
#1
Apr2712, 07:52 AM

P: 55

How would one go about proving ℝ^{n} is not homeomorphic to ℝ^{m} for m≠n. This isn't a homework question, I was told, but we weren't shown the proof.Is there somewhere with a proof I can see, or can someone outline it briefly?




#2
Apr2712, 09:12 AM

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P: 16,583

The proof is really difficult and requires you to know about algebraic topology (and homology in particular).
A free book on the topic can be found here: http://www.math.cornell.edu/~hatcher/AT/ATpage.html The theorem in question is Theorem 2.26 on page 126 



#3
Apr2712, 09:29 AM

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PF Gold
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The "Invariance of Domain" theorem states that a continuous injection F from R^n to R^n is open. In particular, F(R^n) is open in R^n. Assume µ:R^n>R^m is a homeomorphism, for m≠n, and WLOG, assume m<n. Let also i be the inclusion of R^m in R^n as R^m x {(0,...,0)}. Then consider the composition i o µ:R^n>R^n. This is continuous and injective, but the image, R^m x {0}, is not open in R^n, constradicting Invariance of Domain.
The proof of Invariance of Domain uses elementary homology theory. Google gave http://www.math.unl.edu/~mbrittenham...nce.domain.pdf which proves Invariance of Domain as an easy corollary to the intuitively clear (generalized) Jordan separation thm which states that an embedding of S^n in R^{n+1} separates R^{n+1} into 2 connected components: one bounded (the "inside" of the sphere) and one unbounded (the "outside"). 



#4
Apr2712, 11:18 AM

P: 55

Not a homeomorphism R^n to R^m
Thanks for that. It all looks pretty interesting. Now I see why my lecturer wasn't going to show us the proof.




#5
Apr2712, 11:33 AM

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try it yourself for R^1 and R^2. (Hint: use connectivity.)
More generally, if you add one point to R^n at "infinity", you get an nsphere. If the euclidean spaces were homeomorphic, then the spheres would be too. So it suffices to show two spheres of different dimensions are not homeomorphic. A basic tool is "homotopy", or continuous deformation. A continuous (one parameter) family of maps from X to Y, is given by one continuous map from XxI to Y, where I is the unit interval. I.e. we say two maps g,h from X to Y are homotopic if there is a continuous map H:XxI>Y such that for all x in X, g(x) = H(x,0) and h(x) = H(x,1). Homotopy is an equivalence relation. Two maps that are homotopic will remain so after composing with a homeomorphism. So here is a theorem that would do what we want: if n < m, then any two maps from S^n to S^m are homotopic, but the identity map S^m to S^m is not homotopic to a constant map (i.e. there are two maps of S^m to itself that are not homotopic.). Here are some steps in the proof. 0) Every map into R^m is homotopic to a constant, because R^m is a "contractible" space. I.e. if g:X>R^m is any map define H:XxI>R^m as H(x,t) = t.g(x). Then for t=1, we get g, and for t=0 we get the constant map to 0. 1) If a map S^n>S^m is not surjective, then it is homotopic to a constant map. I.e. S^m minus any point is just R^m, and we apply part 0). 2) Here's the part that takes a little work: we can triangulate a sphere into small tetrahedra, so that any map g:S^n>S^m is approximated by a "simplicial map", i.e. one that takes tetrahedra in S^n into tetrahedra in S^m of the same dimension. This approximation can be made so close that it is homotopic to g, but this approximating simplicial map is not surjective, since it does not hit any top dimensional tetrahedra in S^m. It follows that every continuous map S^n>S^m where n<m, is homotopic to a constant map. 3) Now we have to show that there is a map S^m>S^m that is not homotopic to a constant, and in fact the identity map works. To do this we again use simplicial approximation, and define the so called "degree" of a map. Basically the degree of a map is the number of inverse images of a general point, but you have to be careful to cancel out inverse images that go around the target in the opposite direction. I.e. a map from R^1 to R^1 defined by a cubic polynomial essentially has degree one, even though there can be that little up and then down wrinkle in the middle of the graph. I.e. some values of y are hit by three x's but the graph is going up at two of them and down at the third, so with canceling we get only one preimage. So to define degree of a map S^m>S^m, we take a simplicial approximation, then we orient the little tetrahedra, and then count the number of preimages of essentially any small tetrahedron, counting it plus if it preserves orientation, and minus if it reverses it. We must also prove this weighted sum is the same for all choices of tetrahedra in the image. Then, once we know degree makes sense, and is the same for two homotopic maps, we note that the degree of the identity map is one, while the degree of a constant map is zero. voila! 



