Distance in the Poincare Disk model of hyperbolic geometry

Question

I am trying to understand the Poincare Disk model of a hyperbolic geometry and how to measure distances. I found the equation for the distance between two points on the disk as: $d^2 = (dx^2 + dy^2) / (1-x^2-y^2)^2$ Given two points on the disk, I am assuming that $dx$ is the difference in the euclidean $x$ coordinate of the points, and similar for $dy$. So, what are $x$ and $y$ in the formula? Also, I see some formulas online as: $d^2 = 4(dx^2 + dy^2) / (1-x^2-y^2)^2$ I am not sure which one is correct.

Assuming that $X$ and $Y$ are coordinates of one of the points, I have tried a few examples, but can't get the math to work out. For instance, if $A = (0,0)$, $B = (0,.5)$, $C = (0,.75)$, then what are $d(A,B)$, $d(B,C)$, and $d(A,C)$? The value $d(A,B) + d(B,C)$ should equal $d(A,C)$, since they are on the same line, but I can't get this to work out. I get distances of $.666$, $.571$, and $1.714$ respectively.

MathWorld - Hyperbolic Metric

Answer 1

As others have indicated one has to distinguish a metric $(x,y)\mapsto d(x,y)$ which measures distances between points $x$, $y$ in a space $X$ and a Riemannian metric which tells us how lengths of curves $\gamma$ in a manifold $X$ should be computed.

In the complex plane we have the usual euclidean metric $d(z_1,z_2):=|z_2-z_1|$ and at the infinitesimal scale the Riemannian metric $$ds^2:=|dz|^2=dx^2+dy^2\ .$$ The latter formula says that the length of an arbitrary curve $$\gamma: \quad t\mapsto\bigl(x(t),y(t)\bigr)\qquad(a\leq t\leq b)$$ should be computed as $$L(\gamma)=\int_\gamma ds=\int_a^b\sqrt{x'^2(t)+y'^2(t)}\ dt = \int_a^b|z'(t)|\ dt\ .$$ This formula implies that the length of a segment $\sigma$ connecting two points $z_1$ and $z_2$ is just $|z_2-z_1|$.

On the "Poincare disc" $P$ we are given a priori only a Riemannian metric $$ds:= {|dz|\over 1-|z|^2}$$ (resp., $ds^2=\ldots$). This metric allows us to compute the lengths of arbitrary curves in $P$: $$L(\gamma)=\int_a^b{|z'(t)|\over1-|z(t)|^2}\ dt\ .$$ The particular definition of $ds$ is chosen such that this hyperbolic length is invariant under arbitrary conformal movements of $P$ and the curves therein.

A posteriori one can define a metric $d(\cdot,\cdot)$ on $P$ by letting the distance $d(z_1,z_2)$ between two points $z_1$, $z_2\in P$ be the hyperbolic length of the shortest curve $\gamma$ connecting $z_1$ and $z_2$. The actual carrying out of this idea shows that $d(z_1,z_2)$ can be written as an elementary function (using ${\rm artanh}$, etc.) in terms of $z_1$ and $z_2$.

Answer 2

The correct formula that is possible to memorize is for the upper half plane in $\mathbb C.$ The unit speed geodesics there are of just two kinds, parametrized by a letter $t,$ with real constant $B > 0$ and any $A,$ vertical $$ A + i e^t,$$ semicircular $$ A + B \tanh t + i B \; \mbox{sech} \; t.$$ The latter may seem unfamiliar, note $$ \cosh^2 t - \sinh^2 t = 1, $$ $$ \cosh^2 t = \sinh^2 t + 1, $$ divide through by $\cosh^2 t$ to get $$ 1 = \tanh^2 t + \mbox{sech}^2 t. $$ Two points in the upper half plane are either in a vertical line or not. In the latter case, it is necessary to draw the perpendicular bisector of the ordinary segment between them and find out where that hits the real axis to find $A,$ then $B$ next.

Meanwhile, an isometry taking the unit disk to the upper half plane is the Moebius transformation $$ f(z) = \frac{z + i}{i z + 1} .$$ The points you gave were along the imaginary axis, and for real $v$ $$ f(iv) = i \; \left( \frac{1+v}{1-v} \right). $$ Your three specific points gave $$ f(0) = i, \; f(i/2) = 3 i, \; f(3i/4) = 7i. $$ These are all along the vertical geodesic $i e^t,$ so their pairwise distances are just the absolute values of the differences in the necessary values of $t,$ those being $0, \; \log 3, \; \log 7. $ So your $$ d(A,B) = \log 3 \approx 1.0986, $$ $$ d(B,C) = \log 7 - \log 3 \approx 0.8473, $$ $$ d(A,C) = \log 7 \approx 1.9459. $$

Furthermore, rotation of the unit disk is an isometry there, so as long as you have points along a radius of the disk, the distances to the origin (and to each other by subtraction) are quite specific: if a point in the unit disk is at "ordinary" distance $v$ from the origin, with the required $0 \leq v < 1,$ then the ``hyperbolic'' distance from the origin is $$ \log \left( \frac{1+v}{1-v} \right). $$ However, if your two points of interest are not along a radius, you must either find a Moebius transformation of the disk to itself that takes them to a common radius (there are such) or map directly to $\mathbb C$ with my $f(z)$ and work with $\tanh t.$ This latter method is probably less work, and less likely to produce errors, than the former method.

Answer 3

There are 2 related but different ideas which both go by the name "metric". The first is a "metric" as in "metric space". That is it's a function $d:X\times X\rightarrow\mathbb{R}$ satisfying a few axioms (such as the triangle inequality).

On the other hand, a smooth manifold can be equipped with what is called a "Riemannian metric" or, confusingly, just a "metric" for short. Recall that on a manifold, each point has an associated vector space, called the tangent space at that point. A Riemanian metric, then, is a smoothly varying choice of inner product on each tangent space. A Riemannian metric allows us to determine lengths (and angles) of tangent vectors. The notation $$ds^2 = \frac{dx^2 + dy^2}{(1-x^2-y^2)^2}$$ is that of a Riemannian metric. It tells you that the tangent vector $\vec{v} = (v_1, v_2)$ with tail at the point $(x,y)$ has length $\frac{v_1^2 + v_2^2}{(1-x^2-y^2)^2}$.

Now, a Riemannian metric allows us to measure the length of (nice enough) curves. Given a differentiable curve $\gamma$, the length of $\gamma$ is defined as $\int_\gamma |\gamma'(t)|dt$. This, then, leads back to the usual (topological) notion of a metric: We can define the distance between two points to be the infimum distance of all curves between them.

Answer 4

The point is $dx$ and $dy$ in the formula (the one with the 4 in is right btw) don't represent the difference in the $x$ and $y$ co-ordinates, but rather the 'infinitesimal' distance at the point $(x,y)$, so to actually find the distance between two points we have to do some integration.

So the idea is that at the point $(x,y)$ in Euclidean co-ordinates, the length squared, $ds^2$, of an infinitesimally small line is the sum of the infinitesimally small projections of that line onto the $x$ and $y$ axes ($dx^2$ and $dy^2$) multiplied by a scaling factor which depends on $x$ and $y$ ($\frac{4}{(1-x^2-y^2)^2}$).