# Computation of a left-invariant vector field

differential-geometrylie-groupsVector Fields

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.

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 $${\rm d}({\rm L}_{(a,b)})_{(x,y)}$$.