Geometry with an Introduction to Cosmic Topology

Suppose \(u\text{,}\) \(v\text{,}\) and \(w\) are distinct complex numbers. If \(v\) is on the line through \(u\) and \(w\) then this line is the unique cline through the three points. Otherwise, the three points do not lie on a single line, and we may build a circle through these three points as demonstrated in Figure 3.2.5. Construct the perpendicular bisector to segment \(uv\text{,}\) and the perpendicular bisector to segment \(vw\text{.}\) These bisectors will intersect because the three points are not collinear. If we call the point of intersection \(z_0\text{,}\) then the circle centered at \(z_0\) through \(w\) is the unique cline through the three points.

We prove the result in the case of inversion in the unit circle. The general proof will then follow, since any inversion is the composition of this particular inversion together with translations and dilations, which also preserve clines by Theorem 3.1.9.

Suppose the cline \(C\) is described by the cline equation

where \(c, d \in \mathbb{R}\text{,}\) \(\alpha \in \mathbb{C}\text{.}\)

We want to show that the image of this cline under inversion in the unit circle, \(i_{\mathbb{S}^1}(C)\text{,}\) is also a cline. Well, \(i_{\mathbb{S}^1}(C)\) consists of all points \(w = 1/\overline{z}\text{,}\) where \(z\) satisfies the cline equation for \(C\text{.}\) We show that all such \(w\) live on a cline.

If \(z \neq 0\) then we may multiply each side of the cline equation by \(1/(z\cdot \overline{z})\) to obtain

But since \(w = 1/\overline{z}\) and \(\overline{w} = 1/z\text{,}\) this equation reduces to

or

Thus, the image points \(w\) form a cline equation. If \(d = 0\) then the original cline \(C\) passed through the origin, and the image cline is a line. If \(d \neq 0\) then \(C\) did not pass through the origin, and the image cline is a circle. (In fact, we must also check that \(|\alpha |^2 > dc\text{.}\) This is the case because the original cline equation ensures \(|\alpha|^2 > cd\text{.}\))

Suppose the line through \(p\) does not pass through the center of \(C\text{,}\) as in the diagram below. Let \(q\) be the midpoint of segment \(mn\text{,}\) and let \(d = |q - o|\) as in the diagram. Note also that the line through \(q\) and \(o\) is the perpendicular bisector of segment \(mn\text{.}\) In particular, \(|m-q|=|q-n|\text{.}\)

The Pythagorean theorem applied to \(\Delta pqo\) gives

and the Pythagorean theorem applied to \(\Delta nqo\) gives

By subtracting equation (2) from (1), we have

which factors as

Since \(|p-q|-|q-n| = |p-m|\) and \(|p-q| + |q-n| = |p-n|\text{,}\) the result follows.

The case that the line through \(p\) goes through the center of \(C\) is left as an exercise.

Assume \(C\) is the circle of radius \(r\) centered at \(z_0\text{,}\) and \(D\) is a cline through a point \(z \neq z_0\) not on \(C\text{.}\) Let \(z^*\) denote the point symmetric to \(z\) with respect to \(C\text{.}\)

First, suppose \(D\) is a line through \(z\text{.}\) A line through \(z\) passes through \(z^*\) if and only if it passes through the center of \(C\text{,}\) which is true if and only if the line is orthogonal to \(C\text{.}\) Thus, the line \(D\) through \(z\) contains \(z^*\) if and only if it is orthogonal to \(C\text{,}\) and the theorem is proved in this case.

Now assume \(D\) is a circle through \(z\text{.}\) Let \(o\) and \(k\) denote the center and radius of \(D\text{,}\) respectively. Set \(s = |z_o-o|\text{,}\) and let \(t\) denote a point of intersection of \(C\) and \(D\) as pictured below.

