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Geometry

Hilbert geometry

A bounded convex region Ω\Omega becomes a metric space once you define dΩd_\Omega by the cross-ratio of where a line exits its boundary — David Hilbert's 1895 construction, which reduces to the classical Klein model of hyperbolic geometry exactly when Ω\Omega is an ellipse, and to a genuinely new non-Riemannian geometry otherwise.

IntuitionDistances that grow as you approach the edge

Take any bounded convex region in the plane — a disk, an ellipse, a square, a hexagon, anything that does not stretch out to infinity and has no dents. Call it Ω\Omega. Pick two points xx and yy inside Ω\Omega, draw the straight line through them, and let it exit the region at two boundary points aa and bb, so that the four points sit in the order a,x,y,ba, x, y, b along the line. Hilbert's trick is to measure the distance between xx and yy not by a ruler, but by how the two segments ayay and bxbx compare to the two segments axax and byby — a ratio of ratios called a cross-ratio. Concretely, inside the unit disk Ω={x2+y2<1}\Omega = \{x^2+y^2<1\}, take the center x=(0,0)x=(0,0) and a point y=(0.6,0)y=(0.6,0) on the horizontal radius. The line through them meets the circle at a=(−1,0)a=(-1,0) and b=(1,0)b=(1,0). Measuring along that line, ∣ay∣=1.6|ay|=1.6, ∣bx∣=1|bx|=1, ∣ax∣=1|ax|=1, ∣by∣=0.4|by|=0.4, so the cross-ratio is 1.6×11×0.4=4\frac{1.6\times 1}{1\times 0.4}=4, and Hilbert's distance is 12ln⁡4=ln⁡2≈0.693\tfrac12\ln 4=\ln 2\approx 0.693. Now slide yy outward, say to (0.9,0)(0.9,0): ∣ay∣=1.9|ay|=1.9, ∣by∣=0.1|by|=0.1, and the cross-ratio jumps to 1.9×11×0.1=19\frac{1.9\times1}{1\times0.1}=19, giving distance 12ln⁡19≈1.47\tfrac12\ln 19\approx 1.47 — more than double, even though yy only moved a little in ordinary Euclidean terms. The closer a point sits to the boundary of Ω\Omega, the farther away Hilbert's metric pushes it, exactly the behavior that makes the classical Klein model of hyperbolic geometry work: in that model, the boundary circle represents points 'at infinity', infinitely far from the center.

An interactive 3D view of a solid cube that can be rotated and whose faces can be exploded outward, used here as a generic example of a bounded convex body in space.
A solid cube: an example of a bounded convex body in space. Note that Hilbert's construction itself is usually pictured in 2D (a flat convex region such as a disk or polygon) — this 3D cube is not literally a Hilbert-geometry picture, but illustrates the general notion of a bounded convex domain that the construction generalizes to in any dimension.

UndergraduateHilbert's construction: distance from a cross-ratio

Definition: The Hilbert metric

Let Ω\Omega be a bounded, open, convex subset of Rn\mathbb{R}^n (equivalently, an affine chart of a properly convex open set in real projective space RPn\mathbb{RP}^n). For distinct points x,y∈Ωx, y \in \Omega, let the line through xx and yy meet the boundary ∂Ω\partial\Omega at two points aa and bb, labelled so the four points occur in the order a,x,y,ba, x, y, b along the line. David Hilbert defined, in a short 1895 note written as a letter to Felix Klein, the Hilbert distance dΩ(x,y)=12ln⁡[a,x,y,b]=12ln⁡∣ay∣ ∣bx∣∣ax∣ ∣by∣,d_\Omega(x,y) = \frac12 \ln[a,x,y,b] = \frac12 \ln\frac{|ay|\,|bx|}{|ax|\,|by|}, together with dΩ(x,x)=0d_\Omega(x,x)=0, where [a,x,y,b][a,x,y,b] denotes the cross-ratio of the four collinear points and ∣⋅∣|\cdot| denotes ordinary Euclidean length along that line (any affine parametrization gives the same ratio, since a cross-ratio is unchanged by reparametrizing the line).

