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Background and motivation:

Consider the cone $C\subset \mathbb{R}^d$ of vectors with non-negative components, and let $\Delta\subset C$ be the simplex of probability vectors (those for which $\sum v_i = 1$). The cone (and hence the simplex) can be equipped with the Hilbert metric, which has applications to Perron-Frobenius theory, among other things.

In these applications, the following step is important: given a $d\times d$ stochastic matrix $A$ with strictly positive entries, one has $A\Delta\subset \Delta$, and one wishes to estimate the diameter of $A\Delta$ in the Hilbert metric. Using some explicit formulas for the Hilbert metric, it can be shown that $$ (1)\qquad\qquad \mathrm{diam}(A\Delta) = \sup_{v,w\in\Delta} d(Av,Aw) = \max_{i,j} d(Ae_i,Ae_j), $$ where $e_i$ are the standard basis vectors. Geometrically, this can be stated as follows: the diameter of $A\Delta$ is achieved by considering only its extreme points.

The proof of (1) that I know relies on some matrix computations and doesn't feel particularly geometrically informative. It uses the characterisation of the Hilbert metric in terms of a partial order -- there is also a characterisation of the Hilbert metric in terms of a cross-ratio. In the present case it boils down to fixing two points $x,y\in C$, letting $w,z$ be the points at which the line through $x,y$ intersects $\partial C$, and setting $d(x,y)$ to be the log of the cross-ratio of $w,x,y,z$.

I wonder if there is a more geometric proof of (1) using this description of $d$. This motivates the following question, which can be stated without reference to the Hilbert metric but would (1).

Question: (can be read independently of the above)

Let $\Delta,\Delta'$ be simplices of the same dimension and suppose that $\Delta'\subset \Delta$. Let $\ell$ be a line that intersects $\Delta'$ in an interval. Let $x,y\in \Delta'$ be the endpoints of this interval, and let $w,z$ be the points where $\ell$ intersects $\partial \Delta$. Let $\Theta(\ell)$ be the cross-ratio of the points $w,x,y,z$.

Compactness of $\Delta'$ implies that there exists $\ell$ maximising $\Theta$. (We assume that $\Delta'\cap\partial\Delta=0$ so that $\Theta<\infty$.) It can be shown that the supremum is attained when $\ell$ is one of the edges of $\Delta'$, but the proof I know is non-geometric. Is there a geometric proof of this fact?

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The recent book "Nonlinear Perron-Frobenius Theory" by B. Lemmens and R. Nussbaum covers a great many aspects of Hilbert geometries (and does so beautifully, I may add). That would be a good place to look for this if you haven't done so already. –  alvarezpaiva May 18 '13 at 6:32

1 Answer 1

The following is taken from On convex projective manifolds and cusps Adv. Math. 277 (2015), 181–251. If Ω is properly convex, a function $f:\Omega\to{\mathbb R}$ satisfies the maximum principle if for every compact subset $K\subset\Omega$ the restriction $f|K$ attains its maximum at an extreme point of $K$. Corollary 1.9 (Maximum principle). If $C$ is a closed convex set in a properly convex domain $\Omega$, then the distance of a point in $\Omega$ from $C$ satisfies the maximum principle.

The proof is based on (1.8) which uses a symmetry argument.

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