Busemann function explained

In geometric topology, Busemann functions are used to study the large-scale geometry of geodesics in Hadamard spaces and in particular Hadamard manifolds (simply connected complete Riemannian manifolds of nonpositive curvature). They are named after Herbert Busemann, who introduced them; he gave an extensive treatment of the topic in his 1955 book "The geometry of geodesics".

Definition and elementary properties

Let

(X,d) be a metric space. A geodesic ray is a path \gamma : [ 0 , inf ty) \to X which minimizes distance everywhere along its length. i.e., for all t,t' \in [ 0 , inf ty) , $$d\backslash big(\backslash gamma(t),\; \backslash gamma(t\text{'})\; \backslash big)\; =\; \backslash big|\; t\; -\; t\text{'}\; \backslash big|.$$ Equivalently, a ray is an isometry from the "canonical ray" (the set [ 0 , inf ty) equipped with the Euclidean metric) into the metric space X.

Given a ray γ, the Busemann function

B _{\gamma} : X \to R is defined by

$$B\_\backslash gamma(x)=\backslash lim\_\backslash big(d\backslash big(\backslash gamma(t),\; x\; \backslash big)\; -\; t\; \backslash big)$$

Thus, when t is very large, the distance

d ( \gamma(t), x ) is approximately equal to B _{\gamma(x)} + t . Given a ray γ, its Busemann function is always well-defined: indeed the right hand side above F _{t(x)} \stackrel{ def } d\big(\gamma(t), x \big) - t , tends pointwise to the left hand side on compacta, since t - d( \gamma(t), x ) = d(\gamma(t),\gamma( 0 )) - d(\gamma(t),x) is bounded above by d(\gamma( 0 ),x) and non-increasing since, if s\le t ,

$$t-s\; +\; d(x,\backslash gamma(s))\; -\; d(x,\backslash gamma(t))\; \backslash ge\; t\; -s\; -d\; (\backslash gamma(s),\backslash gamma(t))\; =\; 0.$$

It is immediate from the triangle inequality that

$$|B\_\backslash gamma(x)\; -\; B\_\backslash gamma(y)|\backslash le\; d(x,y),$$

so that

B _{\gamma} is uniformly continuous. More specifically, the above estimate above shows that

• Busemann functions are Lipschitz functions with constant 1.

By Dini's theorem, the functions

F _{t(x)=d(x,\gamma(t))} - t tend to B _{\gamma(x)} uniformly on compact sets as t tends to infinity.

Example: Poincaré disk

Let

D be the unit disk in the complex plane with the Poincaré metric

$$ds^2\; =\; .$$

Then, for

|z| < 1 and |\zeta| = 1 , the Busemann function is given by

$$B\_\backslash zeta(z)\; =\; -\; \backslash log\backslash left(\backslash right),$$

where the term in brackets on the right hand side is the Poisson kernel for the unit disk and

\zeta corresponds to the radial geodesic \gamma from the origin towards \zeta ,\gamma(t) = \zeta \tanh(t/ 2 ) . The computation of d(x,y) can be reduced to that of d(z, 0 ) = d(|z|, 0 ) = 2 \operatorname{artanh}(|z|) = log \left( \tfrac{ 1+ |z|}{ 1- |z|} \right) , since the metric is invariant under Möbius transformations in SU( 1 , 1 ) ; the geodesics through 0 have the form \zeta g _{t(0)} where g _{t} is the 1-parameter subgroup of SU( 1 , 1 ) ,

The formula above also completely determines the Busemann function by Möbius invariance.

Busemann functions on a Hadamard space

In a Hadamard space, where any two points are joined by a unique geodesic segment, the function

F = F _{t} is convex, i.e. convex on geodesic segments [x,y] . Explicitly this means that if z(s) is the point which divides [x,y] in the ratio, then F(z(s)) \leq s F(x) + ( 1 - s) F(y) . For fixed a the function d(x,a) is convex and hence so are its translates; in particular, if \gamma is a geodesic ray in X , then F _{t} is convex. Since the Busemann function B _{\gamma} is the pointwise limit of F _{t} ,

• Busemann functions are convex on Hadamard spaces.
• On a Hadamard space, the functions

F _{t(y)=d(y,\gamma(t))} - t converge uniformly to B _{\gamma} uniformly on any bounded subset of X .

Let . Since

\gamma (t) is parametrised by arclength, Alexandrov's first comparison theorem for Hadamard spaces implies that the function is convex. Hence for

$$g(s)\; \backslash le\; (1-)g(0)\; +\; g(t).$$

Thus

$$2sh(s)\backslash le\; (h(s)\; +s)^2\; -\; s^2\; =g(s)\; \backslash le\; (1-)d(x,y)^2\; +\; (2th(t)\; +h(t)^2)\backslash le\; d(x,y)^2\; +\; 2s\; h(t)\; +\; d(x,y)^2,$$

so that

$$|F\_s(y)\; -\; F\_t(y)|=\; |h(s)\; -\; h(t)|\; \backslash le\; (s^\; +\; t^)\; d(x,y)^2.$$

Letting t tend to ∞, it follows that

$$|F\_s(y)\; -\; B\_\backslash gamma(y)|\; \backslash le,$$

so convergence is uniform on bounded sets.

Note that the inequality above for

|F _{s(y)} - F _{t(y)|} (together with its proof) also holds for geodesic segments: if is a geodesic segment starting at and parametrised by arclength then

$$|d(y,\backslash Gamma(s))\; -s\; -d(y,\backslash Gamma(t))\; +t|\; \backslash le\; (s^\; +\; t^)\; d(x,y)^2.$$

Next suppose that are points in a Hadamard space, and let be the geodesic through with and, where . This geodesic cuts the boundary of the closed ball at the point . Thus if, there is a point with such that .