dΩ(x,y)=12ln⁡[a,x,y,b]=12ln⁡∣ay∣ ∣bx∣∣ax∣ ∣by∣d_\Omega(x,y) = \frac12 \ln[a,x,y,b] = \frac12 \ln\frac{|ay|\,|bx|}{|ax|\,|by|}

For Ω\Omega as above, dΩd_\Omega is a metric on Ω\Omega: it is symmetric (dΩ(x,y)=dΩ(y,x)d_\Omega(x,y)=d_\Omega(y,x)), non-negative, and dΩ(x,y)=0d_\Omega(x,y)=0 if and only if x=yx=y. Moreover, for any x,z∈Ωx,z\in\Omega and any point yy on the straight segment [x,z][x,z], dΩ(x,y)+dΩ(y,z)=dΩ(x,z)d_\Omega(x,y)+d_\Omega(y,z)=d_\Omega(x,z); consequently the ordinary Euclidean straight-line segment [x,z][x,z] (traversed at the appropriate non-uniform speed) is a geodesic of (Ω,dΩ)(\Omega,d_\Omega), and dΩd_\Omega satisfies the triangle inequality with equality exactly along such segments.

Why is it true?

The formula dΩ(x,y)=12ln⁡[a,x,y,b]d_\Omega(x,y) = \frac12\ln[a,x,y,b] is symmetric under swapping the roles of xx and yy together with aa and bb, because relabelling turns ∣ay∣∣bx∣∣ax∣∣by∣\frac{|ay||bx|}{|ax||by|} into itself; and because a<x<y<ba<x<y<b forces ∣ay∣>∣ax∣|ay|>|ax| and ∣bx∣>∣by∣|bx|>|by|, the cross-ratio is a product of two numbers each greater than 11, so it exceeds 11 exactly when x≠yx\neq y, making dΩd_\Omega strictly positive off the diagonal. The additivity along a fixed line is really a statement about ordinary numbers on the real line: cross-ratios computed from the same pair of boundary points a,ba,b telescope multiplicatively as a third collinear point is inserted, so a logarithm turns that telescoping product into a sum — the same phenomenon you can check numerically on any three points of a segment.

Proof

Symmetry and positivity. Write [a,x,y,b]=∣ay∣∣ax∣⋅∣bx∣∣by∣[a,x,y,b]=\frac{|ay|}{|ax|}\cdot\frac{|bx|}{|by|}. Since a<x<y<ba<x<y<b on the line, ∣ay∣=∣ax∣+∣xy∣>∣ax∣|ay|=|ax|+|xy|>|ax| and ∣bx∣=∣by∣+∣xy∣>∣by∣|bx|=|by|+|xy|>|by|, so both factors exceed 11 when x≠yx\ne y, hence [a,x,y,b]>1[a,x,y,b]>1 and dΩ(x,y)>0d_\Omega(x,y)>0; when x=yx=y both factors equal 11. Relabelling (x,y,a,b)↦(y,x,b,a)(x,y,a,b)\mapsto(y,x,b,a) sends ∣ay∣∣bx∣∣ax∣∣by∣\frac{|ay||bx|}{|ax||by|} to ∣bx∣∣ay∣∣by∣∣ax∣\frac{|bx||ay|}{|by||ax|}, the identical number, so dΩ(x,y)=dΩ(y,x)d_\Omega(x,y)=d_\Omega(y,x).