#6
Apr2712, 11:47 AM

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another clever way to show the identity map of a 2  sphere say, is at least not homotopic to a constant by a differentiable homotopy, is to note that the solid angle form has ≠0 integral over the sphere. It is easy to show by Stokes theorem that two differentiably homotopic maps will pull back a closed form to have the same integral. Since a constant map pulls back any form to have integral zero, this would do it too.
These two ideas, homotopy and integrals of closed forms, become the basis for a whole theory of studying spaces, called homotopy and homology. There is a way to actually "multiply" two maps from an n sphere into the same space, and make the set of homotopy classes of maps into a group, the nth homotopy group. It is even easier to make the space of closed n forms into a vector space. When modded out by the exact forms one gets a space called the nth (deRham) cohomology space. Then a really short answer to the question above is this: S^n and S^m cannot be homeomorphic for n < m, because both the nth homotopy and cohomology groups of S^m are zero, whereas for S^n they are both non zero. You learn the homotopy theory in a first course in algebraic topology, but some people teach the cohomology theory in advanced calculus, e.g. Ted Shifrin's book, or Michael Spivak's Calculus on manifolds. Notice the proofs above reduce down to the easy theorem that 0 ≠ 1. Using the more subtle fact that 1 ≠ 1, we can prove that a 2sphere does not have any never zero vector fields, since that would make the identity map homotopic, not to a constant map, but to the antipodal map, which has degree 1. The reason you can teach this stuff before algebraic topology and without doing homology theory first, is that for a sphere the cohomology group is generated by the solid angle (or volume) form, so you just write it down. On a 1  sphere that's "dtheta". I.e. the whole ball of wax boils down to showing the identity map of a sphere is not homotopic to a constant. If it were (in dimension one) you would get zero when you integrate dtheta around a circle, but you don't, you get 2pi. The analogous argument works in all dimensions. I.e. an equivalent statement is that there always exists on every sphere, a closed form that is not exact, and you can just write it down. see e.g. http://en.wikipedia.org/wiki/Nsphere for the "volume form". the part about getting zero when you integrate an m form around the n sphere is even easier since the way differential forms work, pulling an m  form back to n  space gives a form which is identically zero when n < m! 



#7
Apr2712, 04:32 PM

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Here is a half formed thought on a different way to prove it  probably wrong. If the map from the lower dimensional sphere were differentiable then the pull back of the volume form of the higher dimensional sphere would be zero. For the standard n sphere, the volume form is given by contracting the volume form of Euclidean space by the unit normal to the sphere. So there can not be a differentiable homeomorphism from a sphere to a higher dimensional standard sphere. So if we give these two topological spheres their standard smooth structures, the map from the lower dimensional sphere can not be differentiable. Maybe there is an approximation theorem which says that there are smooth maps that are arbitrarily close (under some metric) to a given homeomorphism. One might then show that if the smooth map is close enough it must itself be a homeomorphism. Is there a proof like this? 



#8
Apr2712, 04:56 PM

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Lavinia, I agree your idea to deduce the continuous case from the differentiable case by approximation, can likely be made to work. nice!
In fact, Milnor's beautiful book Topology from the differentiable viewpoint contains such an argument for Brouwer's fix point theorem. I.e. if there were a continuous self map of the disc with no fix point, then every point would move at least e>0. If we approximate this map within e/2 by a smooth map, then every point still moves a positive amount. Thus the fact every smooth self map has a fix point implies that every continuous one does too. 



#9
Apr2812, 12:04 AM

P: 55

Thanks for taking time to reply, you've all given me some interesting reading.




#10
Apr2812, 06:09 AM

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#11
Apr2912, 12:30 AM

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P: 1,168

I would say it is moreorless intuitively clear from the perspective of Lebesgue Dimension:
http://en.wikipedia.org/wiki/Lebesgue_dimension That the Lebesgue dimension of R^n is n, and that the Lebesgue dimension is a homeomorphism invariant. For specific values of n,m, you can use the fact that , given a homeomorphism f:X>Y, cutsets C ,of a space X sets whose removal disconnects Xare preserved by homeomorphisms, i.e., if C disconnects X, then f(C) disconnects Y (e.g., if X\C=A\/B , then f(A)\/f(B) are a disconnection of Y'ust use the properties of injections, continuity, etc.). So, for R, you have that a cutset is a point, i.e., the removal of any point disconnects R, but this is not so for R^2, for which you need to remove a line, or a closed curve to disconnect. I think it is clear that the dimension of a cutset for R^n is n1, but I cannot think of a proof. 


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