(Just so we're clear: that the Lie group of planar translations $R^2$ is isomorphic to a quotient of the 2D Euclidean Lie group $E(2)$ and the circle group $U(1)$.)
I am trying to prove that $R^2 = E(2)/U(1)$ directly but not getting very far. It seems as if it should be true, and trivially so, but I am new to Lie groups and am not quite sure what implies what yet. For instance, it is known that $E(2)$ is the semidirect product of $R^2$ and $U(1)$ (or $O(2)$ if you like); and I reckon that implies that E(2) is a principal fibre bundle with base $R^2$ and fibre $U(1)$, is any of that true? And does it imply there is a quotient relationship?
If so - can you give explicit homomorphisms $f$ and $g$ such that the following sequence is exact:
$1\rightarrow U(1)\xrightarrow f E(2)\xrightarrow g R^2\rightarrow1$?
(i.e. Im $f$ = Ker $g$)
I thought that $f$ is the map that takes each element of $U(1)$ to the element of $E(2)$ with the same rotation but $\underline{0}$ translation - i.e. if a general element of E(2) is expressed as $(M,\underline{v})$ for a rotation matrix $M$ and translation vector $\underline{v}$ [with $(M,\underline{a}).(N,\underline{b})=(MN,\underline{a}+M\underline{b})]$ then $f$ embeds $U(1)$ in E(2) as
$f:R\mapsto (R, \underline{0})$
for $R\in U(1)$, $R$ a rotation matrix. But presumably that would mean that $g$ would map the whole of $E(2)$ to $R^2$ by taking $(R,\underline{v})\mapsto (0,\underline{v})$ (Thaking Im $F$, the pure rotations, to the identity element $e_{R^2}=\underline{0}$ - unfortunately, this doesn't appear to be a homomorhpism.