I have the matrix Lie-group $G$ consisting of matrices of the form
$$X = \begin{bmatrix}1&x&y\\ 0&1&z\\ 0&0&1\end{bmatrix}$$
I have to prove that the set $\{\frac{\partial }{\partial x}, \frac{\partial }{\partial y},x\frac{\partial }{\partial y} + \frac{\partial }{\partial z} \}$
is a basis for the space of left invariant vector fields $X_L(g)$.
From my course I know that the map that takes any $$v \mapsto L^v, v \in T_eG, L^v \in X_L(g)$$
where
$$L^v(g) = T_eL_g(v) , L_g \text{ being the left translation,}$$
is a linear isomorphism.
I have found a basis for $$T_eG,$$ $$ v_1=\begin{bmatrix}0&1&0\\ 0&0&0\\ 0&0&0\end{bmatrix} , v_2=\begin{bmatrix}0&0&1\\ 0&0&0\\ 0&0&0\end{bmatrix}, v_3=\begin{bmatrix}0&0&0\\ 0&0&1\\ 0&0&0\end{bmatrix}$$ which I regard as $$(T_e\theta)^{-1}(e_i), \text{where } \theta : X \mapsto (x,y,z) \text{is a global chart}$$.
Now I know $\{ L^{v_1},L^{v_2},L^{v_3} \}$ will be a basis for $X_L(g)$.
It's not clear to me though how I get from the matrix $v_1$ to $\{\frac{\partial }{\partial x} \}$.
Also, after I clarify this, is it enough of an argument that since $T_eG$ is a vector space, then $X_L(G)$ is also a vector space by the isomorphism described above, and since $\{\frac{\partial }{\partial x}, \frac{\partial }{\partial y},x\frac{\partial }{\partial y} + \frac{\partial }{\partial z} \}$ can be obtained from $\{\frac{\partial }{\partial x}, \frac{\partial }{\partial y},\frac{\partial }{\partial z} \}$, it is also a basis since the isomorphism is linear ?
Also, if $\theta(g)=(x(g),y(g),z(g))$, then $x\frac{\partial }{\partial y}|_g =x(g)\frac{\partial}{\partial y}|_g$ ?
I follow a course which defines tangent spaces as sets of equivalence classes of curves. Any help would be appreciated, sorry for any mistakes but this is my first post.
Best Answer
Yes so I know I'm still bad at this but, if you prove that all the vector fields in the set are left-invariant and that they are independently linear (by the bijection I've described there is a v.s. structure on the set of left-invariant vector fields OR because you know there is one on the space of normal vector fields), since the space of normal vector fields has dimension 3, then the set is indeed a basis for the space.