Let {M} be a compact Riemannian manifold of (strictly) negative curvature, so that {M} is a {K(\pi_1 M, 1)}. In the previous post, we saw that the group {\pi_1 M} was significantly restricted: for example, every solvable subgroup of {\pi_1 M} had to be infinite cyclic. The goal of this post and the next is to understand the result of Milnor that the group {\pi_1 M} is of exponential growth in an arithmetic sense.

Milnor’s (wonderful) idea is to translate this into a problem in geometry: that is, to relate the growth of the group {\pi_1 M} to the volume growth of expanding balls in the universal cover {\widetilde{M}}. As I understand, this idea has proved enormously influential on future work on the fundamental groups of Riemannian manifolds with restricted curvature. Note that Milnor’s result also highlights the difference between positive and negative curvature: in positive curvature, the fundamental group of every compact manifold is finite. Most of this material is from Chavel’s Riemannian geometry: a modern introduction.

1. Volume growth

To begin with, let’s say something about volume growth. Let {M} be a complete, simply connected Riemannian manifold whose sectional curvatures are {\leq c < 0}. If we choose {p \in M}, we know that the exponential map

\displaystyle \exp_p : T_p M \rightarrow M

is a diffeomorphism. Note that {\exp_p} sends the euclidean ball of radius {r} diffeomorphically onto the (metric) ball of radius {r} in {M}.

Our goal is to prove:

Theorem 13 The function {r \mapsto \mathrm{vol} ( B_M(p, r))} which sends {r} to the volume of the ball in {M} of radius {r} centered at {p} grows exponentially.

This theorem also highlights the sense in which negative curvature corresponds to the “spreading” of geodesics: the geodesics spread so much that the volumes of linearly expanding balls actually grow exponentially.

In order to prove this result, we need to understand the pullback of the metric {g} on {M} to {T_p M} via {\exp_p}. Without loss of generality, we can ignore zero, and we work in polar coordinates. We have a diffeomorphism

\displaystyle \mathbb{R}_+ \times S(T_p M) \rightarrow T_p M \setminus \left\{0\right\}

where {S(T_p M)} is the unit sphere in {T_p M}. We thus get “coordinates” {(r, \xi)} on {T_p M} where {r} is the radius and {\xi \in S(T_p M)}. Our goal is to express the metric {\exp_p^* g} on {T_p M} as

\displaystyle dr^2 + (M(r, \xi) d \xi)^2 ,

where the (abusive) notation {(M(r, \xi) d\xi)^2} refers to a positive definite quadratic form {M(r, \xi)} on {\xi^{\perp}} (under the polar identification of the tangent space at {(r, \xi)} with {\mathbb{R} dr \oplus \xi^{\perp}}). The existence of such a decomposition of the metric is a consequence of the Gauss lemma.

Now, our goal is to understand {M(r, \xi)}. In other words, given {v \in T_p M} perpendicular to {\xi}, we would like to understand

\displaystyle M(r, \xi) (v, v) \stackrel{\mathrm{def}}{=} g( d( \exp_p)_{r \xi *} (rv), d( \exp_p)_{r \xi *} (rv)) .

If we understand these values, then we understand the pulled back metric {\exp^* g}. However, we can understand these values in terms of geodesic deviation or Jacobi fields. We recall that the value

\displaystyle d ( \exp_p)_{r \xi *} (rv)

is exactly the value (at {r}) of a Jacobi field {J} along the geodesic {\gamma(t) = \exp_p(t\xi)} with {J(0) =0 }, {J'(0) \stackrel{\mathrm{def}}{=} \frac{D}{dt}|_{t = 0} J = v}.

The claim is that, using the assumed upper bounds for curvature, we can bound below the quadratic form {M(r, \xi)}, and thus the volume. This will require a special case of the Rauch comparison theorem.

