# Introduction to Groups and Geometries

## Section3.5Projective geometry

Early motivation for the development of projective geometry came from artists trying to solve practical problems in perspective drawing and painting. In this section, we present a modern Kleinian version of projective geometry.
Throughout this section, $$\F$$ is a field, $$V$$ is a vector space over $$\F\text{,}$$ $$\Proj(V)=(V\setminus \!\{0\})/\F^\ast$$ is the projective space, and $$PGL(V)=GL(V)/\F^\ast$$ is the projective transformation group. See Exercise 2.5.3.6 for definitions and details. We will write $$[T]$$ for the projective transformation that is the equivalence class of the linear transformation $$T$$ of $$V\text{.}$$

### Subsection3.5.1Projective points, lines, and flats

Points in projective space correspond bijectively to 1-dimensional subspaces of $$V$$ via
\begin{equation*} [v] \leftrightarrow \{\alpha v\colon \alpha\in\F\}. \end{equation*}
The set of 1-dimensional subspaces in $$V\text{,}$$ denoted $$G(1,V)\text{,}$$ is an alternative model space for projective geometry. We will usually denote points in projective space using capital letters, such as $$P\text{,}$$ $$Q\text{,}$$ etc.
A line in projective space is a set of the form
\begin{equation*} \ell_\Pi=\{[v]\colon v\in \Pi\setminus\{0\}\} \end{equation*}
for some 2-dimensional subspace $$\Pi$$ in $$V\text{.}$$ Thus, projective lines correspond bijectively to 2-dimensional subspaces of $$V$$ via
\begin{equation*} \ell_{\Pi} \leftrightarrow \Pi. \end{equation*}
The set of 2-dimensional subspaces in $$V$$ is denoted $$G(2,V)\text{.}$$ Points in projective space are called collinear if they lie together on a projective line. We will usually denote projective lines using lower case letters, such as $$\ell\text{,}$$ $$m\text{,}$$ etc.
There is an offset by 1 in the use of the word "dimension" in regards to subsets of $$\Proj(V)$$ and the corresponding subspace in $$V\text{.}$$ In general, a $$k$$-dimensional flat in $$\Proj(V)$$ is a set of the form $$\{[v]\colon v\in G(k+1,V)\}\text{,}$$ where $$G(d,V)$$ denotes the set of $$d$$-dimensional subspaces of $$V\text{.}$$ 1  Flats are also called subspaces in projective space, even though projective space is not a vector space.
Points $$P_1=[v_1],P_2=[v_2],\ldots,P_k=[v_k]$$ are said to be in general position if the vectors $$v_1,v_2,\ldots,v_k$$ are independent in $$V\text{.}$$

### Subsection3.5.2Coordinates

For the remainder of this section, we consider $$V=\F^{n+1}\text{.}$$ For readability, we will write $$P=[v]=[x_0,x_1,x_2,\ldots,x_{n}]$$ (rather than the more cumbersome $$[(x_0,x_1,x_2,\ldots,x_n)]$$) to denote the point in projective space that is the projective equivalence class of the point $$v=(x_0,x_1,x_2,\ldots,x_{n})$$ in $$\F^{n+1}\text{.}$$ The entries $$x_i$$ are called homogeneous coordinates of $$P\text{.}$$ If $$x_0\neq 0\text{,}$$ then
\begin{equation*} P=[x_0,x_1,x_2,\ldots,x_n]=\left[1,\frac{x_1}{x_0},\frac{x_2}{x_0},\ldots,\frac{x_n}{x_0}\right]. \end{equation*}
The numbers $$x_i/x_0$$ for $$1\leq i\leq n$$ are called inhomogeneous coordinates for $$P\text{.}$$ The $$n$$ degrees of freedom that are apparent in inhomogeneous coordinates explain why $$\Proj(\F^{n+1})$$ is called $$n$$-dimensional. Many texts write $$\F\Proj(n)\text{,}$$ $$\F\Proj_n\text{,}$$ or simply $$\Proj_n$$ when $$\F$$ is understood, to denote $$\Proj(\F^{n+1})\text{.}$$

