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Section5.3Measurement in Hyperbolic Geometry

In this section we develop a notion of distance in the hyperbolic plane. If someone is standing at point \(p\) and wants to get to point \(q\text{,}\) he or she should be able to say how far it is to get there, whatever the route taken.

The distance formula is derived following the approach given in Section 30 of Boas' text [2]. We first list features our distance function ought to have, and use the notation that \(d_H(p,q)\) represents the hyperbolic distance from \(p\) to \(q\) in the hyperbolic plane \(\mathbb{D}\text{.}\)

  1. The distance between 2 distinct points should be positive.
  2. The shortest path between 2 points should be on the hyperbolic line connecting them.

  3. If \(p,q,\) and \(r\) are three points on a hyperbolic line with \(q\) between the other two then \(d_H(p,q) + d_H(q,r) = d_H(p,r)\text{.}\)

  4. Distance should be preserved by transformations in \(\cal H\text{.}\) (A lunch pail shouldn't shrink if it is moved to another table.) In other words, the distance formula should satisfy \begin{equation*} d_H(p,q) = d_H(T(p),T(q)) \end{equation*} for any points \(p\) and \(q\) in \(\mathbb{D}\text{,}\) and any transformation \(T\) in \({\cal H}\text{.}\)

  5. In the limit for small distances, hyperbolic distance should be proportional to Euclidean distance.

Perhaps the least obvious of the features listed is the last one. One theme of this text is that locally, on small scales, non-Euclidean geometry behaves much like Euclidean geometry. A small segment in the hyperbolic plane is approximated to the first order by a Euclidean segment. Small hyperbolic triangles look like Euclidean triangles and hyperbolic angles correspond to Euclidean angles; the hyperbolic distance formula will fit with this theme.

To find the distance function, start with a point's distance from the origin. Given a point \(z\) in \(\mathbb{D}\text{,}\) rotate about 0 so that \(z\) gets sent to the point \(x = |z|\) on the positive real axis.

We may find a hyperbolic line \(L\) about which \(x\) gets inverted to the origin. Such a hyperbolic line is constructed in the proof of Theorem 5.1.3. Recall, the line \(L\) is centered at \(x^*\text{,}\) the point symmetric to \(x\) with respect to the unit circle \(\mathbb{S}^1_\infty\text{,}\) and it goes through the points at which \(\mathbb{S}^1_\infty\) intersects the circle with diameter \(0x^*\text{.}\) Let \(x + h\) be a point near \(x\) on the positive real axis, and suppose \(x + h\) gets inverted to the point \(w\text{,}\) as depicted in Figure 5.3.1. One can show (in Exercise 5.3.1) that \begin{equation*} w = \frac{-h}{1 - x^2 - hx}. \end{equation*}

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Figure5.3.1Reflection of the hyperbolic plane sending \(x\) to 0 and \(x+h\) to \(w\text{.}\)

If distance is to be preserved by transformations in \({\cal H}\text{,}\) \begin{equation*} d_H(x,x+h) = d_H(0,w). \tag{1} \end{equation*}

Also, \(0, x,\) and \(x+h\) are all on the same hyperbolic line (the real axis), so assuming \(h > 0\) \begin{equation*} d_H(0,x)+d_H(x,x+h) = d_H(0,x+h). \tag{2} \end{equation*}

For \(0 \lt x \lt 1\) define the function \(d(x) = d_H(0,x)\) which is the hyperbolic distance of \(x\) to the origin. Then (2) and (1) may be combined to give \begin{equation*} d(x+h)-d(x)=d(w). \end{equation*}

Divide both sides by \(h\) to get \begin{equation*} \frac{d(x+h)-d(x)}{h}=\frac{d(w)}{h}. \end{equation*}

As \(h \to 0\) we obtain \begin{equation*} d^\prime(x) = \lim_{h \to 0} \frac{d(w)}{h}. \end{equation*}

