## Lecture 2: Wednesday, August 21, 2013

Today, we give an overview of Projective Geometry. We define affine $n$-space $\mathbb A^n(K)$ and projective $n$-space $\mathbb P^n(K)$, and spend time discussing the geometry of the projective line $\mathbb P^1(K)$. In particular, we discuss the differences when $K = \mathbb R$ and $K = \mathbb C$.

### Preliminaries

Let $G$ and $H$ be sets. We write “$a \in G$” to mean “$a$ is an element of $G$”; and we write “$G \subseteq H$” to mean “every element $a$ of $G$ is also an element of $H$.” Types of sets are $\mathbb Z$, the integers; $\mathbb Q$, the rational numbers; $\mathbb R$, the real numbers; and $\mathbb C$, the complex numbers. We write $\mathbb Z$ because zahlen is German for “number”; and we write $\mathbb Q$ for quotient since each rational number is the ratio of two integers. We will denote the end of the proof by “$\square$”, which is called a tombstone. More notation can be found here.

Let $\phi: G \to H$ be a map. If $\alpha = \phi(a)$ as an element in $H$ for $a$ in $G$, we write $a \mapsto \alpha$. We say that $\phi$ is well-defined if $\phi(a) = \phi(b)$ in $H$ whenever $a = b$ in $G$. For example, consider the map $\phi: \mathbb R \to \mathbb R$ defined by $\phi(x) = x^2$; this is a well-defined map. As another example, consider the map $\phi: \mathbb Q \to \mathbb Z$ defined by $\phi( a/b ) = a$; this is not a well-defined map. Indeed, we have $\dfrac 22 = 1 = \dfrac 11$ so that $2 = \phi \left( \dfrac 22 \right) \neq \phi \left( \dfrac 11 \right) = 1$.

Given a well-defined map $\phi: G \to H$, we say $\phi$ is injective if $\phi(a) \neq \phi(b)$ in $H$ whenever $a \neq b$ in $G$. We say $\phi$ is surjective if given $\alpha \in H$ there exists $a \in G$ such that $\phi(a) = \alpha$. We say $\phi$ is bijective if $\phi$ is both injective and surjective. For example, the trivial map $\phi: \mathbb R \to \mathbb R$ defined by $\phi(x) = x$ is a bijective. As another example, the map $\phi: \mathbb R \to \mathbb R$ defined by $\phi(x) = x^2$ is neither injective ($\phi(-1) = \phi(1)$ yet $-1 \neq 1$) nor surjective (there does not exist a real number $x$ such that $\phi(x) = -1$).

### Equivalence

Let $G$ be a set. Any relation $\sim$ among the elements of $G$ which satisfies the following three axioms is said to be an equivalence relation:

• Reflexivity: For all $a \in G$ we have $a \sim a$.

• Symmetry: For all $a, \, b \in G$, if $a \sim b$ then $b \sim a$.

• Transitivity: For all $a, \, b, \, c \in G$, if $a \sim b$ and $b \sim c$, then $a \sim c$.

If $a \sim b$, we say “$a$ is equivalent to $b$.” The most natural equivalence relation is equality: if $a = b$ and $b = c$ then $a = c$. As another example, let $\phi: G \to H$ be a well-defined map. We define an relation $\sim$ on $G$ as follows: the notation $a \sim b$ means $\phi(a) = \phi(b)$. To see why this is an equivalence relation, note (Reflexivity) is satisfied because $\phi(a) = \phi(a)$; (Symmetry) is satisfied because $\phi(a) = \phi(b)$ if and only if $\phi(b) = \phi(a)$; and (Transitivity) is satisfied because if $\phi(a) = \phi(b)$ and $\phi(b) = \phi(c)$ then $\phi(a) = \phi(c)$.

Given an equivalence relation $\sim$ on $G$, the set $[a] = \{ \, b \in G \, | \, a \sim b \, \}$ is called the equivalence class of the element $a \in G$. Note that $a \in [a]$ by property (Reflexivity). We denote by $G / \sim$ the collection $\{ \, [a] \, | \, a \in G \, \}$ of all equivalence classes. As an example, consider the map $\phi: \mathbb R \to \mathbb R$ defined by $\phi(x) = x^2$. This induces an equivalence $\sim$ on $\mathbb R$ by saying $a \sim b$ precisely when $a^2 = b^2$. Hence $[x] = \{ x, \, -x \}$ is the equivalence class of $x$. The collection of equivalence classes $\mathbb R / \sim$ may be identified with the nonnegative real numbers.

