Lecture 10: Properly Convex Automorphisms Part II

We continue our work classifying the transformations of a properly convex domain. Let $\Omega$ be open and properly convex in $\mathbb{R} P^n$, and suppose $A \in \text{SL}(\Omega)$ is not elliptic. Then as we saw last time, $A$ has an attracting fixed point $x_+ \in \mathbb{P}(E(A)) \cap \overline{\Omega} \subseteq \partial \Omega$. We also saw that $x_+ = [v_+]$ where $v_+$ is a $r(A)$-eigenvector of $A$ (where $r(A)$ is the spectral radius, i.e. the absolute value of the largest eigenvalue of $A$.)

We break our analysis into two cases: * $r(A) > 1$ (hence $A$ is hyperbolic) * $r(A) = 1$ (hence $A$ is parabolic)

Let’s consider the first case. Let $\lambda_+ = r(A)$. We note that $r(A^{-1}) > 1$ also, and we can find an attracting fixed point $x_- = [v_-]$ of $A^{-1}$, where $v_-$ is an $r(A^{-1})$ eigenvector of $A^{-1}$. Then it is clear that $x_-$ is a fixed point of $A$, and that $v_-$ is a $1/r(A^{-1}) = \lambda_-$ eigenvector of $A$.

Now let $\ell_{x_+ x_-} = [x_+, x_-] \cap \overline{\Omega}$ be the projective line in $\overline{\Omega}$ through $x_+$ and $x_-$. Any point $p$ on this line is moved to another point on the line, but how far does it move? To answer this, we view the line as a copy of $\mathbb{R}P^1$ and apply a Mobius transformation to set $x_- = 0$, $p = 1$, and $x_+ = \infty$. Then a simple calculation shows that $Ap = \lambda_+ / \lambda_-$.

We have just seen that for any $p\in \ell_{x_+ x_-}$,

So we set $\tau(A) = \log(\lambda_+/\lambda_-)$ and call this the translation length of $A$.

As we saw in a previous lecture, $A$ also acts on the dual $\Omega^*$, and we can find attracting and repelling fixed points; call them $\varphi_+$ and $\varphi_-$ respectively. In fact, these fixed points correspond to supporting hyperplanes of $\Omega$, written $H_{\varphi_+}$ and $H_{\varphi_-}$. We foliate $\Omega$ by a pencil of hyperplanes with core $H_{\varphi_+} \cap H_{\varphi_-}$. Next we note that any four points from the same four hyperplanes in the foliation have the same cross ratio.
Thus, projecting along these hyperplanes onto the line $\ell_{x_+ x_-}$ is a distance decreasing function, as points on the boundary of $\Omega$ must be closer together than projections of $x_+$ and $x_-$.

Theorem: Supposes that $\Omega$ is strictly convex.

1 If $A\in \text{SL}(\Omega)$ a hyperbolic trasformation, then there exist a unique attractor $x_+$ and repeller $x_-$.

2 The line connecting $x_+$ and $x_-$, called the axis of $A$, consists of all points realizing the translation length, that is, all points $p$ such that $d_{\Omega}(p,Ap) = \tau(A)$.

3 The Jordan blocks of $x_+$ and $x_-$ have size 1.

Proof.

1 As noted above, we know that $x_+ \in \mathbb{P}(E(A)) \cap \overline{\Omega}$ and that $x_+ \in \partial \Omega$, but since $\Omega$ is strictly convex the hyperplane $\mathbb{P}(E(A))$ meets its boundary at just one point. A similar argument applies to $x_-$.

2 It was already shown above that all the points on this line realize the translation length. Additionally, projecting along the pencil of hyperplanes as described just before this theorem is a strictly distance decreasing transformation in the strictly convex case.

3 To show this, we must find an $A$-invariant complement to the space generated by $x_+$ and $x_-$, but the core of our pencil of hyperplanes is just such a complement. QED.

We now consider the case where $r(A) = 1$, when we have a parabolic transformation. Then there is an attracting fixed point $x \in \partial \overline{\Omega}$.

We define the index of $A$, $i(A)$ to be the size of its most powerful Jordan block. If $A \in \text{SL}(\Omega)$ is parabolic, then $i(A)$ is odd.

To prove this, first suppose $A$ has a single Jordan block, so $i(A) = n+1$. Then we can choose coordinates such that $[e_{n+1}] \in \Omega$ and $A = I + N$ where $N$ is a matrix with 1’s above the diagonal and 0 elsewhere. Note that $N^n \neq 0$ and $N^{n+1} = 0$. So,

and for large $p$, the $e_1$ term dominates. So as we repeatedly apply $A$ (or $A^{-1}$) our point approaches $e_1$. Now if $n$ is odd, the sign of the $e_1$ term is the sign of $p$. This implies that $\pm e_1 \in \partial \Omega$, contradicting the fact that $\Omega$ is properly convex. Thus $n$ is even and $i(A)$ is odd.

Now we consider the general case. Suppose $[v] \in \Omega$ and let $W$ be the cyclic $\mathbb{R}[A]$-module generated by $v$. $A|_W$ has a single Jordan block. If we assume $v$ generates the largest Jordan block, then we consider $\Omega' = \mathbb{P}_+(W) \cap \Omega$ and reduce to the previous case.

Theorem: Let $A \in \text{SL}(\Omega)$ be parabolic and suppose $\Omega$ is strictly convex. Then $A$ has a unique attracting fixed point in $\partial \Omega$.

Finally, here’s an interesting unsolved question. Fix $r\in \mathbb{R}$, and consider the set of points moved distance less than $r$ by an isometry of a properly convex set. Is this set connected?

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