We now interrupt this derivation with an important point. In the limit for small \(w\text{,}\) the hyperbolic distance of \(w\) from 0, \(d(w)\text{,}\) is proportional to the Euclidean distance \(|w - 0| = |w|\text{.}\) Since \(w\) is the image of \(x+h\) under the inversion and \(x\) gets inverted to 0, it follows that \(w \to 0\) as \(h \to 0\text{.}\) So, we assume that \begin{equation*} \lim_{h \to 0} \frac{d(w)}{|w|} = k \end{equation*} for some constant \(k\text{.}\) Following convention, we set the constant of proportionality to \(k = 2\text{,}\) as this makes length and area formulas look very nice later on. Now, back to the derivation. \begin{align*} d^\prime(x) \amp = \lim_{h \to 0} \frac{d(w)}{h}\\ \amp = \lim_{h \to 0} \frac{d(w)}{|w|}\frac{|w|}{h}\\ \amp = \lim_{h \to 0} \bigg[ 2 \cdot \frac{h}{(1-x^2-hx)h} \bigg]\\ \amp = \frac{2}{1-x^2}. \end{align*}

To get back to the distance function \(d(x)\) we integrate: \begin{align*} \int \frac{2}{1-x^2} ~dx \amp = \int \bigg( \frac{1}{1-x} + \frac{1}{1+x}\bigg)~dx \tag{partial~fractions}\\ \amp = -\ln(1-x) + \ln(1+x)\\ d(x) \amp = \ln\bigg(\frac{1+x}{1-x}\bigg). \end{align*}

To summarize, the hyperbolic distance from 0 to a point \(z\) in \(\mathbb{D}\) is \begin{equation*} d_H(0,z) = \ln\bigg(\frac{1+|z|}{1-|z|}\bigg). \end{equation*}

Notice that if \(z\) inches its way in \(\mathbb{D}\) out toward the circle at infinity (i.e., \(|z| \to 1\)), the hyperbolic distance from \(0\) to \(z\) approaches \(\infty\text{.}\) This is a good thing. Thinking of Euclid's postulates, this notion of distance satisfies one of our fundamental requirements: One can produce a hyperbolic segment to any finite length.

To arrive at a general distance formula \(d_H(p,q)\text{,}\) observe something curious. The hyperbolic line through \(0\) and \(x\) has ideal points -1 and 1. Furthermore, the expression \((1+x)/(1-x)\) corresponds to the cross ratio of the points \(0\text{,}\) \(x\text{,}\) 1, and -1. In particular, \begin{equation*} (0,x;1,-1) = \frac{0 - 1}{0 + 1}\cdot \frac{x + 1}{x - 1} = \frac{1 + x}{1 - x}. \end{equation*}

Thus, \begin{equation*} d_H(0,x) = \ln((0,x;1,-1)). \end{equation*}

We can now derive a general distance formula, assuming the invariance of distance under transformations in \({\cal H}\text{.}\) There is a transformation \(T\) in \({\cal H}\) that takes \(p\) to the origin and \(q\) to some spot on the positive real-axis, call this spot \(x\) (see Figure 5.3.2). Thus, \begin{align*} d_H(p,q) \amp = d_H(T(p),T(q)) \tag{invariance of distance}\\ \amp = d_H(0,x)\\ \amp = \ln((0,x;1,-1))\\ \amp = \ln((p,q;u,v)),\tag{invariance of cross ratio} \end{align*}

where \(u\) and \(v\) are the ideal points of the hyperbolic line through \(p\) and \(q\text{.}\) To be precise, \(u\) is the ideal point you would head toward as you went from \(p\) to \(q\text{,}\) and \(v\) is the ideal point you would head toward as you went from \(q\) to \(p\text{.}\)

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Figure5.3.2To find the distance between \(p\) and \(q\text{,}\) we may first transform \(p\) to \(0\) and \(q\) to the positive real axis.

SubsectionA working formula for \(\mathbf{d_H(p,q)}\)

One may compute the hyperbolic distance between \(p\) and \(q\) by first finding the ideal points \(u\) and \(v\) of the hyperbolic line through \(p\) and \(q\) and then using the formula \(d_H(p,q)=\ln((p,q;u,v))\text{.}\) In practice, finding coordinates for these ideal points can be a difficult task, and it is often simpler to compute the distance between points by first moving one of them to the origin. (This simpler approach uses the fact that hyperbolic distance is preserved under transformations in \(\cal H\text{.}\) This fact will be proved shortly.)

One transformation in \({\cal H}\) that sends \(p\) to 0 has the form \begin{equation*} T(z) = \frac{z-p}{1-\overline{p}z}. \end{equation*}

The map \(T\) sends \(q\) to some other point, \(T(q)\text{,}\) in \(\mathbb{D}\text{.}\) Assuming again that \(T\) preserves distance, it follows that \(d_H(p,q) = d_H(0,T(q))\text{,}\) and \begin{equation*} d_H(p,q) = \ln\bigg(\frac{1+|T(q)|}{1-|T(q)|}\bigg). \end{equation*}

Making the substitution \(\displaystyle T(q) = \frac{q-p}{1-\overline{p}q}\) provides us with the following working formula for the hyperbolic distance between two points.