This condition persists for Busemann functions. The statement and proof of the property for Busemann functions relies on a fundamental theorem on closed convex subsets of a Hadamard space, which generalises orthogonal projection in a Hilbert space: if is a closed convex set in a Hadamard space, then every point in has a unique closest point in and ; moreover is uniquely determined by the property that, for in,

$$d(x,y)^2\; \backslash ge\; d(x,a)\; ^2\; +\; d(a,y)^2,$$

so that the angle at in the Euclidean comparison triangle for is greater than or equal to .

• If is a Busemann function on a Hadamard space, then, given in and, there is a unique point with such that . For fixed, the point is the closest point of to the closed convex set of points such that and therefore depends continuously on .

Let be the closest point to in . Then and so is minimised by in where is the unique point where is minimised. By the Lipschitz condition . To prove the assertion, it suffices to show that, i.e. . On the other hand, is the uniform limit on any closed ball of functions . On, these are minimised by points with . Hence the infimum of on is and tends to . Thus with and tending towards . Let be the closest point to with . Let . Then, and, by the Lipschitz condition on, . In particular tends to . Passing to a subsequence if necessary it can be assumed that and are both increasing (to ). The inequality for convex optimisation implies that for .

$$d(u\_n,u\_m)^2\; \backslash le\; R\_n^2\; -\; R\_m^2\backslash le\; 2r|R\_n\; -R\_m|,$$

so that is a Cauchy sequence. If is its limit, then and . By uniqueness it follows that and hence, as required.

Uniform limits. The above argument proves more generally that if tends to infinity and the functions tend uniformly on bounded sets to, then is convex, Lipschitz with Lipschitz constant 1 and, given in and, there is a unique point with such that . If on the other hand the sequence is bounded, then the terms all lie in some closed ball and uniform convergence there implies that is a Cauchy sequence so converges to some in . So tends uniformly to, a function of the same form. The same argument also shows that the class of functions which satisfy the same three conditions (being convex, Lipschitz and having minima on closed balls) is closed under taking uniform limits on bounded sets.

Comment. Note that, since any closed convex subset of a Hadamard subset of a Hadamard space is also a Hadamard space, any closed ball in a Hadamard space is a Hadamard space. In particular it need not be the case that every geodesic segment is contained in a geodesic defined on the whole of or even a semi-infinite interval . The closed unit ball of a Hilbert space gives an explicit example which is not a proper metric space.

• If is a convex function, Lipschitz with constant 1 and assumes its minimum on any closed ball centred on with radius at a unique point on the boundary with, then for each in there is a unique geodesic ray such that and cuts each closed convex set with at, so that . In particular this holds for each Busemann function.

The third condition implies that is the closest point to in the closed convex set of points

u such that . Let for be the geodesic joining to . Then is a convex Lipschitz function on with Lipschitz constant 1 satisfying and and . So vanishes everywhere, since if and . Hence . By uniqueness it follows that is the closest point to in and that it is the unique point minimising in . Uniqueness implies that these geodesics segments coincide for arbitrary and therefore that extends to a geodesic ray with the stated property.

• If, then the geodesic ray starting at satisfies

\sup d(\gamma(t),\delta(t)) < inf ty . If is another ray starting at with \sup d(\delta(t),\delta _{1(t))} < inf ty then .

To prove the first assertion, it is enough to check this for sufficiently large. In that case and are the projections of and onto the closed convex set . Therefore, . Hence . The second assertion follows because is convex and bounded on, so, if it vanishes at, must vanish everywhere.

• Suppose that is a continuous convex function and for each in there is a unique geodesic ray such that and cuts each closed convex set at, so that ; then is a Busemann function. is a constant function.

Let be the closed convex set of points with . Since is a Hadamard space for every point in there is a unique closest point to in . It depends continuously on and if lies outside, then lies on the hypersurface —the boundary ∂ of —and satisfies the inequality of convex optimisation. Let be the geodesic ray starting at .

Fix in . Let be the geodesic ray starting at . Let, the Busemann function for with base point . In particular . It suffices to show that . Now take with and let be the geodesic ray starting at corresponding to . Then

$$d(x,y)\; \backslash ge\; d(\backslash gamma(t),\backslash delta(t)),\backslash ,\backslash ,\backslash ,\; d(x,\backslash delta(t))^2\; \backslash ge\; d(x,\backslash gamma(t))^2+d(\backslash gamma(t),\backslash delta(t))^2,\backslash ,\backslash ,\backslash ,d(y,\backslash gamma(t))^2\; \backslash ge\; d(y,\backslash delta(t))^2+d(\backslash gamma(t),\backslash delta(t))^2.$$

On the other hand, for any four points , , , in a Hadamard space, the following quadrilateral inequality of Reshetnyak holds:

$$|d(a,c)^2\; +\; d(b,d)^2\; -\; d(a,d)^2\; -\; d(b,d)^2|\backslash le\; 2\; d(a,b)d(c,d).$$

Setting,,,, it follows that

$$|d(y,\backslash gamma(t))^2\; -\; d(x,\backslash delta(t))^2|\backslash le\; 2d(x,y)^2,$$

so that

$$|d(y,\backslash gamma(t))\; -\; d(x,\backslash gamma(t))|\backslash le\; 2\; \backslash le\; .$$

Hence . Similarly . Hence on the level surface of containing . Now for and in , let the geodesic ray starting at . Then and . Moreover, by boundedness, . The flow can be used to transport this result to all the level surfaces of . For general, if, take such that and set . Then, where . But then, so that . Hence, as required. Similarly if, take such that . Let . Then, so . Hence, as required.

