Vector Fields on Lie Groups in a Coordinate Parameterisation My question is about how to write out vector fields on Lie groups in terms of a coordinate basis when a coordinate parametrisation of the group is given.
Consider the group $E(2)$ of Euclidian transformations. A general element can be written in coordinates as 
$$g = \begin{bmatrix}\cos z & -\sin z & x \\ \sin z & \cos z & y \\ 0 & 0 & 1 \end{bmatrix} $$
where $(x,y,z) \in \mathbb{R}^2 \times S^1$. This gives an identification of the group with $\mathbb{R}^2 \times S^1$. We can get a general left-invariant vector field on the group by multiplying any element of the Lie algebra on the left by the above matrix. The Lie algebra is spanned by
$$e_1 = \begin{bmatrix}0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \ \ e_2 = \begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}, \ \ e_3 = \begin{bmatrix}0 & 0 & 1 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{bmatrix} $$
So for instance there is a vector field 
$$X = ge_1 = \begin{bmatrix}-\sin z & -\cos z & 0 \\ \cos z & -\sin z & 0 \\ 0 & 0 & 0 \end{bmatrix} $$
I would like to picture this vector field by plotting it as a vector field in $\mathbb{R}^2 \times S^1$. How can I write it out in a coordinate basis, i.e. $X = X_x \partial_x + X_y \partial_y + X_z \partial_z$?
 A: Careful: you mean $SE(2)$, the group of proper rigid transformations.
If $g = \begin{bmatrix}\cos c & -\sin c & a \\ \sin c & \cos c & b \\ 0 & 0 & 1 \end{bmatrix}$ is a fixed group element and $h = \begin{bmatrix}\cos z & -\sin z & x \\ \sin z & \cos z & y \\ 0 & 0 & 1 \end{bmatrix}$ then
$$L_g(h) = \begin{bmatrix}\cos (c+z) & -\sin (c+z) & a + x\cos(c) - y\sin(c) \\ \sin (c+z) & \cos (c+z) & b + y\cos(c) + x \sin(c) \\ 0 & 0 & 1 \end{bmatrix},$$
or by identifying the group with $\mathbb R^2 \times S^1$, $$L_g(x,y,z) = (a+x \cos c - y\sin c, b + y\cos c + x \sin c, c+z).$$
Hence by differentiating, we see that the pushforward is $$dL_g = \begin{bmatrix}\cos c & -\sin c & 0
\\ \sin c & \cos c & 0\\ 0 & 0 & 1 \end{bmatrix}.$$
Now your $e_1 = \partial_z$, so
$$(dL_g)_I(e_1) = \begin{bmatrix}\cos c & -\sin c & 0
\\ \sin c & \cos c & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix} = \partial_z.$$
Compare e.g. with $e_2 = \partial_x + \partial_y$, where 
$$\begin{align}(dL_g)_I(e_2) &= \begin{bmatrix}\cos c & -\sin c & 0
\\ \sin c & \cos c & 0\\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix} \\&= \begin{bmatrix} \cos c - \sin c \\ \sin c + \cos c \\ 0\end{bmatrix} \\&= (\cos c - \sin c)\partial_x  + (\sin c + \cos c)\partial_y.\end{align}$$
