## Lecture 3: Friday, August 23, 2013

Euclid of Alexandria wrote a text around 300 B.C. entitled “The Elements” which provided a logical framework for much of modern mathematics. A chapter (which was called a “book” during his time) presented an axiomatic approach of geometry, stating rather obvious postulates such as “any two points can be joined by a line.” In Euclid’s honor, modern geometry is often called Euclidean Geometry.

One of these statements, namely “any two parallel lines never intersect,” came under dispute, and several types of geometry grew out of discussions on whether this parallel postulate was really necessary. For instance, in the 19th century mathematicians formed new geometries, including Hyperbolic Geometry, Elliptic Geometry, and even Projective Geometry. These new constructions are typically called non-Euclidean Geometries.

In the this lecture, we’ll discuss why Projective Geometry is non-Euclidean.

### Projective Lines in the Projective Plane $\mathbb P^2$

We will show that the Parallel Postulate does not hold in projective geometry by showing that there is always a point of intersection — in the projective sense. Say that we have two distinct lines:

\begin{aligned} a_1 \, x + b_1 \, y + c_1 & = 0 \\ a_2 \, x + b_2 \, y + c_2 & = 0 \end{aligned}

We know that there are two possibilities: either the lines are parallel, or they intersect in exactly one point. Since we have an injective map $\mathbb A^2(K) \to \mathbb P^2(K)$ which sends $(x,y) \mapsto (x : y : 1);$ we consider how these lines behave in projective space. The following proposition shows that while two lines might not intersect in at least one affine point, they always intersect in at least one projective point. The idea is that even if the lines are parallel, the lines intersect at a “point at infinity.”

Proposition.
Two (projective) lines always intersect in at least one (projective) point.

Proof: Assuming we have the system of equations as above, make the substitution $x = \tau_1 / \tau_0$ and $y = \tau_2 / \tau_0$, i.e., $\left( x \, : \, y \, : \, 1 \right) = \left( \tau_1 \, : \, \tau_2 \, : \, \tau_0 \right).$ Then the system of equations may be written as two distinct “projective” lines:

\begin{aligned} a_1 \, \tau_1 + b_1 \, \tau_2 + c_1 \, \tau_0 & = 0 \\ a_2 \, \tau_1 + b_2 \, \tau_2 + c_2 \, \tau_0 & = 0 \end{aligned}

Note that we have two equations and three unknowns. One solution is given by the cross product:

$(\tau_1, \tau_2, \tau_0) = (a_1, b_1, c_1 ) \times (a_2, b_2, c_2) = \left( \left| \begin{matrix} b_1 & c_1 \\ b_2 & c_2 \end{matrix} \right|, - \left| \begin{matrix} a_1 & c_1 \\ a_2 & c_2 \end{matrix} \right|, \left| \begin{matrix} a_1 & b_1 \\ a_2 & b_2 \end{matrix} \right| \right).$

It suffices to show that this point is not the origin. We will show that the cross product of two vectors vanishes if and only if the vectors are scalar multiples of each other i.e., if $\vec r_1 = (a_1, b_1, c_1)$ and $\vec r_2 = (a_2, b_2, c_2)$ are two nontrivial lines satisfing $\vec r_1 \times \vec r_2 = \vec 0$ then there is a scalar $\lambda$ such that $\vec r_1 = \lambda \, \vec r_2$. To this end, assume that $\vec r_1 \times \vec r_2 = \vec 0$. We know that $\vec r_1, \, \vec r_2 \neq \vec 0$, so without loss of generality say that $a_2 \neq 0$ and choose $\lambda = a_1 / a_2$. Then we have

\begin{aligned} \vec r_1 - \lambda \, \vec r_2 & = \left( a_1 - \frac {a_1}{a_2} \, a_2, \ b_1 - \frac {a_1}{a_2} \, b_2, \ c_1 - \frac {a_1}{a_2} \, c_2 \right) = \left( 0, \ -\frac 1{a_2} \left| \begin{matrix} a_1 & b_1 \\ a_2 & b_2 \end{matrix} \right|, \ -\frac 1{a_2} \left| \begin{matrix} a_1 & c_1 \\ a_2 & c_2 \end{matrix} \right| \right) \\ & = \left( 0, \, 0, \, 0 \right). \end{aligned}

Hence $\vec r_1 = \lambda \, \vec r_2$ as claimed. But then the lines $a_1 \, x + b_1 \, y + c_1 = 0$ and $a_2 \, x + b_2 \, y + c_2 = 0$ are not distinct, which is a contradiction. Hence the two lines intersect at the unique projective point $(\tau_1: \tau_2: \tau_0) = \left( \left| \begin{matrix} b_1 & c_1 \\ b_2 & c_2 \end{matrix} \right|: - \left| \begin{matrix} a_1 & c_1 \\ a_2 & c_2 \end{matrix} \right| : \left| \begin{matrix} a_1 & b_1 \\ a_2 & b_2 \end{matrix} \right| \right)$. This completes the proof. $\square$

The proof above motivates the following result.