Finally there are necessary and sufficient conditions for two geodesics to define the same Busemann function up to constant:

• On a Hadamard space, the Busemann functions of two geodesic rays

\gamma _{1} and \gamma _{2} differ by a constant if and only if \sup _{t\ge} d(\gamma _{1(t),\gamma} _{2(t))} < inf ty .

Suppose firstly that and are two geodesic rays with Busemann functions differing by a constant. Shifting the argument of one of the geodesics by a constant, it may be assumed that, say. Let be the closed convex set on which . Then and similarly . Then for, the points and have closest points and in , so that . Hence .

Now suppose that . Let be the geodesic ray starting at associated with . Then . Hence . Since and both start at , it follows that . By the previous result and differ by a constant; so and differ by a constant.

To summarise, the above results give the following characterisation of Busemann functions on a Hadamard space:

THEOREM. On a Hadamard space, the following conditions on a function are equivalent:

• is a Busemann function.
• is a convex function, Lipschitz with constant and assumes its minimum on any closed ball centred on with radius at a unique point on the boundary with .
• is a continuous convex function and for each in there is a unique geodesic ray such that and, for any, the ray cuts each closed convex set at .

Bordification of a Hadamard space

In the previous section it was shown that if is a Hadamard space and is a fixed point in then the union of the space of Busemann functions vanishing at and the space of functions is closed under taking uniform limits on bounded sets. This result can be formalised in the notion of bordification of . In this topology, the points tend to a geodesic ray starting at if and only if tends to and for arbitrarily large the sequence obtained by taking the point on each segment at a distance from tends to .

If is a metric space, Gromov's bordification can be defined as follows. Fix a point in and let . Let be the space of Lipschitz continuous functions on, i.e. those for which for some constant . The space can be topologised by the seminorms, the topology of uniform convergence on bounded sets. The seminorms are finite by the Lipschitz conditions. This is the topology induced by the natural map of into the direct product of the Banach spaces of continuous bounded functions on . It is give by the metric .

The space is embedded into by sending to the function . Let be the closure of in . Then is metrisable, since is, and contains as an open subset; moreover bordifications arising from different choices of basepoint are naturally homeomorphic. Let . Then lies in . It is non-zero on and vanishes only at . Hence it extends to a continuous function on with zero set . It follows that is closed in, as required. To check that is independent of the basepoint, it suffices to show that extends to a continuous function on . But, so, for in, . Hence the correspondence between the compactifications for and is given by sending in to in .

When is a Hadamard space, Gromov's ideal boundary can be realised explicitly as "asymptotic limits" of geodesic rays using Busemann functions. If is an unbounded sequence in with tending to in, then vanishes at, is convex, Lipschitz with Lipschitz constant and has minimum on any closed ball . Hence is a Busemann function corresponding to a unique geodesic ray starting at .

On the other hand, tends to uniformly on bounded sets if and only if tends to and for arbitrarily large the sequence obtained by taking the point on each segment at a distance from tends to . For, let be the point in with . Suppose first that tends to uniformly on . Then for,. This is a convex function. It vanishes as and hence is increasing. So it is maximised at . So for each, tends towards 0. Let, and . Then is close to with large. Hence in the Euclidean comparison triangle is close to with large. So the angle at is small. So the point on at the same distance as lies close to . Hence, by the first comparison theorem for geodesic triangles, is small. Conversely suppose that for fixed and sufficiently large tends to 0. Then from the above satisfies

$$|F\_s(y)\; -\; B\_\backslash gamma(y)|\; \backslash le,$$

so it suffices show that on any bounded set is uniformly close to for sufficiently large.

For a fixed ball, fix so that . The claim is then an immediate consequence of the inequality for geodesic segments in a Hadamard space, since

$$|d(y,x\_n)\; -d(y,x\_0)\; -\; d(y,x\_n(s))\; +\; s\; |\backslash le\; \backslash le\; \backslash varepsilon.$$

Hence, if in and is sufficiently large that, then

$$|h\_n(y)-B\_\backslash gamma(y)|=|d(y,x\_n)\; -d(y,x\_0)\; -\; B\_\backslash gamma(y)|\; \backslash le\; |d(y,x\_n)\; -d(y,x\_0)\; -\; d(y,x\_n(s))\; +\; s\; |\; +\; d(x\_n(s),\backslash gamma(s))\; +\; |F\_s(y)\; -\; B\_\backslash gamma(y)|\; \backslash le\; 3\backslash varepsilon.$$

Busemann functions on a Hadamard manifold

Suppose that are points in a Hadamard manifold and let be the geodesic through with . This geodesic cuts the boundary of the closed ball at the two points . Thus if, there are points with such that . By continuity this condition persists for Busemann functions:

• If is a Busemann function on a Hadamard manifold, then, given in and, there are unique points, with such that and . For fixed, the points and depend continuously on .

Taking a sequence tending to and, there are points and which satisfy these conditions for for sufficiently large. Passing to a subsequence if necessary, it can be assumed that and tend to and . By continuity these points satisfy the conditions for . To prove uniqueness, note that by compactness assumes its maximum and minimum on . The Lipschitz condition shows that the values of there differ by at most . Hence is minimized at and maximized at . On the other hand, and for and the points and are the unique points in maximizing this distance. The Lipschitz condition on then immediately implies and must be the unique points in maximizing and minimizing . Now suppose that tends to . Then the corresponding points and lie in a closed ball so admit convergent subsequences. But by uniqueness of and any such subsequences must tend to and, so that and must tend to and, establishing continuity.

The above result holds more generally in a Hadamard space.

• If is a Busemann function on a Hadamard manifold, then is continuously differentiable with for all .

