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

$A = \left(\begin{array}{c|c} B & v \\ \hline 0 \cdots 0 & 1 \end{array}\right).$

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

$\mathbb{D} = \mathbb{P}(\mathcal{C}) = \{[x:y:1] \mid x^2+y^2 < 1\}.$

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

$PGL(\mathbb{D}) = \{[A] \in PGL(3,\mathbb{R}) \mid A\mathcal{C} = \mathcal{C}\} = O(2,1) \cong \text{Isom}(\mathbb{H}^2).$

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

$\beta \left(\begin{array}{c} x \\ y \end{array}\right) = \left(\begin{array}{cc} x & y \end{array}\right) \left(\begin{array}{cc} m & n \\ n & p \end{array}\right) \left(\begin{array}{c} x \\ y \end{array}\right) = mx^2 +2nxy + py^2.$

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:

$T_*(\beta) \left(\begin{array}{c} x \\ y \end{array}\right) = \beta\left(T\left(\begin{array}{c} x \\ y \end{array}\right)\right).$

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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