Non-contractible, simply connected subsets of R^n

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flownt
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Non-contractible, simply connected subsets of R^n

Postby flownt » Fri Oct 17, 2014 3:09 pm UTC

(This problem was suggested to me by a friend who told me the first two observations)

The first observation one can make, is that S^2 is non-contractible and simply connected (trivial fundamental group), yielding that for n>=3 such subsets exist.

Secondly, for n=0,1 all connected subsets are immediately simply connected.

This leaves the case n=2. The example I came up with is the topoligists sine, with an extra piece adjoined:

Spoiler:

Code: Select all

+----*|||||-------+
|                 |
+-----------------+


That is, an extra arc is added from some place on graph, to the origin.

In polar coordinates: r(\theta)=2+sin(4\pi^2/ \theta), \theta\in(0,2\pi]; Let X=r((0,2\pi])

It is pretty easy to show that there are no paths going through the origin (or (2,0) for the radial version ) , yielding that these have a trivial fundamental group. But whereas intuition tells me that this should be non-contractible, an actual prove has elluded me.

Then yesterday my friend suggested the following map (in polar coordinates):

Code: Select all

F: X\times [0,1] \to X, 
(\theta, r(\theta), t)\mapsto { (\theta, r(\theta))                            if \theta/2\pi <= 1-t
                                (\theta + (t-(1-\theta/2\pi)) * (2\pi-\theta)/(\theta/2\pi), r(...)       else


Basically, keep a coordinate fixed "until its time has come" and than move it at a constant speed along the curve such that it ends up at \theta=2\pi exactly at t=1. This seemed surprisingly a contraction for a minute until we realized that the same trick than should work for circles as well.

So the question is:
Why doesn't "keep points fixed until they have to move; then move at constant speed" contract the circle? I know it doesn't because the circle has a non trivial fundamental group Z, but that doensn't help in the case I'm trying to study....

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jestingrabbit
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Re: Non-contractible, simply connected subsets of R^n

Postby jestingrabbit » Fri Oct 17, 2014 4:45 pm UTC

I believe its discontinuous at "the join". The image of points that are near one another should be near one another. The points near the join eventually start getting very distant from one another. Try to prove the continuity at time 1 near the join and you'll see what I mean.
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z4lis
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Re: Non-contractible, simply connected subsets of R^n

Postby z4lis » Fri Oct 17, 2014 10:39 pm UTC

Spoiler:
This is one of the tricky point set proofs. Suppose I have a contraction F, and WLOG we can contract to the point P next to the wiggly sine part and assume that P is fixed throughout the homotopy. Draw a tiny open disk D around it P, and call a time crappy if x-intercepts of the sine curve which started arbitrarily close to P are still in that tiny ball at time t, but not in the line segment pointing into the sine curve. We start at a crappy time, end at a not crappy time, and the set of crappy times is open. The trick is to show that the set of crappy times is actually closed, so suppose that we have an interval [0,t) of crappy times, and we try to show that t is crappy, too. By continuity there must be a small neighborhood V around P *and* a small time interval around t such that for points v in V and times s within that window, F(v,s) is in the tiny ball we drew a little bit ago. This means that if you drew the tiny ball small enough, points that are in the "wiggles" of the sine curve at times right before t cannot leave D until some time after t, and so t must be crappy.
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jestingrabbit
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Re: Non-contractible, simply connected subsets of R^n

Postby jestingrabbit » Wed Oct 22, 2014 9:18 am UTC

I was thinking about this again, and I think the cleanest argument that this can't be continuous is as follows. Firstly, recall that for a continuous function, f: X \to Y, \lim f(x_i) = f(\lim x_i), where x_i is a convergent sequence. Now, your retraction, F, is such that F_t, defined by F_t(x) = F(x, t), is a surjection onto X. So, for a given x_0 in X and t\in [0, 1], there is an x_t such that F_t(x_t) = x_0. I claim that we can therefore pick a sequence {(x_i, t_i)} such that \lim (x_i, t_i) = (the join, 1), and \lim f(x_i, t_i) = \lim x_0 = x_0. Therefore, F is not continuous.

The hard part in using this argument in general is showing that the intermediate functions must be onto.
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MartianInvader
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Re: Non-contractible, simply connected subsets of R^n

Postby MartianInvader » Thu Oct 23, 2014 8:49 pm UTC

I'm not sure I'm reading your notation correctly. Are you just ripping the end off the topologist's sine curve and pulling it around the circle?

If so, keep in mind that at every stage of the homotopy, the image of every convergent sequence should converge to the image of its limit. For example, take the sequence of points at the local maxima of the sine curve (so points where r = 3). If I'm reading your notation correctly, these converge to the point (0, 3) in your space. That means that at every stage, the sequence of the images of these points needs to converge to the image of (0, 3), otherwise the contraction's not continuous. I think that's where your problem is.
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