Is there a manifold structure on a space of conformal maps? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T00:20:12Z http://mathoverflow.net/feeds/question/53360 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53360/is-there-a-manifold-structure-on-a-space-of-conformal-maps Is there a manifold structure on a space of conformal maps? Thomas K 2011-01-26T15:32:48Z 2011-10-21T06:22:12Z <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#53367 Answer by Simon Rose for Is there a manifold structure on a space of conformal maps? Simon Rose 2011-01-26T16:00:29Z 2011-01-26T16:00:29Z <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>