Fundamental Group of the special Euclidean matrix group of the plane

Question

How do you do this? Compute the fundamental group of the special Euclidean group of the plane, that is, all matrices of the form: $ \left( \begin{array}{ccc} \cos(z) & \sin(z) & x \\ -\sin(z) & \cos(z)& y \\ 0 & 0 & 1\end{array} \right)$

I don't know how to start. Please give hints.

Answer 1

Assuming we take matrices under the usual norm, i.e. for $A \in \mathbb{R}^{n \times n}$: $$ \|A\| = \sqrt{\sum_{i,j = 1}^n |A_{ij}|^2} $$ we can show that the Euclidean matrix group is homeomorphic to $\Bbb R \times \Bbb R \times S^1$. This form should be easier to work with.

In other words, show that $$ \left( \begin{array}{ccc} \cos(z) & \sin(z) & x \\ -\sin(z) & \cos(z)& y \\ 0 & 0 & 1\end{array} \right) \mapsto (x,y,(\cos z,\sin z)) $$ Defines a homeomorphism from the matrix group to $\Bbb R \times \Bbb R \times S^1$.

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Showing that the above is a continuous map: in fact, it's easier to use the (topologically equivalent) norms $$ \|A\| =\sum_{i,j = 1}^n |A_{ij}|; \qquad \|(a,b,(c,d))\| = |a| + |b| + |c| + |d| $$

Fix $\epsilon > 0$. Let $A_1$ and $A_2$ be two matrices with $\|A_1 - A_2\| < \epsilon$. That is, $$ |x_1 - x_2| + |y_1 - y_2| + 2 |\cos(z_1) - \cos(z_2)| + 2|\sin(z_1) - \sin(z_2)| < \epsilon $$ We then have $$ \|f(A_1) - f(A_2)\| = \\ |x_1 - x_2| + |y_1 - y_2| + |\cos(z_1) - \cos(z_2)| + |\sin(z_1) - \sin(z_2)| <\\ |x_1 - x_2| + |y_1 - y_2| + 2 |\cos(z_1) - \cos(z_2)| + 2|\sin(z_1) - \sin(z_2)| < \epsilon $$ So, $\|A_1 - A_2\| < \epsilon \implies \|f(A_1) - f(A_2)\| < \epsilon$. Thus, $f$ is a (uniformly) continuous function.

Answer 2

The upper left $2 \times 2$-block parameterizes a familiar space, call it $X$. The $(x,y)$-entries give a copy of $\mathbb{R}^2$. The group is homeomorphic to $X \times \mathbb{R}^2$ and therefore the fundamental group of the Euclidean group is isomorphic to $\pi_1(X)$.