Additivity on a line. Let x,y,zx,y,z be collinear with yy between xx and zz, and let a,ba,b be the (common) boundary intersections of that line. Writing all four one-dimensional distances along the line, [a,x,y,b]⋅[a,y,z,b]=∣ay∣∣ax∣∣bx∣∣by∣⋅∣az∣∣ay∣∣by∣∣bz∣=∣az∣∣ax∣∣bx∣∣bz∣=[a,x,z,b],[a,x,y,b]\cdot[a,y,z,b]=\frac{|ay|}{|ax|}\frac{|bx|}{|by|}\cdot\frac{|az|}{|ay|}\frac{|by|}{|bz|}=\frac{|az|}{|ax|}\frac{|bx|}{|bz|}=[a,x,z,b], since the factors ∣ay∣|ay| and ∣by∣|by| cancel. Taking 12ln⁡\frac12\ln of both sides gives dΩ(x,y)+dΩ(y,z)=dΩ(x,z)d_\Omega(x,y)+d_\Omega(y,z)=d_\Omega(x,z), exactly the equality case of the triangle inequality, so the straight segment [x,z][x,z] realizes the Hilbert distance and is a geodesic.

Triangle inequality off a line. For x,y,zx,y,z not collinear, convexity of Ω\Omega is what is needed: projecting the pair (x,z)(x,z) through yy only ever enlarges the relevant cross-ratio compared with going straight from xx to zz, because the boundary points seen from yy lie no closer to xx or zz than the boundary points seen directly along xzxz — this monotonicity, proved carefully using the convexity of Ω\Omega, gives dΩ(x,z)≤dΩ(x,y)+dΩ(y,z)d_\Omega(x,z)\le d_\Omega(x,y)+d_\Omega(y,z) in general (see Hilbert 1895; a full derivation is given in the Handbook of Hilbert Geometry, Ch. 1).

Example: Computing a Hilbert distance from the definition

Let Ω\Omega be the open unit disk {x2+y2<1}\{x^2+y^2<1\}, and let x=(−0.5,0)x=(-0.5,0) and y=(0.5,0)y=(0.5,0). Compute dΩ(x,y)d_\Omega(x,y) directly from the definition dΩ(x,y)=12ln⁡[a,x,y,b]d_\Omega(x,y)=\frac12\ln[a,x,y,b].

Solution

The horizontal line through xx and yy meets the unit circle at a=(−1,0)a=(-1,0) and b=(1,0)b=(1,0), in the order a,x,y,ba,x,y,b. Along that line, ∣ay∣=1.5|ay|=1.5, ∣bx∣=1.5|bx|=1.5, ∣ax∣=0.5|ax|=0.5, ∣by∣=0.5|by|=0.5, so [a,x,y,b]=1.5×1.50.5×0.5=2.250.25=9[a,x,y,b]=\frac{1.5\times1.5}{0.5\times0.5}=\frac{2.25}{0.25}=9. Hence dΩ(x,y)=12ln⁡9=ln⁡3≈1.099d_\Omega(x,y)=\frac12\ln 9=\ln 3\approx 1.099. (Note the pleasant coincidence that ∣ay∣=∣bx∣|ay|=|bx| here, because xx and yy are symmetric about the center — that symmetry is what makes the cross-ratio a perfect square, 9=329=3^2.)

An interactive ellipse $x^2/a^2+y^2/b^2=1$ with adjustable semi-axes $a$ and $b$, shown as the convex domain on which the Hilbert metric coincides exactly with the Klein model of hyperbolic geometry.
The special ellipse case x2/a2+y2/b2=1x^2/a^2+y^2/b^2=1: here the Hilbert metric dΩd_\Omega is not merely analogous to hyperbolic geometry, it is the classical Klein (projective) model of the hyperbolic plane, transported onto this particular ellipse by a linear change of coordinates.

