# 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. $$X = X_x \partial_x + X_y \partial_y + X_z \partial_z$$?

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}