Obtaining Non-Euclidean Geometries

David Mwakima

Fall 2016

1 Introduction

Hyperbolic geometry is the geometry obtained from the metric (first fundamental form) of the pseudosphere. It is a non-Euclidean geometry, which means that the models of hyperbolic geometry satisfy some but not all of the axioms of Euclidean geometry. In particular, in this paper I shall show that the so-called “parallel axiom/postulate” of Euclidean geometry does not hold in the model of hyperbolic geometry I shall be considering, which is the upper plane model. Moreover, I shall show that in hyperbolic geometry the sum of internal angles of a hyperbolic triangle is strictly less than π contrary to the well-known result of Euclidean geometry, viz. the sum of internal angles of a Euclidean triangle is equal to π. An interesting fact also emerges from hyperbolic geometry, which is that the angles of a hyperbolic polygon determine its area, which is a result we do not have in Euclidean geometry. (Cf. Walkden (2016) p. 42 for an interesting discussion of these facts.) In section 2 I derive the metric of the upper plane model of hyperbolic geometry and show, using this metric, covariant derivatives and Christoffel symbols; that the Gaussian Curvature K of the pseudosphere is in fact -1. In section 3 I identify the geodesics of the upper-plane model and show that the parallel axiom does not hold in hyperbolic geometry. Finally in section 4 I show how to compute distances and area in hyperbolic geometry and show that the sum of interior angles of a hyperbolic triangle is less than π and the more general result relating the area of a hyperbolic polygon with the sum of its interior angles.

1 2 The upper plane model of hyperbolic geometry

Following the discussion on surfaces of constant Gaussian curvature in Pressley (2010) section 8.3. pp. 196ff, we can obtain the pseudosphere given by the coordinate patch √ σ(w, v) = (ewcos(v), ewsin(v), 1 − e2w − cosh−1(e−w))

We obtain this coordinate patch by considering the case where K < 0 of the surface of − f¨ revolution σ(u, v) = (f(u)cos(v), f(u)sin(v), g(u)), which has K = f and choosing K = −1

In his discussion on p. 270, Pressley (2010) does not say why we must have w > 1 for coordinate patch he considers to be smooth and well-defined. But by considering the following derivation we can see why this is the case.

Given σ(w, v) and setting u = e−w gives the reparametrized surface

( √ ) 1 1 1 σ˜(u, v) = cosv, sinv, 1 − − cosh−1u u u u2

Writing u˜ = v....(1) and v˜ = u = e−w....(2) we can get the reparametrization map Φ given by:

{ w = −ln(˜v) by solving for w in (2) Φ = v =u ˜ iff J(Φ) is invertible, where J(Φ) is the Jacobian matrix of the map Φ

( ) ( ) ∂w ∂w − 1 ∂u˜ ∂v˜ 0 v˜ J(Φ) = ∂v ∂v = ∂u˜ ∂v˜ 1 0 detJ(Φ) ≠ 0 provided u =v ˜ > 1. So, it is because of this that Pressley (2010) p. 270 says w > 1. For otherwise the Jacobian tells us that the reparametrization map is not invertible.

2 We know the first fundamental form g of surface of revolution is given by: g = du2 + f(u)2dv2 (Cf. Pressley (2010) p. 123)

In our case with σ(w, v) it is:

2 2w 2 gw,v = dw + e dv T and by taking J (gw,v)J we get du2 + dv2 g = u,v u2 for the reparametrization σ˜(u, v)

The upper half-plane model of hyperbolic geometry is the surface H = {(v, u) ∈ R2|u > 0} equipped with the metric gu,v.

Following Pressley (2010) p. 270, we can identify the upper half of the vu-plane described above with the upper half of the complex plane. So that we have: H = {z ∈ C|Im(z) > 0}

Given our model of hyperbolic geometry, we can then compute distance, areas and angles on our surface. But first we need to know what the length minimizing curves i.e. geodesics of hyperbolic geometry are. The length minimizing curves are the analogues of the familiar straight lines from Euclidean geometry. Using the geodesics of our model of hyperbolic geometry, I will show that the “parallel axiom” does not hold in hyperbolic geometry.

