Does normalized Ricci flow on surfaces yield a bundle?

As is well known, the normalized Ricci flow is defined for all $t>0$ on compact surfaces, and every metric on a compact surfaces converges to a metric constant curvature if $X \neq S^2$ (at least I can't find a reference that asserts that same result for $X=S^2$; B. Chow's "Ricci flow on the 2-sphere" only shows that metrics of positive Gaußian curvature converge to constant curvature metrics). This is somewhat related to this.

One thus has a map $\mathcal{R}(X) \rightarrow T_X$ from the Riemannian moduli space to the Teichmüller space $T_X$ of constant curvature metrics associating to $g\in \mathcal{R}(X)$ its limit $g^\ast$ under the normalized Ricci flow. Note that the fibres of this map are convex.

Question: Is this map a fibre bundle?

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The fact that one has convergence to the round metric on the sphere follows from Chow's work, maybe not the paper you mention though. Precise references are to be found in Chow and Knopff book, "The Ricci flow, an introduction". – Thomas Richard Mar 29 2012 at 22:17
You're wrong, the paper of Chow that you quote proves convergence for all metrics on $S^2$, see his Corollary 1.3. – YangMills Mar 30 2012 at 0:45
Note that even to show that your map is continuous requires some highly nontrivial analysis! – YangMills Mar 30 2012 at 2:08

2 Answers

The formal reference for the result is C. Earle and J. Eells, "A fibre bundle description of Teichmüller theory", J. Differential Geom. Volume 3, Number 1-2 (1969), 19-43. The upshot is that there are two fibrations: One is the fibration of the space of Riemannian metrics $R(S)$ over the space $C(S)$ of conformal structures (where the fiber is the space of positive functions), the second is the principal fibration of the space of conformal structures over the Teichmüller space $T(S)$ of the surface $S$ (where the fiber is $Diff_0(S)$). Combining these two one gets the fibration $R(S)\to T(S)$. To see that the first is a fibration one usually fixes a reference Riemannian metric $g_0$, then for $g\in R(S)$ the function $Vol(g)/Vol(g_0)$ determines a trivialization of $R(S)\to C(S)$. Doing this using uniformization is much harder, and requires some machinery which became fully available only by 1950s. Note that the existence of a constant curvature metric gives only a set-theoretic section. One also needs continuity which is harder and does not follow from some proofs of the Uniformization Theorem. (For instance, it does not follow from the Koebe's or Caratheodory's proofs, or the proof using Green's functions on the universal cover.)

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 Yes, this is a much better answer than mine. – Deane Yang Mar 31 2012 at 12:56

The Ricci flow preserves the conformal class of the metric, and there is a unique metric with constant curvature -1, 0, or 1 in each conformal class. So, forgetting about the Ricci flow, isn't the space of smooth Riemannian metrics over a closed Riemann surface obviously a smooth trivial bundle over the space of constant curvature metrics, where the fiber consists of smooth positive functions?

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@Deane Yang: Deane, actually, triviality of the bundle was not obvious until (I think) 1950s when Ahlfors, Bers and, independently, Bojarsky, established that solution of Beltrami equation depends continuously/smoothly/analytically on the Beltrami differential. (You need to know that metric of constant curvature depends nicely on the conformal structure in order to construct a section. Maybe in the smooth setting this was known before 1950s though...) – Misha Mar 30 2012 at 12:40
Misha, I don't know enough of the history to be able to confirm what you say, but you make a good point. What I said is "obvious", only if you know that in each conformal class there is a unique metric with constant curvature -1, 0, or 1. – Deane Yang Mar 30 2012 at 13:15