# Geometry ## with an Introduction to Cosmic Topology

#### Errata

The bound copy available for purchase essentially matches the current, freely available version of the text. What follows is a list of errata in the bound copy that have been corrected in the online version of the text. I am grateful to readers for pointing out errors.

• pp. 124-5, in the proof of Theorem 5.4.19 the passage describing the construction of the common perpendicular has been corrected to the following (changes emphasized in blue):

Then construct the common perpendicular of $C$ and the imaginary axis, call this perpendicular $D\text{.}$ We construct this common perpendicular as follows. First find the two points $p$ and $q$ symmetric to both $C$ and the imaginary axis (see Theorem 3.2.16). The points $p$ and $q$ will live on the circle at infinity. The common perpendicular of $C$ and the imaginary axis will be the cline through $p$ and $q$ that is also a hyperbolic line (i.e., orthogonal to the circle at infinity).

If $C$ and the imaginary axis intersect, no such perpendicular exists (think triangle angles), so drag $v_2$ away from $v_1$ until these lines do not intersect. Then construct $D$ as in the preceeding paragraph. Let $v_4$ and $v_5$ be the points of intersection of $D$ with $C$ and the imaginary axis, respectively.

• p. 128, Exercise 5.4.11 should read \begin{equation*} \cosh(c) = \frac{(1+|p|^2)(1+|q|^2)-4\text{Re}(p\overline{q})}{(1-|p|^2)(1-|q|^2)} \end{equation*}
• p. 130, Exercise 5.4.14 a) should read

In particular, show that $\cosh(d_H(0,p)) = \cosh^2(a)\text{.}$

• p. 130, Exercise 5.4.14 d) should read

Show that $\cosh(d_H(p,q)) = \cosh^4(a)[1-\sin(2\theta)]+\sin(2\theta)\text{.}$

• p. 215, the last full sentence on the page should read (change is in blue):

It turns out that every surface can be viewed as a quotient space of the form $M/G$, where $M$ is either the Euclidean plane $\mathbb{C}$, the hyperbolic plane $\mathbb{D}$, or the sphere $\mathbb{S}^2$, and $G$ is a group of isometries in Euclidean geometry, hyperbolic geometry, or elliptic geometry, respectively.