[Math] Lie algebra of the two dimensional torus

Question

In "An Introduction To Manifold (2nd Edition)" Remark 16.15, it says " For the torus $\mathbb{R}^2 / \mathbb{Z}^2$, the Lie algebra $\mathfrak{g}$ has $\mathbb{R}^2$ as the underlying vector space and the one-dimensional Lie subalgebras are all the lines through the origin".

My questions are:

1. What's the bracket of $\mathfrak{g}$ which has $\mathbb{R}^2$ as underlying vector space? To make $\mathbb{R}^2$ as a Lie algebra, we can define $[x,y] = 0, x,y\in \mathbb{R}^2$, or $[a,a]=[b,b]=0$ and $[a,b]=a$ where $a$ and $b$ are two basis of $\mathbb{R}^2$. Are both bracket definition valid for 2-torus?

2. Are there possible other lines than those passing through the origin that are one-dimensional Lie subalgebras with bracket definitions other than above two?

Thanks

Answer 1

The projection $\pi \colon \mathbb{R}^2 \rightarrow \mathbb{R}^2 / \mathbb{Z}^2$ between $\mathbb{R}^2$ and the torus is a group homomorphism that is also a local diffeomorphism and so it induces a lie algebra isomorphism $d \pi_{(0,0)} \colon \mathfrak{g} \rightarrow \mathfrak{h}$ between the lie algebra $\mathfrak{g}$ of $\mathbb{R}^2$ and the lie algebra $\mathfrak{h}$ of the torus and $\mathfrak{g}$ can be identified with $T_{(0,0)} \mathbb{R}^2 \cong \mathbb{R}^2$. Since both groups are abelian, their lie algebras are also abelian and so the lie bracket is trivial for both of them.

A one-dimensional subspace (line passing through the origin) of a lie algebra is always a lie algebra. The bracket acts on each line trivially.