Lecture 5: Properly Convex Geometry

In this lecture, we will discuss the geometry of properly convex subsets of $\mathbb{R} P^n$. We begin by discussing an important invariant from 1-dimensional projective geoemtry that will be the basis for a metric on properly convex domains.

The Cross Ratio

Recall that we can identify $\mathbb R P^1$ with $\mathbb R \cup \{\infty\}$. Suppose that $a,x,y,b\in \mathbb R P^1$ are a 4 distinct points, then we can define the cross ratio as

where a limit is taken if one of the points is $\infty$. The cross ratio has many useful properties which we now outline.

1. If the points occur in order (with respect to the orientation on $\mathbb R P^1$) then $[a:x:y:b]>1$.
2. If $A\in PGL_2(\mathbb R)$ then $[Aa:Ax:Ay:Ab]=[a:x:y:b]$
3. If $[a:x:y:b]=[a':x':y':b']$ then there is a unique $A\in PGL_2(\mathbb R)$ such that $A\cdot(a,x,y,b)=(a',x',y',b')$
4. If $(a,x,z,y,b)$ is an ordered quintuple of points in $\mathbb R P^1$ then $[a:x:y:b]=[a:x:z:b][a:z:y:b]$

All of the above properties can be verified by elementary computations. Property 2 shows that the cross ratio is a projective invariant of quadruples of points and property 3 can be thought of as saying that this invariant uniquely determines quadruples of points in $\mathbb R P^1$.

Property 4 can be viewed as a metric property. Specifically, the cross ratio allows us to put a metric on the interval $(a,b)$ using the formula $d_{ab}(x,y)=\log([a:x:y:b])$ for $x,y\in (a,b)$. Property 4 ensures that this metric satisfies the triangle inequality. We now discuss some comparison properties of this metric

Let $a,x,y,b\in \mathbb{R}P^1$ be ordered, then we can think of $[a:x:x+t:b]$ and $[a:x:y:a+t]$ as functions of $t$. I simple computation involving derivatives shows that

1. $[a:x:x+t:b]$ is increasing for $t\in (0,b-x)$
2. $[a:x:y:a+t]$ is decreasing for $t\in (y-1,\infty)$

The first property shows that as points move away from each other on $(a,b)$ that the their distance increases, and the second property shows that the metric gets smaller as the interval gets larger.

The following cross ratio identity in $\mathbb{R}P^2$ will also be of much use to us. For $1\leq i\leq 4$, let $\ell_i$ be four lines passing through a common point, $p\in \mathbb{R}P^2$ and $a_i$ and $b_i$ be quadruples of concurrent points, as in the image below.

Then $[a_1:a_2:a_3:a_4]=[b_1:b_2:b_3:b_4]$. This can be seen as follows: there is a projective transformation fixing $p$ and mapping $a_i$ to $b_i$ for $1\leq i\leq 3$. Such a transformation leaves the lines $\ell_1,\ell_2$, and $\ell_3$ invariant. However, it is easy to see that such a projective transformation must leave $\ell_4$ invariant as well (the $\ell_i$ correspond to 4 concurrent points in $\mathbb{R}P^{1\ast}$), and as such it must map $a_4$ to $b_4$. It should be remarked that the quantity we are calculating is really just the cross ratio of the concurrent lines $l_1,l_2,l_3,l_4$ when regarded as element of $\mathbb{R}P^{2\ast}$.

The Hilbert Metric

Let $\Omega\subset \mathbb{R}P^n$ be an open properly convex domain and let $x,y\in \Omega$. Since $\Omega$ is properly convex the projective line connecting $x$ and $y$ intersects $\partial \Omega$ in two points $a$ and $b$ and we define the Hilbert distance betwee these points as

It is easy to see that $d_\Omega(x,y)=0$ if and only if $x=y$ and that $d_\Omega(x,y)=d_\Omega(y,x)$. Furthermore, since projective maps preserve the cross ratio, we see that elements of $PGL(\Omega)$ are isometries. We now discuss why $d_\Omega$ satisfies the triangle inequality. The triangle inequality follows from a comparison argument with a triangle $\Delta\subset \mathbb{R}P^2$, and so for the time being we assume that $d_\Delta$ satisfies the triangle inequality.

