Variant of the Riemann Mapping Theorem for $Conf(\mathbb H^2)$?

Question

According to the Riemann mapping theorem it is possible to map a simply connected open subset $B \subset \mathbb C$ into any other $B' \subset \mathbb C$ by a (bi-)holomorphic mapping. Moreover, such a mapping is unique.

Now I'm interested in the related case of the infinite dimensional Lie group $Conf(\mathbb H^2)$ that maps the upper half plane to itself. The Lie group is locally isomorphic to $Diff(S^1)$ (I think... some details may vary), so it should be clear that there are subsets $B' \subset \mathbb H^2$ that cannot be mapped from $B$ by a $Conf(\mathbb H^2)$ transformation.

Specifically, I'm interested in some kind of intuition about the possible forms of allowed mappings as above, i.e. some "number" of possible images of such mappings. Clearly the answer will involve $Diff(S^1)$ in some important way...

In short:

-$Diff(\mathbb C)$ maps $B$ to any $B'$ in infinite number of possible ways

-$Conf(\mathbb C)$ maps $B$ to any $B'$ in a unique way

-$Conf(\mathbb H^2)$ maps $B$ to some subsets $B'$... but what kind of $B'$s??

I actually have in mind a way to "number" such maps, but there are probably much better mathematical expositions/proofs lying around somewhere...

EDIT: OK I just knew I should've stuck with infinite dimensional Lie algebras instead of Lie groups (as per Robert Bryant's comment below)... so here are some corrections (I'll leave the above stuff intact for the sake of my own public humiliation):

1) Yes indeed I meant a proper subset which is connected and simply connected

2) OK so the Riemann Mapping is unique up to $PSL(2;\mathbb R)$... didn't realize that!

3) So instead of "$Conf(\mathbb H^2)$" let's think about holomorphic vector fields on $\mathbb H^2$, e.g.

$V = \xi(z) \partial_z + \xi(\bar z) \partial_{\bar z}$.

As $Im(z) \to 0$, these tend to $\xi(x) \partial_x \in Vect(S^1)$ (the boundary of $\mathbb H^2$ is $S^1$). I think it's safe to say that there are flows at least for some $\xi$ such that the resulting mapping is a conformal transformation? That's what I meant by $Conf(\mathbb H^2)$, which was probably wrong in many ways...

So the question then applies to flows of $V$... I hope it's clearer now!

Answer 1

A cultural remark to begin: The comments asking for clarification of your question may have sounded a bit rough, but please understand that they weren't meant personally. In the culture of mathematics, definitions are very important because we have found, time and again, that questions are often based on some misunderstanding, and clarifying the question is usually the first step in seriously concentrating on finding a solution. Training students to ask clear questions is a major part of training young mathematicians, and, for most of us who do this for a living, it becomes second nature to begin by asking the questioner to define the terms of discussion more carefully and/or pointing out that there's some confusion going on. It doesn't necessarily stop happening to you when you finish graduate school either. When I was a postdoc at the Institute for Advanced Study, my first position after graduate school, I thought I had the best opportunity in the world to learn more about Lie groups because Armand Borel was there. I found, though, that whenever I went to ask him a question, he would invariably respond "What do you mean??" in what I then thought was a stern, almost indignant, voice, as though I had just revealed how foolish and ignorant I really was. It took me some time to realize that this was almost always his response, even if he thought it was an excellent question. After I did realize that, though, we got along fine and, indeed, I learned an enormous amount from him.

Anyway, to get to your question: It seems that, by $Conf(\mathbb{H}^2)$, you mean the (real) vector fields on $\mathbb{H}^2$ whose (local) flows are holomorphic. Such a vector field $X$ is the real part of a unique holomorphic vector field $Z$ of the form $$ Z = h(z)\ \frac{\partial\ }{\partial z} = h(x+iy)\ \frac12\left(\frac{\partial\ }{\partial x}-i\frac{\partial\ }{\partial y}\right) $$ where $h$ is holomorphic in the upper half plane. People frequently write $Z$ when they mean $X = Z +\bar Z$, which is why some folks were confused by your expression for an element of $Conf(\mathbb{H}^2)$.

The problem with thinking of this infinite dimensional vector space as the Lie algebra of a Lie group is that most of these vector fields only define local flows on $\mathbb{H}^2$, not global ones, so they don't really generate automorphisms of $\mathbb{H}^2$. In fact, it's a theorem that the only ones that do are the ones for which
$$ h(z) = a + bz + cz^2 $$ where $a$, $b$, and $c$ are real numbers, and this is a Lie algebra isomorphic to ${\frak{sl}}(2,\mathbb{R})$. The flows that are generated in this way generate the Lie group of linear fractional transformations that carry $\mathbb{H}^2$ into itself, and this happens to be the group of automorphisms of $\mathbb{H}^2$ as a complex manifold.

Given this, it seems that, instead, you want $Conf(\mathbb{H}^2)$ to mean something else, namely the Lie algebra of vector fields of the above form in which $h$ is holomorphic on the entire complex plane $\mathbb{C}$ and real-valued on $\mathbb{R}$. In other words, $h$ should have its power series in $z$ have all real coefficients and have infinite radius of convergence. Then, indeed, $Conf(\mathbb{H}^2)$ injects into (but not onto) the Lie algebra $\frak{X}(\mathbb{R})$ of real analytic vector fields on $\mathbb{R}$.

By the way, if $B\subset\mathbb{C}$ is a connected and simply connected open subset, then the set of all holomorphic vector fields on $B$ will still be of the above form for some set of $h\in\mathcal{O}(B)$ (the holomorphic functions on $B$), but picking out the $3$-dimensional subalgebra whose flows generate the automorphisms of $B$ (which, by the Riemann Mapping Theorem, is a $3$-dimensional Lie subalgebra of this space) is, generally, a very difficult thing to do.