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Section 8.6 Lunes

A lune is the region formed between two great circles on a sphere, as illustrated in Figure 8.6.1.  1  Analogously, a lune is the region between two elliptic lines in the Klein disk, as illustrated in Figure 8.6.2.  2 

More precisely, a spherical lune is the region between two elliptic line segments connecting two antipodal points.
Recall that antipodal points are identified, so the Klein lune only has a single vertex!
Figure 8.6.1. A spherical lune with vertices at the poles.
Figure 8.6.2. A Klein lune with vertex on the equator.
Activity 8.6.1.

What does a Klein lune look like if its vertex isn't on the equator?

Hint

Don't forget the wraparound feature of the Klein disk!

Solution
Figure 8.6.3. A Klein lune with vertex not on the equator.

Since we know that the surface area of a sphere is \(4\pi r^2\text{,}\) the area of a lune with angle \(\alpha\) is obtained by proportional reasoning as

\begin{equation*} A(\alpha) = \frac{\alpha}{2\pi}4\pi r^2 = 2\alpha r^2 . \end{equation*}

Since the Klein disk can be thought of as the Northern Hemisphere, its area must be \(2\pi r^2\text{,}\) but, due to the wraparound feature, the entire disk is covered when \(\alpha=\pi\text{.}\) Thus, the area of a Klein lune is given by

\begin{equation*} A(\alpha) = \frac{\alpha}{\pi} 2\pi r^2 = 2\alpha r^2 , \end{equation*}

which is the same as the area of the corresponding spherical lune.

Figure 8.6.4. A triangle covered by three Klein lunes. (The applet may have trouble determing the interior of the triangle for certain configurations.)

Imagine now an elliptic triangle. Construct a lune at each vertex using the angles of the triangle. On the Klein disk, these three lunes already cover the entire disk, as shown in Figure 8.6.4. And more, as the triangle itself is contained in each lune. Letting \(T\) denote the area of the triangle, we therefore have

\begin{equation*} A(\alpha)+A(\beta)+A(\gamma) = 2\pi r^2 + 2 T \end{equation*}

or equivalently

\begin{equation*} T = 2(\alpha+\beta+\gamma)r^2 - 2\pi r^2 = 2r^2 (S-\pi) , \end{equation*}

where \(S\) denotes the angle sum of the triangle. We have therefore shown that the area of a triangle in the Klein disk is a constant (\(2r^2\)) times the angle excess of the triangle.

The same argument works on a sphere. However, lunes on a sphere always come in pairs, since the great circles continue past their intersection points. The construction above therefore yields six lunes, but also six copies of the triangle. The computation now reads

\begin{equation*} 2\bigl(A(\alpha)+A(\beta)+A(\gamma)\bigr) = 4\pi r^2 + 4 T \end{equation*}

which yields the same conclusion

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