Computation of a left-invariant vector field Consider $\mathbb{R}^3\times\mathbb{R}^3$ as a Lie group endowed with the group operation $(x,y).(u,v)=(x+u,y+v+x\times u) $ where ( $x\times u $ is the cross product of the $\mathbb{R}^3$ vectors $x$ and $u$).
I need to explicitly calculate the unique left-invariant vector field $X$ such that $X_{(0,0)}=(u,v)\in T_{(0,0)}(\mathbb{R}^3\times\mathbb{R}^3)$
Now using $X_{(a,b)}=DL_{(a,b)}(X_{(0,0)})$ after some computations I'm getting $X_{(a,b)}=(u,v+a\times u)$ . But the answer seems to be $X_{(a,b)}=(a+u,b+v+a\times u)$.
So can you please give me an explicit computation for $X_{(a,b)}$ so that I know which one of the above is correct.
 A: If $G$ is any Lie group and $X$ is a left-invariant vector field, then $X_g = {\rm d}({\rm L}_g)_e(X_e)$ for every $g\in G$.
If $G = \Bbb R^6$ with this operation, we have that $X_{(a,b)} ={\rm d}({\rm L}_{(a,b)})_{(0,0)}(X_{(0,0)})$. We're requiring $X_{(0,0)}$ to be equal to $(u,v)$. This means that we may just compute the total derivative of ${\rm L}_{(a,b)}$ at the point $(0,0)$, and evaluate it at $(u,v)$. The decomposition $\Bbb R^6 = \Bbb R^3 \times \Bbb R^3$ allows us to compute ${\rm d}({\rm L}_{(a,b)})_{(0,0)}$ as a (block) matrix. First of all, write $${\rm L}_{(a,b)}(x,y) = (x+a,y+b+a\times x).$$Thus $${\rm d}({\rm L}_{(a,b)})_{(0,0)} = \begin{pmatrix} {\rm Id}_{\Bbb R^3} & 0 \\ a\times \_ & {\rm Id}_{\Bbb R^3}\end{pmatrix},$$by setting the "gradients" of the entries of ${\rm L}_{(a,b)}$ in "rows" and using that, given $a$, $x\mapsto a\times x$ is a linear map (and so its derivative is itself). Hence $${\rm d}({\rm L}_{(a,b)})_{(0,0)}(u,v) = \begin{pmatrix} {\rm Id}_{\Bbb R^3} & 0 \\ a\times \_ & {\rm Id}_{\Bbb R^3}\end{pmatrix} \begin{pmatrix} u \\ v \end{pmatrix} = \begin{pmatrix} u \\ v+ a\times u \end{pmatrix}. $$Abandoning matrices, the conclusion is that $X_{(a,b)} = (u,v+a\times u)$.
Your conclusion is correct. Taking the derivative of ${\rm L}_{(a,b)}$ eliminates $a$ from the first entry and $b$ from the second, because derivatives of translations are the identity. The map ${\rm L}_{(a,b)}$ acts on points, its derivative acts on vectors. Don't be misled by the natural isomorphisms $T_{(x,y)}(\Bbb R^3\times \Bbb R^3) = \Bbb R^3 \times \Bbb R^3$: they do not intertwine ${\rm L}_{(a,b)}$ and ${\rm d}({\rm L}_{(a,b)})_{(x,y)}$.