Example5.3.4The distance between two points

For instance, suppose \(p = \frac{1}{2}i\text{,}\) \(q = \frac{1}{2} + \frac{1}{2}i\text{,}\) \(z = .95e^{i5\pi/6}\) and \(w=-.95\text{.}\) Then \(d_H(p,q) \approx 1.49\) units, while \(d_H(z,w) \approx 4.64 \) units.

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Figure5.3.5Some distances in \((\mathbb{D},{\cal H})\text{.}\)

SubsectionThe arc-length differential

Now that we can compute the distance between two points in the hyperbolic plane, we turn our attention to measuring the length of any path that takes us from \(p\) to \(q\text{.}\)

Definition5.3.6

A smooth curve is a differentiable map from an interval of real numbers to the plane \begin{equation*} \vec{r}: [a,b] \to \mathbb{C} \end{equation*} such that \(\vec{r}\hskip.04in^\prime(t)\) exists for all \(t\) and never equals \(\vec{0}\text{.}\)

In the spirit of this text, we write \(\vec{r}(t) = x(t) + i y(t)\text{,}\) in which case \(\vec{r}\hskip.04in^\prime(t) = x^\prime(t) + iy^\prime(t)\text{.}\)

Recall that in calculus we first approximate the Euclidean length of a given smooth curve \(\vec{r}(t) = x(t) + iy(t)\) by summing the contributions of small straight line segments having Euclidean length \begin{align*} \Delta s \amp = |\vec{r}(t+\Delta t) - \vec{r}(t)|\\ \amp = \sqrt{[x(t + \Delta t) - x(t)]^2 + [y(t + \Delta t) - y(t)]^2}\\ \amp = \sqrt{\bigg[ \frac{x(t + \Delta t) - x(t)}{\Delta t}\bigg]^2 + \bigg[ \frac{y(t + \Delta t) - y(t)}{\Delta t}\bigg]^2} |\Delta t|. \end{align*}

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As \(\Delta t \to 0\) we obtain the Euclidean arc-length differential \(ds = \sqrt{(dx/dt)^2 + (dy/dt)^2}~dt\text{,}\) which may be expressed as \begin{equation*} ds = |\vec{r}\hskip.04in^\prime(t)|~dt. \end{equation*}

For instance, consider \(\vec{r}: [0, 2\pi] \to \mathbb{C}\) by \(\vec{r}(t) = R \cos(t) + i R \sin(t)\text{.}\) This map traces a circle of radius \(R\) centered at the origin. To find the length of this curve, which we denote as \({\cal L}(\vec{r})\text{,}\) compute the integral \begin{align*} {\cal L}(\vec{r}) \amp = \int_0^{2\pi} |\vec{r}\hskip.04in^\prime(t)|~dt\\ \amp = \int_0^{2\pi} |-R\sin(t) + i R \cos(t)|~dt\\ \amp = \int_0^{2\pi}\sqrt{R^2 \sin^2(t) + R^2 \cos^2(t)}~dt\\ \amp = \int_0^{2\pi} R ~dt\\ \amp = 2\pi R. \end{align*}

In the hyperbolic plane, we may deduce the arc-length differential by a similar argument. Suppose \(\vec{r}\) is a smooth curve in \(\mathbb{D}\) given by \(\vec{r}(t) = x(t) + iy(t)\text{,}\) for \(a \leq t \leq b\text{.}\) One may approximate the length of a tiny portion of the curve, say from \(\vec{r}(t)\) to \(\vec{r}(t + \Delta t)\text{,}\) by the hyperbolic distance between these two points, \(d_H(\vec{r}(t), \vec{r}(t + \Delta t))\text{.}\) To compute this distance, we first send the point \(\vec{r}(t)\) to 0 by the transformation \begin{equation*} T(z) = \frac{z-\vec{r}(t)}{1-\overline{\vec{r}(t)}z}, \end{equation*} so that \begin{equation*} d_H(\vec{r}(t),\vec{r}(t+\Delta t)) = \ln[1 + |T(\vec{r}(t+\Delta t))|]-\ln[1 - |T(\vec{r}(t+\Delta t))|]. \end{equation*}