### Subsection3.5.3Freedom in projective transformations

In an $$n$$-dimensional vector space, any $$n$$ independent vectors can be mapped to any other set of $$n$$ independent vectors by a linear transformation. Therefore it seems a little surprising that in $$n$$-dimensional projective space $$\F\Proj_n=\Proj(\F^{n+1})\text{,}$$ it is possible to map any set of $$n+2$$ points to any other set of $$n+2$$ points, provided both sets of points meet sufficient "independence" conditions. This subsection gives the details of this result, called the Fundamental Theorem of Projective Geometry.
Let $$e_1,e_2,\ldots, e_n,e_{n+1}$$ denote the standard basis vectors for $$\F^{n+1}$$ and let $$e_0=\sum_{i=1}^{n+1}e_i\text{.}$$ Let $$v_1,v_2,\ldots,v_{n+1}$$ be another basis for $$\F^{n+1}$$ and let $$c_1,c_2,\ldots,c_{n+1}$$ be nonzero scalars. Let $$T$$ be the linear transformation $$T$$ of $$\F^{n+1}$$ given by $$e_i\to c_iv_i$$ for $$1\leq i\leq n+1\text{.}$$ Projectively, $$[T]$$ sends $$[e_i]\to [v_i]$$ and $$[e_0]\to [\sum_i c_iv_i]\text{.}$$
Now suppose there is another map $$[S]$$ that agrees with $$[T]$$ on the $$n+2$$ points $$[e_0],[e_1],[e_2],\ldots,[e_{n+1}]\text{.}$$ Then $$[U]:=[S]^{-1}\circ [T]$$ fixes all the points $$[e_0],[e_1],[e_2],\ldots,[e_{n+1}]\text{.}$$ This means that $$Ue_i=k_ie_i$$ for some nonzero scalars $$k_1,k_2,\ldots,k_{n+1}$$ and that $$Ue_0=k'e_0$$ for some $$k'\neq 0\text{.}$$ This implies
\begin{equation*} (k_1,k_2,\ldots,k_{n+1})=(k',k',\ldots,k') \end{equation*}
so we have $$k'=k_1=k+2=\cdots k_{n+1}\text{.}$$ Therefore $$[U]$$ is the identity transformation, so $$[S]=[T]\text{.}$$ We have just proved the following existence and uniqueness lemma.

### Subsection3.5.4The real projective plane

The remainder of this section is devoted to the planar geometry $$\Proj(\R^3)=\R\Proj_2$$ called the real projective plane. It is of historical interest because of its early practical use by artists. Lines through the origin in $$\R^3$$ model sight lines in the real world as seen from an eye placed at the origin. A plane that does not pass through the origin models the "picture plane" of the artist’s canvas. Figure 3.5.3 shows a woodcut by Albrecht Dürer that illustrates a "perspective machine" gadget used by 16th century artists to put the projective model into practice for image making.
A two dimensional subspace $$\Pi$$ in $$\R^3$$ is specified by a normal vector $$n=(n_1,n_2,n_3)$$ via the equation $$n\cdot v=0\text{,}$$ that is, a point $$v=(x,y,z)$$ lies on $$\Pi$$ with normal vector $$n$$ if and only if $$n\cdot v=n_1x+n_2y+n_3z=0\text{.}$$ Any nonzero multiple of $$n$$ is also a normal vector for $$\Pi\text{,}$$ so the set $$G(2,\R)$$ of 2-dimensional subspaces in $$\R^3$$ is in one-to-one correspondence $$\R^3/\R^\ast\text{.}$$ We will write $$\ell=[n]=[n_1,n_2,n_3]$$ to denote the projective line $$\ell$$ whose corresponding 2-dimensional subspace in $$\R^3$$ has normal vectors proportional to $$(n_1,n_2,n_3)\text{.}$$ Beware the overloaded notation! Whether the equivalence class $$[v]$$ of a vector $$v$$ in $$\R^3$$ denotes a projective point or a projective line has to be specified.
The equation $$n\cdot v=0$$ makes sense projectively. This means that if $$n\cdot v=0$$ for vectors $$n,v\text{,}$$ then
$$(\alpha n)\cdot (\beta v)=0 \;\;\text{for all}\;\; \alpha,\beta\in \F^\ast\text{,}\tag{3.5.1}$$
even though the value of the dot product is not well-defined for projective equivalence classes! Thus we will write $$\ell\cdot P=[n_1,n_2,n_3]\cdot [x,y,z]=0$$ for a projective line $$\ell=[n_1,n_2,n_3]$$ and a projective point $$P=[x,y,z]\text{,}$$ to mean (3.5.1), and we make the following interpretation of the dot product as an incidence relation in $$\R\Proj_2\text{.}$$
$$\ell \cdot P = 0 \;\; \Leftrightarrow \;\;P\;\;\text{lies on}\;\;\ell\;\;\Leftrightarrow \;\;\ell\;\;\text{contains}\;\;P. \tag{3.5.2}$$
Given two independent vectors $$v,w$$ in $$\R^3\text{,}$$ their cross product $$v\times w$$ is a normal vector for the 2-dimensional space spanned by $$v,w\text{.}$$ Given two 2-dimensional subspaces $$\Pi,\Sigma$$ in $$\R^3$$ with normal vectors $$n,m\text{,}$$ the cross product $$n\times m$$ is a vector that lies along the 1-dimensional subspace $$\Pi\cap\Sigma\text{.}$$ The bilinearity of cross product implies that cross product is well-defined on projective classes, i.e., we can write $$[u]\times [v]:=[u\times v]\text{.}$$ Thus we have the following.

