9.6: The Upper Half-Plane

The Mystery Metric

In the previous section, we applied the Gauss-Bonnet Formula to geodesic triangles on a surface with constant positive curvature, namely the unit sphere. Here we investigate a surface with constant negative curvature. Consider the metric coefficients g11(u,v) = 1/v2 = g22(u,v) and g12(u,v) = 0, defined on the upper half-plane, where v > 0. Note that only the intrinsic properties of this surface are given. Since we have no explicit equation to graph, we refer to this surface as the mystery surface. When working with curves and regions on the mystery surface, we draw them in the domain, which is the upper half-plane.


Finding Lengths and Areas

Given the metric coefficients, we can calculate the lengths of curves and the area of regions on a surface. Recall that:

Area = ∫∫√(g11g22 - g122)dudv,           and    
Length = ∫√(g11u'(t)2 + 2g12u'(t)v'(t) + g22v'(t)2)dt

Demonstration 1

In this demonstration, we draw a euclidean triangle in the upper half-plane whose shape and location can be changed using the three hotspots labeled A, B, and C. In the control panel, we use the metric coefficients for hyperbolic space to calculate the hyperbolic area of the triangle as well as the line integral of geodesic curvature around the boundary of the triangle. Applying the Gauss-Bonnet Formula to the triangle, we see that the hyperbolic area and the line integral must be equal:

∫T∫KdA + ∂∫κgds + ∑ext.angles = 2π

Since the sum of the exterior angles of a euclidean triangle is always equal to 2π, we get

∫T∫KdA + ∂∫κgds = 0

Furthermore, the Gaussian curvature for the hyperbolic metric is constant, Κ = -1. This means that

∂T∫κgds = Area(T)

The demonstration calculates the both the line integral of the geodesic curvature and the hyperbolic area of the triangle. Since the area blows up for regions close to the horizontal axis, the triangle is restricted to the region above the line v=½

Exercises

    1. For the metric g11(u,v) = 1/v2 = g22(u,v), g12(u,v) = 0, on the upper half-plane with v > 0, find the geodesic curvature of the portion of the line u(t) = t, v(t) = mt lying in the upper half-plane, where m is a constant.

    2. Explicitly evaluate the line integral     ∂D∫κgds over the boundary of the region D bounded by three straight lines in the parameter domain, from (1,1) to (2,2) to (1,2) and back to (1,1). Without additional calculations, show how to deduce ∫D∫KdA. Evaluate this double integral explicitly, using the fact that K = -1 for this metric.

Geodesics on the Mystery Surface

Geodesics are curves on a surface along which the geodesic curvature is always zero (κg = 0). Geodesic curvature is an intrinsic quantity, and can therefore be written in terms of the metric coefficients:

κgs'(t)3 = u'(t)√(g11/g22)[(-1/2)(∂g11/∂v)u'(t)2 + (∂g22/∂u)u'(t)v'(t) + (1/2)(∂g22/∂v)v'(t)2]
                - v'(t)√(g22/g11)[(1/2)(∂g11/∂u)u'(t)2 + (∂g11/∂v)u'(t)v'(t) - (1/2)(∂g22/∂u)v'(t)2]
+ √(g11g22)[u'(t)v''(t) - v'(t)u''(t)]                                                      

In the derivation of this equation, which can be seen here, we assume that the partial derivative vectors are perpendicular (i.e. g12 = 0). When working with the metric of the mystery surface, we can substitute g11 = g22 = 1/v2. This gives us:

κg(t)s'(t)3 = [u'(t)v'(t)2 + u'(t)3 + v(t)(u'(t)v''(t) - v'(t)u''(t))] / v(t)3

The length of the velocity vector, s'(t), is given by:

s'(t)2 = |X'(t)|2 = (X'(t) · X'(t)) = g11u'(t)2 + 2g12u'(t)v'(t) + g22v'(t)2

Then, using the metric coefficients of the mystery metric, we get:

s'(t) = (1/v(t))√(u'(t)2 + v'(t)2)

Combining this result with our equation for κgs'(t)3, we arrive at the formula for geodesic curvature on the upper half-plane along the curve (u(t), v(t)) in the domain.

κg(t) = [u'(t)v'(t)2 + u'(t)3 + v(t)(u'(t)v''(t) - v'(t)u''(t))] / (u'(t)2 + v'(t)2)3/2

Let C = (C1, C2) be a point in the upper half-plane and R > 0. Consider the following curve (u(t), v(t)):

u(t) = C1 + Rcos(t)
v(t) = C2 + Rsin(t)

Then,

u'(t) = -Rsin(t)         v'(t) = Rcos(t)
u''(t) = -Rcos(t)       v''(t) = -Rsin(t)

The curve (u(t), v(t)) describes a circle of radius R centered at a point C in the domain. After plugging this into our formula for geodesic curvature on the upper half-plane and simplifying, we find that

κg(t) = C2/R

And so, κg(t) = 0 whenever C2 = 0. This means that any circular arc in the domain whose center of curvature lies on the horizontal axis, v = 0, will map to a geodesic on the mystery surface. Note that κg(t) will also equal zero whenever u'(t) = 0. This means that vertical lines in the domain, for which u(t) is a constant, will also map to geodesics on the mystery surface.

Exercises

    1. Let (u(t), v(t)) = (t, c), where a ≤ t ≤ b. Calculate the geodesic curvature along this horizontal line.

    2. Using the metric g11(u,v) = 1/v2 = g22(u,v), g12(u,v) = 0, calculate the area above the curve u(t) = R cos(t), v(t) = R sin(t) above the interval on the u-axis from Rcos(θ) to Rcos(φ). What happens as θ goes to 0 and as φ goes to π?

    3. Since g12 = 0 for the hyperbolic metric, we can use the intrinsic formula that we derived for Gaussian curvature Κ(u,v). Use this formula to calculate the Gaussian curvature of the surface with metric coefficients g11 = g22 = 1/v2.


Applying the Gauss-Bonnet Theorem in Hyperbolic Space

Just as we used the Gauss-Bonnet Theorem to find the area of geodesic triangles drawn on the unit sphere, we can also use it to find the area of geodesic triangles on the mystery surface. In one of the preceding exercises, we calculated the Gaussian curvature of a surface having metric coefficients g11 = g22 = 1/v2 and found that the mystery surface has constant negative Gaussian curvature Κ = -1.

Demonstration 2

In this demonstration, we choose three points, (A, B, and C), in the upper half-plane. The geodesics connecting these points form a triangle with circular sides. From a previous exercise, we know that the hyperbolic area above a geodesic circular arc drawn in the upper half-plane is Area = (φ - θ), such that φ is the angle of the left endpoint of the circular arc and θ is the angle of the right endpoint of the circular arc, where the angles are measured with respect to the center point of the circular arc. The demo colors the regions above each of the geodesics and then calculates the area of each of these regions. Note that the calculation of area assumes that the colored regions extend up to infinity, but the picture only shows them up to a height of h. Then, calculating the area of the geodesic triangle is simply a matter of adding up the areas of the red, green, and yellow regions, and then subtracting off the part where they overlap. The area of the geodesic triangle is calculated using this method.

Exercises

    1. Using the demo, what is the hyperbolic area of the geodesic triangle when all three points lie on the horizontal axis?

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