Is $SO(2,\mathbb R)$ connected

Question

Is $SO(2,\mathbb R)=\{ A\in O(2,\mathbb R): det A=1\}$ connected?Why?

Answer 1

There is a surjective morphism from $(\mathbb{R}, +) \to SO(2,\mathbb{R})$, $\ t \mapsto \left( \begin{array}{cc} \cos t &- \sin t\\ \sin t & \cos t \end{array} \right) $

$\bf{Added:}$. $SO(n,\mathbb{R})$ is connected for all $n$. There are many ways to prove that. One way would be this: any matrix $g$ in $SO(n, \mathbb{R})$ is conjugate ( in $SO(n, \mathbb{R}) $ ) to a block diagonal matrix $h g h^{-1}$ of the form \begin{equation*} h g h^{-1} = \left( \begin{array}{cccc} \cos t_1 & - \sin t_1 & 0 & 0 & \ldots \\ \sin t_1 & \cos t_1 & 0 & 0 & \ldots \\ 0 & 0 & \cos t_2 &- \sin t_2 & \ldots\\ 0 & 0 & \sin t_2 & \cos t_2 & \ldots\\ 0 & 0 & 0 & 0 & \ddots \end{array} \right) \end{equation*} with the last block(s) \begin{equation*} \left( \begin{array}{ccc} \cos t_{ \lfloor{\frac{n}{2}}\rfloor } &-\sin t_{ \lfloor{\frac{n}{2}}\rfloor } \\ \sin t_{ \lfloor{\frac{n}{2}}\rfloor }& \cos t_{ \lfloor{\frac{n}{2}}\rfloor } \end{array} \right) \end{equation*}

if $n$ even or \begin{equation*} \left( \begin{array}{ccc} \cos t_{ \lfloor{\frac{n}{2}}\rfloor } &-\sin t_{ \lfloor{\frac{n}{2}}\rfloor } &0\\ \sin t_{ \lfloor{\frac{n}{2}}\rfloor }& \cos t_{ \lfloor{\frac{n}{2}}\rfloor } &0 \\ 0&0&1 \end{array} \right) \end{equation*} if $n$ is odd.

Thus $SO(n, \mathbb{R})$ is a union of conjugates of a subgroup isomorphic to $SO(2, \mathbb{R})^{\lfloor{\frac{n}{2}\rfloor}}$, so a union of connected subgroup and hence also connected ( the subgroups all have a common point $e$ ).