Lecture 2: Introduction to Projective Geometry Part II

Last time, we saw a theorem stating that the subgroup of transformations in $PGL(n+1,\mathbb{R})$ that preserve an affine patch is isomorphic to the affine transformations in $n$-dimensional Euclidean space. To see this in general, we first introduce the idea of projective coordinates.

If our vector space $V = \mathbb{R}^{n+1}$ has a basis $v_1,\ldots,v_{n+1}$ then we write \[ [t_1:\cdots:t_{n+1}] := [t_1 v_1 + \cdots + t_{n+1}v_{n+1}]. \]

Then we consider the affine patch of points where $t_{n+1} \neq 0$. In this affine patch, we can rewrite the projective coordinates so the last coordinate is 1.

Now if $[A] \in PGL(n+1,\mathbb{R})$ preserves this affine patch, we find

So when we apply $A$ to the vector $(x_1,\ldots,x_n,1)$ we get a vector of the form $(Bx + v, 1)$. The set of all transformations taking a vector $x$ to the vector $Bx + v$ is called the affine group.

Example Affine Manifold

We consider the space $\mathbb{R}P^2$. We let $v_1,v_2,v_3$ be a basis for $\mathbb{R}^3$. Then the points $[v_1],[v_2],[v_3]$ with the lines between each pair divide $\mathbb{R}P^2$ into 4 triangles. We label the middle triangle $\Delta$.

Next, we let $\Gamma = \langle [A],[B] \rangle \cong \mathbb{Z}^2$ where $A$ and $B$ are diagonal matrices with entries $2,1,1$ and $1,3,1$ respectively. Since both transformations have all positive coordinates, $\Gamma$ preserves our 4 triangles in $\mathbb{R}P^2$.

Then $\Delta / \Gamma$ is an affine torus, which is homeomorphic to a torus. To see this we note that in the affine patch $[x:y:1]$, the rectangle $X = [1,2]\times [1,3]$ is a fundamental domain of $\Gamma$ which is moved by transformations that stretch or compress the rectangle. Applying a log transformation to the first quadrant gives a homeomorphism to $\mathbb{R}^2$, where $\Gamma$ now acts by translation and hence the quotient by $\Gamma$ is clearly a torus.

For the next theorem it is useful to note that because linear maps send vector subspaces of $V$ to other vector subspaces, transformations in $PGL(V)$ similarly preserve projective subspaces (eg projective lines) in $\mathbb{P}(V)$.

(A) Fundamental Theorem of Projective Geometry

If $n\ge 2$, $U$ and $V$ are open sets in $\mathbb{R}P^n$, $T$ is a continuous map $T:U \to V$, and $T$ sends $U \cap$(line) to $V \cap$(line) for all projective lines, then $T$ is the restriction of a projective transformation.

This can be interpreted as saying if $T$ preserves “lines” (the intersections of lines with U), then $T$ also preserves “projective subspaces.”

Proof: There exists a projective basis $\{p_1,\ldots,p_{n+2}\} = B$ contained in $U$ such that $T(B)$ is another projective basis. Then there exists a unique projective transformation $S$ such that $S\mid_B = T\mid_B$ (considering $B$ as an ordered basis). Then the map $S^{-1}\circ T$ preserves “lines” and $S^{-1} \circ T$ fixes $B$. We now want to show $S^{-1} \circ T$ is the identity map.

So, we may assume $T$ is a transformation from $U$ to itself that fixes the projcetive basis $B$. We proceed by induction on the dimension of $\mathbb{R}P^n$.

First, if $n=2$ then our projective basis has 4 points. Let $L_0$ denote the six “lines” (lines intersected with $U$) that contain 2 points in $B$. Since $T$ sends each “line” to itself, $T$ fixes a set of points $P_1$ composed of $B$ together with the 3 additional points lying on 2 lines in $L_0$. Next we let $L_1$ be the set of line containing 2 points in $P_1$, obtain $P_2$ from $P_1$ by adding points contained in two lines in $L_1$, and continue in this fashion. The union of all sets $P_n$ will be a dense set in $U$ that is fixed by $T$. Since $T$ is continuous, this shows that $T$ is the identity.

We show how the induction works by sketching the $n=3$ case. Now $B$ is a projective basis of $\mathbb{R}P^3$, and we know that $T$ is the identity on $B$. Then $T$ preserves each “plane” containing 3 points in $B$. By the $n=2$ step, we have that $T$ is the identity on each of these planes. Now we use a similar construction to the $n=2$ step to again get a dense set of points fixed by $T$.

The Klein (or Projective) Model of $\mathbb{H}^2$

We can model the hyperbolic plane as a disc in the “light cone” in $\mathbb{R}^3$ defined by $z= x^2 + y^2$. We let $\mathcal{C}$ denote the unit disc inside this cone at height $z=1$. Then the disc in $\mathbb{R}P^2$ is

The set of projective transformations that fix $\mathbb{D}$ is

There exist homomorphisms from $SL(2,\mathbb{R})$ to $SL(3,\mathbb{R})$:

• the trivial homomorphism $A \mapsto \text{Id}$;

• the homomorpism $A \mapsto \bigl(\begin{smallmatrix} A& \\ &1 \end{smallmatrix} \bigr)$, a reducible homomorphism;

• there also exists an irreducible homomorphism defined as follows:

$SL(2,\mathbb{R})$ acts on the vector space $\mathbb{R}^3$ considered as the vector space of quadratic forms on $\mathbb{R}^2$, that is, polynomials of the form $ax^2 + bxy + cy^2$.

Let $\beta:\mathbb{R}^2 \to \mathbb{R}^1$ be a quadratic form, written as

We then identify $\beta$ with the vector $(m,n,p) \in \mathbb{R}^3$. Now let $T:\mathbb{R}^2 \to \mathbb{R}^2$ be a linear transformation. Then define $T_* (\beta)$ as the following quadratic form:

It is easy to calculate that $(S\circ T)_* = T_* \circ S_*$, so we have defined an anti-homomorphism. So we must fix this define our homomorphism $\theta: GL(2,\mathbb{R}) \to GL(3,\mathbb{R})$ by $\theta(T) = (T^{-1})_*$. This is our irreducible homomorphism, and it restricts to a homomorphism taking $SL(2,\mathbb{R})$ to $SL(3,\mathbb{R})$.

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