### Affine and Projective $n$-Space

Fix a positive integer $n$, and let $K$ denote either $\mathbb Q$, $\mathbb R$, or $\mathbb C$. We define affine $n$-space as $\mathbb A^n(K) = \left \{ \left( \tau_1, \, \tau_2, \dots, \, \tau_n \right) \, \biggl \vert \, \tau_i \in K \right \}$. Sometimes this is denoted as $K^n$, especially in the sense of $\mathbb R^2$ or $\mathbb R^3$ as 2-space and 3-space, respectively. For example, $\mathbb A^1(K) = K$. In general, elements $P \in \mathbb A^n(K)$ are called $K$-rational affine points.

Proposition.
Let $K$ denote either $\mathbb Q$, $\mathbb R$, or $\mathbb C$. Denote $G = \mathbb A^{n+1}(K) - \{ (0,0,\dots,0) \}$ as affine $(n+1)$-space $\mathbb A^{n+1}(K)$ minus the origin $(0,0,\dots,0)$. Two $K$-rational affine points $P = (\tau_1, \dots, \tau_n, \tau_0)$ and $Q = (\omega_1, \dots, \omega_n, \omega_0)$ in $G$ are said to be equivalent if $P = \lambda \cdot Q$ for some nonzero number $\lambda \in K^\times$. This defines an equivalence relation $\sim$ on $G$.

Proof: We must show properties (Reflexivity), (Symmetry), and (Transitivity) are satisfied. The relation $P \sim Q$ means $P = \lambda \cdot Q$ for some $\lambda \in K^\times$. This is reflexive, that is, $P \sim P$ because $P = 1 \cdot P$ as $1 \in K^\times$. We now show it is symmetric. Say $P \sim Q$. Then $P = \lambda \cdot Q$, and so $Q = (1/\lambda) \cdot P$ because $\lambda$ in invertible in $K$. Hence $Q \sim P$. Finally, we show it is transitive. Say $P \sim Q$ and $Q \sim R$. Then $P = \lambda \cdot Q$ and $Q = \mu \cdot R$ for some $\lambda, \, \mu \in K^\times$. Then $P = (\lambda \, \mu) \cdot R$ where $\lambda \cdot \mu \in K^\times$ is also nonzero, so $P \sim R$ as desired. $\square$

Given any $K$-rational point $P$ as above, denote its equivalence class as $\bigl( \tau_1 \, : \, \cdots \, : \, \tau_n \, : \, \tau_0 \bigr) = \biggl[ \, (\tau_1, \, \dots, \, \tau_n, \, \tau_0) \, \biggr] = \left \{ \lambda \cdot \bigl(\tau_1, \, \dots, \, \tau_n, \, \tau_0 \bigr) \, \biggl| \, \lambda \in K^\times \right \}$. We define projective $n$-space as the collection of these equivalence classes: $\mathbb P^n(K) = \left \{ \bigl( \tau_1 \, : \, \cdots \, : \, \tau_n \, : \, \tau_0 \bigr) \, \biggr| \, \tau_i \in K \ \text{but} \ \bigl(\tau_1, \, \dots, \, \tau_n, \, \tau_0 \bigr) \neq \bigl(0, \, \dots, \, 0, \, 0 \bigr) \right \}$. For example, when $n = 1$ we have

\displaystyle \begin{aligned} \mathbb P^1(K) & = \left \{ \bigl(\tau_1 \, : \, \tau_0) \, \biggl \vert \, \tau_1, \, \tau_0 \in K \ \text{but} \ \bigl(\tau_1, \, \tau_0 \bigr) \neq \bigl(0, \, 0 \bigr) \right \} \\ & = \left \{ \bigl( \tau_1 \, : \, \tau_0 \bigr) \, \biggl \vert \, \tau_0 \neq 0 \right \} \cup \left \{ \bigl(\tau_1 \, : \, \tau_0 \bigr) \, \biggl \vert \, \tau_0 = 0 \right \} \\ & = \left \{ \bigl(\tau \, : \, 1 \bigr) \, \biggl \vert \, \tau \in K \right \} \cup \biggl \{ \bigl(1 \, : \, 0 \bigr) \biggr \}. \end{aligned}

Note that the first set is essentially $\mathbb A^1(K) = K$, while the second set consists of one equivalence class, which we call the “point at infinity.” In general, elements $P \in \mathbb P^n(K)$ are called $K$-rational projective points.

