# Is there a manifold structure on a space of conformal maps?

I would be very grateful for any information or pointers for the following:

1) Fix an open subset $U$ of $\mathbb{CP}^1$. a) Does the set of all holomorphic maps from $U$ to $\mathbb{C}$ (with the compact-open topology) have the structure of a manifold in any sense? b) Is there even a notion of a differentiable structure, and what is the tangent space at a typical point (e.g. at the identity)? Does the subset of maps that are conformal on $U$ (i.e. have non-vanishing derivative there) inherit any sensible structure?

2) Is it possible to allow the domain $U$ to vary, e.g. is it possible to consider a collection of all maps from all possible domains (say simply connected ones)?

(I am coming across these maps in the context of conformal loop ensembles (CLEs), which are random families of (countably many, a.s.) loops in $U$, and in order to express certain constructions on these CLEs it appears that one should consider "differentiating" in the space of conformal maps.)

Many thanks!

Update. Maybe some further thoughts: If I fix $U$ to be, say, the open unit disk, then the space of holomorphic maps on $U$ certainly forms a topological vector space. Let's call it $H$. Is this a manifold in any sense (Frechet, I suppose)? Is it smooth (under which notion of differentiability)?

Next, if I restrict to those maps which are conformal on $U$, let's call this $A$, I don't seem to get a vector space; though I think $A$ is a closed subset of $H$ (in the compact-open topology), not being conformal at a point in $U$ is an open condition(?). But what can be said about the topology of $A$? Does $A$ contain a subspace which is an affine space modeled on some space of holomorphic functions? (I.e. "conformal + holomorphic = conformal"?)

• On the question of a manifold structure on holomorphic maps, I recommend taking a look at Kriegl and Michor's book "A convenient setting for global analysis" (downloadable free from the AMS bookstore). I don't remember the details, but they do some work with spaces of holomorphic and analytic maps. – Andrew Stacey Jan 26 '11 at 15:58
• Formally speaking, the tangent space at a typical point should be the space of holomorphic tangent fields on $U$, which is noncanonically isomorphic to the space of entire functions on $U$. You might need to cut that down a bit with growth conditions to get compatibility with whatever differentiable structure you end up imposing on the total space. Also, conformality at a point is an open condition, since it concerns nonvanishing of a derivative. – S. Carnahan Feb 10 '11 at 18:53
• Yes, for any open set $U \subseteq \mathbb{C}^n$, the set $\mathrm{Hol}(U)$ of all holomorphic functions on $U$ is a Frechet space; just take the seminorms $p_n$ to be $p_n(f) = \sup_{K_n} |f|$, where $\{ K_n \}$ is any compact exhaustion of $U$, i.e. a sequence of compact sets $K_1, K_2, K_3, \ldots$ whose union is the whole of $U$, with $K_n \subset \mathrm{int}(K_{n+1})$. – Zen Harper Mar 11 '11 at 6:49
• ...sorry, by "conformal", do you mean locally injective (which is equivalent to $f'(z) \ne 0$ at every point)? Or do you mean globally injective (which is much harder, I think). Assuming $U$ to be connected, the locally uniform limit of locally injective holomorphic functions is either locally injective or constant. So, I think $A$ is not closed unless you add in the constant functions also. P.S. I'm only considering the one-variable case here; I'm not very familiar with Several Complex Variables. – Zen Harper Mar 11 '11 at 6:56
• ...sorry, I've just noticed you're talking about domains in $\mathbb{C}\mathbb{P}^1$, not $\mathbb{C}$. But presumably it is not too different? – Zen Harper Mar 11 '11 at 6:58

A quick comment: I assume you want $U$ to be "non-trivial" i.e. not equal to $\mathbb{C}$ itself; if it were, then the collection of such maps should be infinite dimensional (in particular, it would contain every polynomial).

So assume that $U$ is non-trivial. I'll also assume that $U$ is simply connected, though I'm pretty sure that you can do away with this assumption. Thus $U$ is biholomorphic to the unit disc in $\mathbb{C}$, so we will assume it is the unit disc.

The holomorphic self-maps of the unit disc contain the group $G = PSL_2(\mathbb{R})$ (this is its group of automorphisms, actually). This is a real 3-manifold, so if you restrict yourself to biholomorphisms, you're good.

However, it also contains the maps $z \mapsto z^k$, and so all conjugates of these maps by $G$. There might be something more you can say about this, but I'm not at the moment sure what.

• Well, if $U$ is all of $\mathbb{CP}^1$, then the space of maps is just $\mathbb{C}$, which is not so interesting, but if $U = \mathbb{C}$, then we get the space of entire functions, which I'd also like to include. I certainly don't just want self-biholomorphisms, since as you say those form a finite-dimensional manifold. Rather, I'd like <i>all</i> holomorphic maps to $\mathbb{C}$. – Thomas K Jan 26 '11 at 16:36
• What you're writing doesn't make much sense to me: you say you exclude the space of all entire functions because it is infinite-dimensional, but then you consider the space of all holomorphic functions on the unit disc, which is certainly much bigger (by analytic continuation, the space of entire functions sits inside the latter). The original question is surely about giving the space of holomorphic maps the structure of an infinite-dimensional manifold. – Dan Petersen Jan 26 '11 at 16:45
• No, sorry, I'm not excluding anything. I was merely responding to the question of whether $U$ has to be "non-trivial" in any sense. The original question is about giving the space of holomorphic maps on an arbitrary domain $U$ (let's say simply-connected) the structure of a manifold. I suppose it suffices to consider the cases where $U = \mathbb{CP}^1, \mathbb{C}$ and the open unit disk. Then, having built this manifold (smooth?) I would like to investigate the subspace of conformal maps. – Thomas K Jan 26 '11 at 19:59
• Sorry for the confusion. My comment was directed at Simon Rose. – Dan Petersen Jan 27 '11 at 9:37
• Just a comment: the whole thing really depends on the choice of the domains and targets: for example the space $Hol_{d}(\mathbb{CP}^{1},\mathbb{CP}^{1})$ of degree $d$ rational maps is a $(2d+1)$-dimensional manifold. But Milnor in his book "Dynamics in one complex variable" also mentions an example of a genus 5 surface $S$ so that the space of meromorphic functions $Hol(S,\mathbb{CP}^{1})$ has singularities... – Sylvain Bonnot Jun 17 '11 at 4:23