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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?

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

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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.

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