# Lecture 1: Introduction to Projective Geometry

Let \(V\) be a finite dimensional vector space over a field \(\mathbb{F}\).

The *projectiviziation* of \(\)V\(\) is
\[\mathbb{P}(V) = (V\backslash {0})/\mathbb{F}^\times = (V\backslash{0})/\sim\]
where we say \(v \sim w\) if \(w = \lambda v\) for some nonzero \(\lambda \in \mathbb{F}\).

If \(V = \mathbb{F}^{n+1}\) then we write \(\mathbb{P}(\mathbb{F}^{n+1}) = \mathbb{F} P^n\).

One way to understand the projectivization of \(V\) is as the space of 1-dimensional subspaces of \(V\).

If we take \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{F} = \mathbb{C}\), then we can consider \(\mathbb{F} P^1\) to be the field \(\mathbb{F}\) together with a point at infinity via the following map:

\[\theta: [(x_0,x_1)] \mapsto \begin{cases} \frac{x_0}{x_1} & x_1 \neq 0 \\ \infty & x_1 = 0\end{cases}\]A transformation \(T \in GL(V)\) sends 1-dimensional subspaces to 1-dimensional
subspaces. Thus \(T\) induces a map \([T]:\mathbb{P}(V) \to \mathbb{P}(V)\), which we call a
*projective transformation*.

We now consider the case of \(\mathbb{F} P^1\) further. Suppose \(T\in GL(\mathbb{F}^2)\), and write \(T = \begin{pmatrix} a&b \\ c&d \end{pmatrix}\). Then \[[T]\left[\left(\begin{array}{c} x_0, \ x_1 \ \end{array}\right)\right] = \left[\left(\begin{array}{c} ax_0 + bx_1, \ cx_0 + dx_1 \ \end{array}\right)\right]. \]

Applying the map \(\theta\) above, we find \(\theta([(x_0,x_1)]) = x_0/x_1 = y\) and that \(\theta([T][(x_0,x_1)]) = (ay+b)/(cy+d)\). We conclude that the group of projective transformations is isomorphic to the Mobius transformations.

We now note that \(T \in GL(V)\) acts trivially on \(\mathbb{P}(V)\) if and only if \(T\) fixes every 1-dimensional subspace of \(V\), which occurs if and only if \(T = \lambda \cdot \text{Id}\) for some nonzero \(\lambda\).

**Definition:**
The *projective general linear group* is
\[PGL(V) = GL(V)/{\lambda \cdot \text{Id} : \lambda \neq 0}.\]

*Exercise:* The set of all nonzero multiples of the identity is the center of
\(GL(V)\).

**Definition:**
If \(U \neq 0\) is a vector subspace of \(V\), then \(\mathbb{P}(U)\) is a *projective
subspace* of \(\mathbb{P}(V)\).

- If \(\dim U = 1\) then \(\mathbb{P}(U)\) is one point.
- If \(\dim U = 2\) then \(\mathbb{P}(U)\) is a
*projective line*. - If \(\dim U = \dim V - 1\) then \(\mathbb{P}(U)\) is a
*projective hyperplane*.

The following definition is necessary to define a projective basis.

**Definition:**
A set of points \(S=\{p_i : I\in I\}\) in \(\mathbb{P}(V)\) is in *general position*
if for all \(k \leq \dim(V)\), every subset of \(k\) points in \(S\) is not contained
in a projective subspace of dimension \(k-2\).

So for example, any subset of 3 points of \(S\) must not lie on a projective line.

A *projective basis* of \(\mathbb{F} P^n\) is a set of \(n+2\) points in general position.

**Theorem:**
There exists a unique projective transformation taking one ordered projective
basis to another.

*Proof:*
Fix a basis \(v_1,\ldots,v_n\) of \(V\). Then
\([v_1],[v_2],\ldots,[v_n],[v_1 + \cdots + v_n] = [v_{n+1}]\) is a projective
basis for \(\mathbb{P}(V)\). Given another projective basis \([w_1],\ldots,[w_{n+1}]\),
it will suffice to show that there is a unique projective transformation taking this
basis to our chosen one.

Now, since \(w_1,\ldots,w_n\) is a basis for \(V\), there exists a vector space isomorphism \(T\) sending \(v_i\) to \(w_i\) for all \(1\le i \le n\). Note that \([T]\) is not the unique transformation taking \([v_i]\) to \([w_i]\) because of scaling.

