Affine transformation Group $\operatorname{Aff}(1)$ is the only connected, non-abelian, 2D Lie group

Question

I'm reading Stillwell's Naive Lie Theory. On page 88 he says,

(The affine group of the line) $\operatorname{Aff}(1)$ is in fact the only connected, nonabelian two-dimensional Lie group.

May I ask how to prove this?

Answer 1

Here is one way to do it. It is just a straightforward application of general facts about Lie theory although a simpler argument more specific to the $2$-dimensional case may also be possible. Let $\mathfrak{g}$ be a nonabelian $2$-dimensional Lie algebra. Let $X, Y \in \mathfrak{g}$ be any basis. By hypothesis the commutator $[X, Y] = aX + bY$ where $a, b$ are not both zero. WLOG $b \neq 0$. This gives

$$\left[ \frac{X}{b}, aX + bY \right] = aX + bY$$

so changing basis to $Z = \frac{X}{b}, W = aX + bY$ we get $\boxed{ [Z, W] = W }$. This determines the unique nonabelian $2$-dimensional Lie algebra, over any field. It is the semidirect product $K \rtimes K$ of two copies of the $1$-dimensional Lie algebra, where the action of $K$ on $K$ is the unique nontrivial one.

Over $\mathbb{R}$, $\mathfrak{g}$ has a unique corresponding simply connected Lie group $G$, which is the semidirect product $\mathbb{R} \rtimes \mathbb{R}$ where the action of $\mathbb{R}$ on $\mathbb{R}$ is given by $r \mapsto e^r$ (we could take any other positive base for the exponential as well). Equivalently, this is the semidirect product $\mathbb{R} \rtimes \mathbb{R}_{+}^{\times}$, which means strictly speaking it is an index $2$ subgroup in the full affine group $AGL_1(\mathbb{R}) \cong \mathbb{R} \rtimes \mathbb{R}^{\times}$, which is disconnected.

Any other Lie group with Lie algebra $\mathfrak{g}$ must be covered by $G$, so must be a quotient of $G$ by a discrete subgroup of its center. But $G$ has trivial center. We can do this by a direct calculation but it's cleaner to argue geometrically: an element of this group has the form $x \mapsto ax + b$ where $a > 0$. If $a \neq 1$ then this map is a scaling around some point, and conjugating by translation changes the location of that point, so it does not commute with translation. If $a = 1$ then this map is a translation, and as we just saw conjugating by translation changes the location of scalings, so it does not commute with scalings.

So $Z(G) = 1$ and $G$ covers no other Lie groups. The conclusion is that $\mathbb{R} \rtimes \mathbb{R} \cong \mathbb{R} \rtimes \mathbb{R}_{+}^{\times}$ is the unique $2$-dimensional connected nonabelian Lie group.