Before doing this, however, I show how using covariant derivatives and Christoffel symbols we can confirm that the Gaussian Curvature K is indeed −1

The covariant derivatives with respect to v and u of some vector a = (α, β) are given by: ∇ v v u u va = (∂vα + Γvvα + Γvuβ, ∂vβ + Γvvα + Γvuβ) ∇ v v u u ua = (∂uα + Γuvα + Γuuβ, ∂uβ + Γuvα + Γuuβ)

3 We know that the Gaussian Curvature K measures the failure of ∇v and ∇u to commute. So we obtain the expression.

2 1 ⊥ (∇v∇u − ∇u∇v)a = K(EG − F ) 2 a (1) · · · ⊥ − where E =σ ˜u σ˜u, G =σ ˜v σ˜v and F =σ ˜u σ˜v are the entries of g(u,v) and a = (β, α) if a = (α, β)

du2+dv2 Given our metric g(u,v) = u2 , we can compute the Christoffel symbols Γ. After doing so we find that the only non-zero Christoffel symbols are:

1 1 Γu = Γv = Γv = − and Γu = uu uv vu u vv u

Picking a = (1, 0) we compute the left hand side of (1) as follows: ( ) ( ) 1 1 1 ∇ v u − va = (∂v(1) + Γvu(0), ∂v(0) + Γvv(1)) = 0 (0), 0 + (1) = 0, ( u u ) ( u ) 1 1 1 ∇ v u − − − ua = (∂u(1) + Γuv(1), ∂u(0) + Γuu(0) = 0 (1), 0 (0) = , 0 ( ) ( ( ))u ( u ) u ∇ ∇ ∇ − 1 − 1 1 − 1 − 1 v( ua) = v , 0 = 0 (0), 0 + = 0, 2 ( u) ( u ( )u (u )) ( u ) ∇ ∇ ∇ 1 − 1 1 − 1 1 − 2 u( va) = u 0, = 0 (0), ∂u = 0, 2 ( u) ( u ) ( u ) u u u 1 2 1 ∇ (∇ a) − ∇ (∇ a) = 0, − − 0, − = 0, v u u v u2 u2 u2 √ 1 1 1 ⊥ − − 2 2 − The right hand side of (1) is: K(0, u2 ) since (EG F ) = u4 and a = (0, 1)

Therefore, comparing the left and right hand side we find that K = −1, which is what we were to show.

3 Geodesics of hyperbolic geometry and the “Par- allel Axiom”

Let γ(t) = (x(t), (y(t)) be a unit speed curve in H

4 By proposition 9.2.3 we say γ(t) is a geodesic iff it satisfies the geodesic equations:

u 2 u u 2 u¨ + Γuuu˙ + 2Γuvu˙v˙ + Γvvv˙ = 0 (2) v 2 v v 2 v¨ + Γuuu˙ + 2Γuvu˙v˙ + Γvvv˙ = 0 (3)

du2+dv2 u v Given our metric g(u,v) = u2 , we see that the only non-zero Γ’s are: Γuu = Γuv = − 1 u 1 H u and Γvv = u . So, we demand that γ(t) satisfy the geodesic equations of given by: 1 1 =u ¨ − u˙ 2 + v˙ 2 = 0 (4) u u 2 =v ¨ − u˙v˙ = 0 (5) u

But we also know by Theorem 9.2.1. and equation (9.2) in Pressley (2010) p. 220 and proposition 9.2.3 on p. 223 that (5) implies that if γ(t) is a geodesic

d (F u˙ + Gv˙) = 0 dt

d v˙ v˙ 2 So, dt ( u2 ) = 0 since F = 0 and u2 = c where c is a constant and v˙ = cu ...(τ). There are two cases:

Case 1: c = 0

Then v˙ = 0 and v = k =⇒ the geodesics are vertical straight lines in the vu - plane.

