# Topology of the space of conformal classes of metrics

Definitions: Consider on a fixed smooth manifold $M$ the space $\text{Met}(M)$ of Riemannian metrics on $M.$ This lives inside an infinite dimensional topological vector space (in fact, in is a Frechet space).

Two metrics $h,g \in \text{Met}(M)$ are said to be conformally equivalent if there exists a nonvanishing (a fortiori positive) smooth function $f$ such that $g(-,-)=fh(-,-).$ This defines an equivalence relation on $Met(M).$ We define the quotient space $\text{Conf}(M):= \text{Met}(M)/\{\text{conformal equivalence}\}$ and endow it with the quotient topology.

My question: It's clear that $\text{Met}(M)$ is contractible since any two metrics can be joined by a straight-line homotopy. Is it true that $\text{Conf}(M)$ is also contractible?

Note that we have the fiber sequence $\{\text{positive functions on M}\} \to \text{Met}(M) \to \text{Conf}(M).$ Since the first two spaces are contractible by straight-line homotopies, it follows from the long exact sequence on homotopy groups that $\text{Conf}(M)$ has vanishing homotopy groups. If this space had the homotopy type of a CW complex, it would be contractible by Whitehead's theorem. However I don't see why it would have the homotopy type of a CW complex...

• It suffices to show that this is a metrizable manifold by a theorem of Palais. But I think this is straightforward, as it's locally modeled on a Frechet space modded out by a closed subspace.
– user98602
Oct 3 '16 at 3:28
• You should state which topology you are using on $Met(M)$ for the question to make sense. Also, $Met(M)$ is not a vector space; the natural structure is the one of an open convex subset of a metric space. Oct 5 '16 at 22:55

Since you did not specify the topology on the space of conformal classes, I will make up my own. Namely, the set of conformal structures on $M^n$ can be identified with the set of reductions of the frame bundle to the bundle whose structure group is the conformal group $CO(n)\cong R_+\times O(n)$. In other words, this is the set of sections of the bundle $E$ over $M$ whose fibers are copies of $F=GL(n,R)/CO(n)$, which is a contractible manifold. I will therefore equip $Conf(M)$ with the $C^\infty$-compact-open topology on the space of sections of $E\to M$. Now my answer to this question proves that $Conf(M)$ is contractible.
• Thanks for your answer. Do you mind explaining why $F$ is contractible? Oct 10 '16 at 14:24
• @MichaelAlbanese: There are two ways to see this. One is to observe that this space is diffeomorphic to $SL(n,R)/SO(n)$, which is the symmetric space for the group $SL(n,R)$ and, hence, has nonpositive curvature and hence is contractible by Cartan-Hadamard theorem. The second is to not that $GL(n,R)/O(n)$ is the convex cone of positive definite bilinear forms and hence, contractible. The multiplicative group $R_+$ acts on this space by scaling making it a principal fiber bundle which has to be trivial since it admits a section, coming from symmetric matrices with unit determinant. Oct 10 '16 at 15:07