2. The case of constant curvature

Let’s see how the computation of {M(r, \xi)} plays out when we are working with a space of constant curvature {c}. In this case, if we keep the same notation as above, we need to understand the Jacobi field {J} along the geodesic {\exp_p(t \xi)} with {J(0) = 0, J'(0) = v}. This satisfies the Jacobi equation

\displaystyle J''(t) = - R(J(t), \gamma'(t)) \gamma'(t). \ \ \ \ \ (3)

Now, {J(t)} started out orthogonal to {\gamma'(t)} (as {J'(0) =0}), so this is always true by the following lemma.

Lemma 14 Let {J} be a Jacobi field along a geodesic {\gamma} with {J(0), J'(0)} perpendicular to {\gamma'(0)}. Then {J(t)} is always perpendicular to {\gamma'(t)}.

Proof: Indeed, we have

\displaystyle \frac{d^2}{dt^2}\left \langle J(t), \gamma'(t)\right\rangle \equiv 0

by the Jacobi equation and the antisymmetry of the curvature tensor. So if {\left \langle J(t), \gamma'(t)\right\rangle} and its derivative vanish at {t =0 }, it must vanish identically. \Box

Anyway, this means that

\displaystyle J''(t) = - c J(t).

If {e_1, \dots, e_{n-1}} is a parallel orthonormal basis along {\gamma} of {(\dot{\gamma})^{\perp}}, then the basis of solutions of this second-order equation are

\displaystyle J_i(t) = \frac{1}{\sqrt{-c}}\sinh( \sqrt{-c} t) e_i(t).

Taking {t = r}, it follows that the quadratic form {M(r, \xi)} on {\xi^{\perp}} is exactly { \frac{1}{\sqrt{-c}}\sinh( \sqrt{-c} r)} — or rather, that times the given (euclidean) metric on {T_p M}. In other words, we find that {\exp_p^* (g)} in polar coordinates is

\displaystyle d r^2 + \left( \frac{\sinh( \sqrt{-c} r)}{\sqrt{-c}} \right)^2 d \sigma_{n-1}^2,

where {d \sigma_{n-1}^2} is the metric on {S^{n-1}}. This is the polar expression of hyperbolic space.

Now, we can get from here the exponential growth of the volume. The volume form associated to {\exp_p^* g} is given by

\displaystyle d \mathrm{vol} = \left( \frac{\sinh( \sqrt{-c} r)}{\sqrt{-c}} \right)^{n-1} dr \wedge d \mathrm{vol}_{S^{n-1}},

where {d \mathrm{vol}} is the volume form on {S^{n-1}}. Then

\displaystyle \mathrm{Vol}( B(_Mp, R)) = \int_{r \leq R}\left( \frac{\sinh( \sqrt{-c} r)}{\sqrt{-c}} \right)^{n-1} dr \wedge d \mathrm{vol}_{S^{n-1}}

is an exponentially growing function on {R}. (If {c =0}, we would see polynomial growth instead.)

 3. The general case

The goal is to generalize the above analysis in the case of a space of constant negative curvature to allow the curvature to vary. The principle is that since the curvature is less than {c < 0}, geodesics should deviate (i.e., Jacobi fields should grow) more than they do in the space of constant curvature {c} — which we have seen is already exponential. The Rauch comparison theorem provides a formalization of this principle, but we will only need a fairly limited analysis.

Let {M} be as before, except we now assume its sectional curvatures are {\leq c < 0}. Consider the geodesic {\gamma(t) = \exp_p(t \xi)} as before and let {v \in \xi^{\perp} \subset T_p M} be a unit vector, and let {J(t)} be the Jacobi field along {\gamma} with {J(0) = 0, J'(0) = v}. We note that {J(t)} is always perpendicular to {\gamma'(t)}, as before, and we have that

\displaystyle \left \langle J''(t), J(t) \right\rangle = - \left \langle R( J(t), \gamma'(t)) \gamma'(t), J(t)\right\rangle \geq -c \left \lVert J(t) \right \rVert^2.

