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 $$p\in \ell_{x_+ x_-}$$,

$d_{\Omega}(p,Ap) = \log [x_-:p:Ap:x_+] = \log(\lambda_+/\lambda_-).$

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,

$(I+N)^p[e_{n+1}] = [e_{n+1}] + {p \choose 1} [e_n] + \cdots + {p \choose n} [e_1]$

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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