Isometry groups of Riemannian submersions with totally geodesic fibers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:36:01Z http://mathoverflow.net/feeds/question/79186 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79186/isometry-groups-of-riemannian-submersions-with-totally-geodesic-fibers Isometry groups of Riemannian submersions with totally geodesic fibers Renato G Bettiol 2011-10-26T19:49:00Z 2012-01-25T07:22:13Z <p>Suppose $F\to M\stackrel{\pi}{\to} B$ is a Riemannian submersion with totally geodesic fibers, all manifolds compact. In general, unless $M=B\times F$ is a Riemannian product, the isometry groups of these manifolds is not related in any obvious way. In this direction:</p> <p><strong>Question 1.</strong> Give sufficient conditions so that the isometry group $Iso(B)$ of the base also acts by isometries on the total space $M$.</p> <p><strong>Question 2.</strong> Same for the isometry group $Iso(F)$ of the fiber.</p> <p>A few remarks:</p> <p>(i) In general, an isometry of the base (or of the fiber) does not extend to an isometry of the total space. Consider e.g. a Klein bottle $K$ as a $S^1$-bundle and note that any rotation of a fiber $S^1$ cannot extend to a continuous map of $K$.</p> <p>(iii) By a result of Hermann (see e.g. Besse's book on Einstein mflds), since the fibers are assumeed totally geodesic, the holonomy group is a subgroup of the isometry group of the fiber.</p> <hr> <p>[edit: Please disregard the following remark; see V Kapovitch's answer below.]</p> <p>(ii) A possible answer for (Q1), ignoring the compactness requirement, is when $\sec_B\leq 0$ and $\pi_1(B)=1$. Note this implies $B$ diffeomorphic to the Euclidean space, hence noncompact. With these assumptions it is not hard to see that an isometry $f:B\to B$ can be lifted to an isometry $\tilde f:M\to M$ by letting $\tilde f(x)$ be the endpoint of the lift of the unique minimal geodesic from $p=\pi(x)\in B$ to $f(p)\in B$. Non-positive curvature is used to guarantee uniqueness of the minimal geodesic from $p$ to $f(p)$ and simply- connectedness is used to guarantee continuity of the lifted map. [incorrect -- see Kapovitch's answer below -- I left the text only so readers can see this previous mistake.]</p> http://mathoverflow.net/questions/79186/isometry-groups-of-riemannian-submersions-with-totally-geodesic-fibers/79724#79724 Answer by Vitali Kapovitch for Isometry groups of Riemannian submersions with totally geodesic fibers Vitali Kapovitch 2011-11-01T15:42:53Z 2011-11-01T15:42:53Z <p>I'm not aware of any natural conditions that would ensure existence of the lifts (extensions) as you want. But I want to point out that your statement in (ii) is wrong. There is no reason to expect that the lift you construct this way will be isometric as it need not preserve the horizontal distribution. you can always construct such lifts locally so the assumption of $sec_B\le 0$ is irrelevant for local Lie group actions. In that case what you say amounts to claiming that a horizontal lift of a Killing vector field on the base remains Killing on the total space. It's easy to see that when the fibers are totally geodesic this is equivalent to the vanishing of the $A$-tensor on that vector field. This is certainly a rather strong condition which is very rarely satisfied. in particular, if the whole $A$-tensor is zero then the whole submersion is flat (locally a Riemannian product) since the $T$-tensor is also zero by your assumptions.</p>