## Three hyperbolic metrics

Up to a constant factor, there is just one conformally invariant Riemannian metric ${\rho}$ on the disk ${\mathbb{D}=\{z\in {\mathbb C}\colon |z|<1\}}$. Indeed, on the tangent space at ${0}$ the metric must be a multiple of the Euclidean one, due to rotational invariance. Normalizing so that at ${0}$ both metrics coincide, we can use the invariance under conformal automorphisms (Möbius transformations)

$\displaystyle \psi_a(z) = \frac{z+a}{1+\bar a z},\quad |a|<1$

to find that on the tangent space at ${a}$ the metric ${\rho}$ differs from Euclidean only by the factor of ${|\psi'_a(a)| (1-|a|^2)^{-1}}$. This can be written as ${d\rho(z) = (1-|z|^2)^{-1}|dz|}$, indicating what we integrate to find the length of curves with respect to ${\rho}$. This is the Poincaré disk model of the hyperbolic plane. The geodesics of ${\rho}$ are precisely the circles orthogonal to ${\partial\mathbb{D}}$, and diameters.

We could model the hyperbolic plane on any other proper simply-connected domain ${\Omega\subset {\mathbb C}}$, just by pulling back ${\rho}$ under the Riemann map ${\phi\colon\Omega\rightarrow\mathbb{D}}$. Explicitly, ${d\rho_{\Omega}(z)=(1-|\phi(z)|^{-2})\,|\phi'(z)|\,|dz|}$. But is this really explicit? We have no closed form for ${\phi}$ except for very special domains ${\Omega}$. The density ${\rho_\Omega}$ can be quite tricky: see this MathOverflow question.

On the disk ${\mathbb{D}}$, the density of ${\rho}$ is roughly the reciprocal of the distance to the boundary. The Koebe distortion theorem yields the same for all simply connected domains:

$\displaystyle \mathrm{dist}\,(z,\partial \Omega)^{-1} \le \frac{|d\rho_\Omega(z)|}{|dz|}\le 4\, \mathrm{dist}\,(z,\partial \Omega)^{-1}$

This suggests replacing the hyperbolic metric ${\rho_\Omega}$ with the quasihyperbolic metric ${d\delta_\Omega(z)=\mathrm{dist}\,(z,\partial \Omega)^{-1}\,|dz|}$. The density of ${\delta_\Omega}$ is about as simple as we could wish for, but the shape of its geodesics is not as obvious; indeed, a number of papers were written on this subject in the last 35 years.

Can we have a hyperbolic-type metric with explicit geodesics and explicit density? The Hilbert metric delivers both, at least in bounded convex domains. The idea is simple: instead of dividing the length of each tangent vector ${v\in T_a}$ by ${\mathrm{dist}\,(a,\partial \Omega)}$, we divide it by ${\mathrm{dist}_v\,(a,\partial \Omega)}$, which is the distance measured in the direction of ${v}$. In other words, this is how long you could walk in the direction ${v}$ before hitting the boundary. It is clear that the length of any curve going to the boundary is infinite due to the divergence of ${\int_{0^+} \frac{dx}{x}}$, same as for hyperbolic and quasihyperbolic metrics. Since it is awkward to have a “metric” that is not symmetric (the lengths of ${v}$ and ${-v}$ are not the same), we symmetrize:

$\displaystyle \|v\|_{\Omega,a} =\frac12 \left(\frac{1}{\mathrm{dist}_v\,(a,\partial \Omega)} + \frac{1}{\mathrm{dist}_{-v}\,(a,\partial \Omega)} \right) |v|_{{\mathbb R}^2}$

Now the distances between points are defined in the usual way, as the infimum of lengths of connecting curves ${\gamma \colon [a,b]\rightarrow\Omega}$,

$\displaystyle L(\gamma) = \int_a^b \|\gamma'(t)\|_{\Omega,\gamma(t)}\,dt$

This is the Hilbert metric ${d_\Omega}$. Despite all its non-Euclideanness, the geodesics of ${d_\Omega}$ are line segments. This (nontrivial) fact makes it easy to calculate the distance between two points ${a,b\in\Omega}$: besides the points themselves, one only needs to consider the pair of points ${a',b'}$ where the line through ${a,b}$ meets ${\partial \Omega}$. The integral ends up being (half of) the logarithm of the cross-ratio of these four points. Assuming ${a',a,b,b'}$ are situated in the listed order, the relevant cross-ratio is

$\displaystyle \frac{|a'-b|\, |a-b'|}{|a'-a|\,|b-b'|}$

which is greater than ${1}$, making the logarithm positive.

I personally find the integral of reciprocals of distances more intuitively accessible than the logarithm of cross-ratio. Either version of the definition makes it clear that the Hilbert metric is invariant under projective transformations, a property not shared by the metrics ${\rho}$ and ${\delta}$.

It turns out that ${d_{\mathbb D}}$ is also a model of the hyperbolic plane, but with geodesics being line segments rather than circular arcs. Along each diameter ${d_{\mathbb D}}$ coincides with ${\rho}$, because at Euclidean distance ${r}$ from the center it scales vectors by

$\displaystyle \frac{1}{2}\left(\frac{1}{1-r}+\frac{1}{1+r}\right) = \frac{1}{1-r^2}$

Behold the magic of partial fractions!

