# Lie bracket of left invariant vector fields on Torus

I think that if $$X,Y$$ are two left invariant vector fields on the $$n$$-torus $$S^1\times ... \times S^1$$, then their Lie bracket is zero. But I don’t know how to prove it. How should I start?

• This follows from the fact that the group $S^1 \times \cdots \times S^1$ is abelian. Commented May 2, 2019 at 16:30
• Do you know how to write down explicitly the vector fields? Commented May 2, 2019 at 16:32
• @Travis so I must prove every abelian Lie group has this property
– user555729
Commented May 2, 2019 at 16:54
• @ArcticChar Yes I can but I didn’t write them because I think it has a more general solution
– user555729
Commented May 2, 2019 at 16:55
• But we you write that down correctly, then it is pretty obvious that the lie bracket is zero. Commented May 2, 2019 at 17:10

If $$G$$ is Abelian, then $$\mathrm{inv}: G\to G$$ given by $$g \mapsto g^{-1}$$ is a homomorphism. Therefore, its differential at identity gives us a Lie algebra homomorphism $$D_I\,\mathrm{inv}$$ such that $$G \mapsto -G$$. If $$X,Y \in \mathfrak{g}$$ then
$$[X,Y] = [-X,-Y]=[D_I \mathrm{inv}(X),D_I \mathrm{inv}(Y)]=D_I\mathrm{inv}[X,Y]=-[X,Y]$$
Since $$S^1 \times \cdots \times S^1$$ is an Abelian group, your idea that the Lie bracket on left invariant vector fields on $$n$$-torus vanishes is correct.