Let B={x12+⋯+xn2<1}\mathbb{B}=\{x_1^2+\dots+x_n^2<1\} be the open unit ball. Then (B,dB)(\mathbb{B},d_{\mathbb{B}}) is isometric to nn-dimensional hyperbolic space of constant curvature −1-1, via the classical Beltrami–Klein model. More generally, if E⊂RnE\subset\mathbb{R}^n is any ellipsoid (the image of B\mathbb{B} under an invertible affine map TT), then (E,dE)(E,d_E) is isometric to (B,dB)(\mathbb{B},d_{\mathbb{B}}) via TT, hence again to hyperbolic space; no other choice of bounded convex Ω\Omega (not affinely equivalent to an ellipsoid) gives a Hilbert geometry isometric to a constant-curvature space.

Why is it true?

Hilbert's cross-ratio construction is literally the same one Arthur Cayley and Felix Klein used decades earlier to build the projective model of hyperbolic geometry inside a conic — Hilbert's 1895 letter to Klein pointed out that convexity of Ω\Omega, not the special quadratic shape of a conic, is what makes the cross-ratio formula define a metric at all. Specializing his general construction back to Ω=B\Omega=\mathbb{B} therefore reproduces Cayley and Klein's construction exactly, which is precisely why an ellipse is the one shape for which Hilbert geometry is not merely 'hyperbolic-like' but literally hyperbolic.

Proof

Along a diameter. Take x=0x=0 and y=(r,0,…,0)y=(r,0,\dots,0) with 0<r<10<r<1. The line through them meets ∂B\partial\mathbb{B} at a=(−1,0,…,0)a=(-1,0,\dots,0) and b=(1,0,…,0)b=(1,0,\dots,0), so ∣ay∣=1+r|ay|=1+r, ∣ax∣=1|ax|=1, ∣bx∣=1|bx|=1, ∣by∣=1−r|by|=1-r, giving dB(0,y)=12ln⁡1+r1−r=artanh⁡(r)d_{\mathbb{B}}(0,y)=\frac12\ln\frac{1+r}{1-r}=\operatorname{artanh}(r). This matches exactly the standard formula for hyperbolic distance from the center to a point at Euclidean radius rr in the Beltrami–Klein model of curvature −1-1.

Off a diameter. For general x,y∈Bx,y\in\mathbb{B}, choose a rotation R∈O(n)R\in O(n) (an isometry of the round ball fixing 00) carrying the line through x,yx,y to a coordinate axis; since RR preserves B\mathbb{B} and maps the four points a,x,y,ba,x,y,b to another quadruple in the same cross-ratio, dB(x,y)=dB(Rx,Ry)d_{\mathbb{B}}(x,y)=d_{\mathbb{B}}(Rx,Ry), reducing to the diametral case worked out above after also translating one point to 00 via a hyperbolic isometry of B\mathbb{B} (a Möbius transformation preserving B\mathbb{B}, which likewise preserves cross-ratios). Since the formula obtained this way is exactly the Beltrami–Klein distance formula, (B,dB)(\mathbb{B},d_{\mathbb{B}}) is isometric to hyperbolic nn-space.

General ellipsoids. Let T:Rn→RnT:\mathbb{R}^n\to\mathbb{R}^n be an invertible linear (or affine) map with T(B)=ET(\mathbb{B})=E. An affine map is a projective transformation of RPn\mathbb{RP}^n that fixes the hyperplane at infinity, and cross-ratios of collinear points are invariant under every projective transformation; since TT sends the line through x,y∈Bx,y\in\mathbb{B} to the line through Tx,Ty∈ETx,Ty\in E, and sends the boundary pair a,ba,b to the boundary pair Ta,TbTa,Tb of EE, it sends the defining quadruple to the defining quadruple, so [a,x,y,b]=[Ta,Tx,Ty,Tb][a,x,y,b]=[Ta,Tx,Ty,Tb] and hence dE(Tx,Ty)=dB(x,y)d_E(Tx,Ty)=d_{\mathbb{B}}(x,y) for all x,yx,y. Thus T:(B,dB)→(E,dE)T:(\mathbb{B},d_{\mathbb{B}})\to(E,d_E) is an isometry, so (E,dE)(E,d_E) is again hyperbolic nn-space.