Now we replace \(w_i\) by \(T^{-1} w_i\). It remains to show that there exists a projective transformation that fixes each \([v_i]\) for \(i\le n\) and sends \([v_{n+1}]\) to \([w_{n+1}]\).

We know \([w_{n+1}] = [\lambda_1 v_1 + \lambda_2 v_2 + \cdots + \lambda_n v_n]\), and we know that \(\lambda_i \neq 0\) for all \(i\) because our vectors are in general position. Thus the diagonal matrix with diagonal entries \(\lambda_i\) is the desired transformation.

To show uniqueness, we can suppose we have 2 transformations taking one basis to another, then compose one with the inverse of another. The result will be a projective transformation that fixes a projective basis. So it will suffice to show that if a projective transformation fixes every point in a projective basis, then it is \([\lambda \cdot \text{Id}] = [\text{Id}]\).

So suppose that \([T e_i] = [e_i]\) for all \(i\le n\) and that \([T(e_1 + \cdots + e_n)] = [e_1 + \cdots + e_n]\). So \(Te_i = \lambda_i e_i\) for all \(i\le n\). Then since \([\lambda_1 e_1 + \lambda_2 e_2 + \cdots + \lambda_n e_n] = [e_1 + \cdots + e_n]\) there is some scalar \(\lambda\) with \(\lambda_i = \lambda\) for all \(i\). So \(T = \lambda \cdot \text{Id}\) as desired.

One way to understand \(\mathbb{R} P^n\) is as the union of an affine patch \(\mathbb{R}^n\) with a projective hyperplane \(\mathbb{R} P^{n-1}\) at infinity. This conception seems to suggest that there are two distinct kinds of points in projective space, but there is no actual difference.

**Definition:**
An *affine patch* \(\mathbb{A}^n\) in \(\mathbb{R} P^n\) is the complement of a hyperplane \(H\).

Since you can find a projective transformation sending any hyperplane to any other hyperplane, you can also send any affine patch to any other affine patch.

**Definition:**
Let \(V\) be a vector space over \(\mathbb{F}\). A map \(\alpha:V\to V\) is an
*affine map* if there exist a linear map \(\beta:V\to V\) and
a vector \(v_0 \in V\) such that \(\alpha(x) = v_0 + \beta(x)\) for all \(x\in V\).

In other words, an affine map is a linear map plus a constant vector.

**Definition:**
An *affine geometry* is a pair \((V, \text{Aff}(V))\) where \(V\) is a vector space
and \(\text{Aff}(V)\) is the group of affine isomorphisms of \(V\).

**Proposition:**
\(PGL(\mathbb{R} P^n \setminus \mathbb{R} P^{n-1}) \cong \text{Aff}(\mathbb{A}^n)\).

Before proving the proposition we give some useful definitions.

**Definition:**
Let \(V\) be a real vector space. Then the *positive projective space* over \(V\)
is \(\mathbb{S}(V) = (V\setminus 0) /( v \cong \lambda v): \lambda > 0\).

So, \(\mathbb{S}(\mathbb{R}^{n+1}) = S^n\) and \(\mathbb{P}(V)\) is equal to \(\mathbb{S}(V) /( [v] \cong [-v])\).

**Definition:**
\(PGL^+ (V) = GL(V)/ \lambda \cdot \text{Id} : \lambda > 0\).

Now we prove the proposition in the case of 2 projective dimensions. When we make \(\mathbb{R}P^2\) out of equivalence classes of Euclidean 3-space, we can represent any equivalence classes not in the \(xy\) plane by points in the plane \(z=1\) by writing \([x_0:x_1:x_2] = [x_0/x_2:x_1/x_2: 1]\). The points not accounted for here are of the form \([x_0:x_1:0]\) which may be considered a copy of \(\mathbb{R} P^1\). So the points in the \(z=1\) plane are a copy of \(\mathbb{A}^2\).

Now if we consider a transformation in \(PGL(\mathbb{R} P^3 - \mathbb{R} P^{2})\), we need a transformation that preserves \(\langle e_1,e_2 \rangle\). So our transformation \([T]\) may be written

\[T = \begin{pmatrix} a & b & e \\ c & d & f \\ 0 & 0 & 1 \end{pmatrix}\]Where scaling allows us to set the bottom corner entry to 1. Then \(T\) applied to an arbitrary \([x_0:x_1:1]\) will be a vector with 1 in the first position and the second two being an arbitrary affine transformation of \((x_0,x_1)\).

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