### Exercises3.5.5Exercises

Essential exercises: 1, 2, 3, 4, 5
At least discuss the idea: 6
May be omitted: 7

#### Instructor’s solution for Exercise 3.5.5.1 (proof of the Fundamental Theorem of Projective Geometry).

Let $$P_0,P_1,P_2,\ldots,P_{n+1}$$ be a set of $$n+2$$ points in $$\Proj(\F^{n+1})$$ such that all subsets of size $$n+1$$ are in general position. Let $$v_0,v_1,v_2,\ldots,v_{n+1}$$ be vectors in $$\F^{n+1}$$ such that $$P_i=[v_i]\text{.}$$ By assumption, the points $$P_1,P_2,\ldots,P_{n+1}$$ are in general position, so the vectors $$v_1,v_2,\ldots,v_{n+1}$$ are independent and form a basis of $$\F^{n+1}\text{.}$$ Let $$c_1,c_2,\ldots,c_{n+1}$$ be scalars such that $$v_0=\sum_{i=1}^{n+1}c_iv_i\text{.}$$ By Lemma 3.5.1, there is a unique projective transformation $$T$$ that maps $$[e_i]\to P_i\text{,}$$ $$0\leq i\leq n+1\text{.}$$ By the same argument, there is a unique projective transformation $$S$$ that maps $$[e_i]\to Q_i\text{,}$$ $$0\leq i\leq n+1\text{.}$$ It follows that the projective transformation $$S\circ T^{-1}$$ maps $$P_i\to Q_i\text{,}$$ $$0\leq i\leq n+1\text{.}$$ For uniqueness, suppose there are two projective maps, say $$U,V$$ that map $$P_i\to Q_i\text{,}$$ $$0\leq i\leq n+1\text{.}$$ It then follows that $$U\circ T, V\circ T$$ both map $$[e_i]\to Q_i\text{,}$$ $$0\leq i\leq n+1\text{.}$$ By the uniqueness statement in Lemma 3.5.1, we have $$U\circ T=V\circ T\text{.}$$ Thus we have
\begin{equation*} U=(U\circ T)\circ T^{-1} = (V\circ T)\circ T^{-1}=V. \end{equation*}
Alternatively, we could have stated and proved a separate lemma that says that any projective transformation that fixes $$n+2$$ points in $$\Proj(\F^{n+1})$$ must be the identity. [The special case of this statement, that we already know, is that a Möbius transformation with three fixed points is the identity.]