Proposition.
A line $L$ going through two distinct affine points $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ can be expressed in terms of a $3 \times 3$ determinant $L: \qquad \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x & y & 1 \end{matrix} \right| = 0$.

Proof: Using the embedding $\mathbb A^2(K) \to \mathbb P^2(K)$ which sends $(x,y) \mapsto (x:y:1)$, express the two affine points as $P_1 = (x_1 : y_1 : 1) = (a_1 : b_1 : c_1)$ and $P_2 = (x_2 : y_2 : 1) = (a_2 : b_2 : c_2)$. Consider the following $3 \times 3$ determinant expressed as a sum of three $2 \times 2$ minors:

$F(\tau_1, \tau_2, \tau_0) = \left| \begin{matrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ \tau_1 & \tau_2 & \tau_0 \end{matrix} \right| = \left| \begin{matrix} b_1 & c_1 \\ b_2 & c_2 \end{matrix} \right| \tau_1 - \left| \begin{matrix} a_1 & c_1 \\ a_2 & c_2 \end{matrix} \right| \tau_2 + \left| \begin{matrix} a_1 & b_1 \\ a_2 & b_2 \end{matrix} \right| \tau_0.$

This is a polynomial of degree $d = 1$. As the rows will be linearly dependent when $(\tau_1 : \tau_2 : \tau_0) = P_1$ or $P_2$, we see that $F(P_1) = F(P_2) = 0$. Hence $L: \ F(\tau_1, \tau_2, \tau_0) = 0$ is indeed a line which goes through both $P_1$ and $P_2$. $\square$

We give an example. Say that we want to find the point of intersection for the two lines $y = x + 1$ and $y = x + 3$. We homogenize these equations by substituting $x = \tau_1 / \tau_0$ and $y = \tau_2 / \tau_0$. We find the equations

\begin{aligned} \tau_1 & - \tau_2 + \tau_0 = 0 \\[5pt] \tau_1 & - \tau_2 + 3 \, \tau_0 = 0 \end{aligned} \qquad \implies \qquad \begin{aligned} \tau_1 & = \lambda \\[5pt] \tau_2 & = \lambda \\[5pt] \tau_0 & = 0 \end{aligned}

for any $\lambda \in K^\times$. Hence $(\tau_1 : \tau_2 : \tau_0) = (1:1:0)$ is the unique projective point of intersection.

### Projective Geometry and Manifolds

Let $X$ be a subset of a topological space. We say $X$ is a manifold of dimension $n$ if there exist

• Open Cover: a collection of open sets $U_\alpha \subset X$ such that $X = \bigcup_{\alpha} U_\alpha$, and

• Atlas: continuous maps $\phi_\alpha: U_\alpha \to \mathbb A^n(\mathbb R)$ which are bijections.

There are variants on this definition for differentiable manifolds and compact manifolds, but we will content ourselves with this rather naive definition. For example, often there is a compatibility condition for intersections $U_\alpha \cap U_\beta$, but we will ignore this for now. We discuss how $\mathbb P^n(\mathbb R)$ is a manifold by considering specific values of $n$. We will discuss this idea in more detail later, but for now we discuss an example.

### The Torus and $\mathbb P^1(\mathbb R) \times \mathbb P^1(\mathbb R)$

Fix real numbers $0 < r < R$. A torus is a manifold which is given by the equation $\mathbb T: \left( R - \sqrt{u^2 + v^2} \right)^2 + w^2 = r^2$. To be more precise, we have the set

$\mathbb T(\mathbb R) = \left \{ (u,v,w) \in \mathbb A^3(\mathbb R) \ \biggl| \ \left( R - \sqrt{u^2 + v^2} \right)^2 + w^2 = r^2 \right \}.$

Geometrically, this is a small circle of radius $r$ which is rotated around a larger circle of radius $R$. A torus is sometimes called a donut. More information can be found here.