From the previous properties of, for each there is a unique geodesic γ(t) parametrised by arclength with such that . It has the property that it cuts at : in the previous notation and . The vector field defined by the unit vector

• \gamma ( 0 ) at is continuous, because is a continuous function of and the map sending to is a diffeomorphism from onto by the Cartan-Hadamard theorem. Let be another geodesic parametrised by arclength through with . Then • ( \delta ( 0 ), • \gamma ( 0 )) . Indeed, let, so that . Then

$$|H(\backslash delta(s))\; -\; H(x)|\; \backslash le\; d(\backslash delta(s),x).$$

Applying this with and, it follows that for

$$(r-\; d(\backslash delta(s),u))/s\; \backslash le\; (h(\backslash delta(s))\; -\; h(y))/s\; \backslash le\; (d(\backslash delta(s),v)\; -r)/s.$$

The outer terms tend to

• ( \delta ( 0 ), • \gamma ( 0 )) as tends to 0, so the middle term has the same limit, as claimed. A similar argument applies for .

The assertion on the outer terms follows from the first variation formula for arclength, but can be deduced directly as follows. Let

• a = \delta ( 0 ) and • b = \gamma ( 0 ) , both unit vectors. Then for tangent vectors and at in the unit ball^[1]

$$d(\backslash exp\_y\; p,\backslash exp\_y\; q)\; =\; \backslash |p-q\backslash |\; +\; \backslash varepsilon\; \backslash max\; \backslash |p\backslash |^2,\; \backslash |q\backslash |^2$$ with uniformly bounded. Let and . Then

$$(d(\backslash delta(s),v)\; -r)/s=\; (d(\backslash exp\_y(t^3\; a),\backslash exp\_y(-t^2\; b))\; -\; t^2)/t^3\; =\; (\backslash |t^3a\; +t^2b\backslash |\; -\; t^2)/t^3\; +\; \backslash varepsilon\; |t|=\; (\backslash |ta\; +b\backslash |\; -1)/t\; +\; \backslash varepsilon\; |t|.$$

The right hand side here tends to as tends to 0 since

$$\backslash |b\; +\; t\; a\backslash ||\_=\; \backslash ,\; \backslash |b\; +\; t\; a\backslash |^2|\_\; =\; (a,b).$$

The same method works for the other terms.

Hence it follows that is a function with dual to the vector field, so that . The vector field is thus the gradient vector field for . The geodesics through any point are the flow lines for the flow for, so that is the gradient flow for .

THEOREM. On a Hadamard manifold the following conditions on a continuous function are equivalent:

1. is a Busemann function.
2. is a convex, Lipschitz function with constant 1, and for each in there are points at a distance from such that .
3. is a convex function with .

It has already been proved that (1) implies (2).

The arguments above show mutatis mutandi that (2) implies (3).

It therefore remains to show that (3) implies (1). Fix in . Let be the gradient flow for . It follows that and that is a geodesic through parametrised by arclength with . Indeed, if, then

$$|s-t|=|h(\backslash alpha\_s(x))-h(\backslash alpha\_t(x))|\; \backslash le\; d(\backslash alpha\_s(x),\backslash alpha\_t(x))\backslash le\; \backslash int\_^t\; \backslash |d\backslash alpha\_\backslash tau(x)/d\backslash tau\backslash |\; \backslash ,\; d\backslash tau\; =\backslash int\_s^t\; \backslash |dh(\backslash alpha\_\backslash tau(x))\backslash |\; \backslash ,\; d\backslash tau\; =\; |s-t|,$$

so that . Let, the Busemann function for with base point . In particular . To prove (1), it suffices to show that .

Let be the closed convex set of points with . Since is a Hadamard space for every point in there is a unique closest point to in . It depends continuously on and if lies outside, then lies on the hypersurface —the boundary of —and the geodesic from to is orthogonal to . In this case the geodesic is just . Indeed, the fact that is the gradient flow of and the conditions imply that the flow lines are geodesics parametrised by arclength and cut the level curves of orthogonally. Taking with and,

$$d(x,y)\; \backslash ge\; d(\backslash alpha\_t(x),\backslash alpha\_t(y)),\backslash ,\backslash ,\backslash ,\; d(x,\backslash alpha\_t(y))^2\; \backslash ge\; d(x,\backslash alpha\_t(x))^2+d(\backslash alpha\_t(x),\backslash alpha\_t(y))^2,\backslash ,\backslash ,\backslash ,d(y,\backslash alpha\_t(x))^2\; \backslash ge\; d(y,\backslash alpha\_t(y))^2+d(\backslash alpha\_t(x),\backslash alpha\_t(y))^2.$$

On the other hand, for any four points,,, in a Hadamard space, the following quadrilateral inequality of Reshetnyak holds:

$$|d(a,c)^2\; +\; d(b,d)^2\; -\; d(a,d)^2\; -\; d(b,d)^2|\backslash le\; 2\; d(a,b)d(c,d).$$

Setting,,,, it follows that

$$|d(y,\backslash alpha\_t(x))^2\; -\; d(x,\backslash alpha\_t(x))^2|\backslash le\; 2d(x,y)^2,$$

so that

$$|d(y,\backslash alpha\_t(x))\; -\; d(x,\backslash alpha\_t(x))|\backslash le\; 2\; \backslash le\; .$$

Hence on the level surface of containing . The flow can be used to transport this result to all the level surfaces of . For general take such that and set . Then, where . But then, so that . Hence, as required.

Note that this argument could be shortened using the fact that two Busemann functions and differ by a constant if and only if the corresponding geodesic rays satisfy . Indeed, all the geodesics defined by the flow satisfy the latter condition, so differ by constants. Since along any of these geodesics is linear with derivative 1, must differ from these Busemann functions by constants.

