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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