We must argue that \(C\) and \(D\) are orthogonal if and only if \(z^*\) is on \(D\text{.}\) Now, \(C\) and \(D\) are orthogonal if and only if \(\angle otz_0\) is right, which is the case if and only if \(r^2 = s^2-k^2\) by the Pythagorean theorem. Applying Lemma 3.2.7 to the point \(z_0\) (which is outside \(D\)) and the line through \(z_0\) and \(z\text{,}\) we see that

where \(w\) is the second point of intersection of the line with circle \(D\text{.}\)

Note also that as symmetric points, \(z\) and \(z^*\) satisfy the equation

Thus, if we assume \(z^*\) is on \(D\text{,}\) then it must be equal to the point \(w\text{,}\) in which case equations (1) and (2) above tell us \(s^2-k^2 = r^2\text{.}\) It follows that \(D\) is orthogonal to \(C\text{.}\) Conversely, if \(D\) is orthogonal to \(C\text{,}\) then \(s^2-k^2=r^2\text{,}\) so \(|z_0-w|=|z_0-z^*|\text{.}\) Since \(z^*\) and \(w\) are both on the ray \(\overrightarrow{z_0z}\) it must be that \(z^* = w\text{.}\) In other words, \(z^*\) is on \(D\text{.}\)

The result was stated for lines in Theorem 3.1.19. Here we assume \(C\) is a circle of inversion. Consider two curves \(\boldsymbol{r_1}\) and \(\boldsymbol{r_2}\) that intersect at a point \(z\) that is not on \(C\) or at the center of \(C\text{.}\) Recall, \(\angle(\boldsymbol{r_1},\boldsymbol{r_2}) = \angle(L_1,L_2)\) where \(L_i\) is the line tangent to curve \(\boldsymbol{r_i}\) at \(z\text{,}\) for \(i = 1,2\text{.}\) We may describe this angle with two circles \(C_1\) and \(C_2\) tangent to the tangent lines \(L_1\) and \(L_2\text{,}\) respectively, with the additional feature that the circles meet the circle of inversion \(C\) at right angles, as in Figure 3.2.11. Indeed, \(C_1\) is the circle through \(z\) and \(z^*\) whose center is at the intersection of lines \(m_1\) and \(k\text{,}\) where \(m_1\) is the line through \(z\) that is perpendicular to \(L_1\text{,}\) and \(k\) is the perpendicular bisector of segment \(zz^*\text{.}\) Circle \(C_2\) also goes through \(z\) and \(z^*\text{,}\) and its center is on the intersection of \(k\) and the line \(m_2\) through \(z\) that is perpendicular to \(L_2\text{.}\)

The advantage to describing \(\angle(L_1,L_2)\) with these circles is that the image of the angle, \(\angle(i_C(L_1),i_C(L_2))\text{,}\) is also described by these two circles, at their other intersection point \(z^*\text{.}\) Notice that these angles will have opposite signs. For instance, in Figure 3.2.11, our initial angle is negative, described by sweeping arc \(C_1\) clockwise onto \(C_2\text{,}\) but in the image, we sweep \(i_C(C_1)\) counterclockwise onto \(i_C(C_2)\text{.}\) We leave it as an exercise for the reader to check that the angle of intersection of \(C_1\) and \(C_2\) at \(z^*\) is the same magnitude as the angle between \(C_1\) and \(C_2\) at \(z\text{.}\)

Now we show that inversion preserves angle magnitudes for angles that occur on the circle \(C\) (i.e., \(z\) is on \(C\)). Let \(C^\prime\) be a concentric circle to \(C\text{.}\) Then \(i_C(z) = S\circ i_{C^\prime}\) where \(S\) is a dilation of \(\mathbb{C}\) whose fixed point is the common center of circles \(C\) and \(C^\prime\) (see Exercise 3.2.12). Since our angle is not on circle \(C^\prime\text{,}\) \(i_{C^\prime}\) preserves the magnitude of the angle by reason of the preceding argument. The dilation \(S\) preserves angles according to Theorem 3.1.12. Thus \(i_C\) preserves angle magnitudes as well. We leave the case of the angle occurring at the origin to the next section. Bearing that exception in mind, this completes the proof.

Assume \(C\) is the cline of inversion, and assume \(p\) and \(q\) are symmetric with respect to a cline \(D\) as in Figure 3.2.13 (where \(C\) and \(D\) are represented as circles).