Compactification of a proper Hadamard space

defined a compactification of a Hadamard manifold which uses Busemann functions. Their construction, which can be extended more generally to proper (i.e. locally compact) Hadamard spaces, gives an explicit geometric realisation of a compactification defined by Gromov—by adding an "ideal boundary"—for the more general class of proper metric spaces, those for which every closed ball is compact. Note that, since any Cauchy sequence is contained in a closed ball, any proper metric space is automatically complete.^[2] The ideal boundary is a special case of the ideal boundary for a metric space. In the case of Hadamard spaces, this agrees with the space of geodesic rays emanating from any fixed point described using Busemann functions in the bordification of the space.

If is a proper metric space, Gromov's compactification can be defined as follows. Fix a point in and let . Let be the space of Lipschitz continuous functions on, .e. those for which for some constant . The space can be topologised by the seminorms, the topology of uniform convergence on compacta. This is the topology induced by the natural map of C(X) into the direct product of the Banach spaces . It is give by the metric .

The space is embedded into by sending to the function . Let be the closure of in . Then is compact (metrisable) and contains as an open subset; moreover compactifications arising from different choices of basepoint are naturally homeomorphic. Compactness follows from the Arzelà–Ascoli theorem since the image in is equicontinuous and uniformly bounded in norm by . Let be a sequence in tending to in . Then all but finitely many terms must lie outside since is compact, so that any subsequence would converge to a point in ; so the sequence must be unbounded in . Let . Then lies in . It is non-zero on and vanishes only at . Hence it extends to a continuous function on with zero set . It follows that is closed in, as required. To check that the compactification is independent of the basepoint, it suffices to show that extends to a continuous function on . But, so, for in, . Hence the correspondence between the compactifications for and is given by sending in to in .

When is a Hadamard manifold (or more generally a proper Hadamard space), Gromov's ideal boundary can be realised explicitly as "asymptotic limits" of geodesics by using Busemann functions. Fixing a base point, there is a unique geodesic parametrised by arclength such that and

• \gamma ( 0 ) is a given unit vector. If is the corresponding Busemann function, then lies in and induces a homeomorphism of the unit -sphere onto, sending • \gamma ( 0 ) to .

Quasigeodesics in the Poincaré disk, CAT(-1) and hyperbolic spaces

Morse–Mostow lemma

In the case of spaces of negative curvature, such as the Poincaré disk, CAT(-1) and hyperbolic spaces, there is a metric structure on their Gromov boundary. This structure is preserved by the group of quasi-isometries which carry geodesics rays to quasigeodesic rays. Quasigeodesics were first studied for negatively curved surfaces—in particular the hyperbolic upper halfplane and unit disk—by Morse and generalised to negatively curved symmetric spaces by Mostow, for his work on the rigidity of discrete groups. The basic result is the Morse–Mostow lemma on the stability of geodesics.

By definition a quasigeodesic Γ defined on an interval with is a map into a metric space, not necessarily continuous, for which there are constants and such that for all and :

$$\backslash lambda^\; |s-t|\; -\; \backslash varepsilon\; \backslash le\; d(\backslash Gamma(s),\backslash Gamma(t))\; \backslash le\; \backslash lambda\; |s-t|\; +\; \backslash varepsilon.$$

The following result is essentially due to Marston Morse (1924).

Morse's lemma on stability of geodesics. In the hyperbolic disk there is a constant depending on and such that any quasigeodesic segment defined on a finite interval is within a Hausdorff distance of the geodesic segment .

Classical proof for Poincaré disk

The classical proof of Morse's lemma for the Poincaré unit disk or upper halfplane proceeds more directly by using orthogonal projection onto the geodesic segment.

• It can be assumed that Γ satisfies the stronger "pseudo-geodesic" condition:

$$\backslash lambda^|s-t|\; -\backslash varepsilon\; \backslash le\; d(\backslash Gamma(s),\backslash Gamma(t))\; \backslash le\; \backslash lambda\; |s-t|.$$

can be replaced by a continuous piecewise geodesic curve Δ with the same endpoints lying at a finite Hausdorff distance from less than : break up the interval on which is defined into equal subintervals of length and take the geodesics between the images under of the endpoints of the subintervals. Since is piecewise geodesic, is Lipschitz continuous with constant,, where . The lower bound is automatic at the endpoints of intervals. By construction the other values differ from these by a uniformly bounded depending only on and ; the lower bound inequality holds by increasing ε by adding on twice this uniform bound.

• If is a piecewise smooth curve segment lying outside an -neighbourhood of a geodesic line and is the orthogonal projection onto the geodesic line then:^[3]

$$\backslash ell(P\; \backslash circ\; \backslash gamma)\; \backslash le\; .$$

Applying an isometry in the upper half plane, it may be assumed that the geodesic line is the positive imaginary axis in which case the orthogonal projection onto it is given by and . Hence the hypothesis implies, so that

$$\backslash ell(P\backslash circ\backslash gamma)\; =\; \backslash int\_a^b$$

\over

\gamma

\le \int_a^b

\over \cosh(s)\, \Im \gamma

=.

• There is a constant depending only on and such that lies within an -neighbourhood of the geodesic segment .^[4]

Let be the geodesic line containing the geodesic segment . Then there is a constant depending only on and such that -neighbourhood lies within an -neighbourhood of . Indeed for any, the subset of for which lies outside the closure of the -neighbourhood of is open, so a countable union of open intervals . Then

$$\backslash ell(\backslash Gamma|\_)\; \backslash le\; s\_1\; \backslash equiv\; \backslash lambda^2\; (2s\; +\backslash varepsilon)\; \backslash left(1\; -\backslash right),$$

since the left hand side is less than or equal to and

$$$$

\over \lambda

-\varepsilon \le d(\Gamma(c),\Gamma(d)) \le 2s + d(P \circ \Gamma|_) \le 2s + .

