To find a Coordinate Patch About a Point in Euclidean Subspace. 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 $\mathbb R^6$.
So should be able to find a coordinate patch about $(0,p,q)$ on $M$.
I think the function $\alpha: \mathbb R^2\times (-\epsilon,\epsilon)\to M$ defined as $\alpha(x,\theta)=(x,x+R_\theta(p), x+ R_\theta(q))$ is a coordinate patch for sufficiently small (and positive) $\epsilon$.
I can show that $\alpha$ is bijective and has a constant rank 3 for sufficiently small and positive $\epsilon$.
But I am not able to show that $\alpha^{-1}$ is continuous, that is, $\alpha(G)$ is open in $M$ whenever $G$ is open in $\mathbb R^2\times (-\epsilon,\epsilon)$.
Can anybody see how to do that, or else, find some other coordinate patch about $(0,p,q)$ on $M$.
Thanks.
 A: (I address the question of showing $M$ is a manifold) By a linear change of coordinates, you may as well assume $p,q$ are the standard basis vectors $e_1, e_2$. 
Then $M$ is given as set $(x, x + (\cos \theta, \sin \theta), x + (-\sin \theta, \cos \theta))$. Applying the diffeomorphism of $\mathbb{R}^6$ given by $(v, w, z) \mapsto (v, w - v, z - v)$ (with inverse $(v, w, z) \mapsto (v, w + v, z + v)$), it now looks like $$(x, (\cos \theta, \sin \theta), (-\sin \theta, \cos \theta) ) \subseteq \mathbb{R}^6$$I leave as an exercise the details of showing this is $\cong \mathbb{R}^2 \times S^1$. 
A: I've been difficult enough that I should just post a full solution at this point ;-).
The goal is to show that $\alpha:\mathbb{R}^2\times (\epsilon,\epsilon)\rightarrow M$, when restricted to a suitable open subset, is an open map.  You have already shown that $\alpha$, when restricted to a suitable open subset is continuous, and bijective onto its image, and of constant rank 3.  To that end, let $V$ be any open subset of $\mathbb{R}^2\times (-\epsilon,\epsilon)$.  Our goal is to show that $\alpha(V)$ is open in $M$.  Note that do this, it's enough to show that for any $p\in V$, there is an open set $U$ with $p\in U\subseteq V$ and $\alpha(U)$ open in $M$.  So, fix a $p\in V$.
Using the constant rank theorem applied to $\alpha:\mathbb{R}^2\times (-\epsilon, \epsilon)\rightarrow \mathbb{R}^6$ and the point $p$, we get open subsets $U_1,U_2\subseteq \mathbb{R}^2\times (\epsilon,\epsilon)$ with $p\in U_1$, an open subset $U_3\subseteq \mathbb{R}^6$, and diffeomorphisms $\phi:U_1\rightarrow U_2$ and $\psi:U_3\rightarrow U_3$ for which $(\psi \circ \alpha \circ \phi^{-1})(y) = (y_1,y_2,y_3,0,0,0)$.  Since $p\in U_1\cap V$, we may, by shrinking $U_1$ and $U_2$, assume wlog that $U_1\subseteq V$.  We will show that $\alpha(U_1)$ is open in $M$.
Now we know that $\psi(\alpha(U_1)) = \psi(\alpha(\phi^{-1}(U_2)) =  U_2\times \{(0,0,0)\}\subseteq U_3$.  Let $W = U_2\times (-\epsilon_1,\epsilon_1)^3$ where $\epsilon_1$ is chosen small enough that $W\subseteq U_3$.  Note that $W$ is an open subset of $\mathbb{R}^6$.
Finally, I claim that $\psi^{-1}(W)\cap M = \alpha(U_1)$.  Since $\psi$ is continuous, $\psi^{-1}(W)$ is open in $\mathbb{R}^6$, so $\alpha(U_1)$ is open in $M$.
