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?
Lie bracket of left invariant vector fields on Torus
differential-geometrylie-algebraslie-groupssmooth-manifoldsVector Fields
Best Answer
We know that the set of left-invariant vector fields on a Lie group under Lie bracket is isomorphic to the Lie algebra of the group. On the other hand, the bracket operation in the Lie algebra of an Abelian group vanishes. Here's why:
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.