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How would one go about such a classification and is there anything interesting that can be said about their corresponding Lie groups?

user560483
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    There is only one nontrivial one (up to isomorphism). It has bracket $[x,y] = y$. Try to prove that for any 2-dimensional algebra with bracket $[x,y] = ax+by$, that you can change basis so that $[x,y] = y$. – Nick Apr 30 '19 at 17:23

2 Answers2

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There are exactly two different Lie algebras of dimension $2$ over an arbitrary field. Let $(x,y)$ be a basis. One Lie algebra is the abelian Lie algebra $K^2$, with $[x,y]=0$. The second Lie algebra is the non-abelian solvable Lie algebra $\mathfrak{r}_2(K)$, with Lie bracket $[x,y]=x$. It is easy to see that every $2$-dimensional Lie algebra with bracket $[x,y]=ax+by$ is isomorphic to one of the two. A corresponding Lie group is not considered for arbitrary fields, but rather for real and complex numbers (or $p$-adic numbers).

References:

Two Dimensional Lie Algebra

Classsifying 1- and 2- dimensional Algebras, up to Isomorphism

Dietrich Burde
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There are two and (up to isomorphism) only two real $2$-dimensinal Lie algebras:

  • The abelian $2$-dimensional Lie algebra $\mathfrak g$. In this case, $(\forall X,Y\in\mathfrak g):[X,Y]=0$. It is the Lie algebra of three connected Lie groups. $(\mathbb R^2,+)$, $(\mathbb R,+)\times(S^1,\cdot)$ and $(S^1,\cdot)\times(S^1,\cdot)$;
  • The $2$-dimensional Lie algebra $\mathfrak g$ with a basis $\{e_1,e_2\}$ such that $[e_1,e_2]=e_2$. It is the Lie algebra of one connected Lie group: the Lie group of the matrices of the form $\left[\begin{smallmatrix}a&b\\0&a^{-1}\end{smallmatrix}\right]$, with $a>0$ and $b\in\mathbb R$.
José Carlos Santos
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