Last time: $$\theta\colon PSL(2,R)\hookrightarrow SL(3,R)$$. If $$V$$ is a real vector space, let $$S^2(V)$$ be the collection of quadratic forms $$\beta\colon V\to R$$. These are functions of the form $$\ell x^2+2mxy+ny^2$$ for real numbers $$\ell,m,n$$. This can be expressed as a symmetric matrix $$\beta=\begin{pmatrix} \ell & m \\ m & n\end{pmatrix}$$. There is an action of $$SL(2,R)$$ on $$S^2(R^2)$$, given by, for $$A\in SL(2,R)$$, setting $$\theta(A)\colon S^2(R^2)\to S^2(R^2)$$ to be $$(A^{-1})^t\begin{pmatrix} \ell & m \\ m & n\end{pmatrix}A^{-1}$$. Concretely, if $$A=\begin{pmatrix}a & b \\ c & d\end{pmatrix}$$, we have $$\theta(A)\beta=\begin{pmatrix}d^2\ell -2dcm +c^2n & -db\ell + (ad+bc)m-acn \\ -db\ell +(ad+bc)m-acn & b^2\ell-2abm+a^2n\end{pmatrix}$$.

By identifying $$S^2(\mathbb{R}^2)$$ with $$\mathbb{R}^3$$ via $$\beta\mapsto (\ell,m,n)$$, this gives $$\theta(A)=\begin{pmatrix}d^2 & -2dc & c^2 \\ -2b\ell & ad+bc & -2ac \\ b^2 & -2ab & a^2\end{pmatrix}$$.

Claim: $$\text{Im}(\theta)=SO(1,2)=\text{Isom}_+(J)$$, where $$J=\begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0\end{pmatrix}$$.

This is because the determinant map which sends $$\beta$$ to $$\ell n-m^2$$ is invariant under the action of $$SL(2,R)$$, namely since $$\theta(A)(\beta)=(A^{-1})^tP(A^{-1})$$, and $$\det(A)=1$$, we have $$\det(\theta(A)(\beta))=\det(\beta)$$. Therefore, the image of $$\theta$$ is a subspace of $$\text{Isom}_+(J)$$, and by comparing dimensions and using the fact that $$\text{Isom}_+(J)$$ is connected, we get equality.

Fact: $$SL(2,R)/SO(2)\cong H^2$$. Therefore $$\text{Im}(\theta)/\theta(SO(2))=SO(1,2)/\theta(SO(2))\cong H^2$$.

$$SO(1,2)$$ acting on $$R^3$$. By mapping $$(\ell,m,n)\mapsto(x_3-x_1,x_2,x_1+x_3)$$, we change $$\ell n-m^2$$ to $$x_3^2-x_1^2-x_2^2$$. Given an $$M\in SO(2,1)$$, this acts on the new coordinates by conjugation with the coordinate map, and preserves the quadratic form. In particular, the 1-locus is a hyperboloid of two sheets which is preserved under the action of $$SO(2,1)$$, and we can metrize the positive (meaning $$x_3>0$$) sheet of this hyperboloid $$H$$ with a hyperbolic metric.

So the action of $$SO(1,2)$$ on $$H$$ is conjugate by a diffeomorphism to action of $$PSL(2,R)$$ on $$H^2=\{x+iy\mid y>0\}$$.

We may also projectivize $$R^3$$, in which case both sheets of the hyperboloid are identified with the unit disc $$\mathbb{D}$$ in the affine patch $$[x_1:x_2:1]$$, with $$x_1^2+x_2^1<1$$. Then the action of $$PSO(1,2)$$ on $$\mathbb{D}$$ is conjugate by a diffeomorphism to an action of $$PSL(2,R)$$ on the upper half plane $$H^2$$. Then $$(\mathbb{D},PO(1,2))$$ is called the Klein model of $$H^2$$.

Definition: A subset $$\Omega\subseteq R P^n$$ is convex if for every line $$\ell$$ in $$R P^n$$, $$\Omega \cap \ell$$ is connected (possibly empty). If also the closure $$\bar{\Omega}$$ is disjoint from some $$R P^{n-1}$$, then $$\Omega$$ is properly convex. For example, consider $$\mathbb{D}\subseteq R P^2$$.

Definition: A properly convex projective orbifold is $$M^n=\Omega/\Gamma$$, where $$\Omega$$ is a properly convex open set in $$R P^n$$ and $$\Gamma$$ is a discrete subgroup of $$PGL(\Omega)=\{\alpha\in PGL(n+1,R)\mid \alpha(\Omega)=\Omega\}$$. If $$\Gamma$$ has no elements of finite order, or if $$\Gamma$$ acts freely on $$\Omega$$, this is a manifold.

Example: A hyperbolic manifold is a properly convex projective orbifold.

By taking $$\Omega$$ to be the interior of a triangle, and $$\Gamma =\langle \begin{pmatrix} 2 & & \\ & 1 & \\ & & 1\end{pmatrix},\begin{pmatrix} 1 & & \\ &2 & \\ & & 1\end{pmatrix}\rangle$$, we obtain a torus.

Duality: Let $$V^*=\text{Hom}(V,R)$$, and consider $$P(V^*)$$. A $$T\in GL(V)$$ acts on $$V^*$$ by $$T_*(\phi)=\phi\circ T^{-1}$$.

If $$\beta$$ is a basis of $$V$$, and $$\beta^*$$ is the dual basis of $$V^*$$, then the matrices are related by $$[T_*]_{\beta^*}=([T]_\beta^{-1})^t$$. The automorphism of $$GL(n,\mathbb{R})$$ given by $$A\mapsto (A^{-1})^t$$ is called a global Cartan involution.

Let $$\Omega$$ be a subset of an affine patch $$A^n$$ in $$R P^n$$. Define $$\mathscr{C}(\Omega)=\{tv\mid v\in \Omega, t>0\}$$. For example, if $$\Omega=\mathbb{D}$$, then $$\mathscr{C}\mathbb{D}$$ is the interior of the light cone, the set $$\{(x_1,x_2,x_3)\mid x_3^2>x_1^2+x_2^2,x_3>0\}$$.

Let $$\mathscr{C}^*\Omega=\{\phi\in V^*\mid \phi(x)>0\forall x\in\overline{\mathscr{C}\Omega}\}$$. This is an open convex cone, because if $$\phi_1,\phi_2\in\mathscr{C}^*$$ and $$\lambda\in [0,1]$$, then $$(\lambda\phi_1+(1-\lambda)\phi_2)(x)>0$$ for all $$x\in \overline{\mathscr{C}\Omega}$$.

Definition: The dual domain of $$\Omega$$ is $$\Omega^*=P(\mathscr{C}^*\Omega)$$, which is a convex set in $$P(V^*)$$. It is open and properly convex.

Example: Take $$\Omega$$ to be the image of a spherical triangle in $$S^2$$ in $$R P^2$$, with angles $$\alpha,\beta,\gamma$$ and opposing side lengths $a,b,c$. The standard inner product of $$V=R^3$$ gives a canonical isomorphism from $$V\to V^*$$ by mapping $$v$$ to $$\langle v,-\rangle$$ (linear functionals are representable). This identifies $$P(V)$$ with $$P(V^*)$$, and in this case, $$\Omega^*$$ is a triangle with side lengths $$\alpha,\beta,\gamma$$ and angles $$a,b,c$$.

$$\Omega$$ is strictly convex if it is properly convex and its boundary contains no line segments. For example, a triangle is properly but not strictly convex.

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