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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