Is there a manifold structure on a space of conformal maps? - MathOverflow most recent 30 from http://mathoverflow.net2013-06-20T00:20:12Zhttp://mathoverflow.net/feeds/question/53360http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/53360/is-there-a-manifold-structure-on-a-space-of-conformal-mapsIs there a manifold structure on a space of conformal maps?Thomas K2011-01-26T15:32:48Z2011-10-21T06:22:12Z
<p>I would be very grateful for any information or pointers for the following:</p>
<p>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?</p>
<p>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)?</p>
<p>(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.)</p>
<p>Many thanks!</p>
<p><i>Update.</i> 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)?</p>
<p>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), <i>not</i> 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"?)</p>
http://mathoverflow.net/questions/53360/is-there-a-manifold-structure-on-a-space-of-conformal-maps/53367#53367Answer by Simon Rose for Is there a manifold structure on a space of conformal maps?Simon Rose2011-01-26T16:00:29Z2011-01-26T16:00:29Z<p>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).</p>
<p>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.</p>
<p>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.</p>
<p>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.</p>