Case 21: c ≠ 0

Here we use the fact that γ(t) is unit speed and so we write:

u˙ 2 +v ˙ 2 = 1 (6) u2 u˙ 2 = u2 − v˙ 2 (7) √ u˙ = u (1 − c2u2) plugging in v˙ = cu2....(τ) in (7) and simplying. (8)

1I thank Cliff Taubes for making this clear tome

5 We can divide (τ) and (8) to get: dv cu2 = √ du u (1 − c2u2) cu = √ (1 − c2u2) 1 d √ = ∓ (1 − c2u2) c du Rearranging gives: ( ) d 1√ v (1 − c2u2) = 0 du c which implies 1√ v (1 − c2u2) = v where v is some constant c 0 0 1√ v − v = ∓ (1 − c2u2) 0 c 1 (v − v )2 = (1 − c2u2) 0 c2 1 (v − v )2 + u2 = (9) 0 c2

But (9) is the equation of a semi-circle with center (v0, 0) in the vu - plane.

Having exhausted all the cases, it follows that there are two length minimizing curves in H. The geodesics v = k for some constant k i.e. vertical half-lines orthogonal to the real axis (u = 0) and the semi-circles with centers on the real axis.

These results explain Proposition 11.1.2 in Pressley (2010) p. 271, which we then use to state and explain Proposition 11.1.3 in (Ibid.).

Proposition 11.1.3

(i) There is a unique geodesic passing through any two distinct points of H

(ii) Suppose γ is a geodesic in H and that p is a point in the upper-half plane that is not in γ. Then there are infinitely many geodesics though p that do not interesect γ

Proof

6 (i) Let α, β ∈ H and α ≠ β. If the line passing though α and β is parallel to the imaginary axis, then the unique geodesic passing through the points α, β is the half-vertical line containing them. If the line through α, β is not parallel to the imaginary axis, its perpendicular bisector intersects the real axis at some point ζ and the unique hyperbolic line passing through α, β is the semi-circle with center ζ and radius |α − ζ| = |β − ζ|

(ii) Let p be in H and let γ = v = β be the vertical line orthogonal to the real axis at β. And suppose that Re(p) > β. If the line through p is not parallel to γ, then by (i) we can get the semi-circle through p with center ζ on the real axis and radius |p − ζ| that does not intersect γ provided |p − ζ| ≤ |ζ − β| Also, by (i) above, we can get the vertical line parallel to γ, which then clearly does not intersect γ, when we look at the limiting case when ζ → ∞

Significance

In his Elements, Euclid had what he called the “Fifth Postulate”:

If a straight line falling on two straight lines makes the interior angles on the same side less than two right-angles, then the two straight lines, if extended indefinitely, meet on the side on which the angles are less than two right-angles.

Figure and quote adopted from Walkden (2016) p. 10

7 But in Definition 232, Euclid defines ‘parallel lines’ as follows:

Parallel lines are straight-lines which, being in the same plane, and being produced to infinity in each direction, meet with one another in neither (of these directions)

So, the “Fifth Postulate” has been understood to be equivalent to the definition of parallel lines and some mathematicians following David Hilbert3 refer to the “Parallel Axiom” which states:

If p is a point that is not on a straight line l, there is a unique straight line passing through p that does not intersect l i.e. which is ‘parallel’ to l.

In Euclidean geometry, it is known that a straight line is the length minimizing curve that joins two points. Since geodesics are locally length minimizing curves, this is equivalent to saying that (in Euclidean and Non-Euclidean geometry) the length minimizing curve between any two points is the geodesic that joins them.

But if this is the case, we have shown in Proposition 11.1.2 that there are infinitely many geodesics through a point p, which do not intersect the geodesic γ. So, the “Parallel Axiom” does not hold in models of hyperbolic geometry, which then implies that the “Fifth Postulate” does not hold either.

Figure adopted from Hitchin (2013) p. 90

2See http://farside.ph.utexas.edu/Books/Euclid/Elements.pdf 3See Hitchin p. 88 - 90

8 4 Angles, distance and area

(a) Angles

Proposition

The angles measured in models of Euclidean Geometry are the same as the angles measured in models of Hyperbolic Geometry.

