simply connected curve


by saravanan13
Tags: connected, curve, simply
saravanan13
saravanan13 is offline
#1
Apr22-11, 01:01 AM
P: 56
Dear Friends

In the complex functions, I completely understand the simply connected region but not the multiply connected region?
An apple is a simply connected region but No. 8 is multiply connected. How?
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lavinia
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#2
Apr22-11, 03:38 AM
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Quote Quote by saravanan13 View Post
Dear Friends

In the complex functions, I completely understand the simply connected region but not the multiply connected region?
An apple is a simply connected region but No. 8 is multiply connected. How?
Simply connected means that every closed curve can be shrunk to a point. On the figure eight there are infinitely many curves that can not be shrunk to a point. Also on a circle.

A disk minus a point is not simply connected. For instance look at a branch of the logarithm in the unit disk minus the origin.
saravanan13
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#3
Apr22-11, 07:08 AM
P: 56
Quote Quote by lavinia View Post
Simply connected means that every closed curve can be shrunk to a point. On the figure eight there are infinitely many curves that can not be shrunk to a point. Also on a circle.

A disk minus a point is not simply connected. For instance look at a branch of the logarithm in the unit disk minus the origin.
Thank
I could not follow your Second statement. Especially " branch of logarithm". Why cant a circle be simply connected?

lavinia
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#4
Apr22-11, 08:34 AM
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simply connected curve


Quote Quote by saravanan13 View Post
Thank
I could not follow your Second statement. Especially " branch of logarithm". Why cant a circle be simply connected?
The angle function on a circle is defined only locally but its exterior derivative is globally defined. Therefore it is a closed 1 form that is not the exterior derivative of a function.

The integral of the derivative of the complex logarithm around a circle centered at the origin is the same as the integral of the angle function.

Another way to look at this is - suppose the curve that loops around the circle once were null homotopic. Then there would be a map from a disk to the circle that was equal to this curve on the boundary of the disk. Stokes Theorem then tells you that the integral of the exterior derivative of the angle function over this loop must be zero. But the intergral is not zero. It is 2pi.

I suggest that you look at the covering space argument that also proves that the circle is not simply connected. This avoids homology and uses purely topological arguments.

Intuitively, a null homotopic loop on the circle would have to retrace its path and return to its end point in the opposite direction that it came in. The loop that goes around once does not do this.


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