#### 2.Coordinate charts and inhomogeneous coordinates.

To facilitate thinking about the interplay between the projective geometry $$\Proj(\F^{n+1})=\F\Proj_n$$ and the geometry of $$\F^{n}$$ (rather than $$\F^{n+1}\text{!}$$) it is useful to have a careful definition for "taking inhomogeneous coordinates in position $$i$$". Here it is: Let $$U_i$$ be the subset of $$\F\Proj_n$$ of points whose homogeneous coordinate $$x_i$$ is nonzero. Let $$\pi_i\colon U_i\to \F^n$$ be given by
\begin{equation*} [x_0,x_1,x_2,\ldots x_{i-1},x_i,x_{i+1},\ldots,x_n] \to \left(\frac{x_0}{x_i},\frac{x_1}{x_i},\frac{x_2}{x_i},\ldots \frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},\ldots,\frac{x_n}{x_i}\right). \end{equation*}
The one-sided inverse $$\F^n\to \F\Proj_n$$ given by
\begin{equation*} (x_0,x_1,\ldots x_{i-1},\widehat{x_i},x_{i+1},\ldots,x_n) \to [x_0,x_1,\ldots x_{i-1},1,x_{i+1},\ldots,x_n] \end{equation*}
(where the circumflex hat indicates a deleted item from a sequence) is called the $$i$$-th coordinate chart for $$\F\Proj_n\text{.}$$ What is the map that results from applying the $$0$$-th coordinate chart $$\C\to \C\Proj_2$$ followed by taking homogeneous coordinates in position 1?

#### Instructor’s solution for Exercise 3.5.5.2.

The map $$\pi_1\circ \text{chart}_0$$ is inversion, as follows.
\begin{equation*} z\stackrel{\text{chart}_0}{\to} [1,z] \stackrel{\pi_1}{\to} 1/z \end{equation*}

#### 3.Möbius geometry is projective geometry.

Show that Möbius geometry $$(\extC,\M)$$ and the projective geometry $$(\Proj(\C^2),PGL(2))$$ are equivalent via the map $$\mu\colon \Proj(\C^2) \to \extC$$ given by
$$\mu([\alpha,\beta]) = \left\{ \begin{array}{cc} \alpha/\beta \amp \beta\neq 0\\ \infty \amp \beta=0 \end{array} \right..\tag{3.5.3}$$
Comment: Observe that $$\mu$$ is an extension of $$\pi_1\colon U_1\to \C$$ given by $$\pi_1([x_0,x_1])=\frac{x_0}{x_1}$$ (defined in Exercise 3.5.5.2).

#### Instructor’s solution for Exercise 3.5.5.3.

It is clear that $$\mu$$ is a bijection. Let $$T\in \M$$ be given by $$Tz=\frac{az+b}{cz+d}\text{.}$$ Let $$M\in GL(2,C)$$ be given by $$M=\twotwo{a}{b}{c}{d}\text{.}$$ If $$\beta\neq 0\text{,}$$ we have
\begin{align*} \mu([M][\alpha,\beta])\amp =\mu\left([a\alpha+b\beta,c\alpha+d\beta]\right)\\ \amp = \frac{a\alpha+b\beta}{c\alpha+d\beta}\\ \amp = \frac{a\frac{\alpha}{\beta}+b}{c\frac{\alpha}{\beta}+d}\\ \amp = T(\alpha/\beta)\\ \amp = T(\mu([\alpha,\beta]). \end{align*}
If $$\beta=0\text{,}$$ we have
\begin{equation*} \mu([M][\alpha,0])= \mu([a\alpha,c\alpha])=a/c=T(\infty) = T(\mu([\alpha,0]). \end{equation*}
Thus we have $$\mu\circ T\circ \mu{-1} = [M]\in PGL(2,\C)$$ and $$\mu^{-1}\circ [M]\circ \mu = T\in \M\text{.}$$

#### 4.Cross ratio.

The projective space $$\Proj_1=\Proj(\F^2)$$ is called the projective line). The map $$\mu\colon \Proj_1\to \extF\text{,}$$ given by $$\mu([x_0,x_1])=\frac{x_0}{x_1}$$ (defined in Exercise 3.5.5.3, but where $$\F$$ may be any field, with $$\extF=\F\cup \{\infty\}$$) takes the points
\begin{equation*} [e_0]=[1,1],[e_2]=[0,1],[e_1]=[1,0] \end{equation*}
in $$\Proj_1$$ to the points $$1,0,\infty$$ in $$\extF\text{,}$$ respectively. Let $$(\cdot,P_1,P_2,P_3)$$ denote the unique projective transformation $$[T]$$ that takes $$P_1,P_2,P_3$$ to $$[e_0],[e_2],[e_1]\text{.}$$ The cross ratio $$(P_0,P_1,P_2,P_3)$$ is defined to be $$\mu([T](P_0))\text{.}$$
1. Show that this definition of cross ratio in projective geometry corresponds to the cross ratio of Möbius geometry for the case $$\F=\C\text{,}$$ via the map $$\mu\text{,}$$ that is, show that the following holds.
\begin{equation*} (P_0,P_1,P_2,P_3)=(\mu(P_0),\mu(P_1),\mu(P_2),\mu(P_3)) \end{equation*}
2. Show that
\begin{equation*} (P_0,P_1,P_2,P_3)=\frac{\det(P_0P_2)\det(P_1P_3)}{\det(P_1P_2)\det(P_0P_3)} \end{equation*}
where $$\det(P_iP_j)$$ is the determinant of the matrix $$\twotwo{a_i}{a_j}{b_i}{b_j}\text{,}$$ where $$P_i=[a_i,b_i]\text{.}$$

