Math-1778 · mathematics · Topology

Math-1778

Surface quotient, bundle normal forms, and a WNN escape condition

Status Partially solved · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedPartially solvedOpen

Statement

Definitions and mathematical statement

Let $G$ be either the Lie group $S^3$ of unit quaternions or the rotation group $\operatorname{SO}(3)=\{A\in\mathbb R^{3\times3}:A^TA=I,\det A=1\}$. Both have real dimension three. A smooth action of $G$ on a manifold is free if the stabilizer of each point is the identity subgroup. It is isometric for a Riemannian metric $g$ if every group element acts by an isometry of that same metric.

Does there exist a closed connected smooth five-dimensional manifold $M$, a Riemannian metric $g$ with strictly positive sectional curvature on every tangent two-plane, and a free isometric action of one of these two groups on $(M,g)$? The question can be separated into the two existence cases $G=S^3$ and $G=\operatorname{SO}(3)$. The quotient $B=M/G$ is then a smooth closed surface, and the orbit map is a principal $G$-bundle and a Riemannian submersion when $B$ is given its quotient metric. No particular diffeomorphism type for $M$ or $B$ is prescribed in advance. An action with finite nontrivial isotropy is not free and therefore does not answer this question. Likewise, a positive-curvature metric and a free action considered separately do not suffice: the action must preserve the positive-curvature metric. The fibers are not additionally assumed totally geodesic or mutually isometric as intrinsic metrics.

Short English statement

Can a positively curved closed five-manifold admit a free isometric action by the unit quaternions or by the three-dimensional rotation group?

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