I would like to know what the simply-connected Lie groups of dimension $2$ are.
It is well-known that for every Lie algebra, there is exactly one simply-connected Lie group having it as its Lie algebra. I know that there are two two-dimensional Lie algebras, namely the abelian Lie algebra and another Lie algebra with basis ${x, y}$ such that $[x, y]=x$. The simply-connected Lie group corresponding to the abelian Lie algebra is $\mathbb{R}^2$. Can you tell me, what the Lie group corresponding to the other Lie algebra is?
[Math] What are the simply-connected two-dimensional Lie groups
lie-groups
Best Answer
The group $\textrm{Aff}^+(\Bbb R) \cong \Bbb R^+ \!\ltimes \Bbb R$ of affine, oriented transformations of the real line (under composition) has dimension $2$ and is nonabelian, so by your classification its Lie algebra is the nonabelian algebra you mention (for an appropriate choice of basis). The above isomorphism identifies the usual composition of affine functions $t \mapsto a t + b$ and $t \mapsto a' t + b'$ with the operation $$(a, b) \cdot (a', b') := (a a', a b' + b).$$
One can realize $\textrm{Aff}^+(\Bbb R)$ as the matrix group $$\left\{\pmatrix{a&b\\0&1} : a, b \in \Bbb R; a > 0 \right\} < \textrm{GL}(2, \Bbb R)$$ via $(t \mapsto a t + b) \leftrightarrow \pmatrix{a&b\\0&1} .$
The induced Lie algebra is $$\mathfrak{aff}^+ = \left\{\pmatrix{a&b\\0&0} : a, b \in \Bbb R\right\} < \frak{gl}(2, \Bbb R),$$ and the basis $(x, y)$ given by $$x = \pmatrix{0&1\\0&0}, \quad y = \pmatrix{-1&0\\0&0}$$ satisfies the given relation $[x, y] = x$.
Finally, we can just as well realize this group structure on the set $\Bbb R^2$ with a nonstandard multiplication $\ast$: Conjugating the group operation $\cdot$ given above by the diffeomorphism $\Bbb R^2 \to \Bbb R^+ \!\ltimes \Bbb R$ given by $(r, s) \mapsto (e^r, s)$ yields the rule $$(c, d) \ast (c', d') = (c + c', d + e^c d').$$