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Topic: Curve Homotopic to a Point Contour (Read 2691 times) |
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JP05
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Show there exists at least one closed curve C in the complex plane that is homotopic to the point contour z = 0.
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| « Last Edit: Apr 9th, 2006, 1:29pm by JP05 » |
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Icarus
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Re: Curve Homotopic to a Point Contour
« Reply #1 on: Apr 9th, 2006, 6:41pm » |
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z=sin(t)
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JP05
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Re: Curve Homotopic to a Point Contour
« Reply #2 on: Apr 10th, 2006, 11:58am » |
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Care to elaborate?
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Obob
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Re: Curve Homotopic to a Point Contour
« Reply #3 on: Apr 10th, 2006, 1:52pm » |
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C is simply connected. Therefore every closed curve is homotopic to the point contour z=0.
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Icarus
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Re: Curve Homotopic to a Point Contour
« Reply #4 on: Apr 10th, 2006, 3:42pm » |
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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).
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| « Last Edit: Apr 10th, 2006, 4:20pm by Icarus » |
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Michael Dagg
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Re: Curve Homotopic to a Point Contour
« Reply #5 on: Apr 10th, 2006, 4:24pm » |
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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].
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| « Last Edit: Apr 30th, 2006, 11:12pm by Michael Dagg » |
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Regards,
Michael Dagg
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JP05
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Re: Curve Homotopic to a Point Contour
« Reply #6 on: Apr 10th, 2006, 6:09pm » |
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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.
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| « Last Edit: Apr 10th, 2006, 6:33pm by JP05 » |
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Icarus
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Re: Curve Homotopic to a Point Contour
« Reply #7 on: Apr 11th, 2006, 4:29pm » |
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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, 2 ].
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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| « Last Edit: Apr 11th, 2006, 5:33pm by Icarus » |
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"Pi goes on and on and on ...
And e is just as cursed.
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When their digits are reversed? " - Anonymous
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