## Quasi-isometries and stability of quasi-geodesics

Continuation of expository series on Gromov hyperbolicity. Recall that a map ${f\colon X\rightarrow Y}$ is a quasi-isometry if there are constants ${L,M}$ such that ${L^{-1}|x_1x_2|-M\le |f(x_1)f(x_2)|\le L|x_1x_2|+M}$ for all ${x_1,x_2\in X}$. This is a coarse version of the bi-Lipschitz condition. Surprisingly, Gromov hyperbolicity is preserved under quasi-isometries of geodesic spaces. The surprising part is that the multiplicative constant ${L}$ does not kill the additive constant ${\delta}$.

Theorem. Suppose ${X}$ and ${Y}$ are geodesic metric spaces, and ${Y}$ is Gromov hyperbolic. If there exists a quasi-isometry ${f\colon X\rightarrow Y}$, then ${X}$ is also Gromov hyperbolic.

Proof goes like this. Assuming that ${X}$ contains a fat geodesic triangle ${a,b,c}$, we consider the geodesic triangle in ${Y}$ with vertices ${f(a),f(b),f(c)}$, and want to prove that it is also fat. Since ${f}$ is a quasi-isometry, it follows that the images of geodesics ${ab}$, ${bc}$ and ${ac}$ form a roughly-triangular shape which has the fatness property: there is a point on one of the sides that is far away from the other two sides. The problem reduces to showing that this roughly-triangular shape lies within a certain distance ${R}$ (independent of ${a,b,c}$) from the actual geodesic triangle with vertices ${f(a),f(b),f(c)}$. This is known as stability of quasi-geodesics. A quasi-geodesic is a quasi-isometric image of a line segment, similar to how a geodesic is a (locally) isometric image of a segment.

By the way, quasi-geodesic stability fails in ${\mathbb R^2}$. We can connect the points ${(-n,0)}$ and ${(n,0)}$ by the quasi-geodesic ${y=n-|x|}$, which is at distance ${n}$ from the true geodesic between these points.

I’ll prove a more specialized and weaker statement, which however contains the essence of the full result. Namely, let $\mathbb H^2$ denote the hyperbolic plane and assume that ${f\colon [a,b]\rightarrow \mathbb H^2}$ is bi-Lipschitz: ${L^{-1}|x_1-x_2|\le |f(x_1)f(x_2)|\le L|x_1-x_2|}$ for all ${x_1,x_2\in [a,b]}$. The claim is that the image of ${f}$ lies in the ${R}$-neighborhood of the geodesic through ${f(a)}$ and ${f(b)}$, where ${R}$ depends only on ${L}$.

There are three standard models of hyperbolic plane: disk, halfplane and infinite strip. I’ll use the last one, because it’s the only model in which a geodesic is represented by Euclidean line. Specifically, ${\mathbb H^2}$ is identified with the infinite strip ${\{x+iy\in\mathbb C\colon |y|<\pi/2\}}$ equipped with the metric ${|dz|/\cos y}$. (To see where the metric comes from, apply $z\mapsto i e^{iz}$ to map the strip onto upper halfplane and pull back the hyperbolic metric $|dw|/\mathrm{Im}\,w$.)

The hyperbolic and Euclidean metrics coincide on the real line, which is where we place ${f(a)}$ and ${f(b)}$ with the help of some hyperbolic isometry. Let ${\Gamma=f[a,b]}$ be our quasi-geodesic. Being a bi-Lipschitz image of a line segment, ${\Gamma}$ satisfies the chord-arc condition: the length of any subarc of ${\Gamma}$ does not exceed ${L^2}$ times the distance between its endpoints. Pick ${y_0\in (0,\pi/2)}$ such that ${1/\cos y_0=2L^2}$. Let ${D}$ be the hyperbolic distance between the lines ${y=y_0}$ and ${y=0}$. This distance could be calculated as ${\int_0^{y_0}\sec y\,dy}$, but I’d rather keep this integral as an exquisite Calculus II torture device.

The problem facing us is that quasigeodesic may be $L^2$ times longer than the distance between its endpoints, which seems to allows it to wander far off the straight path. However, it turns out there is a uniform bound on the length of any subarc ${\Gamma'}$ of ${\Gamma}$ that lies within the substrip ${\{y \ge y_0\}}$. We lose no generality in assuming that the endpoints of $\Gamma'$ are on the line $y=y_0$; they will be denoted ${x_j+iy_0}$, ${j=1,2}$. The key point is that connecting these two points within $\{y\ge y_0\}$ is rather inefficient, and such inefficiency is controlled by the chord-arc property.

The hyperbolic distance between ${x_j+iy_0}$ is at most $|x_1-x_2|+2D$, because we can go from ${x_1+iy_0}$ to ${x_1}$ (distance ${D}$), then from ${x_1}$ to ${x_2}$ (distance ${|x_1-x_2|}$), and finally from ${x_2}$ to ${x_2+iy_0}$ (distance ${D}$). On the other hand, the length of ${\Gamma'}$ is at least ${|x_1-x_2|/\cos y_0 = 2L^2|x_1-x_2|}$ because the density of hyperbolic metric is at least $1/\cos y_0$ where $\Gamma'$ lives. The chord-arc property yields ${2L^2 |x_1-x_2| \le L^2 (|x_1-x_2|+2D)}$, which simplifies to ${|x_1-x_2| \le 2D}$. Hence, the distance between the endpoints of ${\Gamma'}$ is at most ${4D}$, and another application of the chord-arc property bounds the length of ${\Gamma'}$ by ${4DL^2}$.

In conclusion, the claimed stability result holds with ${R= D+2DL^2}$.

Complete proofs can be found in many books, for example Metric Spaces of Non-Positive Curvature by Bridson and Haefliger or Elements of Asymptotic Geometry by Buyalo and Schroeder. I used Schroeder’s lecture notes An introduction to asymptotic geometry.