Metric products

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  • #26
WWGD
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You can prove this (##\mathbb{R}^{2n+1}\ncong X\times X##) with algebraic topology (homology)- do you have a different argument in mind?
Yes, there's this interesting one, (not originally mine). It uses the degree of a map, specifically, homeomor-
phisms have degree ## \pm 1## and satisfy ##def(f \circ g)=deg(f) \cdot deg(g)##, so that for any homeomorphism f, we have : ##deg(f \circ f) ==1 ##.
Assume then there is a homeo h ## X^2 \rightarrow \mathbb R^3 ##. Then we have a homeo
from ## X^4 \rightarrow \mathbb R^6 ##. Consider the homeo ## h: X^4 \rightarrow X^4 : (a,b,c,d) \rightarrow (d,a,b,c) ## with ## h \circ h : (a,b,c,d) \rightarrow (c,d,a,b) ## which gives us the map ##h' \circ h' \mathbb R^6 \rightarrow \mathbb R^6 (a,b,c,d,e,f) \rightarrow ##
If ##X## is a smooth manifold then wouldn't ##X^2## be even dimensional?
Yes, I dont know if this applies to any other type of space though. And it seems like a novel proof.
 
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  • #27
WWGD
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Well, clearly a 2-manifold cannot be even locally diffeomorphic to ##\mathbb R^3##, I just thought it was a cool proof, did not use Invariance of dimension, etc.
 
  • #28
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It is a cool proof, and it doesn't assume that ##X## is a (even topological) manifold, so is stronger than an argument from invariance of domain, etc.

A more mundane proof would be to remove a point and use a Kunneth formula on ##X\times X=\mathbb{R}^{2n+1}##.
 
  • #29
lavinia
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It is a cool proof, and it doesn't assume that ##X## is a (even topological) manifold, so is stronger than an argument from invariance of domain, etc.

A more mundane proof would be to remove a point and use a Kunneth formula on ##X\times X=\mathbb{R}^{2n+1}##.
If it does not assume that ##X## is a topological manifold then how does one make the argument about orientation reversal?

For instance, how does ##f## possibly have degree ##-1## for an arbitrary topological space? Suppose that ##X## is a non-orientable manifold.
 
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  • #30
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If it does not assume that ##X## is a topological manifold then how does one make the argument about orientation reversal?

For instance, how does ##f## possibly have degree ##-1## for an arbitrary topological space? Suppose that ##X## is a non-orientable manifold.
The proof does not consider the degree of a map ##X\to X##, but of a map ##X^4\to X^4##. This makes sense because ##X^4\cong\mathbb{R}^6##. I think @WWGD had a typo of ##X^6## for ##\mathbb{R}^6## in his post.
 
  • #31
lavinia
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The proof does not consider the degree of a map ##X\to X##, but of a map ##X^4\to X^4##. This makes sense because ##X^4\cong\mathbb{R}^6##. I think @WWGD had a typo of ##X^6## for ##\mathbb{R}^6## in his post.
Not really my point. The underlying intuition was that the iterated mapping ##h^2## on ##X^4## is orientation preserving. So how is ##X^4## orientable in the first place? What does that mean?E.g.if ##X## is a non-orientable manifold then ##X^4## is also non-orientable. For instance the four fold Cartesian product of the real projective plane with itself is not orientable. In fact is ##w_{i}## is the generator of its first ##Z_3## cohomology then the first Stiefel-Whitney class of the four fouldCartesian product is ##w_1+w_2+w_3+w_4##.

BTW: I am curious to see your Kunneth formula proof.
 
  • #32
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Sorry I must be missing something. ##\mathbb{R}^6## is orientable, and ##X^4## is homeomorphic to ##\mathbb{R}^6##, so it is also orientable. If you're asking what definition of orientable works here, I think you can use the usual definition of being able to consistently choose generators for the local homology groups ##H_6(X^4,X^4-\text{point})##.
 
  • #33
lavinia
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Sorry I must be missing something. ##\mathbb{R}^6## is orientable, and ##X^4## is homeomorphic to ##\mathbb{R}^6##, so it is also orientable. If you're asking what definition of orientable works here, I think you can use the usual definition of being able to consistently choose generators for the local homology groups ##H_6(X^4,X^4-\text{point})##.
Right. Now I get it. Nice.

There is a theorem here that is implicitly assumed which is that the square of any homeomorphism of an orientable manifold is orientation preserving. This is clear for a compact manifold but for a non-compact manifold, how does one prove it?
 
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  • #36
WWGD
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The proof does not consider the degree of a map ##X\to X##, but of a map ##X^4\to X^4##. This makes sense because ##X^4\cong\mathbb{R}^6##. I think @WWGD had a typo of ##X^6## for ##\mathbb{R}^6## in his post.
Myb bad, thanks for pointing it out , for links and followup. I am editing as we speak.
 
  • #37
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There is a theorem here that is implicitly assumed which is that the square of any homeomorphism of an orientable manifold is orientation preserving. This is clear for a compact manifold but for a non-compact manifold, how does one prove it?
I think the point is the same as in the closed case. If ##X## is orientable, then all of the local homology groups ##H_n(X,X-\text{point})## are naturally isomorphic (identify the preferred generators), so a homeomorphism ##f:X\to X## with ##f(x_1)=x_2## gives an isomorphism ##H_n(X,X-x_1)\to H_n(X,X-x_2)\cong H_n(X,X-x_1),## and then ##f\circ f## has to induce the identity, so ##f\circ f## preserves orientation.

The only difference from the closed case is that the generators are not induced by a homology class of ##X##.
 
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  • #38
lavinia
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I think the point is the same as in the closed case. If ##X## is orientable, then all of the local homology groups ##H_n(X,X-\text{point})## are naturally isomorphic (identify the preferred generators), so a homeomorphism ##f:X\to X## with ##f(x_1)=x_2## gives an isomorphism ##H_n(X,X-x_1)\to H_n(X,X-x_2)\cong H_n(X,X-x_1),## and then ##f\circ f## has to induce the identity, so ##f\circ f## preserves orientation.

The only difference from the closed case is that the generators are not induced by a homology class of ##X##.
Right. Two oriented charts are connected by a path of overlapping oriented charts.
 

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