Hence every point lies at a distance less than or equal to of . To deduce the assertion, note that the subset of for which lies outside the closure of the -neighbourhood of is open, so a union of intervals with and both at a distance from either or . Then

$$\backslash ell(\backslash Gamma|\_)\; \backslash le\; s\_2\; \backslash equiv\; \backslash lambda^2(2(s\; +\; s\_1)\; +\; \backslash varepsilon),$$

since

$$$$

\over \lambda

-\varepsilon \le d(\Gamma(c),\Gamma(d)) \le 2(s+s_1).

Hence the assertion follows taking any greater than .

• There is a constant depending only on and such that the geodesic segment lies within an -neighbourhood of .

Every point of lies within a distance of . Thus orthogonal projection carries each point of onto a point in the closed convex set at a distance less than . Since is continuous and connected, the map must be onto since the image contains the endpoints of . But then every point of is within a distance of a point of .

Gromov's proof for Poincaré disk

The generalisation of Morse's lemma to CAT(-1) spaces is often referred to as the Morse–Mostow lemma and can be proved by a straightforward generalisation of the classical proof. There is also a generalisation for the more general class of hyperbolic metric spaces due to Gromov. Gromov's proof is given below for the Poincaré unit disk; the properties of hyperbolic metric spaces are developed in the course of the proof, so that it applies mutatis mutandi to CAT(-1) or hyperbolic metric spaces.

Since this is a large-scale phenomenon, it is enough to check that any maps from for any to the disk satisfying the inequalities is within a Hausdorff distance of the geodesic segment . For then translating it may be assumed without loss of generality is defined on with and then, taking (the integer part of), the result can be applied to defined by . The Hausdorff distance between the images of and is evidently bounded by a constant depending only on and .

Now the incircle of a geodesic triangle has diameter less than where ; indeed it is strictly maximised by that of an ideal triangle where it equals . In particular, since the incircle breaks the triangle breaks the triangle into three isosceles triangles with the third side opposite the vertex of the original triangle having length less than, it follows that every side of a geodesic triangle is contained in a -neighbourhood of the other two sides. A simple induction argument shows that a geodesic polygon with vertices for has each side within a neighbourhood of the other sides (such a polygon is made by combining two geodesic polygons with sides along a common side). Hence if, the same estimate holds for a polygon with sides.

For let, the largest radius for a closed ball centred on which contains no in its interior. This is a continuous function non-zero on so attains its maximum at some point in this segment. Then lies within an -neighbourhood of the image of for any . It therefore suffices to find an upper bound for independent of .

Choose and in the segment before and after with and (or an endpoint if it within a distance of less than from). Then there are with, . Hence, so that . By the triangle inequality all points on the segments and are at a distance from . Thus there is a finite sequence of points starting at and ending at, lying first on the segment, then proceeding through the points, before taking the segment . The successive points are separated by a distance no greater than and successive points on the geodesic segments can also be chosen to satisfy this condition. The minimum number of points in such a sequence satisfies . These points form a geodesic polygon, with as one of the sides. Take, so that the -neighbourhood of does not contain all the other sides of the polygon. Hence, from the result above, it follows that . Hence

$$3\; +\; 2(\backslash lambda+\backslash varepsilon)^h\; +\; 6\backslash lambda\; h\; +\; \backslash varepsilon>\; 2^\; +\; 2.$$

This inequality implies that is uniformly bounded, independently of, as claimed.

If all points lie within of the, the result follows. Otherwise the points which do not fall into maximal subsets with . Thus points in have a point with in the complement of within a distance of . But the complement of, a disjoint union with and . Connectivity of implies there is a point in the segment which is within a distance of points and with and . But then, so . Hence the points for in lie within a distance from of less than .

Extension to quasigeodesic rays and lines

Recall that in a Hadamard space if and are two geodesic segments and the intermediate points and divide them in the ratio, then is a convex function of . In particular if and are geodesic segments of unit speed defined on starting at the same point then

$$d(\backslash Gamma\_1(t),\backslash Gamma\_2(t))\; \backslash le\; d(\backslash Gamma\_1(R),\backslash Gamma\_2(R)).$$

In particular this implies the following:

• In a CAT(–1) space, there is a constant depending only on and such that any quasi-geodesic ray is within a bounded Hausdorff distance of a geodesic ray. A similar result holds for quasigeodesic and geodesic lines.

If is a geodesic say with constant and, let be the unit speed geodesic for the segment . The estimate above shows that for fixed and sufficiently large, is a Cauchy sequence in with the uniform metric. Thus tends to a geodesic ray uniformly on compacta the bound on the Hausdorff distances between and the segments applies also to the limiting geodesic . The assertion for quasigeodesic lines follows by taking corresponding to the geodesic segment .

