7
$\begingroup$

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...

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – user98602
    Oct 3 '16 at 3:28
  • $\begingroup$ 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. $\endgroup$ Oct 5 '16 at 22:55
4
+50
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Thanks for your answer. Do you mind explaining why $F$ is contractible? $\endgroup$ Oct 10 '16 at 14:24
  • $\begingroup$ @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. $\endgroup$ Oct 10 '16 at 15:07
  • $\begingroup$ Hence, this principal fibre bundle is trivial. Thus, its base is contractible. $\endgroup$ Oct 10 '16 at 15:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.