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- Mar 10, 2012

- 835

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.