Differential Geometry: Comparing Metric Tensors

[lavinia]

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]

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]

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?

 

[Infrared]

I found a MO thread (https://mathoverflow.net/questions/60375/is-mathbb-r3-the-square-of-some-topological-space) that has both the Kunneth argument (first answer) and @WWGD's orientation solution (link in second answer)

I worry that we've hijacked the thread...

 

[lavinia]

I found a SO thread (https://mathoverflow.net/questions/60375/is-mathbb-r3-the-square-of-some-topological-space) that has both the Kunneth argument (first answer) and @WWGD's orientation solution (link in second answer)

I worry that we've hijacked the thread...

yes.

 

[WWGD]

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]

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 $X$.

 

[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 $X$.

Right. Two oriented charts are connected by a path of overlapping oriented charts.