Can the group of holomorphic automorphisms of an open subset of the complex plane be isomorphic to the additive group of real numbers?

Question

We can construct open sets in the complex plane $\mathbb{C}$ whose automorphism group is isomorphic to $\mathbb{Z}$, but is there an open set whose automorphism group is isomorphic to $\mathbb{R}$?

Answer 1

No, we cannot.

Let $S$ be a connected noncompact Riemann surface whose group of conformal automorphisms is nondiscrete. Then (this follows from the uniformization theorem) $S$ is either conformal to the cylinder or to the annulus or to the once punctured disk or to the disk. Now, it is elementary to observe that the (connected components of the identity, which have index at most 2 in the full automorphism group) groups of conformal automorphisms of these surfaces are respectively ${\mathbb C}^\times$, $S^1$, $S^1$ and $PSL(2, {\mathbb R})$.

Edit. In the case $S$ is disconnected one can argue as follows. If two components of $S$ are conformal to each other then $Aut(S)$ contains order 2 elements, while ${\mathbb R}$ is torsion-free. Now, assume that none of the components are conformal to each other. You have the connected components $S_j, j\in J$. Then $$ Aut(S)\cong \prod_{j\in J} Aut(S_j).$$ Then, none of the exceptional cases $Aut_0(S_j)\cong {\mathbb C}^\times, \cong S^1, \cong PSL(2,R)$, can occur (since they all contain nontrivial elements of finite order), hence, each $S_j$ is a hyperbolic surface with nonabelian fundamental group. Then $Aut(S_j)$ is discrete and, moreover, contains no infinitely divisible infinite order elements, i.e. infinite order elements $g$ such that $g^{1/n}$ exists for infinitely many natural numbers $n$ (for this you need to know a little bit about Fuchsian groups, I can explain if you are interested). This means that $Aut(S)$ also has no infinitely divisible elements. However, the entire group ${\mathbb R}$ is divisible. A contradiction.

Answer 2

It cannot be $R$. Consider two cases:

a) Your region $D$ is hyperbolic Then $D=H/G$, where $H$ is the upper half-plane, and $G$ a discrete group. Let $\Gamma$ be the pullback of your group of automorphisms. Then we have $\gamma g=\beta(g)\gamma$ for every $\gamma\in\Gamma$ and every $g\in G$ and some automorphism $\beta:G\to G$. As $\Gamma$ is not discrete, but $G$ is discrete, we conclude that $\beta=\mathrm{id}$, so $G$ and $\Gamma$ commute. But the centralizer of any element $h$ of $SL(2,R)$ is the one-parametric subgroup passing through $h$. We conclude that $G$ is contained in a one-parametric group. As $G$ is discrete, and infinite (unless it is trivial), it must be isomorphic to $Z$, or be trivial. Then $D$ is a punctured disc or a ring or the disk if $G$ is trivial. In all three cases the full group of automorphisms is not isomorphic to $R$.

b) If $D$ is parabolic, it is either $C$ or $C^*$. In both cases the full group of automorphisms is not isomorphic to $R$.