Abstract. In this paper we study submanifold with nonpositive extrinsic curvature in a positively curved manifold. Among other things we prove that, if K ⊂ (S n , g) is a totally geodesic submanifold in a Riemannian sphere with positive sectional curvature where n ≥ 5, then K is homeomorphic to S n−2 and the fundamental group of the knot complement π 1 (S n − K) ∼ = Z.
IntroductionIn [Re] the author constructed nontrivial torus knots in S 3 which are totally geodesic with respect to some Riemannian metric with positive curvature. Inspired by this work, it is interesting to ask the following Problem 1: Let (S n , g) be a Riemannian sphere with positive sectional curvature and let i : K → (S n , g) be a codimension 2 totally geodesic submanifold. Could i(K) be a nontrivial knot if n ≥ 2?The problem emerges naturally in the study of transformation group theory acting on manifolds. The famous Smith conjecture asserts that if a cyclic group acting on S 3 has 1-dimensional fixed point set, then the fixed point set must be an unknot. The proof of the conjecture was finally given in 1979 depended on several major advances in 3-manifold theory, in particular the work of William Thurston on hyperbolic structures on 3-manifolds, results by William Meeks and ShingTung Yau on minimal surfaces in 3-manifolds, and work by Hyman Bass on finitely generated subgroups of GL(2, C) (cf. [MB]). The wellknown Thurston's conjecture for 3-orbifold, proved in [BLP], implies that any such an action is topologically conjugate to a linear action. Clearly, every fixed point component of a linear action is a trivial knot and so the latter assertion implies the Smith conjecture.2000 Mathematics Subject Classification. Primary 53C42; Secondary 53C22.