Example: Affine invariance: the same distance, transported onto an ellipse

Let E={x12/4+x22/1.44<1}E=\{x_1^2/4+x_2^2/1.44<1\} be the ellipse with semi-axes a=2a=2, b=1.2b=1.2, the image of the unit disk B\mathbb{B} under the linear map T(x1,x2)=(2x1,1.2x2)T(x_1,x_2)=(2x_1,1.2x_2). Using the disk computation from the intuition section (x=(0,0)x=(0,0), y=(0.6,0)y=(0.6,0), giving dB(x,y)=ln⁡2d_{\mathbb{B}}(x,y)=\ln 2), compute dE(Tx,Ty)d_E(Tx,Ty) directly from the definition on EE, and check it against Theorem 'The ellipse case is exactly the Klein model of hyperbolic geometry'.

Solution

TT sends x=(0,0)x=(0,0) to x′=(0,0)x'=(0,0) and y=(0.6,0)y=(0.6,0) to y′=(1.2,0)y'=(1.2,0). The horizontal line through x′,y′x',y' meets ∂E\partial E where x12/4=1x_1^2/4=1, i.e. at a′=(−2,0)a'=(-2,0) and b′=(2,0)b'=(2,0). Along that line, ∣a′y′∣=1.2+2=3.2|a'y'|=1.2+2=3.2, ∣b′x′∣=2−0=2|b'x'|=2-0=2, ∣a′x′∣=0−(−2)=2|a'x'|=0-(-2)=2, ∣b′y′∣=2−1.2=0.8|b'y'|=2-1.2=0.8, so [a′,x′,y′,b′]=3.2×22×0.8=6.41.6=4[a',x',y',b']=\frac{3.2\times 2}{2\times 0.8}=\frac{6.4}{1.6}=4, giving dE(x′,y′)=12ln⁡4=ln⁡2d_E(x',y')=\frac12\ln 4=\ln 2. This is exactly dB(x,y)=ln⁡2d_{\mathbb{B}}(x,y)=\ln 2, confirming the theorem's claim that the linear map TT is an isometry from (B,dB)(\mathbb{B},d_{\mathbb{B}}) onto (E,dE)(E,d_E): no new computation was really needed, because TT is affine and the cross-ratio never notices affine reparametrizations of the line.

AdvancedBeyond the ellipse: Hilbert's fourth problem and Finsler geometry

Hilbert's 1895 letter to Klein was not an isolated curiosity: it fed directly into Problem IV on his famous list of 23 problems presented at the 1900 International Congress of Mathematicians in Paris, which asks for a characterization of all metrics on a region of real projective space whose geodesics are exactly the ordinary straight line segments. By Theorem 'The Hilbert distance is a genuine metric, with straight segments as geodesics' above, every Hilbert geometry built from a convex domain Ω\Omega is automatically an example. Georg Hamel solved the smooth ('regular') case in 1901, showing that every sufficiently smooth such metric is locally a Minkowski (translation-invariant, norm-induced) metric; the problem without any smoothness assumption was substantially resolved through Herbert Busemann's integral-geometric approach — representing admissible metrics via measures on the space of hyperplanes, in the spirit of the Crofton formula — and Aleksei Pogorelov's 1973 general solution built on it, though the two-dimensional case has continued to attract refinements since. A single fact governs exactly how 'curved' a Hilbert geometry (Ω,dΩ)(\Omega,d_\Omega) can be: dΩd_\Omega comes from a genuine Riemannian metric of constant curvature −1-1 precisely when Ω\Omega is an ellipsoid (the theorem above); for every other bounded convex Ω\Omega, dΩd_\Omega is still a perfectly good metric with straight-line geodesics, but it arises from a non-Riemannian Finsler structure — the 'unit ball' of directions at each point is a rescaled copy of Ω\Omega itself rather than a round ellipsoid, so lengths depend on direction in a way no Riemannian metric can reproduce. Infinitesimally, if t+(x,v)t_+(x,v) and t−(x,v)t_-(x,v) denote how far one can travel from x∈Ωx\in\Omega along +v+v and −v-v before exiting Ω\Omega, the Hilbert distance is generated by the Finsler norm below. When Ω\Omega is centrally symmetric (Ω=−Ω\Omega=-\Omega, e.g. a square or a regular hexagon centered at the origin), this Finsler structure becomes translation-invariant and reduces, near the center, to the geometry of a normed vector space — a second classical family of symmetric convex domains alongside the ellipsoids, genuinely Riemannian only when that norm happens to be Euclidean.

