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Simply connected curve 
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#1
Apr2211, 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? 


#2
Apr2211, 03:38 AM

Sci Advisor
P: 1,716

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. 


#3
Apr2211, 07:08 AM

P: 56

I could not follow your Second statement. Especially " branch of logarithm". Why cant a circle be simply connected? 


#4
Apr2211, 08:34 AM

Sci Advisor
P: 1,716

Simply connected curve
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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