Let be a complete Riemannian manifold. In the previous post, we saw that the condition that have nonpositive sectional curvature had a reformulation in terms of the “spreading” of geodesics: that is, nearby geodesics in nonpositive curvature spread at least as much as they would in euclidean space. One consequence of this philosophy was the Cartan-Hadamard theorem: the universal cover of is diffeomorphic to . In fact, for a simply connected (and complete of nonpositive curvature), the exponential map

is actually a diffeomorphism.

This suggests that without the simple connectivity assumption, the “only reason” for two geodesics to meet is the fundamental group. In particular, geodesics have no conjugate points, meaning that the exponential map is always nonsingular. Moreover, again by passing to the universal cover, it follows that for any and (relative to the basepoint ), there is a unique geodesic loop at representing . In this post, I’d like to discuss some of the topological consequences of having a metric of nonpositive sectional curvature. Most of this material is from Do Carmo’s *Riemannian Geometry. *

**1. Torsion-freeness of the fundamental group**

Let’s explore some consequences of these assertions.

Theorem 8If is complete of nonpositive curvature, then any has infinite order.

To see this, note that if had finite order, it would define a free action of a finite group on the universal cover , which is diffeomorphic to . But there is a homotopical obstruction to this: no nontrivial finite group can act freely on . In fact, then would be a finite-dimensional model for a (see for instance this MO thread). It is known that no nontrivial finite group can admit a finite-dimensional : in fact, the group cohomology

is nonzero in infinitely many dimensions.

In particular, the fundamental group of any nonpositively curved manifold (say, a compact Riemann surface of positive genus) can have no torsion (although it is usually highly non-abelian).

It is also possible to prove the above theorem by geometric rather than topological means. In fact, one can prove the stronger result:

Theorem 9 (Cartan)Let be a simply connected complete manifold of nonpositive curvature. Let be a compact group acting on by isometries. Then has a fixed point.

The proof of this result uses the notion of the *center of mass* of a compact subset , which is an extension of the notion in euclidean space to any Riemannian structure on of nonpositive curvature. Given , there is a unique which minimizes

this is called the center of mass. (The measure on is the restriction of the Riemannian measure on .) More generally, this notion makes sense whenever one has a compact measure space with a continuous map

If acts on , then for any , the center of mass of the map

is a fixed point for the -action.

**2. Preissman’s theorem**

In strictly negative curvature, one can say a little more about the fundamental group: it is highly non-abelian.

Theorem 10Let be a compact Riemannian manifold of strictly negative sectional curvature. Then any abelian subgroup of is cyclic.

In particular, this provides a strong topological restriction on negatively curved manifolds, which rules out negatively curved metrics on tori. Note that itself can’t be cyclic, because then would have the homotopy type of .

Example 1No compact product manifold can have negative curvature. In fact, if is a product of two compact manifolds, then any commute with each other in . Therefore if had negative curvature, one of would have to be contractible, which is a contradiction unless one of is a point.

Let be the universal cover of , so that is diffeomorphic to via the exponential map at any point. The proof of Preissman’s theorem is based on the following fact: given a geodesic triangle in between points , the sum of the angles is strictly less than . This is a consequence of the fact that the corresponding triangle in euclidean space of lengths has larger angles. In particular, for any geodesic quadrilateral in , the sum of the angles must be less than .

With this in mind, we will prove that any fixed-point-free isometry of has a unique invariant geodesic, called an **axis**.

Lemma 11Given an isometry without fixed points, there is a unique geodesic of left invariant by .

*Proof: *For uniqueness, the idea is that if there were two axes for , we would be able to construct a quadrilateral of total angle . Namely, observe first that don’t intersect: if they intersected at , they’d intersect at as well and two geodesics in that intersect twice are the same. Thus, choose a point on , and draw a geodesic from to some point on . Then one has a quadrilateral (based on the geodesics ) whose angles sum to from the construction, a contradiction.

For existence, the idea is to choose such that is minimized. We can do this for the following reason: the function

is invariant under and descends to a function on the compact manifold . Indeed, if is the projection, then is the length of the shortest loop at in the homotopy class . So there is a point which minimizes this distance.

Now let be the geodesic segment from to . The claim is that preserves the entire geodesic defined by and that is therefore an axis. To see this, it suffices to show that the broken geodesic path from to given by and is actually a geodesic segment. But if it is not a segment — if there is a twist in the path — then for close to , we would get a path from to of length which is not a geodesic (it has a twist) and which can be shortened, a contradiction.

With the lemma proved, the proof of Preissman’s theorem is now rather straightforward. Consider , acting on the universal cover , and suppose that they *commute*. We know that has an axis such that . Since commute, it follows that is also an axis for and therefore . More generally, given any abelian subgroup , this reasoning shows that there is a single axis for all of .

But if is an abelian subgroup of with axis , then there is an injective homomorphism

because any must act on by translation by a constant along the geodesic. However, the only finitely generated abelian group that imbeds in is , so .

Corollary 12 (Byers)Under the above hypotheses, any solvable subgroup of is infinite cyclic.

*Proof:* It suffices to show that if is a subgroup with the property that has a subgroup with abelian and , then .

To see this, choose a geodesic in (as in the previous proof) invariant under . For , one has

so that, since leaves invariant,

It follows that since axes are unique, and we find that itself leaves the geodesic invariant. The same reasoning as above now shows that has to be infinite cyclic.

A somewhat related manifestation of the high degree of non-abelianness of comes from the theorem (of Milnor) that is a group of exponential growth under these hypotheses.

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