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Consider the Lie group $G = \left\{ \ \left[ \begin{matrix} \cos(\theta) & \sin(\theta) & 0 & \alpha \\ -\sin(\theta) & \cos(\theta) & 0 & \beta \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right] \in M_{4 \times 4}(\mathbb{R}) \ \bigg| \ \alpha, \beta, \theta \in \mathbb{R} \ \right\}$.

I'm supposed to show that $G$ is diffeomorphic to $\mathbb{R}^{3}$.

Obviously I would need to find some diffeomorphism $\phi : \mathbb{R}^{3} \to G$. I'm thinking I should pick $\phi( \alpha, \beta, \theta )$ to map to a matrix of the above form. I know I'd have to show that $\phi$ is a bijection, is smooth and has a smooth inverse.

My issue is about the sector consisting of the $\theta$ stuff in matrices $M \in G$. I feel like that sector is diffeomorphic to $S^{1}$ and not $\mathbb{R}$, which then seems to imply that I shouldn't be able to define a diffeomorphism to $\mathbb{R}^{3}$, but rather to $S^{1} \times \mathbb{R}^{2}$. Is my thinking correct? Or am I missing out on a detail?

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Yes, the map $$S^1\times\mathbb{R}^2\to G,\quad (\theta,\alpha,\beta)\mapsto\left[ \begin{matrix} \cos(\theta) & \sin(\theta) & 0 & \alpha \\ -\sin(\theta) & \cos(\theta) & 0 & \beta \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} \right] \ $$ is a diffeomorphism and hence $G$ is not diffeomorphic to $\mathbb{R}^3$. However, $S^1\times\mathbb{R}^2$ (and hence $G$) is diffeomorphic to $\mathbb{R}^3-\{z\text{-axis}\}$, as can be seen by rotating an open half plane around the $z$-axis.

To see that $\mathbb{R}^3$ and $S^1\times\mathbb{R}^2$ are not diffeomorphic, you can look at their fundamental group: $\pi_1(\mathbb{R}^3)=1$ while $\pi_1(S^1\times\mathbb{R}^2)=\pi_1(S^1)\times\pi_1(\mathbb{R}^2)=\mathbb{Z}\times 1=\mathbb{Z}$.

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