Let $x,y,z\in \Omega$. Any three such points are contained in a projective 2-plane that intersects $\Omega$, thus the proof can be reduced to the two dimensional case. The key point is that we can always find a triangle $\Delta$ such that $d_\Omega(x,z)=d_\Delta(x,z)$, $d_\Omega(z,y)=d_\Delta(z,y)$, and $d_\Omega(x,y)\leq d_\Delta(x,y)$. This is illustrated in the following picture.

As a result we see that

We now verify that the triangle inequality is satisfied for triangular domains in $\mathbb{R}P^2$. This will be done by showing that for a triangle $\Delta$, $(\Delta,d_\Delta)$ is isometric to $(\mathbb{R}^2,d_{Hex})$, where $d_{Hex}$ is the metric on $\mathbb{R}^2$ coming from the norm whose unit ball is a regular hexagon.

We can realize a projective triangle as

If we realize $\mathbb{R}^2$ as the plane in $\mathbb{R}^3$ where the coordinates sum to zero, then there is a map $L:\Delta\to \mathbb{R}^2$ given by $[e^{c_1},e^{c_2},e^{c_3}]\mapsto (c_1,c_2,c_3)$. The group of diagonal matrices in $SL_3(\mathbb{R})$ with positive entries acts as a transitive group of isometries on $\Delta$ with respect to the Hilbert metric. For any such diagonal matrix $A$, a simple computation shows that $L\circ A$ is a translation of $\mathbb{R}^3$ that preserves $\mathbb{R}^2$, and is thus a $d_{Hex}$ isometry. It thus suffices to show that metric balls in $\Delta$ centered at $p_0=[1,1,1]$ are mapped by $L$ to metric balls in $\mathbb{R}^2$ centered at $(0,0,0)$ of the same radius.

Let $d>0$, for $1\leq i\leq 3$ let $\ell_i$ be the line connecting $[e_i]$ to the midpoint of the opposite side of $\Delta$, and let $p_i^{\pm}$ be the points on $\ell_i$ distance $d$ from $p_0$ (where $p_i^+$ is the point closer to $[e_i]$). Let $\ell_i^\pm$ be the line starting at $[e_i]$ and passing through $p_{i\pm 1}^+$, where the indices are taken mod 3.

The line $\ell_1$ can be parameterized as $[e^{2t/3},e^{-t/3},e^{-t/3}]$, and a simple computation shows that this is an isometric embedding of $\mathbb{R}$ to $\ell_1$. Consequently, the points $p_{1}^\pm=[e^{\pm 2d/3},e^{\mp d/3},e^{\mp d/3}]$

By construction, the points $p_i^\pm$ are on the boundary of the unit ball centered at $p_0$. These points span a hexagon $H$ (shown in blue above) in $\Delta$ and using the above identity for cross ratios we see that the points on the boudnary of $H$ are distance d from $p_0$ and thus $H$ is the unit ball centered at $p_0$. Furthermore, the six sides of this $H$ are contained in the lines $\ell_i^\pm$.

Consider the line

This line has $[e_2]$ as one of its endpoints and its a simple computation shows that its intersection with $\ell_1$ is $p_1^-$, and so we see that $\ell=\ell_2^+$. A similar argument shows that

As a result, we see that The images of $\ell_i^{\pm}$ under $L$ are lines in $\mathbb{R}^2$ that span a hexagon in $\mathbb{R}^2$. Furthermore, the $L$-image of the $d$-ball centered at $p_0$ is obtained by scaling the $L$-image of the 1-ball centered at $p_0$ by a factor of $d$. As a result we see that $L$ is an isometry. Consequently, $d_\Delta$ satisfies the triangle inequality, and thus $(\Delta, d_\Delta)$ is a metric space.

An interesting and fun fact that can be easily verified from the above picture is that for this metric $\pi$ (i.e. the ratio of the circumfrence and the diameter of a circle) is 3.

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