Math-1780 · mathematics · Topology

Math-1780

The maximal symmetry-rank conjecture and a sharp product defect formula

Status Partially solved · unsolvedmath.comLLM status Open unverified

Top Open Problems Not reported cataloguedPartially solvedOpen

Statement

Definitions and mathematical statement

Let $(M^n,g)$ be a closed connected simply connected Riemannian manifold of dimension $n\geq2$ with nonnegative sectional curvature: $\sec_g(\sigma)\geq0$ for every tangent two-plane $\sigma$. Its isometry group $\operatorname{Isom}(M,g)$ is a compact Lie group. Define its symmetry rank to be the dimension of a maximal torus in the identity component of this group. Equivalently, it is the largest integer $r$ for which the torus $T^r=(S^1)^r$ acts effectively and isometrically on $(M,g)$. Effective means that no nonidentity group element fixes every point of the manifold.

The proposed bound asks whether every such manifold satisfies \[ \operatorname{rank}\operatorname{Isom}(M,g) \leq\left\lfloor\frac{2n}{3}\right\rfloor. \] The integer floor is simply the integral form of the source inequality $r\leq2n/3$. The metric is arbitrary within the nonnegative-curvature class, and the bound is to depend only on dimension. No assumption that the torus action is free, has a fixed point, or has a specified orbit type is imposed. The problem concerns the maximal dimension of a torus subgroup, not the dimension of the entire isometry group, which is a different invariant. Simple connectivity is a hypothesis on the underlying manifold and cannot be dropped merely because some orbits are connected. A counterexample would consist of a single such metric with an effective isometric torus action exceeding the displayed rank.

Short English statement

Is the symmetry rank of a simply connected, nonnegatively curved closed manifold at most two thirds of its dimension?

Sources

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