- Two spaces are said to be homotopically equivalent if there are continuous maps $H:M →N$ and $G:N→M$ such that $HG$ and $GH$ are both homotopic to the identity map. Homotopically equivalent is not the same as homeomorphic. A contractible space is homotopically equivalent to a point but not in general, homeomorphic to a point. An annulus is homtopically equivalent to a circle.

- In some spaces, every closed curve is null homotopic. Such a space is said to be "simply connected". For instance, the sphere of any dimension greater than one, is simply connected. A simply connected space need not be contractible though. No sphere is contractible.

- In Algebraic Topology one measures the failure of a space to be simply connected by its fundamental group. By definition, a space is simply connected if its fundamental group is trivial. The fundamental group of a point is trivial. But the fundamental group of a circle is the integers.

- If two spaces are homotopically equivalent, then they have isomorphic fundamental groups. Since Eulcidean space is homotopically equivalent to a point, its fundamental group is trivial. A circle is not homotopically equivalent to a point since its fundamental group is not trivial.