FΩ(x,v)=12 ∥v∥(1t+(x,v)+1t−(x,v))F_\Omega(x,v) = \frac12\,\|v\|\left(\frac{1}{t_+(x,v)}+\frac{1}{t_-(x,v)}\right)

AdvancedBridges forward: convex projective structures and their limits

Once a compact manifold MM is presented as Ω/Γ\Omega/\Gamma for a properly convex domain Ω⊂RPn\Omega\subset\mathbb{RP}^n and a discrete group Γ\Gamma of projective transformations acting freely and cocompactly on Ω\Omega, it carries a convex projective structure, and the Hilbert metric dΩd_\Omega descends to a genuine (Finsler, generally non-Riemannian) Riemannian-like metric on MM itself. Studying how such structures deform — their curvature, their geodesic flows, their moduli — pulls in the tools of smooth differential geometry (Higgs bundles, harmonic maps, connections) even though the underlying metric is only Finsler; this is one path from Hilbert geometry towards the broader landscape of differential geometry on manifolds. In the opposite direction, letting a strictly convex domain Ωt\Omega_t degenerate — its boundary flattening toward a polytope as a parameter tt moves to an extreme — makes the Hilbert metric increasingly polyhedral: distances become governed by which facet of the limiting polytope a geodesic runs closest to, in a 'max/min of linear functions' way that echoes, as an analogy rather than a literal identity, how tropical geometry replaces curved algebraic varieties by piecewise-linear polyhedral complexes under logarithmic degeneration. Both directions matter for later chapters of this library: the differential-geometric direction towards higher Teichmüller theory and Anosov representations, and the degenerating, piecewise-linear direction towards tropical geometry.

ResearchConvex divisible domains and higher-rank rigidity

In the unit disk Ω={x2+y2<1}\Omega=\{x^2+y^2<1\}, let x=(0,0)x=(0,0) and y=(0.6,0)y=(0.6,0). The line through them meets the circle at a=(−1,0)a=(-1,0) and b=(1,0)b=(1,0), giving ∣ay∣=1.6|ay|=1.6, ∣bx∣=1|bx|=1, ∣ax∣=1|ax|=1, ∣by∣=0.4|by|=0.4. What is dΩ(x,y)d_\Omega(x,y)?

For which shape of bounded convex domain Ω\Omega does the Hilbert metric dΩd_\Omega coincide exactly with the classical Klein (Beltrami–Klein) model of hyperbolic geometry?

Which statement correctly describes the Hilbert metric dΩd_\Omega on a general bounded convex domain Ω\Omega that is not an ellipsoid (say, a square or a triangle)?

Hilbert's fourth problem (1900) asks for a characterization of all metrics on a region of real projective space whose geodesics are exactly the ordinary ___.

References

  1. Athanase Papadopoulos, Marc Troyanov (eds.) (2014). Handbook of Hilbert Geometry · DOI:10.4171/147
  2. Athanase Papadopoulos, Marc Troyanov (2014). From Funk to Hilbert Geometry · arXiv:1406.6983
  3. David Hilbert (1895). Über die gerade Linie als kürzeste Verbindung zweier Punkte · DOI:10.1007/BF02096204