# Metric products

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WWGD
Gold Member
2019 Award
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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WWGD
Gold Member
2019 Award
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.

Infrared
Gold Member
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 $X\times X=\mathbb{R}^{2n+1}$.

lavinia
Gold Member
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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Infrared
Gold Member
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.

lavinia
Gold Member
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.

Infrared
Gold Member
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 $H_6(X^4,X^4-\text{point})$.

lavinia
Gold Member
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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WWGD
Gold Member
2019 Award
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.

Infrared
Gold Member
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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lavinia
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 $X$.