To arrive at an arc-length differential, we want to let \(\Delta t\) approach 0. As this happens, \(T(\vec{r}(t+\Delta t))\) approaches \(T(\vec{r}(t))\text{,}\) which is 0. From calculus we also know that \(\ln(1+x) \approx x\) for \(x\) very close to 0. Thus, for small \(\Delta t\text{,}\) we have \begin{align*} d_H(\vec{r}(t),\vec{r}(t+\Delta t)) \amp \approx |T(\vec{r}(t+\Delta t))| + |T(\vec{r}(t+\Delta t))|\\ \amp =2\cdot \bigg| \frac{\vec{r}(t+\Delta t) - \vec{r}(t)}{1 - \overline{\vec{r}(t)}\vec{r}(t+\Delta t)}\bigg |\\ \amp =2\cdot \frac{\big | \frac{\vec{r}(t+\Delta t) - \vec{r}(t)}{\Delta t}\big |}{1 - \overline{\vec{r}(t)}\vec{r}(t+\Delta t)}\cdot | \Delta t |. \end{align*}

Now, as \(\Delta t \to 0\text{,}\) the numerator in the above quotient goes to \(|\vec{r}\hskip.04in^\prime(t)|\) and the denominator goes to \(1 - |\vec{r}(t)|^2\text{,}\) and we arrive at the following hyperbolic arc-length differential.

Definition5.3.7

If \(\vec{r}: [a,b] \to \mathbb{D}\) is a smooth curve in the hyperbolic plane, define the length of \(\vec{r}\) to be \begin{equation*} {\cal L}(\vec{r}) = \int_a^b \frac{2}{1 - |\vec{r}(t)|^2}~|\vec{r}\hskip.04in^\prime(t)|dt. \end{equation*}

One can immediately check that the hyperbolic distance between two points in \(\mathbb{D}\) corresponds to the length of the hyperbolic line segment connecting them.

The proof of this theorem is left as an exercise. One can prove that hyperbolic reflections preserve arc-length as well. This should come as no surprise, given the construction of the distance formula at the start of this section. Still, one can prove this fact from our definition of arc-length (Exercise 5.3.6). Thus, all hyperbolic reflections and all transformations in \(\cal H\) are hyperbolic isometries: they preserve the hyperbolic distance between points in \(\mathbb{D}\text{.}\)

Another consequence of the invariance of distance, when applied to hyperbolic rotations, is the following:

We are now in a position to argue that in the hyperbolic plane, the shortest path (geodesic) connecting two points \(p\) and \(q\) is along the hyperbolic line through them.

Proof Sketch: We first argue that the geodesic from 0 to a point \(c\) on the positive real axis is the real axis itself.

Suppose \(\vec{r}(t) = x(t) + iy(t)\) for \(a \leq t \leq b\text{,}\) is an arbitrary smooth curve from 0 to \(c\) (so \(\vec{r}(a) = 0\) and \(\vec{r}(b) = c\)).

Suppose further that \(x(t)\) is nondecreasing (if our path backtracks in the \(x\) direction, we claim the path cannot possibly be a geodesic). Then \begin{equation*} {\cal L}(\vec{r}) = \int_a^b\frac{2}{1-[(x(t))^2 + (y(t))^2]}\sqrt{(x^\prime(t))^2 + (y^\prime(t))^2}~dt. \end{equation*}

The hyperbolic line segment from 0 to \(c\) can be parameterized by \(\vec{r}_0(t) = x(t) + 0i\) for \(a \leq t \leq b\text{,}\) which has length \begin{equation*} {\cal L}(\vec{r}_0) = \int_a^b\frac{2}{1-[x(t)]^2}\sqrt{(x^\prime(t))^2}~dt. \end{equation*}

The curve \(\vec{r}_0\) is essentially the shadow of \(\vec{r}\) on the real axis.

One can compare the integrands directly to see that \({\cal L}(\vec{r}) \geq {\cal L}(\vec{r}_0).\)

Since transformations in \({\cal H}\) preserve arc-length and hyperbolic lines, it follows that the shortest path between any two points in \(\mathbb{D}\) is along the hyperbolic line through them.

Example5.3.12Two paths from \(\boldsymbol{p}\) to \(\boldsymbol{q}\)

Figure 5.3.13 shows two paths from \(p = .5i\) to \(q = .5+.5i\text{:}\) the (solid) hyperbolic segment from \(p\) to \(q\text{,}\) and the (dashed) path \(\vec{r}\) that looks like a Euclidean segment. Which path is shorter?

