### Hyperbolic Triangles and the Gauss-Bonnet Theorem

Poincaré’s half-plane model for hyperbolic geometry comprises the upper half plane ${\mathbf{H} = \{(x,y): y>0\}}$ together with a metric

$\displaystyle d s^2 = \frac { d x^2 + d y^2 } { y^2 } \,.$

It is remarkable that the entire structure of the space ${(\mathbf{H},ds)}$ follows from the metric.

The “straight lines” or geodesics in this model are semi-circles with centres on the x-axis and vertical half-lines. In the following figure, the two shaded regions are bounded by geodesics, so they are triangles. Visually, one is large and one small. However, distances depend on the vertical coordinate. In fact, since the two triangles are similar, that is, they have three corresponding angles equal, they are also equal in area. This is a consequence of the Gauss-Bonnet Theorem. (See earlier posts [3] and [4] on the Poincaré’ Half-plane).

Geodesic triangles in the Poincare Half-plane. The two shaded triangles are similar and therefore have the same area.

The Gauss-Bonnet Theorem

There is a beautiful result that links the total curvature of a surface, the curvature of its boundary and and the Euler Number of the surface. Suppose that ${M}$ is a two-dimensional manifold. The Gauss-Bonnet Theorem then states that

$\displaystyle \int\!\!\!\int_{M} K\,da + \oint_{\partial M} k_g\,ds = 2\pi\chi(M) \,. \ \ \ \ \ \ \ \ \ \ \ (1)$

Here ${K(M)}$ is the Gaussian curvature of ${M}$, ${k_g}$ is the geodesic curvature of the boundary ${\partial M}$, and ${\chi(M)}$ is the Euler Number of ${M}$ [2].

For a (geodesic) triangle ${T}$ in ${(\mathbf{H},ds)}$, the Euler Number ${\chi = V-E+F}$ is equal to 1 (here ${V}$, ${E}$ and ${F}$ are respectively the number of vertices, edges and faces of ${T}$). So the right hand side of (1) is just ${2\pi}$.

Let the triangle ${T}$ have angles ${\alpha}$, ${\beta}$ and ${\gamma}$. To evaluate the line integral, we replace each vertex by a small circular arc of radius ${\epsilon}$ and curvature ${1/\epsilon}$. The straight segments contribute nothing. Each arc-length is ${\epsilon}$ times  the supplement of the angle, so the line integral is

$\displaystyle \oint_{\partial M} k_g\,ds = (\pi-\alpha)+(\pi-\beta)+(\pi-\gamma) = 2\pi + \delta$

where ${\delta = \pi - (\alpha+\beta+\gamma)}$ is the angle deficit. Finally, since the Gaussian curvature is constant, the area integral in (1) is just ${KA}$, the negative of the area of T.

The Gauss-Bonnet formula now becomes

$\displaystyle KA + 2\pi + \delta = 2\pi \qquad\mbox{or, since K=-1, }\qquad A = \delta$

This is often called the Local Gauss-Bonnet Theorem [1]: “The total curvature of a geodesic triangle on a surface equals the angle excess of the triangle”. In the present case, the curvature is negative so the “excess” becomes the deficit ${\delta}$.

Since the two triangles in the figure above are similar, their angle deficits are equal and therefore they have equal areas.

Other Examples

A consequence of the Gauss-Bonnet Theorem is that no triangle in ${(\mathbf{H},ds)}$ can have an angle greater than ${\pi}$, and only triangles with all angles equal to zero have this area. In the next figure we show three triangles. In each of these, two of the angles are zero and the third is a right angle. Therefore, in each case, ${\delta = \pi/2}$ and the areas of all three are also ${A = \pi/2}$.

Geodesic triangles with angle deficit π/2.  All three have area A=π/2.

Finally, the figure below shows three triangles for which all the angles vanish. For these triangles the angle sum is zero so the deficit is ${\pi}$. They are triangles of maximal area and, despite appearances, all have equal area.

Geodesic triangles with angle deficit δ=π. All three have area A=π.

Sources

[1] Richeson, D. S., 2008: Euler’s Gem: The Polyhedron Formula and the Birth of Topology. Princeton Univ. Press, 317pp. ISBN: 978-0-691-12677-7.

[2] Gauss-Bonnet Theorem. WIkipedia article:  http://en.wikipedia.org/wiki/Gauss-bonnet

[3] Poincaré’s half-plane model: This blog

[4] Poincaré’s half-plane model (bis): This blog