#### Instructor’s solution for Exercise 3.5.5.4.

1. Let $$S=(\cdot,\mu(P_1),\mu(P_2),\mu(P_3))\text{.}$$ By Exercise 3.5.5.3, the map $$\mu^{-1}\circ S\circ \mu$$ is a projective transformation, and it is easy to see that $$\mu^{-1}\circ S\circ \mu$$ takes $$P_1,P_2,P_3$$ to $$[e_0], [e_2],[e_1]\text{,}$$ so by the Fundamental Theorem of Projective Geometry, we must have $$\mu^{-1}\circ S\circ \mu = [T]\text{.}$$ Evaluating both sides of this equation at $$P_0\text{,}$$ we have
\begin{equation*} (P_0,P_1,P_2,P_3)=(\mu(P_0),\mu(P_1),\mu(P_2),\mu(P_3)). \end{equation*}
2. It is easy to verify that
\begin{equation*} (\mu(P_0),\mu(P_1),\mu(P_2),\mu(P_3))= \frac{\det(P_0P_2)\det(P_1P_3)}{\det(P_1P_2)\det(P_0P_3)}\text{.} \end{equation*}

#### 5.Condition for collinearity in $$\R\Proj_2$$.

Let $$u=(u_1,u_2,u_3),v=(v_1,v_2,v_3),w=(w_1,w_2,w_3)$$ be vectors in $$\R^3\text{,}$$ and let $$M$$ be the matrix $$M=\left[\begin{array}{ccc} u_1 \amp v_1 \amp w_1\\ u_2 \amp v_2 \amp w_2\\ u_3 \amp v_3 \amp w_3\\\end{array}\right]$$ Show that $$[u],[v],[w]$$ are collinear in $$\R\Proj_2$$ if and only if $$\det M=0\text{.}$$

#### Instructor’s solution for Exercise 3.5.5.5.

A line in projective space $$\R\Proj_2$$ the set of equivalence classes of all vectors in a 2-dimensional subspace of $$\R^3\text{.}$$ The three vectors $$u,v,w$$ lie in a 2-dimensional subspace together if and only if the three vectors are linearly dependent. This holds if and only if the determinant of the given $$3\times 3$$ matrix is zero.

#### 6.

The following is a famous theorem of classical geometry.
##### Pappus’ Theorem.
Let $$A,B,C$$ be three distinct collinear points in $$\R\Proj_2\text{.}$$ Let $$A',B',C'$$ be another three distinct collinear points on a different line. Let $$P,Q,R$$ be the intersection points $$P=BC'\cap B'C\text{,}$$ $$Q=AC'\cap A'C\text{,}$$ $$R=AB'\cap A'B\text{.}$$ Then points $$P,Q,R$$ are collinear. See Figure 3.5.5.
Follow the outline below to prove Pappus’ Theorem under the additional assumption that no three of $$A,A',P,R$$ are collinear. Applying the Fundamental Theorem of Projective Geometry, we may assume $$A=[e_1]\text{,}$$ $$A'=[e_2]\text{,}$$ $$P=[e_3]\text{,}$$ and $$R=[e_0]\text{.}$$
• Check that $$AR=[0,-1,1]$$ and $$A'R=[1,0,-1]\text{.}$$
• Explain why it follows that $$B'=[r,1,1]$$ and $$B=[1,s,1]$$ for some $$r,s\text{.}$$
• Explain why $$C=[rs,s,1]$$ and $$C'=[r,rs,1]\text{.}$$
• Explain why $$Q=[rs,rs,1]\text{.}$$
• Observe that $$P,Q,R$$ all lie on $$[1,-1,0]\text{.}$$
Hint.
For the second bullet point, use the fact that $$B'=[x,y,z]$$ lies on $$AR$$ to get $$y=z\text{.}$$ For the third bullet point, use known coordinates for $$A,B,B',P$$ to get coordinates for lines $$AB,PB'\text{.}$$ Then $$C= AB\cap PB'\text{.}$$ Use a similar process for $$C'\text{.}$$ Four the fourth bullet point, use $$Q=AC'\times A'C\text{.}$$