### Projective Line $\mathbb P^1$

We give a geometric interpretation of $\mathbb P^1(K)$ when $K$ is either $\mathbb Q$, $\mathbb R$, or $\mathbb C$.

Proposition.
Let $K$ denote either $\mathbb Q$ or $\mathbb R$, and fix a nonzero $r \in K$. Define the collection of $K$-rational points on the circle of radius $r$ as the set $S^1(K) = \left \{ \left( u, \, w \right) \in \mathbb A^2(K) \, \biggr \vert \, u^2 + w^2 = r^2 \right \}$; and consider the map defined by $\phi: \, \mathbb P^1(K) \to S^1(K)$ which sends $\left( \tau_1 : \tau_0 \right) \mapsto \left( \dfrac {2 \, \tau_1 \, \tau_0}{\tau_1^2 + \tau_0^2} \, r, \ \dfrac {\tau_1^2 - \tau_0^2}{\tau_1^2 + \tau_0^2} \, r \right)$. This map is a bijection, that is, $\mathbb P^1(K) \simeq S^1(K)$.

Proof: First, we show that $\phi$ is well-defined. Say that $(\tau_1 : \tau_0) = (\omega_1 : \omega_0)$. Then $\tau_1 = \lambda \, \omega_1$ and $\tau_0 = \lambda \, \omega_0$ for some $\lambda \in K^\times$, so that

\begin{aligned} \phi \bigl( (\tau_1 : \tau_0) \bigr) & = \left( \dfrac {2 \, \tau_1 \, \tau_0}{\tau_1^2 + \tau_0^2} \, r, \ \dfrac {\tau_1^2 - \tau_0^2}{\tau_1^2 + \tau_0^2} \, r \right) = \left( \dfrac {\lambda^2 \cdot 2 \, \omega_1 \, \omega_0}{\lambda^2 \cdot \left(\omega_1^2 + \omega_0^2 \right)} \, r , \ \dfrac {\lambda^2 \cdot \left( \omega_1^2 - \omega_0^2 \right)}{\lambda^2 \cdot \left( \omega_1^2 + \omega_0^2 \right)} \, r \right) \\ & = \left( \dfrac {2 \, \omega_1 \, \omega_0}{\omega_1^2 + \omega_0^2} \, r, \ \dfrac {\omega_1^2 - \omega_0^2}{\omega_1^2 + \omega_0^2} \, r \right) = \phi \bigl( (\omega_1 : \omega_0) \bigr). \end{aligned}

Next, we show that $\phi$ is injective. Say that we have two points $(\tau_1 : \tau_0)$ and $(\omega_1 : \omega_0)$ which map to the same image. Then we have

\begin{aligned} (0,0) & = \phi \left( (\tau_1:\tau_0) \right) - \phi \left( (\omega_1:\omega_0) \right) \\ & = \left( \dfrac {2 \left( \tau_0 \, \omega_0 - \tau_1 \, \omega_1 \right) \left( \tau_1 \, \omega_0 - \tau_0 \, \omega_1 \right)}{\left( \tau_1^2 + \tau_0^2 \right) \left( \omega_1^2 + \omega_0^2 \right)} \, r, \ \dfrac {2 \left( \tau_0 \, \omega_1 + \tau_1 \, \omega_0 \right) \left( \tau_1 \, \omega_0 - \tau_0 \, \omega_1 \right)}{\left( \tau_1^2 + \tau_0^2 \right) \left( \omega_1^2 + \omega_0^2 \right)} \, r \right). \end{aligned}