Efremovich–Tikhomirova theorem

Before discussing CAT(-1) spaces, this section will describe the Efremovich–Tikhomirova theorem for the unit disk with the Poincaré metric. It asserts that quasi-isometries of extend to quasi-Möbius homeomorphisms of the unit disk with the Euclidean metric. The theorem forms the prototype for the more general theory of CAT(-1) spaces. Their original theorem was proved in a slightly less general and less precise form in and applied to bi-Lipschitz homeomorphisms of the unit disk for the Poincaré metric;^[5] earlier, in the posthumous paper, the Japanese mathematician Akira Mori had proved a related result within Teichmüller theory assuring that every quasiconformal homeomorphism of the disk is Hölder continuous and therefore extends continuously to a homeomorphism of the unit circle (it is known that this extension is quasi-Möbius).^[6]

Extension of quasi-isometries to boundary

If is the Poincaré unit disk, or more generally a CAT(-1) space, the Morse lemma on stability of quasigeodesics implies that every quasi-isometry of extends uniquely to the boundary. By definition two self-mappings of are quasi-equivalent if, so that corresponding points are at a uniformly bounded distance of each other. A quasi-isometry of is a self-mapping of, not necessarily continuous, which has a quasi-inverse such that and are quasi-equivalent to the appropriate identity maps and such that there are constants and such that for all in and both mappings

$$\backslash lambda^\; d(x,y)\; -\; \backslash varepsilon\; \backslash le\; d(f\_k(x),f\_k(y))\; \backslash le\; \backslash lambda\; d(x,y)\; +\backslash varepsilon.$$

Note that quasi-inverses are unique up to quasi-equivalence; that equivalent definition could be given using possibly different right and left-quasi inverses, but they would necessarily be quasi-equivalent; that quasi-isometries are closed under composition which up to quasi-equivalence depends only the quasi-equivalence classes; and that, modulo quasi-equivalence, the quasi-isometries form a group.^[7]

Fixing a point in, given a geodesic ray starting at, the image under a quasi-isometry is a quasi-geodesic ray. By the Morse-Mostow lemma it is within a bounded distance of a unique geodesic ray starting at . This defines a mapping on the boundary of, independent of the quasi-equivalence class of, such that . Thus there is a homomorphism of the group of quasi-isometries into the group of self-mappings of .

To check that is continuous, note that if and are geodesic rays that are uniformly close on, within a distance, then and lie within a distance on, so that and lie within a distance ; hence on a smaller interval, and lie within a distance by convexity.

On CAT(-1) spaces, a finer version of continuity asserts that is a quasi-Möbius mapping with respect to a natural class of metric on, the "visual metrics" generalising the Euclidean metric on the unit circle and its transforms under the Möbius group. These visual metrics can be defined in terms of Busemann functions.^[8]

In the case of the unit disk, Teichmüller theory implies that the homomorphism carries quasiconformal homeomorphisms of the disk onto the group of quasi-Möbius homeomorphisms of the circle (using for example the Ahlfors–Beurling or Douady–Earle extension): it follows that the homomorphism from the quasi-isometry group into the quasi-Möbius group is surjective.

In the other direction, it is straightforward to prove that the homomorphism is injective. Suppose that is a quasi-isometry of the unit disk such that is the identity. The assumption and the Morse lemma implies that if is a geodesic line, then lies in an -neighbourhood of . Now take a second geodesic line such that and intersect orthogonally at a given point in . Then lies in the intersection of -neighbourhoods of and . Applying a Möbius transformation, it can be assumed that is at the origin of the unit disk and the geodesics are the real and imaginary axes. By convexity, the -neighbourhoods of these axes intersect in a -neighbourhood of the origin: if lies in both neighbourhoods, let and be the orthogonal projections of onto the - and -axes; then so taking projections onto the -axis, ; hence . Hence, so that is quasi-equivalent to the identity, as claimed.

Cross ratio and distance between non-intersecting geodesic lines

Given two distinct points on the unit circle or real axis there is a unique hyperbolic geodesic joining them. It is given by the circle (or straight line) which cuts the unit circle unit circle or real axis orthogonally at those two points. Given four distinct points in the extended complex plane their cross ratio is defined by

$$(a,b;c,d)\; =.$$

If is a complex Möbius transformation then it leaves the cross ratio invariant: . Since the Möbius group acts simply transitively on triples of points, the cross ratio can alternatively be described as the complex number in such that for a Möbius transformation .

Since,, and all appear in the numerator defining the cross ratio, to understand the behaviour of the cross ratio under permutations of,, and, it suffices to consider permutations that fix, so only permute, and . The cross ratio transforms according to the anharmonic group of order 6 generated by the Möbius transformations sending to and . The other three transformations send to, to and to .^[9]

Now let be points on the unit circle or real axis in that order. Then the geodesics and do not intersect and the distance between these geodesics is well defined: there is a unique geodesic line cutting these two geodesics orthogonally and the distance is given by the length of the geodesic segment between them. It is evidently invariant under real Möbius transformations. To compare the cross ratio and the distance between geodesics, Möbius invariance allows the calculation to be reduced to a symmetric configuration. For, take . Then where . On the other hand, the geodesics and are the semicircles in the upper half plane of radius and . The geodesic which cuts them orthogonally is the positive imaginary axis, so the distance between them is the hyperbolic distance between and, . Let, then, so that there is a constant such that, if, then

$$d([a,d];[b,c])\; -\; C\; \backslash le\; \backslash log\; (a,b;c,d)\; \backslash le\; d([a,d];[b,c])\; +\; C,$$

since is bounded above and below in . Note that are in order around the unit circle if and only if .

A more general and precise geometric interpretation of the cross ratio can be given using projections of ideal points on to a geodesic line; it does not depend on the order of the points on the circle and therefore whether or not geodesic lines intersect.^[10]

• If and are the feet of the perpendiculars from and to the geodesic line, then .

Since both sides are invariant under Möbius transformations, it suffices to check this in the case that,, and . In this case the geodesic line is the positive imaginary axis, right hand side equals, and . So the left hand side equals . Note that and are also the points where the incircles of the ideal triangles and touch .

Proof of theorem

A homeomorphism of the circle is quasisymmetric if there are constants such that

$$\backslash displaystyle$$

It is quasi-Möbius is there are constants such that

$$\backslash displaystyle$$

\le c

(z_1,z_2;z_3,z_4)

^d,

where

$$\backslash displaystyle$$

denotes the cross-ratio.