Proof

By corollary 6.3.4. p. 136 in Pressley (2010), the local diffeomorphism f : R2 → H H 1 is conformal since the metric of is proportional by a factor of u2 to the metric of Euclidean R2.

(b) Distance

Following Pressley (2010) p. 272ff. we prove that the hyperbolic distance between two points z1, z2 ∈ H is given by:

| − | −1 z2 z1 dH = 2tanh (10) |z2 − z¯1| where z¯1 denotes the conjugate complex of z1.

Proof

Pressley (2010) proves it for the case in which z1 and z2 lie on the hyperbolic line which is a semi-circle and leaves the simpler case where z1 and z2 lie on the half-line as an exercise. (Cf. Exercise 11.1.2 in Pressley (2010) p.227) I shall give the actual proof in the case where they lie on the half-line (following his worked on solution for this exercise) and discuss the proof of the second case suggesting what one needs to make it more perspicuous.

Case 1: z1 and z2 lie on the vertical half-lines

Suppose that z1 and z2 lie on a half a vertical half-line, say z1 = r +is and z2 = r +it ∈ R du2+dv2 where r, s, t and t > s. Then given our metric g = u2 and since v is constant

9 (only looking at the vertical lines), the dv term vanishes and we have: ∫ t 1 du2 dH(z , z ) = du by equation (6.1) on page 123 Pressley taking the square root of 1 2 u u2 s( ) t = ln = d where d is the distance between z and z s 1 2

⇒ t d = s = e But from the distance formula (10) above we can write ( ) ( ) ( ) t − s ed − 1 d 2tanh−1 = 2tanh−1 = 2tanh−1 tanh = d t + s ed + 1 2

The first fraction was obtained by considering the u coordinates of z1 and z2, which are s and t respectively, and treating them as the points that go into (10). The second fraction was obtained by multiplying each term in the numerator and denominator 1 t d of the first fraction by s and using s = e . The third fraction was obtained using e2d−1 the identity: tanhd = e2d+1

It follows that ( − ) −1 z2 z1 dH(z1, z2) = 2tanh = d z2 + z1

Case 2: z1 and z2 lie on the semi-circle

The proof for the first case in the book is not immediately clear unless one isfamiliar with the following useful identities, formulas and heuristics4. ∫ 1 − θ θ (i) sinθ dθ = ln(cos( 2 )) + ln(sin( 2 )) + C − a (ii) lna lnb = ln b

e2d−1 (iii) tanhd = e2d+1

(iv) eln(x) = x

(iv) Setting A = tan(ψ/2) and B = tan(φ/2) allows one to see the proof on p. 273 perspicuously. From bottom of p. 272 and using (iii) and (iv), we can write e(ln(A/B))

4I thank Cliff Taubes for fruitful discussion on some of these points

10 = A/B so that the numerator in the second fraction on p. 273 is A/B − 1 and the denominator is A/B + 1. Now multiplying both the numerator and denominator by B gives (A − B)/(A + B), which when we substitute back gives us the third fraction.

(v) The trigonometric identities: cos(a)cos(b) + cos(a)cos(b) = cos(a − b) sin(a)cos(b) + sin(b)cos(a) = sin(a + b) sin(a)cos(b) − sin(b)cos(a) = sin(a − b) cos(2a) = 1 − 2sin2(a)

(c.) Area and angles of hyperbolic polygons

In presenting the following proof, I follow Pressley (2010) p. 273 - 274 with some modifications and comments on what steps in the proof need to be spelt outabit more clearly.

We’ll need the following lemma.

Lemma

Let z1 and z2 be the end points of a segment l of a geodesic in H that forms part of the semi-circle with center c on the real axis and suppose that the radius vectors joining c to z1 and∫ c to z2, make angles φ and ψ, respectively, with the positive dv − real axis. Then B u = φ ψ, which is independent of the radius of the semicircle chosen. The formula is correct even if the geodesic is part of a half-line, for in this case since dv = constant. The integral is equal to 0.