#### Instructor’s solution for Exercise 3.5.5.6 (proof of Pappus’ Theorem).

Following the outline, we have the following.
• We have the following.
\begin{align*} AR\amp = [e_1]\times [e_0] = [1,0,0]\times [1,1,1] = [0,-1,1]\\ A'R\amp = [e_2]\times [e_0] = [0,1,0]\times [1,1,1] = [1,0,-1] \end{align*}
• Since $$B'$$ lies on $$AR\text{,}$$ we have $$B'\cdot AR=0\text{,}$$ so $$B'=[r,x,x]$$ for some $$r,x\text{.}$$ It must be that $$x\neq 0\text{,}$$ for otherwise, we would have $$B'=A\text{.}$$ So we can take $$B'$$ to be $$[r,1,1]\text{.}$$ Similar reasoning yields $$B = [1,s,1]$$ for some $$s\text{.}$$
• Next we have the following.
\begin{align*} AB\amp = [e_1]\times [1,s,1] = [0,-1,s]\\ PB'\amp = [e_3]\times [r,1,1] = [-1,r,0] \end{align*}
Now we have $$C=AB\times PB'=[rs,s,1]\text{.}$$ Similar reasoning applies to $$C'\text{,}$$ which lies on $$A'B'$$ and also on $$BP$$ and yields $$C'=[r,rs,1]\text{.}$$
• Point $$Q$$ lies on $$AC'=[0,-1,rs]$$ and also on $$A'C=[1,0,-rs]\text{,}$$ so must have the form $$AC'\times A'C=[rs,rs,1]\text{.}$$
• We have $$X\cdot [1,-1,0]=0$$ for $$X=P,Q,R\text{,}$$ so we conclude that $$P,Q,R$$ are collinear.

A quadric in $$\Proj(\F^{n+1})$$ is a set of points whose homogeneous coordinates satisfy an equation of the form
$$\sum_{0\leq i\leq j\leq n}c_{ij}x_ix_j=0.\tag{3.5.4}$$
A quadric in $$\R\Proj_2$$ is called a conic.
1. Explain why (3.5.4) is a legitimate definition of a set of points in $$\Proj(\F^{n+1})\text{.}$$
2. Consider the conic $$C$$ given by
\begin{equation*} x_0^2+x_1^2 -x_2^2=0. \end{equation*}
What are the figures in $$\R^2$$ that result from taking inhomogeneous coordinates (see Exercise 3.5.5.2) on $$C$$ in positions $$0,1,2\text{?}$$

#### Instructor’s solution for Exercise 3.5.5.7.

1. Given a vector $$x=(x_0,x_1,\ldots,x_{n+1})\in\F^{n+1}\text{,}$$ let $$c(x)$$ denote the expression on the left side of (3.5.4). We have $$c(kx) = k^2c(x)$$ for any scalar $$k\text{,}$$ so $$c(x)=0$$ if and only if $$c(kx)=0\text{.}$$
2. Taking inhomogeneous coordinates in position 0, let $$u=x_1/x_0,v=x_2/x_0\text{.}$$ The points $$(1,u,v)$$ satisfy $$1+u^2-v^2=0\text{,}$$ which is the equation for a hyperbola in the $$u,v$$-plane. Taking inhomogeneous coordinates in position 1 is also a hyperbola. Taking inhomogeneous coordinates in position 2, with $$u=x_0/x_2,v=x_1/x_2\text{,}$$ we have $$u^2+v^2-1=0\text{,}$$ which is the equation for the unit circle in the $$u,v$$-plane.
The set $$G(d,V)$$ is called the Grassmannian of $$d$$-dimensional subspaces of $$V\text{,}$$ named in honor of Hermann Grassmann.