Dimension of certain subgroup of isometry group of positively curved manifold - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T17:11:57Z http://mathoverflow.net/feeds/question/77714 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77714/dimension-of-certain-subgroup-of-isometry-group-of-positively-curved-manifold Dimension of certain subgroup of isometry group of positively curved manifold unknown (google) 2011-10-10T17:40:15Z 2011-10-14T20:16:00Z <p>Let $M$ be a closed $n$-dimensional Riemannian manifold with positive sectional curvature. Let $G$ be a close subgroup of isometry group ${\rm Iso}(M)$. Suppose the action of $G$ on $M$ is not transitive, hence $M/G$ has dimension at least $1$. </p> <p>By a theorem of Grove and Searle the symmetry rank $${\rm symran}(M)\le [\frac{n+1}{2}]$$ I am wondering is there any upper bound for the dimension of $G$ mentioned above?</p> http://mathoverflow.net/questions/77714/dimension-of-certain-subgroup-of-isometry-group-of-positively-curved-manifold/77729#77729 Answer by Vitali Kapovitch for Dimension of certain subgroup of isometry group of positively curved manifold Vitali Kapovitch 2011-10-10T19:20:21Z 2011-10-10T21:25:36Z <p>Forgetting positive curvature, if $\dim M^n/G=k$ then by looking at the transitive action of $G$ on the principal orbit one gets a trivial bound $\dim G\le \dim O(n-k)=\frac{(n-k+1)(n-k)}{2}$. This bound is realized for $k=1$ on a round $S^n$ and $G=O(n-1)$. As the sphere is positively curved this bound is sharp.</p> <p>Addressing the comment below, the assumption of $M/G$ being a manifold is not a natural one in this context. It hardly ever happens when $M$ has positive curvature. In particular, by a result of Wilking (Lemma 5 in <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&amp;s1=623978&amp;vfpref=html&amp;r=13&amp;mx-pid=2199227" rel="nofollow">"Positively curved manifolds with symmetry"</a>) based on his connectedness principle, if the principal isotropy group $H$ is not trivial and $M/G\ne pt$ then $M/G$ has a boundary. If $H=1$ and $M/G$ is a smooth manifold then the $G$-action is free and hence $rank G\le 1$ by Berger's vanishing theorem.</p> <p>I should add that there is large literature on the subject of isometric group actions on positively curved manifolds (mostly by Wilking, Grove, Ziller, Searle and Rong) and I suggest you study it if you want to pursue these kind of questions. A good place to start is a survey by Wilking <a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=IID&amp;s1=623978&amp;vfpref=html&amp;r=7&amp;mx-pid=2408263" rel="nofollow">"Nonnegatively and positively curved manifolds".</a></p>