OK, having put my foot in my mouth earlier in this conversation with a false alarm, I have realized that I still have a spare foot. Having inspected the illustration carefully, I agree that the curve is not contractible. But I do not understand the claim that removing any of the impediments allows the curve to contract; in fact, I think the claim is wrong. The leftmost of the three punctures in the illustration is outside the curve. The other two points are inside the curve. Surely removing an impediment outside the curve can have no effect on contractibility. And removing one of the two impediments inside the curve leaves the other one to prevent contraction. On Sat, Oct 23, 2010 at 9:58 PM, Bill Thurston <wpthurston@gmail.com> wrote:
Sorry for the terse style --- I should have taken more time to explain it. Bill On Oct 23, 2010, at 8:41 PM, Fred lunnon wrote:
I withdraw the remark about typos; it's just that the style is terse, and requires several passes to parse correctly ... WFL
On 10/23/10, Fred lunnon <fred.lunnon@gmail.com> wrote:
The material appears fascinating --- but beware unfortunate typos ... WFL
On 10/23/10, Bill Thurston <wpt4@cornell.edu> wrote:
A simple curve in the plane is the boundary of a disk. If the disk has no points in its interior, the curve is contractible. The curve has winding number 1 or -1 about any point in the interior of the disk.
The point is that in the plane with 3 or more punctures, there can be very complicated simple curves, even though they automatically have small winding numbers (0, 1 or -1) about all the punctures.
Bill On Oct 22, 2010, at 10:40 PM, James Propp wrote:
What's the simplest example of a simple closed curve in the triply-punctured plane that has winding number 0 around each of the punctures but is not contractible? (Back when I was in grad school in Berkeley in the '80s, there was a painting of one such curve on the wall, along with the associated word in the fundamental group of the surface.) Also, what's the simplest way to prove that no analogous curve exists for the doubly-punctured plane?
Jim Propp
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