It follows that if {f(t) = \left \lVert J(t) \right \rVert = \left \langle J(t), J(t)\right\rangle^{1/2}}, then

Screenshot-52

By Cauchy-Schwarz applied to the last two terms, we find that this is at least

\displaystyle f''(t) \geq \frac{ \left \langle J''(t), J(t)\right\rangle}{\left \lVert J(t)\right \rVert} \geq -c f(t).

So we get the differential inequality:

\displaystyle f''(t) + c f(t) \geq 0 , \ \ \ \ \ (4)

rather than the differential equation that we got before with solutions the {\sinh} function. Moreover, by

\displaystyle f'(t)^2 = \frac{ \left \langle J'(t), J(t)\right\rangle^2}{ \left \langle J(t), J(t)\right\rangle}

and L’Hopital’s rule, we get {f'(0) \geq 1} as {v = J'(0)} is a unit vector.

The idea is that a solution of the differential inequality will have to be at least the solution of the corresponding differential equation, and this is the heart of the comparison with constant sectional curvature. Let’s state this as follows:

Lemma 15 Let {c < 0}. Let {f(t)} be a nonnegative {C^2} function with {f(0) = 0, f'(0)= 1} satisfying the differential inequality (4). Then {f(t) \geq \frac{1}{\sqrt{-c}}\sinh( \sqrt{-c} t)}.

Proof: Consider {h(t) =\frac{f(t)}{\frac{1}{\sqrt{-c}}\sinh( \sqrt{-c}t)}}; we have that {h(0) = 1}. We’d like to show that

\displaystyle h'(t) \geq 0\quad t \geq 0, \ \ \ \ \ (5)

which will imply the result. Write {\widetilde{f}(t) = \frac{1}{\sqrt{-c}}\sinh( \sqrt{-c} t)}. Then (5) follows from

\displaystyle h'(t) = \frac{f' \widetilde{f} - f \widetilde{ f}'}{\widetilde{f}(t)^2}

if we can show that {f' \widetilde{f} - f \widetilde{f}' \geq 0}. To see this, we note that {f' \widetilde{f} - f \widetilde{f}'} takes the value {0} at {t =0 }, so we will be done if we can show that its derivative is nonnegative. But we have

\displaystyle \frac{d}{dt} \left( f' \widetilde{f} - f \widetilde{f}'\right) = f'' \widetilde{f} - f \widetilde{f}'' \geq (-c) f \widetilde{f} - (-c) f \widetilde{f} = 0.

\Box

It follows that the pull-back metric {\exp_p^* g = dr^2 + (M(r, \xi) d \xi)^2} has the property that the quadratic form {M(r, \xi)} on {\xi^{\perp} \subset T_p M} is greater than {\sinh^2(\sqrt{-c} r)} times the euclidean metric on {\xi^{\perp}}. That is, if we interpret {M(r, \xi)} as a linear transformation {\xi^{\perp} \rightarrow \xi^{\perp}}, then {M(r, \xi) \geq \sinh^2( \sqrt{-c} r)}. It follows that

\displaystyle d \mathrm{vol} = \sqrt{ \det M(r, \xi)}dr \wedge d \mathrm{vol}_{S^{n-1}} = \sinh( \sqrt{-c} r)^{n-1}dr \wedge d \mathrm{vol}_{S^{n-1}},

so the volume form we get is bigger than the one we got before in constant curvature.

We have thus proved the following:

Theorem 16 Let {M} be a complete, simply connected Riemannian manifold with sectional curvatures {\leq c < 0}. Let {p \in M}. Let {B_M(p, r)} be the ball of radius {r} in {M} centered at {p}. Then {\mathrm{vol}(B_M(p, r))} grows exponentially. In fact, it grows as fast as the analogous volumes of balls in the constant curvature space of curvature {-c}.

 

In the next post, we’ll see what this has to do with the growth rate of fundamental groups.

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