I Constructing left invariant vector fields on SO(3)

[Mikeey aleex]

hello every one
can one please construct for me left invariant vector field of so(3) rotational algebra using Euler angles ( coordinates ) by using the push-forward of left invariant vector field ? iv'e been searching for a method for over a month , but i did not find any well defined method for my case . i uploaded a pdf the give a direct formula for left invariant vector field , I've been trying to use push-forward method for one parameter subgroup , but i failed

 

[fresh_42]

hello every one
can one please construct for me left invariant vector field of so(3) rotational algebra using Euler angles ( coordinates ) by using the push-forward of left invariant vector field ? iv'e been searching for a method for over a month , but i did not find any well defined method for my case . i uploaded a pdf the give a direct formula for left invariant vector field , I've been trying to use push-forward method for one parameter subgroup , but i failed

You could proceed along the lines
$$
\mathbb{S}^3 \stackrel{\cong_\varphi}{\longrightarrow} \mathbb{U(1,\mathbb{H}}) \stackrel{\cong_\psi}{\longrightarrow} SU(2,\mathbb{C}) \stackrel{Ad}{\longrightarrow} SO(\mathfrak{su(2,\mathbb{C}})) \stackrel{=}{\longrightarrow} SO(3,\mathbb{R})
$$
The first two isomorphisms with
$$
\varphi(x_1+x_2+x_3+x_4)=x_1\mathbf{1}+x_2\mathbf{i}+x_3\mathbf{j}+x_4\mathbf{k}
$$
and
$$
\psi(x_1\mathbf{1}+x_2\mathbf{i}+x_3\mathbf{j}+x_4\mathbf{k}) = \begin{bmatrix}x_1\mathbf{1}+x_2\mathbf{i} & -x_3\mathbf{1}-x_4\mathbf{i} \\x_3\mathbf{1}-x_4\mathbf{i} & x_1\mathbf{1}-x_2\mathbf{i} \end{bmatrix}
$$
are quite easy but the coordinate form of $Ad$ isn't so pleasant, at least in Cartesian coordinates. Maybe it's a better in Euler Coordinates.

 

[Mikeey aleex]

You could proceed along the lines
$$
\mathbb{S}^3 \stackrel{\cong_\varphi}{\longrightarrow} \mathbb{U(1,\mathbb{H}}) \stackrel{\cong_\psi}{\longrightarrow} SU(2,\mathbb{C}) \stackrel{Ad}{\longrightarrow} SO(\mathfrak{su(2,\mathbb{C}})) \stackrel{=}{\longrightarrow} SO(3,\mathbb{R})
$$
The first two isomorphisms with
$$
\varphi(x_1+x_2+x_3+x_4)=x_1\mathbf{1}+x_2\mathbf{i}+x_3\mathbf{j}+x_4\mathbf{k}
$$
and
$$
\psi(x_1\mathbf{1}+x_2\mathbf{i}+x_3\mathbf{j}+x_4\mathbf{k}) = \begin{bmatrix}x_1\mathbf{1}+x_2\mathbf{i} & -x_3\mathbf{1}-x_4\mathbf{i} \\x_3\mathbf{1}-x_4\mathbf{i} & x_1\mathbf{1}-x_2\mathbf{i} \end{bmatrix}
$$
are quite easy but the coordinate form of $Ad$ isn't so pleasant, at least in Cartesian coordinates. Maybe it's a better in Euler Coordinates.

i know this method but it's not working very well , there is a clear method by Dr. frederic P. sculler, using push-forward of left invariant vector field in a chart , but the issue is that i can not project the matrix of ( SO(3) group ) in a proper way on a chart . can i do this method for a one parameter subgroup like SO(3) ?
see the videos.

 

[fresh_42]

For an atlas you should use the two stereographic projections from the north, resp. the south poles onto the equatorial plane.

I don't really like to watch the four hours of lectures now you linked to. Why don't you go the other way around: take a basis of $\mathfrak{so}(3)$ and build the exponential function with them? Or take some curves (1-parameter subgroups) on $SO(3)$ through its generators and differentiate. So, yes, this method (whatever it is) should apply here, too.

Do you want to prove something or just want to know which the basis vectors, resp. generators are? They can easily be written out.
Or simply note that $X^\tau X = 1$ becomes $X^\tau + X = 0$ and $\det X = 1$ becomes $\operatorname{trace} X = 0$.
What is done for $SL(2)$ can also be done for $SU(2)$ and $SO(3)$ in a similar way. Their Dynkin diagrams are all the same, namely $\circ$.