We may compute the length of the hyperbolic segment connecting \(p\) and \(q\) with the distance formula from Theorem 5.3.3. This distance is approximately 1.49 units.

By contrast, consider the path in \(\mathbb{D}\) corresponding to the Euclidean line segment from \(p\) to \(q\text{.}\) This path may be described by \(\vec{r}(t) = t + \frac{1}{2}i\) for \(0 \leq t \leq \frac{1}{2}\text{.}\) Then \(\vec{r}\hskip.04in^\prime(t) = 1\) and \begin{align*} {\cal L}(\vec{r}) \amp = \int_0^{\frac{1}{2}}\frac{2}{1-(t^2+\frac{1}{4})}dt\\ \amp \approx 1.52. \end{align*}

It is no surprise that the hyperbolic segment connecting \(p\) to \(q\) is a shorter path in \((\mathbb{D},{\cal H})\) than the Euclidean line segment connecting them.

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Figure5.3.13Two paths from \(p = .5i\) to \(q = .5+.5i\text{.}\) Which is shorter?
Example5.3.14Constructing a perpendicular bisector

For any two points \(p\) and \(q\) in \(\mathbb{D}\text{,}\) we may construct the perpendicular bisector to hyperbolic segment \(pq\) by following the construction in Euclidean geometry. Construct both the hyperbolic circle centered at \(p\) that goes through \(q\) and the hyperbolic circle centered at \(q\) that goes through \(p\text{.}\) The hyperbolic line through the two points of intersection of these circles is the perpendicular bisector to segment \(pq\text{,}\) labeled \(L\) in Figure 5.3.15.

Hyperbolic reflection about \(L\) maps \(p\) to \(q\) and \(q\) to \(p\text{.}\) Since hyperbolic reflections preserve hyperbolic distances, each point on \(L\) is hyperbolic equidistant from \(p\) and \(q\text{.}\) That is, for each \(z\) on \(L\text{,}\) \(d_H(z,p)=d_H(z,q)\text{.}\)

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Figure5.3.15Constructing the perpendicular bisector of segment \(pq\text{.}\)

In Euclidean geometry one uses perpendicular bisectors to construct the circle through three noncollinear points. This construction can break down in hyperbolic geometry. Consider the three points \(p, q,\) and \(r\) in Figure 5.3.16. The corresponding perpendicular bisectors do not intersect. There is no point in \(\mathbb{D}\) hyperbolic equidistant from all three of these points. In particular, in hyperbolic geometry, there need not be a hyperbolic circle through three noncollinear points.

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Figure5.3.16Three noncollinear points need not determine a circle in hyperbolic geometry.

SubsectionExercises

1

Suppose \(0\lt x \lt 1\) and \(L\) is a hyperbolic line about which \(x\) gets inverted to the origin. (Such an inversion was constructed in Theorem 5.1.3.) For a real number \(h\text{,}\) let \(w\) be the image of \(x+h\) under this inversion. Prove that \(w = \frac{-h}{1-x^2-hx}\text{.}\)

2

Determine a point in \(\mathbb{D}\) whose hyperbolic distance from the origin is 2,003,007.4 units.

3

Suppose \(L\) is any hyperbolic line, and \(C\) is any cline through the ideal points of \(L\text{.}\) For any point \(z\) on \(L\text{,}\) its perpendicular distance to \(C\) is the length of the hyperbolic segment from \(z\) to \(C\) that meets \(C\) at right angles. Prove that the perpendicular distance from \(C\) to \(L\) is the same at every point of \(L\text{.}\) Hint: Use the fact that distance is an invariant of hyperbolic geometry.

4

Determine the hyperbolic distance from the point \(p = 0.5\) to the point \(q = 0.25 + 0.5i\text{.}\)

6

Hyperbolic reflections preserve distance in \((\mathbb{D},{\cal H})\)

a. Use the definition of arc-length to prove that hyperbolic reflection about the real axis preserves arc-length.

b. Use part (a) and Theorem 5.3.8 to argue that hyperbolic reflection about any hyperbolic line preserves arc-length in \((\mathbb{D}, {\cal H})\text{.}\)

7

Suppose \(z_0\) is in the hyperbolic plane and \(r > 0\text{.}\) Prove that the set \(C\) consisting of all points \(z\) in \(\mathbb{D}\) such that \(d_H(z,z_0) = r\) is a Euclidean circle.