## Dimension of certain subgroup of isometry group of positively curved manifold

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 $1$.

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?

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 Why the downvote?! It seems to me a reasonable question. Perhaps it would help to define the "symmetry rank" (=rank of the isometry group). – JosÃ© Figueroa-O'Farrill Oct 10 2011 at 19:20

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.
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 "Positively curved manifolds with symmetry") 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.
 Thanks for the example. I am wondering what if extra assumption "$M/G$ is a manifold" is impossed? – unknown (google) Oct 10 2011 at 19:35