Theorem.
The mapping $\mathbb P^1(\mathbb R) \times \mathbb P^1(\mathbb R) \to \mathbb T(\mathbb R)$ defined by

$\bigl( (\tau_1 : \tau_0), \ (\omega_1 : \omega_0) \bigr) \mapsto \left( \dfrac {\tau_1^2-\tau_0^2}{\tau_1^2+\tau_0^2} \left[ R - \dfrac {\omega_1^2-\omega_0^2}{\omega_1^2+\omega_0^2} \, r \right], \ \dfrac {2 \, \tau_1 \, \tau_0}{\tau_1^2+\tau_0^2} \left[ R - \dfrac {\omega_1^2-\omega_0^2}{\omega_1^2+\omega_0^2} \, r \right], \ \dfrac {2 \, \omega_1 \, \omega_0}{\omega_1^2+\omega_0^2} \, r \right)$
is a bijection. In particular,

• $\mathbb T(\mathbb R) \simeq \mathbb P^1(\mathbb R) \times \mathbb P^1(\mathbb R) \simeq S^1(\mathbb R) \times S^1(\mathbb R)$.
• $\mathbb P^1(\mathbb R) \times \mathbb P^1(\mathbb R)$ is a manifold of dimension 2.
• $\mathbb A^2(\mathbb R) \simeq \mathbb A^1(\mathbb R) \times \mathbb A^1(\mathbb R)$, but $\mathbb P^2(\mathbb R) \not \simeq \mathbb P^1(\mathbb R) \times \mathbb P^1(\mathbb R)$.

Both the projective plane $\mathbb P^2(\mathbb R)$ and the torus $\mathbb T(\mathbb R) \simeq \mathbb P^1(\mathbb R) \times \mathbb P^1(\mathbb R)$ are manifolds of dimension 2, but they are different manifolds.

Proof:We derive the birational equivalence above by rationally parametrizing points on $\mathbb T$. We will draw two surfaces and consider the intersection with the torus. For $(\omega_1 : \omega_0) \in \mathbb P^1(\mathbb R)$, consider the surface

$S_1 = \left \{ (u,v,w) \in \mathbb A^3(\mathbb R) \ \biggl| \ \omega_0^2 \, \bigl( u^2 + v^2 \bigr) = \bigl( \omega_1 \, w - \omega_0 \, (R+r) \bigr)^2 \right \}.$

This is a cone with vertex $(u,v,w) = \bigl( 0, \ 0, \ (\omega_1/\omega_0) \, (R+r) \bigr)$; it intersects the torus at a circle defined by the equations $u^2 + v^2 = \left( R - \dfrac {\omega_1^2 - \omega_0^2}{\omega_1^2 + \omega_0^2} \, r \right)^2$ and $w = \dfrac {2 \, \omega_1 \, \omega_0}{\omega_1^2 + \omega_0^2} \, r$. For $(\tau_1 : \tau_0), (\omega_1 : \omega_0) \in \mathbb P^1(\mathbb R)$, now consider the surface

$S_2 = \left \{ (u,v,w) \in \mathbb A^3(\mathbb R) \ \biggl| \ \tau_0 \, u - \tau_1 \, v + \tau_0 \left( R - \dfrac {\omega_1^2-\omega_0^2}{\omega_1^2+\omega_0^2} \, r \right) = 0 \right \}.$

This is a plane with normal vector $(\tau_1 : \tau_0 : 0)$ which goes through the vertical line generated by the equations $u = (\omega_1^2 - \omega_0^2)/(\omega_1^2 + \omega_0^2) \, r - R$ and $v = 0$. The three surfaces $S_1$, $S_2$, and $\mathbb T$ intersect at four points, but we list only two of them:

\begin{aligned} (u,v,w) = & \left( -R + \frac {\omega_1^2-\omega_0^2}{\omega_1^2+\omega_0^2} \, r, \ 0, \ \frac {2 \, \omega_1 \, \omega_0}{\omega_1^2+\omega_0^2} \, r \right), \\[5pt] & \left( \frac {\tau_1^2-\tau_0^2}{\tau_1^2+\tau_0^2} \left[ R - \frac {\omega_1^2-\omega_0^2}{\omega_1^2+\omega_0^2} \, r \right], \ \frac {2 \, \tau_1 \, \tau_0}{\tau_1^2+\tau_0^2} \left[ R - \frac {\omega_1^2-\omega_0^2}{\omega_1^2+\omega_0^2} \, r \right], \ \frac {2 \, \omega_1 \, \omega_0}{\omega_1^2+\omega_0^2} \, r \right). \end{aligned}

Hence we have a bijection $\mathbb P^1(\mathbb R) \times \mathbb P^1(\mathbb R) \to \mathbb T(\mathbb R)$ given by

$\bigl( (\tau_1 : \tau_2), \ (\omega_1 : \omega_0) \bigr) \mapsto \left( \dfrac {\tau_1^2-\tau_0^2}{\tau_1^2+\tau_0^2} \left[ R - \dfrac {\omega_1^2-\omega_0^2}{\omega_1^2+\omega_0^2} \, r \right], \ \dfrac {2 \, \tau_1 \, \tau_0}{\tau_1^2+\tau_0^2} \left[ R - \dfrac {\omega_1^2-\omega_0^2}{\omega_1^2+\omega_0^2} \, r \right], \ \dfrac {2 \, \omega_1 \, \omega_0}{\omega_1^2+\omega_0^2} \, r \right).$

We saw in the previous lecture that $\mathbb P^1(\mathbb R) \simeq S^1(\mathbb R)$, so this proves (1).

