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Compact manifolds of nonpositive curvature

Let M be a compact C°° riemannian manifold of nonpositive curvature and with fundamental group π. It is well known [8, p. 102] that M is a K(π, 1) and thus completely determined up to homotopy type by π. In light of this fact it is natural to ask to what extent the riemannian structure of M is determined by the structure of π , and the intent of this paper is to demonstrate that rather strong implications of this sort exist. In the case that M has strictly negative curvature, the group π is known to be highly noncommutative. Every abelian, in fact, every solvable, subgroup of π is cyclic [3]. It is therefore a plausible conjecture that in the nonpositive curvature case, π will possess large amounts of commutativity only under special geometric circumstances. We shall show that this is true, that indeed those properties of π which involve commutativity have a dramatic reflection in the riemannian structure of M. Our first theorem concerns abelian subgroups of π, which, since no element of π has finite order [8, p. 103], must be torsion free. As remarked above, when M is negatively curved, every abelian subgroup has rank one. However, when the curvature of M is simply nonpositive, we prove the following. The flat torus theorem. There exists an abelian subgroup of rank k in π if and only if there exists a flat k-torus isometrically and totally geodesically immersed in M. The second theorem concerns the case where π is a product of groups. In particular, we shall prove: The splitting theorem. Let M be real analytic and assume that π has no center. If π can be expressed as a direct product of groups π = srfι X X s/N, then M is isometric to a riemannian product M = Mλ X X MN, where πι(Mk) = ^kfork= 1, . . , N . It is shown in § 4 that in the case that π has a nontrivial center, the splitting theorem, as stated, is not true. However, by a slight weakening of the conclusion, one can obtain a similar theorem for the general case. As one may by now suspect, the appearance of a nontrivial center in π must