We may construct two clines \(E\) and \(F\) that go through \(p\) and \(q\text{.}\) In the figure, cline \(E\) is a circle and cline \(F\) is a line. These clines intersect \(D\) at right angles (Theorem 3.2.8). Since inversion preserves clines and angle magnitudes, we know that \(E^*=i_C(E)\) and \(F^*=i_C(F)\) are clines intersecting the cline \(D^*=i_C(D)\) at right angles. Both \(E^*\) and \(F^*\) contain \(p^*=i_C(p)\text{,}\) so they both contain the point symmetric to \(p^*\) with respect to \(D^*\) (Theorem 3.2.8), but the only other point common to both \(E^*\) and \(F^*\) is \(q^* = i_C(q)\text{.}\) Thus, \(p^*\) and \(q^*\) are symmetric with respect to \(D^*\text{.}\)

If \(k = 1\text{,}\) the set \(D\) is a Euclidean line, according to Theorem 2.4.2, so we assume \(k \neq 1\text{.}\) Let \(C\) be the circle centered at \(p\) with radius 1. Suppose \(z\) is an arbitrary point in the set \(D\text{.}\) Inverting about \(C\text{,}\) let \(z^* = i_C(z)\) and \(q^* = i_C(q)\) as in the following diagram.

Observe first that \(\Delta pz^*q^*\) and \(\Delta pqz\) are similar.

Indeed, \(|p-z|\cdot|p-z^*|=1=|p-q|\cdot|p-q^*|\) by the definition of the inversion transformation, so we have equal side-length ratios

and the included angles are equal, \(\angle q^*pz^* = \angle qpz\text{.}\)

It follows that

from which we derive

Thus, the set \(D\) of all points \(z\) satisfying \(|z-p|=k|z-q|\) has image \(i_C(D)\) under this inversion consisting of all points \(z^*\) on a circle centered at \(q^*\) with radius \((k|p-q|)^{-1}\text{.}\) Since inversion preserves clines and \(p\) is not on \(i_C(D)\text{,}\) it follows that \(D\) itself is a circle.

First, assume one cline is a line \(L\text{,}\) and the other is a circle \(C\) centered at the point \(z_0\) as pictured in Figure 3.2.15. Let \(L_1\) be the line through \(z_0\) that is perpendicular to \(L\text{,}\) and let \(z_1\) be the point of intersection of \(L\) and \(L_1\text{.}\) Next, construct the circle \(C_1\) having the diameter \(z_0z_1\text{.}\) Circle \(C_1\) intersects circle \(C\) at some point, which we call \(t\text{.}\) Notice that \(\angle z_0 t z_1\) is right, and so the circle \(C_2\) centered at \(z_1\) through \(t\) is orthogonal to \(C\text{.}\) Furthermore, the center of \(C_2\text{,}\) \(z_1\text{,}\) lies on line \(L\text{,}\) so \(C_2\) is orthogonal to \(L\text{.}\) Let \(p\) and \(q\) be the two points at which \(C_2\) intersects \(L_1\text{.}\) By construction, and by using Theorem 3.2.8, \(p\) and \(q\) are symmetric to both \(C\) and \(L\text{.}\)

Now, suppose \(C_1\) and \(C_2\) are circles that do not intersect. We may first perform an inversion in a circle \(C\) that maps \(C_1\) to a line \(C_1^*\text{,}\) and \(C_2\) to another circle, \(C_2^*\text{,}\) as suggested in Figure 3.2.17 (any circle \(C\) centered on a point of \(C_1\) works). Then by reason of the preceeding argument, there exist two points \(p\) and \(q\) that are symmetric with respect to \(C_1^*\) and \(C_2^*\text{.}\) Since inversion preserves symmetry points, \(i_C(p)\) and \(i_C(q)\) are symmetric with respect to both \(i_C(C_1^*)\) and \(i_C(C_2^*)\text{.}\) But \(i_C(C_1^*) = C_1\) and \(i_C(C_2^*) = C_2\) so we're found two points symmetric to both \(C_1\) and \(C_2\text{.}\) (In fact, we have one exception. If \(C_1\) and \(C_2\) are concentric circles, this strategy will produce points \(i_C(p)\) and \(i_C(q)\text{,}\) one of which is the center of \(C\text{,}\) and we have not yet extended the notion of inversion to include the center. We do so in the next section in such a way that the theorem applies to this exceptional case as well.)