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The Hyperbolic Plane

The plane of constant negative curvature, drawn in the Poincaré disc. Build a triangle and compare its angles and area with a flat one, set a point beside a line and see which lines through it never meet the line, or measure a circle against 2πr.

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Lesson

The theory — The Hyperbolic Plane

The hyperbolic plane is the complete, simply connected surface whose Gaussian curvature is the same negative number everywhere; here K = −1, so lengths are in units of the curvature radius and areas in its square. A triangle's area is π less its angle sum, and through a point not on a line pass infinitely many lines that never meet it. The Poincaré disc draws the whole infinite plane inside a circle: straight lines become arcs that meet the rim at right angles, angles are drawn true, and lengths shrink towards the rim, which is infinitely far away.

What each symbol means

K
the Gaussian curvature, −1 at every point of this plane
Π(d)
the angle of parallelism: lines through a point at distance d from a line meet it only within Π(d) of the perpendicular

No triangle here holds more than π, however long its sides

Press Equilateral, side 1. The sides are straight lines of this plane, drawn as arcs because the disc is a map, and the angles add to 157.93°. The shortfall from 180° is the area: 180° − 157.93° = 22.07°, which in radians is 0.385. Now press Equilateral, side 10. Each angle has closed to 0.772°, the sum is 2.32°, and the area is 3.101, 98.71% of π, where a flat triangle with sides of 10 holds 43.30. Since the area is π less the angle sum, and angles cannot fall below zero, no triangle in this plane holds more than π, whatever its sides. The table shows the approach: side 5 already reaches 84.41% of π, side 20 reaches 99.99%. Only a triangle whose corners have gone off to infinity, with three angles of 0°, holds π exactly.

Through a point off a line, infinitely many lines never meet it

Press Parallels at distance 1. The black line is a straight line of this plane, and the point is 1 from it. Watch the line turning through the point. It meets the black line only while it is within 40.395° of the perpendicular, on one side or the other. Lobachevsky called that angle the angle of parallelism, Π(d), and in 1840 gave it as tan(Π/2) = e−d. The two lines at exactly Π(d), in pink, close in on the black line towards its ends and never reach it, and every line between them in the remaining 99.21° misses it too. Euclid's geometry has exactly one parallel through the point; this one has infinitely many. Press Parallels at distance 3: Π(d) is down to 5.700°, and 168.60° of directions miss. Bring the point close instead and the fan narrows towards nothing, 55.05° at d = 0.5: near the line, the plane behaves almost as paper does.

A circle's rim grows exponentially, and most of a disc lies near its edge

Press Circle of radius 10. The circumference is 2π sinh 10 = 69,198.18, against 62.83 on paper, 1,101 times as long. The area inside, 2π(cosh 10 − 1) = 69,191.90, is almost the same number, and 63.22% of it lies within 1 of the rim. Far out, every step of 1 multiplies the rim by nearly e. Henderson and Taimina crochet a model of this plane with that rule: in every new row they turn N stitches into N + 1, so each row is the same factor longer than the last, and the fabric, holding more edge than a flat sheet has room for, has to fold and wave. The saddle in Gauss's Theorema Egregium ran long in the same way. Type r = 0.5 here and the ring gives a curvature of −1.0126 (the estimate is exact only as the ring shrinks), where the saddle's ring gave −0.884 at the same radius: this plane is curved −1 everywhere, the saddle only at its centre.

Problem solved in full

  1. A triangle known only by its angles 4 steps

    A triangle in this plane has angles of 30°, 45° and 60°. What is its area, and how long are its sides? On paper the question has no answer: every enlargement of a triangle keeps its angles. Press Equilateral, side 1 first, where the page gives an angle sum and an area.

    Open this problem in the tool

    1. On the preset the area is π less the angle sum, in radians:

      A=π−(A+B+C)=(180∘−157.93∘)×π180∘=0.385\mathcal A=\pi-(A+B+C)=(180^\circ-157{.}93^\circ)\times\dfrac{\pi}{180^\circ}=0{.}385

    2. The same rule gives this triangle's area at once:

      A=(180∘−30∘−45∘−60∘)×π180∘=π4=0.7854\mathcal A=(180^\circ-30^\circ-45^\circ-60^\circ)\times\dfrac{\pi}{180^\circ}=\dfrac{\pi}{4}=0{.}7854

    3. For the sides, this plane has a second law of cosines, which takes angles in and gives a side out:

      cosh⁡a=cos⁡30∘+cos⁡45∘cos⁡60∘sin⁡45∘sin⁡60∘=0.866025+0.3535530.612372=1.991564 ⇒ a=1.3121\cosh a=\dfrac{\cos 30^\circ+\cos 45^\circ\cos 60^\circ}{\sin 45^\circ\sin 60^\circ}=\dfrac{0{.}866025+0{.}353553}{0{.}612372}=1{.}991564\ \Rightarrow\ a=1{.}3121

    4. and the other two the same way, starting from B and from C:

      cosh⁡b=cos⁡B+cos⁡Ccos⁡Asin⁡Csin⁡A ⇒ b=1.6231,cosh⁡c=cos⁡C+cos⁡Acos⁡Bsin⁡Asin⁡B ⇒ c=1.8131\cosh b=\dfrac{\cos B+\cos C\cos A}{\sin C\sin A}\ \Rightarrow\ b=1{.}6231,\qquad \cosh c=\dfrac{\cos C+\cos A\cos B}{\sin A\sin B}\ \Rightarrow\ c=1{.}8131

    Answer

    Area π/4 = 0.7854, sides 1.3121, 1.6231 and 1.8131. No other triangle has these angles: enlarge it and its area grows, so its angle sum has to fall. Similar triangles of different sizes do not exist in this plane, and a triangle's angles carry a length scale with them, the curvature radius.

    A=π4=0.7854,a, b, c=1.3121, 1.6231, 1.8131\mathcal A=\dfrac{\pi}{4}=0{.}7854,\qquad a,\ b,\ c=1{.}3121,\ 1{.}6231,\ 1{.}8131

Learning path

Geometry on a curved surface

References (2)

· Reviewed against published sources. Corrections welcome.