5. Hyperbolic Geometry

Hyperbolic geometry is the geometry you get by assuming all the postulates of Euclid, except the fifth one, which is replaced by its negation.

In hyperbolic geometry there exist a line $ l$ and a point $ P$ not on $ l$ such that at least two distinct lines parallel to $ l$ pass through $ P$.

It is more difficult to imagine, but have in mind the following image (and imagine that the lines never meet $ l$):

Figure 1: Two intersecting lines parallel to $ l$
Image Section51

The first property that we get from this axiom is the following lemma (we omit the proof, which is a bit technical):

Lemma 1   Rectangles don't exist in hyperbolic geometry.

Using this lemma, we can prove the following Universal Hyperbolic Theorem:

Theorem 1   In hyperbolic geometry, for every line $ l$ and every point $ P$ not on $ l$ there pass through $ P$ at least two distinct parallels through $ P$. Moreover there are infinitely many parallels to $ l$ through $ P$.


Drop the perpendicular $ PQ$ to $ l$ and erect a line $ m$ through $ P$ perpendicular to $ PQ$, like in the figure below.

Figure 2: Proof of the universal hyperbolic theorem
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Let $ R$ be another point on $ l$, erect perpendicular $ t$ to $ l$ through $ R$ and drop perpendicular $ PS$ to $ t$. Now $ PS$ is parallel to $ l$, since both are perpendicular to $ t$.

Assume that $ m$ and $ PS$ are the same line (so $ S\in m$). This would mean that $ PQRS$ is a rectangle, which contradicts the lemma above. Hence there are two distinct parallels to $ l$ through $ P$. By varying $ R$, we get infinitely many parallels. $ \square$

Other important consequences of the Lemma above are the following theorems:

Theorem 2   In hyperbolic geometry, all triangles have angle sum $ <180^{\circ}$. (From this it follows immediately that all convex quadrilaterals have angle sum less than $ 360^{\circ}$).

Theorem 3   In hyperbolic geometry if two triangles are similar, they are congruent.

Note: This is totally different than in the Euclidean case. It tells us that it is impossible to magnify or shrink a triangle without distortion.


Assume the contrary: there are triangles $ \triangle ABC$ and $ \triangle A'B'C'$ that are similar (they have the same angles), but are not congruent. The no corresponding sides are congruent (otherwise, they would be congruent, using the principle $ angle-side-angle$). We may assume, without loss of generality, that $ AB>A'B'$ and $ AC>A'C'$.

Figure 3: Similar triangle are congruent.
Image Section53

Then, by definition of $ >$ there exists a point $ B''$ on $ AB$ and a point $ C''$ on $ AC$ such that $ AB''=A'B'$ and $ AC''=A'C'$. Then, since the angles are the same, by $ angle-side-angle$, $ \triangle A'B'C'\cong \triangle AB''C''$. Hence $ \angle AB''C''=\angle B'$ and $ \angle AC''B''=\angle C'$.

But we also have that $ \angle C=\angle C'$ and $ \angle B=\angle B'$, so $ \angle AB''C''=\angle B$ and $ \angle AC''B''=\angle C$. This implies that the lines $ BC$ and $ B'C'$ are parallel, hence the quadrilateral $ BB''CC''$ is convex, and the sum of its angles is exactly $ 360^{\circ}$, which contradicts the theorem above.

Hence similar triangles are congruent. $ \square$

An amazing consequence of this theorem is that in hyperbolic geometry a segment can be determined with the aid of an angle. This is stated more dramatically by saying that hyperbolic geometry has an absolute unit of length!

Defect and area of a triangle

There is also the concept of the defect of a triangle, which is defined to be the quantity that is missing for the sum of the angles to add up to 180 degrees. And in fact the area of a triangle in hyperbolic geometry is proportional to its defect D. So we have that A=kD, where k is a positive number.
Two triangles have the same area if and only if they have the same angle-sum.

Now that you have experienced a flavour of proofs in hyperbolic geometry, Try some exercises!


The following are exercises in hyperbolic geometry. You are to assume the hyperbolic axiom and the theorems above.

  1. Prove that if A'B'BA is a Saccheri quadrilateral (the angles A' and B' are right angles and AA'=BB'), then the summit AB is greater than the base A'B'. (Hint: Join the midpoints M and M')
  2. Suppose that lines l and l' have a common perpendicular MM'. Let A and B be points on l such that M is not the midpoint of AB. Show that A and B are not equidistant froml'.
  3. Suppose that the parallel lines l and l' have a common perpendicular MM'. Prove that MM' is the shortest segment between any point of l and any point of l'.
  4. Suppose that the parallel lines l and l' have a common perpendicular MM'. Let A and B be points on l such that A is between M and B. Drop perpendiculars AA' and BB' to l'. Prove that AA'<BB'.
  5. Is every Saccheri quadrilateral a convex quadrilateral? Why or why not?

Back to Escher

M. C. Escher created four patterns using hyperbolic geometry: Circle Limit I, Circle Limit III, Circle Limit III and Circle Limit IV. What Escher used for his drawings is the Poincaré model for hyperbolic geometry.
"Circle Limit I"
"Circle Limit II"
"Circle Limit III"
"Circle Limit IV"

What does it mean a model? By a model, we mean a choice of an underlying space, together with a choice of how to represent basic geometric objects, such as points and lines, in this underlying space. The need to have models for the hyperbolic plane (or better said, the hyperbolic geometry of the plane) is that it is very difficult to work with an Euclidean representation, but do non-Euclidean geometry. As you saw above, it is difficult to picture the notions that we work with, even if the proofs follow logically from our assumptions.

There are two more popular models for the hyperbolic plane: the upper half-plane model and the Poincaré plane model. We will analyse both of them in the following sections.