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"?)

compact exhaustionof $U$, i.e. a sequence of compact sets $K_1, K_2, K_3, \ldots$ whose union is the whole of $U$, with $K_n \subset \mathrm{int}(K_{n+1})$. $\endgroup$ – Zen Harper Mar 11 '11 at 6:49locallyinjective (which is equivalent to $f'(z) \ne 0$ at every point)? Or do you meangloballyinjective (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. $\endgroup$ – Zen Harper Mar 11 '11 at 6:561more comment