It is immediate that quasisymmetric and quasi-Möbius homeomorphisms are closed under the operations of inversion and composition.

If is quasisymmetric then it is also quasi-Möbius, with and : this follows by multiplying the first inequality for and . Conversely any quasi-Möbius homeomorphism is quasisymmetric. To see this, it can be first be checked that (and hence) is Hölder continuous. Let be the set of cube roots of unity, so that if in, then . To prove a Hölder estimate, it can be assumed that is uniformly small. Then both and are greater than a fixed distance away from in with, so the estimate follows by applying the quasi-Möbius inequality to . To verify that is quasisymmetric, it suffices to find a uniform upper bound for in the case of a triple with, uniformly small. In this case there is a point at a distance greater than 1 from, and . Applying the quasi-Möbius inequality to,, and yields the required upper bound. To summarise:

• A homeomorphism of the circle is quasi-Möbius if and only if it is quasisymmetric. In this case it and its inverse are Hölder continuous. The quasi-Möbius homeomorphisms form a group under composition.

To prove the theorem it suffices to prove that if then there are constants such that for distinct points on the unit circle

$$\backslash displaystyle$$

\le A

(a,b;c,d)

^B.

It has already been checked that (and is inverse) are continuous. Composing, and hence, with complex conjugation if necessary, it can further be assumed that preserves the orientation of the circle. In this case, if are in order on the circle, so too are there images under ; hence both and are real and greater than one. In this case

$$(F(a),F(b);F(c),F(d))\; \backslash le\; A\; (a,b;c,d)^B.$$

To prove this, it suffices to show that . From the previous section it suffices show . This follows from the fact that the images under of and lie within -neighbourhoods of and ; the minimal distance can be estimated using the quasi-isometry constants for applied to the points on and realising .

Adjusting and if necessary, the inequality above applies also to . Replacing,, and by their images under, it follows that

$$A^\; |(a,b;c,d)|^\; \backslash le\; |(F(a),F(b);F(c),F(d))|\; \backslash le\; A\; |(a,b;c,d)|^B$$

if,, and are in order on the unit circle. Hence the same inequalities are valid for the three cyclic of the quadruple . If and are switched then the cross ratios are sent to their inverses, so lie between 0 and 1; similarly if and are switched. If both pairs are switched, the cross ratio remains unaltered. Hence the inequalities are also valid in this case. Finally if and are interchanged, the cross ratio changes from to, which lies between 0 and 1. Hence again the same inequalities are valid. It is easy to check that using these transformations the inequalities are valid for all possible permutations of,, and, so that and its inverse are quasi-Möbius homeomorphisms.

Busemann functions and visual metrics for CAT(-1) spaces

Busemann functions can be used to determine special visual metrics on the class of CAT(-1) spaces. These are complete geodesic metric spaces in which the distances between points on the boundary of a geodesic triangle are less than or equal to the comparison triangle in the hyperbolic upper half plane or equivalently the unit disk with the Poincaré metric. In the case of the unit disk the chordal metric can be recovered directly using Busemann functions and the special theory for the disk generalises completely to any proper CAT(-1) space . The hyperbolic upper half plane is a CAT(0) space, as lengths in a hyperbolic geodesic triangle are less than lengths in the Euclidean comparison triangle: in particular a CAT(-1) space is a CAT(0) space, so the theory of Busemann functions and the Gromov boundary applies. From the theory of the hyperbolic disk, it follows in particular that every geodesic ray in a CAT(-1) space extends to a geodesic line and given two points of the boundary there is a unique geodesic such that has these points as the limits . The theory applies equally well to any CAT space with since these arise by scaling the metric on a CAT(-1) space by . On the hyperbolic unit disk quasi-isometries of induce quasi-Möbius homeomorphisms of the boundary in a functorial way. There is a more general theory of Gromov hyperbolic spaces, a similar statement holds, but with less precise control on the homeomorphisms of the boundary.

Example: Poincaré disk

Applications in percolation theory

See also: percolation theory. More recently Busemann functions have been used by probabilists to study asymptotic properties in models of first-passage percolation and directed last-passage percolation.

References

• • • • • • • • • • , Appendix.
                    • Book: Mostow . G. Daniel . Strong Rigidity of Locally Symmetric Spaces . Annals of Mathematics Studies . 78 . 1973 . Princeton University Press . 978-0-6910-8136-6 . j.ctt1bd6kr9.

Notes and References

1. In geodesic normal coordinates, the metric . By geodesic convexity, a geodesic from to lies in the ball of radius . The straight line segment gives an upper estimate for of the stated form. To obtain a similar lower estimate, observe that if is a smooth path from to, then . (Note that these inequalities can be improved using the sharper estimate).
2. Note that a metric space which is complete and locally compact need not be proper, for example with the metric . On the other hand, by the Hopf–Rinow theorem for metric spaces, if is complete, locally compact and geodesic—every two points and are joined by a geodesic parametrised by arclength—then is proper (see). Indeed if not, there is a point in and a closed ball maximal subject to being compact; then, since by hypothesis is non-compact for each, a diagonal argument shows that there is a sequence with decreasing to but with no convergent subsequence; on the other hand taking on a geodesic joining and, with, compactness of implies, and hence, has a convergent subsequence, a contradiction.
3. , Lemma 4
4. , Lemmas 5–6
5. Bi-Lipschitz homeomorphisms are those for which they and their inverses are Lipschitz continuous
6. See:
7. See:
8. See:
9. Note that there is a natural homomorphism of onto, acting by conjugation on and . Indeed these permutations together with the identity form a normal Abelian subgroup equal to its own centraliser: the action of by conjugation on the non-trivial elements yields the homomorphism onto .
10. See: