Geometry with an Introduction to Cosmic Topology

Rather than derive the arc-length formula here as we did for hyperbolic geometry, we state the following definition and note the single sign difference from the hyperbolic case. This sign difference is consistent with the sign difference in the algebraic descriptions of the transformations in the respective geometries.

The proof of this theorem is left as an exercise, and is essentially the same as the proof that hyperbolic arc-length is an invariant of hyperbolic geometry, from which it follows that area is invariant. One can also prove that the shortest path between two points is along the elliptic line between them. That is, geodesics follows elliptic lines.

However, one must be careful when measuring the shortest path between points in the projective plane. If \(p\) and \(q\) are two points in \(\mathbb{P}^2\) then there is exactly one elliptic line through them both, but we may view this line as consisting of two elliptic segments, both of which connect \(p\) to \(q\text{.}\) That is, if a bug finds herself at point \(p\) and wants to walk along an elliptic line to point \(q\text{,}\) she can do so by proceeding in either direction along the line, as shown in Figure 6.3.3. The elliptic distance between \(p\) and \(q\), which we denote by \(d_S(p,q)\text{,}\) is then the minimum value of the two segment lengths.

With the sphere as our model, we can check our formulas against measurements on the sphere. For instance, \(d_S(0,1)=2\arctan(1)=\pi/2\text{.}\) The elliptic segment from 0 to 1 corresponds via stereographic projection to one-quarter of a great circle on the unit sphere. Any great circle on the unit sphere has circumference \(2\pi\text{,}\) so one-quarter of a great circle has length \(\pi/2\) on the sphere. We also note that the distance formula \(d_S(0,x) = 2\arctan(x)\) applies to spherical geometry \((\mathbb{C}^+,{\cal S})\) for all positive real numbers \(x\text{,}\) and this distance matches the corresponding distances of the points on the unit 2-sphere, see Exercise 6.3.12.

We emphasize that \(\pi/2\) is an upper bound for the distance between two points in \((\mathbb{P}^2,{\cal S})\text{.}\) However, there is no upper bound on how long a journey along an elliptic line can be. If Bormit the bug wants to head out from point \(p\) and travel \(r\) units along any line, Bormit can do it, without obstruction, for any \(r \gt 0\text{.}\) Of course, if \(r\) is large enough, Bormit will do “laps” on this journey. We say a path is a geodesic path if it follows along an elliptic line.

Each transformation \(T\) in \(\cal{S}\) is an elliptic Möbius transformation by Theorem 6.2.4 that fixes two antipodal points, say \(p\) and \(p_a\text{.}\) So \(T\) pushes points along type II clines of \(p\) and \(p_a\text{,}\) and since transformations preserve distance between points, these type II clines of \(p\) and \(p_a\) determine elliptic circles; all points on these type II clines are equidistant from \(p\text{.}\)

Now, suppose \(p\) and \(q\) are any distinct points in \(\mathbb{P}^2\text{.}\) There exists a type II cline of \(p\) and \(p_a\) that goes through \(q\text{.}\) If this cline lives entirely inside the closed unit disk it represents the elliptic circle centered at \(p\) through \(q\text{.}\) Of course, this cline may not live entirely inside the disk, as is the case in Figure 6.3.6. But each point on the type II cline of \(p\) and \(p_a\) through \(q\) has antipodal point on the type II cline of \(p\) and \(p_a\) through \(q_a\text{.}\) So, in \(\mathbb{P}^2\) we may represent the elliptic circle centered at \(p\) through \(q\) by the portions of these two type II clines of \(p\) and \(p_a\) that live in the closed unit disk.

We note also that elliptic circles centered at the origin correpond to Euclidean circles centered at the origin. In particular, a Euclidean circle centered at the origin with Euclidean radius \(a\) (with \(0 \lt a \lt 1\)) corresponds to an elliptic circle centered at the origin with elliptic radius \(2\arctan(a)\text{.}\)

The proof of this theorem is left as an exercise. Circles with elliptic radius greater than or equal to \(\pi/2\) are also investigated in the exercises. They may not look like circles!

The area of a triangle.

We now turn our attention to finding a formula for the area of a triangle in elliptic geometry. We begin with lunes. A lune is the region in \(\mathbb{P}^2\) bounded between two elliptic lines. How do two elliptic lines bound a region? Two lines trap a region because we are identifying antipodal points on the unit circle. A bug living in the shaded region of \(\mathbb{P}^2\) pictured in Figure 6.3.8 would be able to visit all shaded points without crossing the boundary walls determined by the two elliptic lines. The shaded region is a single, connected region bounded by two lines. So what is the area of this region?

Lemma 6.3.9. The area of a lune.

In \((\mathbb{P}^2,{\cal S})\text{,}\) the area of a lune with angle \(\alpha\) is \(2\alpha\text{.}\)

Proof.

To compute the area of a lune, first move the vertex of the lune to the origin in such a way that one leg of the lune lies on the real axis, as in Figure 6.3.10. Then half of the lunar region can be described in polar coordinates by \(0 \leq r \leq 1\) and \(0 \leq \theta \leq \alpha\text{.}\)

So the area of a lune having angle \(\alpha\) is given by

\begin{align*} A \amp = 2\int_0^\alpha\int_0^1 \frac{4r}{(1+r^2)^2}dr d\theta \tag{Let $u = 1+r^2$}\\ \amp = 2\int_0^\alpha \bigg[\frac{-2}{1+r^2} \bigg|_0^1\bigg]~d\theta\\ \amp = 2\int_0^\alpha 1 ~d\theta\\ \amp = 2\alpha\text{.} \end{align*}

Example 6.3.11. Triangle area in \((\mathbb{P}^2,{\cal S})\).

Triangle \(\Delta pqr\) below is formed from 3 elliptic lines. Notice that each corner of the triangle determines a lune, and that the three lunes cover the entire projective plane, with some overlap. In particular, the three lunes in sum cover the triangle three times, so the sum of the three lune areas equals the area of the entire projective plane plus two times the area of the triangle.

The area of the entire projective plane is \(2\pi\) (see Exercise 6.3.9), so we have the following relation:

\begin{align*} 2\pi + 2\cdot A(\Delta pqr) \amp = A(\text{Lune}~ p) + A(\text{Lune}~ q) + A(\text{Lune}~ r)\\ 2\pi + 2\cdot A(\Delta pqr) \amp = (2\alpha + 2\beta + 2\gamma)\\ A(\Delta pqr) \amp = (\alpha + \beta + \gamma) - \pi\text{.} \end{align*}

We summarize the result of this example with the following theorem.

Theorem 6.3.12.

In elliptic geometry \((\mathbb{P}^2,{\cal S})\text{,}\) the area of a triangle with angles \(\alpha, \beta, \gamma\) is

\begin{equation*} A = (\alpha+\beta+\gamma) - \pi\text{.} \end{equation*}

From this theorem it follows that the angles of any triangle in elliptic geometry sum to more than 180\(^\circ\text{.}\)

We close this section with a discussion of trigonometry in elliptic geometry. We derive formulas analogous to those in Theorem 5.4.12 for hyperbolic triangles. We assume here that the triangle determined by distinct points \(p, q\) and \(z\) in \((\mathbb{P}^2,{\cal S})\) is formed by considering the shortest paths connecting these three points. So triangle side lenghths will not exceed \(\pi/2\) in what follows.

a. Position our triangle conveniently, with one corner at the origin, one on the positive real axis at the point \(r\) (\(0 \lt r \leq 1)\text{,}\) and one at the point \(q = ke^{i\gamma}\) (with \(0 \lt k \leq 1)\) as in Figure 6.3.13. Then \(a = d_S(0,r)=2\arctan(r)\text{,}\) so that

\begin{align*} \cos(a) \amp = \cos(2\arctan(r))\\ \amp = \cos^2(\arctan(r))-\sin^2(\arctan(r)) \end{align*}

by the cosine double angle formula. If we set \(\theta = \arctan(r)\) we may use the following right triangle to rewrite the above description of \(\cos(a)\) as follows:

\begin{align*} \cos(a) \amp = \frac{1}{1+r^2}-\frac{r^2}{1+r^2}\\ \amp = \frac{1-r^2}{1+r^2} \end{align*}

On the other hand,

\begin{align*} \sin(a) \amp = \sin(2\arctan(r))\\ \amp = 2\sin(\arctan(r))\cos(\arctan(r))\text{,} \end{align*}

from which it follows that

\begin{equation*} \sin(a) = \frac{2r}{1+r^2}\text{.} \end{equation*}

Turning our attention to the side with length \(b\text{,}\) \(b = d_S(0,q) = 2\arctan(|q|)=2\arctan(k),\) since \(q = ke^{i\gamma}\) with \(k > 0\text{.}\) It follows that

\begin{equation*} \cos(b) = \frac{1-k^2}{1+k^2}~~~\text{and}~~~\sin(b) = \frac{2k}{1+k^2}\text{.} \end{equation*}

And the third side? Theorem 6.3.4 tells us

\begin{equation*} c = d_S(r,ke^{i\gamma}) = 2\arctan\left(\left|\frac{ke^{i\gamma} - r}{1+rke^{i\gamma}}\right|\right)\text{.} \end{equation*}

Since in general \(\cos(2\arctan(x)) = (1-x^2)/(1+x^2)\text{,}\) it follows that

\begin{equation*} \cos(c) = \frac{|1+rke^{i\gamma}|^2 - |ke^{i\gamma}-r|^2}{|1+rke^{i\gamma}|^2 + |ke^{i\gamma}-r|^2}\text{,} \end{equation*}

which can be expanded using \(|z|^2 = z \cdot \overline{z}\text{,}\) and then reduced to obtain

\begin{equation*} \cos(c) = \frac{1 + r^2k^2 - k^2 - r^2+2rk(e^{i\gamma}+e^{-i\gamma})}{1+r^2 + k^2 + r^2k^2}\text{.} \end{equation*}

Now, \(e^{i\gamma}+e^{-i\gamma}=2\cos(\gamma)\text{,}\) so we have

\begin{align*} \cos(c) \amp = \frac{(1-r^2)(1-k^2) + 2rk2\cos(\gamma)}{(1+r^2)(1+k^2)} \\ \amp = \frac{1-r^2}{1+r^2}\cdot\frac{1-k^2}{1+k^2} + \frac{2r}{1+r^2}\cdot \frac{2k}{1+k^2}\cos(\gamma)\\ \amp = \cos(a)\cos(b)+\sin(a)\sin(b)\cos(\gamma)\text{.} \end{align*}

Though we worked out the formula for a conveniently located triangle, it holds for any triangle in elliptic geometry because angles and distances are preserved by transformations in elliptic geometry, and there is a transformation that takes any triangle to this convenient location.

b. To prove the elliptic law of sines, first construct the circle containing side \(c\text{.}\) This circle goes through \(r\) and \(q\) and their antipodal points \(-1/r\) and \(q_a\) as pictured in Figure 6.3.15. We let \(p\) denote the center of the circle, and \(R\) its Euclidean radius.

Let \(m_r\) be the midpoint of the segment connecting \(r\) and \(-1/r\text{,}\) and let \(m_q\) be the midpoint of the segment connecting \(q\) and \(q_a\text{.}\) So,

\begin{equation*} |m_r - r| = \frac{1}{2}\bigg(r+\frac{1}{r}\bigg) = \frac{1+r^2}{2r}\text{,} \end{equation*}

and

\begin{equation*} |m_q - q| = \bigg|\frac{1}{2}\bigg(k+\frac{1}{k}\bigg)e^{i\gamma}\bigg| = \frac{1+k^2}{2k}\text{.} \end{equation*}

Note further that \(\Delta pm_rr\) is right, and \(\angle m_r r p = \pi/2 - \beta\) so that \(\angle rpm_r = \beta\text{.}\) From this right triangle we see \(\sin(\beta) = (1+r^2)/(2rR)\text{.}\)

Similarly, \(\Delta pm_q q\) is right, and we have \(\angle m_q p q = \alpha\text{,}\) and \(\sin(\alpha) = (1+k^2)/(2kR)\text{.}\)

Comparing ratios,

\begin{align*} \frac{\sin(a)}{\sin(\alpha)} \amp = \frac{2r}{1+r^2}\cdot\frac{2kR}{1+k^2}\\ \amp = \frac{2k}{1+k^2}\cdot\frac{2rR}{1+r^2}\\ \amp = \frac{\sin(b)}{\sin(\beta)}\text{.} \end{align*}

To see that the ratio \(\sin(c)/\sin(\gamma)\) must match the preceding common ratio, note that transformations in elliptic geometry preserve distances and angles, so we may transform our triangle above to one in which the length \(c\) is now on the real axis with one end at the origin. The argument above ensures that the ratio \(\sin(c)/\sin(\gamma)\) then matches one of the other ratios, and so all three agree.