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 8 If
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 10 Let
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 1 No 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 11 Given 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.
Leave a Reply