We now prove (2). Choose the set $\alpha = \bigl \{ \bigr( \left( p_1 : p_0 \right), \ \left( q_1 : q_0 \right) \bigr) \bigr \}$ consisting of just one point in $\mathbb P^1(\mathbb R) \times \mathbb P^1(\mathbb R)$, and consider the open set

U_\alpha = \left \{ \left( \left( \tau_1 : \tau_0 \right), \ \left( \omega_1 : \omega_0 \right) \right) \in \mathbb P^1(\mathbb R) \times \mathbb P^1(\mathbb R) \ \left| \ \begin{aligned} p_0 \, \tau_1 - p_1 \, \tau_0 & \neq 0 \\ q_0 \, \omega_1 - q_1 \, \omega_0 & \neq 0 \end{aligned} \right. \right \}.

Note that $p_0 \, \tau_1 - p_1 \, \tau_0 = 0$ if and only if $(\tau_1 : \tau_0) = (p_1 : p_0)$, so $\mathbb P^1(\mathbb R) \times \mathbb P^1(\mathbb R) = \bigcup_\alpha U_\alpha$. Consider the map $\phi_\alpha: U_\alpha \to \mathbb A^2(\mathbb R)$ defined by $\biggl( \left( \tau_1 : \tau_0 \right), \ \left( \omega_1 : \omega_0 \right) \biggr) \mapsto \left( \dfrac {p_1 \, \tau_1 + p_0 \, \tau_0}{p_0 \, \tau_1 - p_1 \, \tau_0}, \ \dfrac {q_1 \, \omega_1 + q_0 \, \omega_0}{q_0 \, \omega_1 - q_1 \, \omega_0} \right)$. This is a well-defined map because the denominators are never zero. It is easy to check that it is also a bijection, so $U_\alpha \simeq \mathbb A^2(\mathbb R)$. Hence $\mathbb P^1(\mathbb R) \times \mathbb P^1(\mathbb R)$ is a manifold of dimension 2.

Finally, we show (3). It is clear that $\mathbb A^2(\mathbb R) \simeq \mathbb A^1(\mathbb R) \times \mathbb A^1(\mathbb R)$. Recall that we have an embedding $\mathbb A^n(\mathbb R) \to \mathbb P^n(\mathbb R)$. In fact, we may write

\begin{aligned} \mathbb P^n(\mathbb R) & = \left \{ \left( \tau_1 : \cdots : \tau_n : \tau_0 \right) \ \biggl| \ \tau_i \in \mathbb R \right \} \\[5pt] & = \left \{ \left( \tau_1 : \cdots : \tau_n : 1 \right) \ \biggl| \ \tau_i \in \mathbb R \right \} \cup \left \{ \left( \tau_1 : \cdots : \tau_n : 0 \right) \ \biggl| \ \tau_i \in \mathbb R \right \} \\[5pt] & \simeq \mathbb A^n(\mathbb R) \cup \mathbb P^{n-1}(\mathbb R). \end{aligned}

In particular, $\mathbb P^1(\mathbb R) \simeq \mathbb A^1(\mathbb R) \cup \{ (1:0) \}$ and $\mathbb P^2(\mathbb R) \simeq \mathbb A^2(\mathbb R) \cup \mathbb P^1(\mathbb R)$. This gives

\begin{aligned} \mathbb P^1(\mathbb R) \times \mathbb P^1(\mathbb R) & \simeq \left( \mathbb A^1(\mathbb R) \cup \{ (1:0) \} \right) \times \left( \mathbb A^1(\mathbb R) \cup \{ (1:0) \} \right) \\[5pt] & \simeq \mathbb A^2(\mathbb R) \cup \mathbb A^1(\mathbb R) \cup \mathbb A^1(\mathbb R) \cup \{ (1:0) \} \\[5pt] & \simeq \mathbb A^1(\mathbb R) \cup \mathbb P^2(\mathbb R). \end{aligned}

Hence $\mathbb P^2(\mathbb R) \not \simeq \mathbb P^1(\mathbb R) \times \mathbb P^1(\mathbb R)$ because the latter is much larger than the former, namely by the affine line $\mathbb A^1(\mathbb R)$. $\square$

The real projective plane $\mathbb P^2(\mathbb R)$ is a very strange object: It is a manifold of dimension 2, but it is actually non-orientable! We will return to this manifold in the homework exercises.