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).

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About Edray Herber Goins, Ph.D.

Edray Herber Goins grew up in South Los Angeles, California. A product of the Los Angeles Unified (LAUSD) public school system, Dr. Goins attended the California Institute of Technology, where he majored in mathematics and physics, and earned his doctorate in mathematics from Stanford University. Dr. Goins is currently an Associate Professor of Mathematics at Purdue University in West Lafayette, Indiana. He works in the field of number theory, as it pertains to the intersection of representation theory and algebraic geometry.
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2 Responses to Lecture 2: Wednesday, August 21, 2013

  1. Pingback: Lecture 10: Wednesday, September 11, 2013 | Lectures on Dessins d'Enfants

  2. Pingback: MA 59800 Course Syllabus | Lectures on Dessins d'Enfants

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