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[SOLVED] To Show a Function is a Coordinate Patch

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hello MHB.
I have been trying to settle this question for a long time now and it is very important for me to solve this.

Let $p, q\in \mathbb R^2$ be points such that $p$ and $q$ are linearly independent (when considered as vectors in $\mathbb R^2$).

For any given $\theta$, write $$R_{\theta}=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$.

Let $M\subseteq (\mathbb R^2)^3$ be defined as $M=\{(x,x+R_\theta(p),x+ R_\theta(q)):x\in\mathbb R^2, \theta\in \mathbb R\}$.

I need to show that $M$ is a $3$-manifold in $\mathbb R^6$.

I think the function $\alpha: \mathbb R^2\times \mathbb R\to M$ defined as $\alpha(x,\theta)=(x,x+R_\theta(p), x+ R_\theta(q))$ is a coordinate patch.

I can show that $\alpha$ is bijective and has a constant rank 3.

But I am not able to show that $\alpha^{-1}$ is continuous, that is, $\alpha$ is an open map.

Can anybody see how to do that, or else, find some other coordinate patch about $(0,p,q)$ on $M$.

Thanks.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Hello MHB.
I have been trying to settle this question for a long time now and it is very important for me to solve this.

Let $p, q\in \mathbb R^2$ be points such that $p$ and $q$ are linearly independent (when considered as vectors in $\mathbb R^2$).

For any given $\theta$, write $$R_{\theta}=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$$.

Let $M\subseteq (\mathbb R^2)^3$ be defined as $M=\{(x,x+R_\theta(p),x+ R_\theta(q)):x\in\mathbb R^2, \theta\in \mathbb R\}$.

I need to show that $M$ is a $3$-manifold in $\mathbb R^6$.

I think the function $\alpha: \mathbb R^2\times \mathbb R\to M$ defined as $\alpha(x,\theta)=(x,x+R_\theta(p), x+ R_\theta(q))$ is a coordinate patch.

I can show that $\alpha$ is bijective and has a constant rank 3.

But I am not able to show that $\alpha^{-1}$ is continuous, that is, $\alpha$ is an open map.

Can anybody see how to do that, or else, find some other coordinate patch about $(0,p,q)$ on $M$.

Thanks.
Just a little quibble to start with: the map $\alpha: \mathbb R^2\times \mathbb R\to M$ is not bijective because $\alpha(x,\theta+2\pi) = \alpha(x,\theta).$ You need to say that the map $\alpha$ has domain $\mathbb R^2\times \mathbb S$, where $\mathbb S$ denotes the unit circle in $\mathbb R^2$.

The map $(x,x+R_\theta(p), x+ R_\theta(q)) \mapsto x$ is clearly continuous. So you only need to show that the map $(R_\theta(p), R_\theta(q)) \mapsto \theta$ is continuous.

There is a theorem (you can find it here) that a continuous bijective map from a compact space to a Hausdorff space is automatically bicontinuous. If you apply that to the map $\beta: \theta \mapsto (R_\theta(p), R_\theta(q))$, it tells you immediately that $\beta^{-1}$ is continuous.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Just a little quibble to start with: the map $\alpha: \mathbb R^2\times \mathbb R\to M$ is not bijective because $\alpha(x,\theta+2\pi) = \alpha(x,\theta).$ You need to say that the map $\alpha$ has domain $\mathbb R^2\times \mathbb S$, where $\mathbb S$ denotes the unit circle in $\mathbb R^2$.
Ooops! My bad. I should have used $(0,2\pi)$ instead of $\mathbb R$. Thanks.

The map $(x,x+R_\theta(p), x+ R_\theta(q)) \mapsto x$ is clearly continuous. So you only need to show that the map $(R_\theta(p), R_\theta(q)) \mapsto \theta$ is continuous.

There is a theorem (you can find it here) that a continuous bijective map from a compact space to a Hausdorff space is automatically bicontinuous. If you apply that to the map $\beta: \theta \mapsto (R_\theta(p), R_\theta(q))$, it tells you immediately that $\beta^{-1}$ is continuous.
Let me work on this. Thanks.

Really helpful.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Ooops! My bad. I should have used $(0,2\pi)$ instead of $\mathbb R$. Thanks.


Let me work on this. Thanks.

Really helpful.
After doing (Headbang) for quite a while I have the following:

Using the above we can show that $\alpha$ is a $C^\infty$ function of constant rank $3$ from $\mathbb R^2\times \mathbb S$ to $M$ and $\alpha$ has a continuous inverse.

Since $\mathbb R^2\times \mathbb S$ is a $3$-manifold (smooth), we conclude, using the map $\alpha$, that $M$ is also a $3$-manifold (smooth).

Am I correct?

EDIT:

Thinking again I realized that since $\alpha$ is defined on $\mathbb R^2\times \mathbb S$ (and not on $\mathbb R^2\times \mathbb R$), I don't what is the derivative of $\alpha$. I have read the concept of defferentiating a function defined from open subsets of Euclidean spaces but not on spaces like $\mathbb R^2\times \mathbb S$.

I am still stuck. :(
 
Last edited:

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834