Does normalized Ricci flow on surfaces yield a bundle? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T10:58:42Z http://mathoverflow.net/feeds/question/92616 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92616/does-normalized-ricci-flow-on-surfaces-yield-a-bundle Does normalized Ricci flow on surfaces yield a bundle? Malte 2012-03-29T22:04:10Z 2012-03-31T14:49:32Z <p>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 <a href="http://mathoverflow.net/questions/47981/ricci-flow-with-surgery-in-dimension-2" rel="nofollow">this</a>.</p> <p>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.</p> <p>Question: Is this map a fibre bundle?</p> http://mathoverflow.net/questions/92616/does-normalized-ricci-flow-on-surfaces-yield-a-bundle/92658#92658 Answer by Deane Yang for Does normalized Ricci flow on surfaces yield a bundle? Deane Yang 2012-03-30T08:05:21Z 2012-03-30T08:05:21Z <p>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?</p> http://mathoverflow.net/questions/92616/does-normalized-ricci-flow-on-surfaces-yield-a-bundle/92745#92745 Answer by Misha for Does normalized Ricci flow on surfaces yield a bundle? Misha 2012-03-31T11:48:22Z 2012-03-31T14:49:32Z <p>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.) </p>