Proof

We can parametrize the geodesic using polar coordinates so that v = rcosθ and u = rsinθ, where r is the radius of the geodetic semi-circle. Then the integral

∫ ∫ ∫ z2 dv ψ −rsinθ ψ = dθ = − dθ = φ − ψ z1 u φ rsinθ φ

Proposition

11 Let R be an n-sided hyperbolic polygon in H with internal angles α1, α2, . . . αn. Then the area of the polygon is given by ∑n A(R) = (n − 2)π − αi i=1

Proof

Let v1, . . . vn be the vertices of R and B its boundary, which consists of n geodesics [a1, a2], [a2, a3],... [an, a1]. Let α1 be the internal angle of R at the vertex a1 and so on.

du2+dv2 Given g(u, v) = w2 , by equation 6.12 in Pressley (2010): ∫∫ dvdu A(R) = 2 R u

By Green’s Theorem, we can compute this area by computing the line integral around the boundary. So we have:

∫ ∫∫ ( ) ∂g ∂f A(R) = fdv + gdu = − dvdu B R ∂v ∂u

Where f and g are smooth curves of (u, v). The equation directly above forces us to 1 take f = u and g = 0. So that we have: ∫∫ ∫ dvdu dv A(R) = 2 = R u B u

Using the lemma above, let φi and ψi be the angles defined in the lemma corre- sponding to the side [zi, zi+1] for 1 ≤ i ≤ n where zn+1 is z1. Since there are n such segments we get ∫∫ ∫ dvdu dv ∑n = = (φ − ψ ) u2 u i i R B i=1

Pressley (2010) p. 276 shows us how this sum can be simplified. The outward pointing normal of the boundary B rotates anticlockwise by an angle of ψi − φi,

12 moving∑ along the segment [zi, zi+1]. Since there are n such segments, this is a total n − − of i=1 ψi φi. But at each vertex the normal turns through an angle of π αi. − There are n vertices, so∑ it turns through n(π αi) going around the boundary. So n − − in total we have nπ + i=1(ψi φi αi) as the total turning angle. By an application of theorem (3.1.4) labeled Hopf’s Umlaufstatz in Pressley (2010) p. 57, which says that any turning angle of a simple closed curve changes by 2π going once around the curve. It follows that:

∑n 2π = nπ + (ψi − φi − αi) i=1 Rearranging gives:

∑n ∑n (n − 2)π − αi = (φi − ψi) = A(R) i=1 i=1

If R is a triangle with internal angles α, β and γ, the above equation tells us that π − (α + β + γ) = A(R). Since α + β + γ is the sum of the internal angles of the hyperbolic triangle, this implies that these internal angles do not add up to π, which is contrary to the familiar result from Euclidean geometry where the sum of a triangle with internal angles α, β and γ is: α + β + γ = π. For spherical geometry, we had the following result in Theorem 6.4.7 Pressley (2010) p.145 for the area of a spherical triangle with internal angles, α, β and γ: (α + β + γ) − π

Remarks. In Euclidean geometry, the angles of a triangle do not determine the triangle’s area (since similar triangles have the same angles but not necessarily the same area). This is not true in hyperbolic geometry, Möbius transformations preserve areas. The only way the area of a hyperbolic triangle in the upper-plane model can be equal to π is if all the internal angles are equal to zero and this is when the vertices of the triangle lie on the circle at infinity. (Cf. Walkden (2016) p. 42)

5 Works cited

(i) Hitchin, Nigel (2013) Geometry of Surfaces Retrieved from: http://people. maths.ox.ac.uk/hitchin/hitchinnotes/geometry_of_surfaces2013.pdf

13 (ii) Pressley, Andrew (2010) Elementary Differential Geometry. London. Springer.

(iii) Walkden, Charles (2016) Hyperbolic Geometry Retrieved from: http://www. maths.manchester.ac.uk/~cwalkden/hyperbolic-geometry/hyperbolic_geometry. pdf