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
\[A = \left(\begin{array}{cc} B & v \\ \hline 0 \cdots 0 & 1 \end{array}\right).\]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 antihomomorphism. 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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