Lecture 8: Projective Isometries

We have a properly convex domain with a Hilbert metric. We define the distance between two points $x,y$ by $d_{\Omega}(x,y) = log([a:x:y:b])$. If we have a projective transformation that preseves cross ratios we call it an isometry of $d_{\Omega}$.

If we restrict ourselves to matrices with determinant plus or minus one $SL^{\pm} (\Omega)$ and then lift to the sphere (since the double cover of a projective space $RP^{n}$ is a sphere $S^n$) and choose one of the preimages $\tilde{\Omega}$ with the property that $\tilde{\Omega}$ maps to itself it becomes much simpler to talk about eigen values and classify isometries.

$A$ is called elliptic if there exists $p \in \Omega$ such that $Ap=p$. Here $p=[v]$ the equivalence class of some vector i.e. $[Av]=[v]$. This implies that $Av = \lambda v$ for some $\lambda > 0$. Hence fixed points coresspond to eigenvectors of the matrix with positve eigenvalues. $A$ is parabolic if every eigenvalue of $A$ has modulus $1$, but $A$ is not elliptic. Finally we say that $A$ is parabolic otherwise.

We define the translation length $t(A)=\inf_{x \in \Omega} d_{\Omega}(x,Ax) \geq 0$ and also define the $minset(A)=\{x \in \Omega \mid d_{\Omega}(x,Ax) = t(A)\}$ of an isometry. In Hyperbolic geometery $H^{n}$ $t(A) > 0$ if and only if $A$ is hyperbolic. In this case points along the geodesic are moved the minimum amount i.e. the minset lies in some geodesic, called the axis. If $t(A)=0$ then $A$ is elliptic if and only if the minset is nonempty and parabolic (along with a unique point $p \in \partial \tilde{\Omega}$ called the parabolic fixed point) if and only if the minset is empty.

Proposition: Let $A \in SL(\Omega)$, then $t(A)=log\vert\lambda / \mu\vert$ where $\lambda$ is an eigenvalue of largest modulus and $\mu$ is an eigenvalue of smallest modulus.

Proposition: $A \in SL(\Omega) \subset SL^{\pm}(n+1,R)$ is elliptic if and only if $A$ is conjugate onto $O(n+1)$

Proof: By the definition of elliptic there is some point in the domain $p \in A$ that is fixed by the elliptic. Here we have a diffeomorphism so $dA$ acts on $T_p \simeq R^{n}$. $d_{\Omega}$ is a norm on $T_p(\Omega)$ specifically a Finsler metric. $dA$ is an isometry of $T_p(\Omega)$, which implies there exists an inner product $\beta$ on $T_p(\Omega)$ such that $dA$ is an isometry of $\beta$. Implies $dA \in Isom(\beta)=O(n)$. (We did not finish proof in lecture, but this implies the conclusion apparently.)

$\bar{\Omega}$ is homeomorphic to some $[0,1]^n$ so any isometry of $\Omega$, $A \in SL(\Omega)$ it preserves the closure $A(\bar{\Omega})=\bar{\Omega}$. The Brower fixed point theorem implies that there exists $p \in \bar{\Omega}$ such that $Ap=p$.

Case1: The fixed point is in the interior $p \in \Omega$ then $A$ is an elliptic.

Case2: The fixed point is on the boundary $p \in \partial \bar{\Omega}$. (In addition it fixes a hyperplane containing that point on the boundary.)

Define a supporting hyper plane $H$ to $\Omega$ at $p \in \partial \bar{\Omega}$ is a projective hyperplane $H=P(V)$ with $dim(V)=n$, such that $p \in H$ and $H \cap \Omega = \emptyset$.

Exercise: Supporting hyperplanes exist.

Proposition: The set of supporting hyperplanes $\{P(ker \phi) : [\phi] \in CC(RP^n)^*\}$ is dual to a nonempty compact convex set in $(RP^n)^*$.

Proof: $\Omega \subset RP^n$ and $\Omega^* \subset RP^n$ a linear map is in the dual domain if and only if $P(ker \phi)$ misses the interior of the domain, $\Omega^* = P(\{\phi \in V^* \mid \phi(\Omega) >0\})$. $\Omega$ must be in $S^n \subset R^{n+1} = V$ for this to make sense, so we have to lift to the double cover ($V$ a vector space).

Corollary: If $A \in SL(\Omega)$ fixes a point in the boundary of the ball $p \in \partial \Omega$ then there exists a supporting hyperplane $H$ to $\Omega$ at $p$ such that $A(H)=H$

Proof: $H$ corresponds to a point in a dual compact convex set $C$. $A$ sends supporting hyperplanes to supporting hyperplanes, so $A: C \rightarrow C$. Then by the Brower Fixed Point Theorem there exists $[\phi] \in C$ such that $A[\phi]=[\phi]$. Therefore $H=ker \phi$ is a supporting hyperplanefixed by $A$.

Structure of fixed set of $A \in SL(\Omega)$: Let $p=[x]$ for $x \in R^{n+1}$, $A$ fixes $p$ if and only if there is a positive number $\lambda$ such that $Ax=\lambda x$. In other word equivalent to $x$ being and eigenvector with a positive eigenvalue. Given an eigenvalue $\lambda>0$ for $A$ we have the eigenspace $E_\lambda$. Define the Fixset $Fix(\lambda, A, \bar{\Omega})= \bar{\Omega} \cap P(E_{\lambda})$ this will be compact and convex as both $\bar{\Omega}$ and $P(E_{\lambda})$ are. The set of all fixed points will be $Fix(A,\Omega) = \bigcup_{\lambda > 0} Fix(\lambda, A, \Omega)$.

Example: Suppose $\Omega = \Delta$ then

Consider $A=\{ \begin{pmatrix} 2&0&0 \\ 0 & 2 & 0 \\ 0&0&1/4 \end{pmatrix}$ then the fixed points we have are $Fix(\lambda =2,A,\Delta)$ and $Fix(\lambda = 1/4,A, \Delta)$.

Dynamics of $[A] \in P_{+}GL(n+1,R)$:

First of all the largest eigenvalues will ‘win’ (i.e. they will blow up the fastest as a matrix is multiplied by itself repeatedly) hence we need only consider the largest eigenvalues. Consider $A=\{ \begin{pmatrix} 2&0&0 \\ 0 & 2 & 2 \\ 0&0&2 \end{pmatrix}$ then $A^n=2^n \{ \begin{pmatrix} 1&0&0 \\ 0 & 1 & n \\ 0&0&1 \end{pmatrix}$. From this we can see that the largest Jordan block will blow up fastest therefore we will just talk about the dynamics of a single $k+1 \times k+1$ Jordan block. The growth of some block is $A^{n}=\lambda^n (I + nN + \frac{n(n-1)}{2}N^2+ ... + (\frac{n}{k}N^k))$ the last term wins as $N \rightarrow \infty$. We define the power of a Jordan blonck with eigenvalue $\lambda \in C$ is $(\vert\lambda\vert,k+1)$ and we use the lexicographical order. We get a partial ordering on Jordan blocks from this.

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