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The following text is an attempt to see Hilbert's fourth problem in a new light.

Definition. A pseudometric $d$ on $\mathbb{R}^n$ is called projective if whenever a point $z$ belongs to a line segment $xy$, it follows that $d(x,z)=d(x,y) + d(y,z)$. In other words, a pseudometric is projective if line segments are geodesic segments.

It is immediate (and important) that the set of all projective pseudometrics in $\mathbb{R}^n$ is a convex cone. This suggest the following problem:

Problem. Construct all projective pseudometrics $d$ that are indecomposable in the sense that if $d$ is a positive linear combination of projective pseudometrics $d_1$ and $d_2$, then both of these metrics are multiples of $d$.

Examples of indecomposable projective pseudometrics. Given a closed half-space $H \subset \mathbb{R}^n$, the pseudometric $\delta_H(x,y)$ that takes the value $1$ when exactly one of the two points lies on $H$ and takes the value $0$ in any other case is an indecomposable projective pseudometric.

Question. Are there other examples when $n = 2$? Find other examples for $n > 2$.

I suspect there are no other indecomposable projective pseudometrics in two dimensions, but I know there have to be some (and more than just some) other such pseudometrics in dimension greater than two. To explain why, I have to motivate the question.

Motivation. Hilbert's fourth problem is somewhat vaguely posed as "find all geometries in which the straight line is the shortest connection between two points". In Wikipedia the problem appears with the note "too vague to be stated resolved or not." In this question I'm trying to get my thoughts straight on a way in which the problem can be stated not only clearly, but also insightfully through Choquet's (integral) representation theory.

Consider the vector space $V_n$ of locally bounded real-valued functions on $\mathbb{R}^n \times \mathbb{R}^n$ that are identically zero on the diagonal. The set of all pseudometrics in $\mathbb{R}^n$ is a convex cone in $V_n$ as is the set of all projective pseudometrics in $\mathbb{R}^n$. I shall call this last cone ${\cal H}_n$ and think of it as the set of weak solutions of Hilbert's fourth problem. The space $V_n$ can be topologized in a number of ways to make it into a locally convex topologically vector space in which ${\cal H}_n$ is a closed convex cone. For example, If $k$ is a positive integer and $\epsilon$ is a positive real number, we can consider the following system of neighborhoods of the origin: $$ U_{k,\epsilon} := \{d \in V_n: d(x,y) < \epsilon \text{ if $x$ and $y$ belong to the Euclidean ball of radius k} \}. $$ However, I'm not convinced this will be the good topology for what I want at the end (namely, the application of Choquet's theory to Hilbert's fourth problem).

In any case, the problem that was stated at the beginning of this questions is simply: describe the set of extreme rays of the cone ${\cal H}_n$.

This question, while interesting by itself, becomes much more interesting if one could set things up so that the following vaguely-stated representation theorem holds: for every projective pseudometric $d$, there exists a probability measure $\mu$ supported on the set of extreme rays so that $d$ is the barycenter of $\mu$.

Why should something like this be true? Well, faith and optimism are unexplainable phenomena, but the Busemann-Pogorelov-Alexander-Ambartzumian solution of Hilbert's fourth problem in two dimensions can be re-interpreted as follows:

Theorem. If $d$ is a continuous projective metric on the plane, there exists a Borel measure $\mu$ defined on the space of closed half-planes such that $$ d(x,y) = \int \delta_H(x,y) \, d\mu(H) . $$

In other words, there is a Choquet representation for continuous projective metrics on the plane.

In higher dimensions, we can construct all sufficiently smooth projective metrics in the same way, but we need to use signed measures. I guess then that there must be other indecomposable projective pseudometrics besides the $\delta_H$.

Apology and conclusion. I'm sorry about this horribly long attempt to get my ideas straight the upshot of which is: if every projective pseudometric in $\mathbb{R}^n$ can be written as the barycenter of some probability measure supported on the set indecomposable projective pseudometrics, then possibly more precise interpretation of Hilbert's fourth problem would be to construct all indecomposable projective pseudometrics in $\mathbb{R}^n$.

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+1: I really like this question; I wish I had something useful to say though! – Suvrit Nov 29 '13 at 19:03

It seems to me that the set of projective semi-distances that you describe in the example (let's call them of type I) can be slightly generalized the following way. In the definition of $d_H$, let $H$ be any open half-plane, plus an open half-line contained in its boundary. In other words, up to an affinity, these other distances (say of type II) are defined by $d(x,y)=0$ or $1$ according whether $x \mathbf{\ge} y$ or not, where $\ge$ denotes the lexicographic order in $\mathbb{R}^2$ (we may also parametrize all of them by the unit tangent bundle of $\mathbb{R}^2$). In contrast, the distance of type I are analogously related to the order $x_1\ge y_1$. These two classes cover all $\{0,1\}$-valued projective semi-distances, and are all indecomposable (even in a slightly more general sense).

Indeed, given a semi-distance $d$ we may consider the equivalence relation $x\sim y\ \mathrm{iff}\ d(x,y)=0$, whose classes are the closures of points. (Let me anticipate a remark here: if $d$ is a semi-distance that is topologically stronger than another semi-distance $d'$, then the partition into closures of points induced by $d $ refines the partition induced by $d'$). If $d$ is projective, the closure of any point is a convex set (a convex combination of a pair of points at zero distance, must be also at zero distance from them).

If $d$ is a $\{0,1\}$-valued projective semi-distance, then the set of points at distance $1$ from any given point $p$ is also a convex set: for, if $d(p,x)=d(p,y)=1$ and $z:=tx+(1-t)y$ for some $0\le t\le 1$, then since $d(x,z)+d(z,y)=d(x,y)\le1$, at least one among $d(x,z)$ and $d(y,z)$ is zero, so that in any case $ d(p,z)= 1$.

But a convex set of $\mathbb{R^2}$ whose complement is also convex is necessarily either an open half-plane, or an closed half plane, or an open hals-plane plus an (either open or closed) half-line contained in its boundary. As a consequence, the partition induced by a (non-zero) $\{0,1\}$-valued projective semi-distance $d$ necessarily consists of two classes, and therefore is of the above types I or II.

A $\{0,1\}$-valued projective semi-distance $d$ that is topologically stronger than a non-zero semi-distance $d'$ refines the partition of the $d'$-closures of points. But this easily implies that $d'$ induces the same partition, and that it can only take two values. Thus it is a positive multiple of $d$. In particular, if $d=d_1+d_2$, both $d_1$ and $d_2$ are multiples of $d$, (even if we do not assume they are projective).

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It seems this situation generalizes to dimension $n$, taking as $H$ the analogous union of half-spaces of decreasing dimension, each in the boundary of the previous one. – Pietro Majer Jan 26 at 18:00
    
To complete the description of all indecomposable projective semi-distances, one should prove that they are two-values functions – Pietro Majer Jan 26 at 18:43
    
Let me think about this and I'll get back to you soon. – alvarezpaiva Feb 5 at 17:59

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