I am new to the Lie Algebra of a Lie Group and I have difficulty showing that the Lie algebra of the Lie Group of the $n$-torus $\mathbb{T}^n = \mathbb{S}^1 \times \cdots \times \mathbb{S}^1$ is isomorphic to $\mathbb{R}^n$.
I know that using angle coordinates $\theta^i$ for each $i$th $\mathbb{S}^1$ factor, $\partial/ \partial \theta^i$, which is the angle coordinate vector field, forms a basis $(\partial/ \partial \theta^1, \dots , \partial/ \partial \theta^n)$ for Lie($\mathbb{T}^n)$.
But how can I show that this is abelian, i.e. for all smooth left invariant vector fields, $X$ and $Y$ on $\mathbb{T}^n$, we have $[X,Y]=0$? I know the $[\partial / \partial \theta^i, \partial/ \partial \theta^j]=0$ for all $i,j$ but how do I use this to show that $[X,Y]=[X^i \frac{\partial}{\partial \theta^i},Y^i \frac{\partial}{\partial \theta^i}]=(X^i \frac{\partial Y^j}{\partial \theta^i}- Y^i \frac{\partial X^j}{\partial \theta^i})\frac{\partial}{\partial \theta^j}=0?$ And why is this isomorphic to $\mathbb{R}^n$?
Best Answer
I'm writing a longer response here rather than in the comments, since it is no place for extended discussion.
If $\Gamma\subset V$ a lattice, the map $V\to V/\Gamma$ is a local diffeomorphism as the associated short exact sequence $0\to\Gamma\to V\to V/\Gamma\to 0$ induces $0\to Lie(\Gamma)\to V\to Lie(V/\Gamma)\to 0$ as a short exact sequence of Lie algebras (here we make the identification $Lie(V)\cong T_eV\cong V$) and since $\dim \Gamma=0$ the map $V\to Lie(V/\Gamma)$ is an isomorphism.
More generally if $\pi:M\to M/G$ where $G$ is a discrete group acting via a covering space action we have an identification $T_q\pi:T_qM\to T_p(M/G)$ for $q\in \pi^{-1}(p)$. Changing between identifications is done by precomposing by $T_{\phi_g^{-1}q}\phi_g$. We can actually think of vector fields on $M/G$ as vector fields on $M$ which are invartiant with respect to $G$, i.e. $\phi_{g*}X=X$.
We can then think of a vector field on $V/\Gamma$ as a vector field on $V$ which is invariant under the $\Gamma$ action. In particular a vector field $X$ on $V$ defines a vector field on $V/\Gamma$ if $X(v+\gamma)=d\phi_{\gamma}(v)X(v)$ and since the differential $d\phi_{\gamma}(v)=id: V\to V$ (remember that $T_vV\cong V$ canonically so the derivative $d\phi_{\gamma}$ is just $T_v\phi_{\gamma}$) we have $X(v+\gamma)=X(v)$ for all $\gamma\in \Gamma$. This means we can identify $\mathfrak{X}(V/\Gamma)$ with $\Gamma$ invariant smooth functions $V\to V$. Now for such a vector field to be left invariant it must have $X(v+u)=X(v)$ for all $u\in V$ since the $V/\Gamma$ action on $\mathfrak{X}(V/\Gamma)$ descends from the action of $V$ on $\mathfrak{X}(V)$, that is, the following diagram commutes $\require{AMScd}$ \begin{CD} \mathfrak{X}(V)^\Gamma @>{\Psi}>> \mathfrak{X}(V/\Gamma)\\ @V{\phi_{\gamma*}}VV @VV{\phi_{[\gamma]*}}V\\ \mathfrak{X}(V)^\gamma @>{\Psi}>>\mathfrak{X}(V/\Gamma) \end{CD} Where $\Psi$ is the map which identifies vector fields on $V/\Gamma$ with $\Gamma$ invariant vector fields on $V$. Of course if $X(v+u)=X(v)$ for all $u\in V$ then $X$ is actually constant and an element of $Lie(V/\Gamma)$ is just an element in $V$.
Per my discussion in my last comment, $\mathbb{R}^n$ is a vector space with basis elements denoted by $e^1=(1,0, \cdots,0), e^2=(0,1,\cdots,0)$ and so on, so the map I described $\partial_{\theta^i}\mapsto e^i$ is just the map which takes the ith basis vector of $Lie(T^n)$ to the ith basis vector of $\mathbb{R}^n$.