This forces $\tau_1 \, \omega_0 - \tau_0 \, \omega_1 = 0$, so that $(\tau_1:\tau_0) = (\omega_1:\omega_0)$. Finally, we show that $\phi$ is surjective. Let $(u, w)$ be a point on the circle. Choose a point $(\tau_1:\tau_0) \in \mathbb P^1(K)$ such that $\tau_1 \left( r - w \right) = \tau_0 \, u$. The equation $u^2 + w^2 = r^2$ implies $u = \bigl( 2 \, \tau_1 \, \tau_0 \bigr) / \bigl( \tau_1^2 + \tau_0^2 \bigr) \cdot r$ and $w = \bigl( \tau_1^2 - \tau_0^2 \bigr) / \bigl( \tau_1^2 + \tau_0^2 \bigr) \cdot r$. $\square$

To make this bijection a bit more concrete, say that $\tau = \tau_1/\tau_0 \in K$. Then the $K$-rational affine point $( u, \, w) = \left( \dfrac {2 \, \tau_1 \, \tau_0}{\tau_1^2 + \tau_0^2} \, r, \ \dfrac {\tau_1^2 - \tau_0^2}{\tau_1^2 + \tau_0^2} \, r \right) = \left( \dfrac {2 \, \tau}{\tau^2 + 1} \, r, \ \dfrac {\tau^2 - 1}{\tau^2 + 1} \, r \right)$ is on the circle. However, the “north pole” $(0,r)$ is not in the image of this map; we need $\tau = \infty$, that is, $(\tau_1:\tau_0) = (1:0)$. This explains why this point is called the “point at infinity.”

Proposition.
Let $K = \mathbb C$, and fix a nonzero $r \in \mathbb R$. Define the collection of real points on the sphere of radius $r$ as the set $S^2(\mathbb R) = \left \{ \left( u, \, v, \, w \right) \in \mathbb A^3(\mathbb R) \, \biggr| \, u^2 + v^2 + w^2 = r^2 \right \}$; and consider the map $\phi: \, \mathbb P^1(\mathbb C) \to S^2(\mathbb R)$ defined by $\left( \tau_1 : \tau_0 \right) \mapsto \left( \dfrac {2 \, \text{Re} \left(\tau_1 \, \overline {\tau_0} \right)}{\vert \tau_1 \vert^2 + \vert \tau_0 \vert^2} \, r, \ \dfrac {2 \, \text{Im} \left( \tau_1 \, \overline {\tau_0} \right)}{\vert \tau_1 \vert^2 + \vert \tau_0 \vert^2} \, r, \ \dfrac {\vert \tau_1 \vert^2 - \vert \tau_0 \vert^2}{\vert \tau_1 \vert^2 + \vert \tau_0 \vert^2} \, r \right)$. This map is a bijection, that is, $\mathbb P^1(\mathbb C) \simeq S^2(\mathbb R)$.

We call $\mathbb P^1(\mathbb C) \simeq S^2(\mathbb R)$ the Riemann Sphere.

Sketch of Proof: It is easy to check that this map is well-defined. To show this map is surjective, let $(u, v, w)$ be a point on the sphere and choose a point $(\tau_1:\tau_0) \in \mathbb P^1(\mathbb C)$ such that $\tau_1 \left( r - w \right) = \tau_0 \left(u + i \, v \right)$. The equation $u^2 + v^2 + w^2 = r^2$ implies $u = \dfrac {2 \, \text{Re} \left(\tau_1 \, \overline {\tau_0} \right)}{\vert \tau_1 \vert^2 + \vert \tau_0 \vert^2} \, r$, $v = \dfrac {2 \, \text{Im} \left( \tau_1 \, \overline {\tau_0} \right)}{\vert \tau_1 \vert^2 + \vert \tau_0 \vert^2} \, r$, and $w = \dfrac {\vert \tau_1 \vert^2 - \vert \tau_0 \vert^2}{\vert \tau_1 \vert^2 + \vert \tau_0 \vert^2} \, r$. (Here $\overline {\tau_0}$ is the complex conjugate of the complex number $\tau_0$.) To show that this map in injective, the proof is essentially the same as above. $\square$

This proof is very closely related to Stereographic Projection. While $\mathbb P^1(\mathbb C) \simeq S^2(\mathbb R)$ is the sphere, note that $\mathbb P^1(\mathbb R) \simeq S^1(\mathbb R)$ is a Great Circle on the sphere. This gives a geometric way to view $\mathbb P^1(\mathbb R)$ as sitting inside $\mathbb P^1(\mathbb C)$.