- #36
wrobel
Science Advisor
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wrobel said:a hypothesis on #9
I think this runs into two problems: first of all, it works on 2 dimensions, but in 3 dimensions, there are many possible sets bounded by the curve, and they are not mapped to themselves. This is essentially the same reason I ran into the problem of holes. I like that you used a line segment though, it's easier to think about that way.mathwonk said:@AndreasC: What about this version of your idea? choose any point p in the plane, and join p to f(p) by a line segment L. Then consider the simple closed curve f(L). If L meets f(L) only at the endpoints, then their union bounds a compact set K homeomorphic to a disc. Then K must be mapped to a compact connected set with the same boundary curve, hence to itself. Apply Brouwer to K.
If L does meet f(L), choose a point q on L such that f(q) is also on L, and the distance from q to f(q) is minimal. Let M be the sub line segment joining q to f(q). Then no interior point of M can map to a point of M, i.e. M and f(M) meet only at endpoints, so apply the earlier argument to the set bounded by M union f(M).
Feel free to look for one, but I think I'm allowed to ask one question using more advanced techniques per thread :)mathwonk said:From there, I need Whitehead and the homology of P^infinity, to finish. Maybe there is a more elementary way?
mathwonk said:By the way, I may be missing something, but this argument does not seem to use fof. = id, just that f is a homeomorphism. Thus it seems to show every homeomorphism of a nice contractible space has a fixed point?
mathwonk said:yes I see that too. trying to see now where I used f^2 = id. oho! thank you!
I also added another idea/speculation above, on extending to the sphere.
it seems the extension of a translation to the n - sphere would have a fixed point of index 2 (or 0?) at infinity. (I am thinking mainly in R^2.)
My argument is about dimensions higher than the plane. Basically you do the same thing you did for the plane, but once more for 3 dimensions, and then again for 4, etc. In 3 dimensions, the set A' is now essentially a 2-disk in 3d space, but f(A') is not the same as A'. However it also is homeomorphic to a 2-disk, and they share a boundary. If they are not the same, their union is essentially a 2-sphere, that is mapped to itself. The interior (A'') is a 3d ball, and because its border is mapped to itself, then A'' is mapped to itself. We do the same for 4 dimensions once more.Infrared said:
How are you getting a (branched) double cover of a sphere? ##f## is a homeomorphism so a degree 1 map. Maybe I misunderstand?mathwonk said:well what about for n=2? it seems one would get a double cover of a surface by the 2-sphere with only one branch point, contradicting riemann-hurwitz.
so maybe one version of the index at infinity is the degree of the cover minus one, i.e. 2-1 = 1 (if orientation preserving).
But why does it have to be disjoint? The whole point is that it is not disjoint, they share a boundary. Do you mean their interiors being disjoint? Well, I think they don't have to be disjoint. I think Brouwer should work even if the union is homeomorphic to a bunch of disks connected by line segments or whatever.mathwonk said:for me the hard part is getting f(A') disjoint from A'
Will it? I think it should be just a bunch of closed loops connected by a single lines in 2d, then sphere-like things connected by lines in 3d, etc.mathwonk said:yes, disjoint except for their boundaries, otherwise the intersection might be quite wild.
wrobel said:What about the case ##f^k=\mathrm{id},\quad k=3,4,\ldots##? I mean continuous ##f##; my trick still holds here.
Apparently yes. Here is a construction of an abelian group ##A## which is isomorphic to ##A\oplus\mathbb{Z}^2## but not isomorphic to ##A\oplus\mathbb{Z}.##mathwonk said:challenge: does an example exist with abelian groups????
Woah I did think free groups but it never occured to me the free group on 3 generators is isomorphic to a subgroup of the free group on 2. It's very counter intuitive.mathwonk said:now for #7: by the Seifert-VanKampen theorem, the fundamental group of a wedge of n circles, is the free group on n generators, and a covering map induces injections of fundamental groups. Now you may be able to see how to construct a 2:1 covering map from a wedge of 3 circles to a wedge of 2 circles; just put the end circle over the end circle, then wrap the middle circle twice around the other circle, and then map the other end circle down to the first circle. This implies the free group on 3 generators is isomorphic to a subgroup of the free group on 2 generators. Now a free Group on n generators is not isomorphic to the free group on m generators unless n=m, since modding out by the commutator subgroup would imply the isomorphism of free abelian groups G,H on n ≠ m generators. That fails by modding out further by the subgroups 2G, 2H, forcing isomorphism of vector spaces G/2G, H/2H, over the field Z/2Z, but of different dimensions.
challenge: does an example exist with abelian groups????