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