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general >> complex analysis >> Curve Homotopic to a Point Contour
(Message started by: JP05 on Apr 9th, 2006, 1:26pm)

Title: Curve Homotopic to a Point Contour
Post by JP05 on Apr 9th, 2006, 1:26pm
Show there exists at least one closed curve C in the complex plane that is homotopic to the point contour z = 0.

Title: Re: Curve Homotopic to a Point Contour
Post by Icarus on Apr 9th, 2006, 6:41pm
z=sin(t)

Title: Re: Curve Homotopic to a Point Contour
Post by JP05 on Apr 10th, 2006, 11:58am
Care to elaborate?

Title: Re: Curve Homotopic to a Point Contour
Post by Obob on Apr 10th, 2006, 1:52pm
C is simply connected.  Therefore every closed curve is homotopic to the point contour z=0.

Title: Re: Curve Homotopic to a Point Contour
Post by Icarus on Apr 10th, 2006, 3:42pm
I assumed he meant the curve itself has to be homotopic to a point, not that there is a homotopy of the ambient space that carries it to a point.

As for elaborating: z(r,t)=sin(rt).

Title: Re: Curve Homotopic to a Point Contour
Post by Michael_Dagg on Apr 10th, 2006, 4:24pm
The first example Icarus gave works: the homotopy between z = sin(t) and z = 0 is X(s,t) = (1 - s)w(t), where w(t) = sin(t), s, t in [0,1].

Title: Re: Curve Homotopic to a Point Contour
Post by JP05 on Apr 10th, 2006, 6:09pm
Yeah, I was just asking how so to get more.

That result with (1-s)sin(t) wrote over [0,1] is in Whitehead's book but it talks about reverse path elimination.

But, sin(rt) works too.

Title: Re: Curve Homotopic to a Point Contour
Post by Icarus on Apr 11th, 2006, 4:29pm
Contours are homotopic to a point if they do not encircle any points in their complement. I.e., the complement has a single component. Effectively, this means they "start" at some point z0, traverse a path to some other point z1, where they reverse direction and retrace the same path back to z0 to complete the loop (I am of course over-simplifying - they can turn around many times). Any path that does this is homotopic to z0, for instance, by "pulling z1 in" - that is, by turning around earlier than z1.

z = sin(t) was simply the easiest example I could come up with. Its image consists of the unit interval [-1, 1], and can be considered a contour by limiting it to [0, 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif].

My z = sin(rt) answer was supposed to be a homotopy of contours, but alas, it falls short, as Michael was too kind to say. However, his trick of X(s,t) = (1 - s)w(t) will work for any radial contour.



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