Collineations of the unit disk

Question

I've been reading a proof that all models of Lobachevskian geometry are isomorphic. As part of the proof, the following statement was made, but no proof or reference was given. (The proof can actually be fixed relatively easily so that it doesn't rely on the statement, but I find the statement interesting in its own right.)

Any bijection of the disk $x^2 + y^2 < 1$ onto itself that maps chords to chords is projective.

This can be restated as follows, viewing the disk as the Klein model of the hyperbolic plane.

Any bijection of the hyperbolic plane onto itself that preserves alignment is an isometry.

The first statement refers of course to the ordinary Euclidean unit disk. A projective mapping means any mapping of the form $x' = (ax + by + c)/(gx + hy + i), y' = (dx + ey + f)/(gx + hy + i)$. If the unit disk is replaced with the whole plane, then the statement is equivalent to the so-called "fundamental theorem of affine geometry."

Without loss of generality, we may assume that the origin is fixed (since it can be shown that projective mappings of the required kind exist that take any point to any point) and that the line $y = 0$ is invariant (applying a rotation). In that case, the problem is to show that only four mappings are possible, namely, the identity mapping, reflections with respect to the coordinate axes, and central symmetry with respect to the origin.

Really, I'd be happier with a reference than with a proof given here, particularly if the proof is long. A reference in another European language would probably be okay.

Answer 1

We can extend any bijection of the disk that maps chord to chords into a unique collineation of the projective plane. The main step is to prove that the collineations of the projective plane are exactly the projectivities, this is the Fundamental theorem of projective geometry. (For a nice proof see e. g. Rey Casse: Projective Geometry - An Introduction, Theorem 4.27.) It means that the collineations of the Cayley-Klein plane are exactly the automorphisms of the unit disks, i.e. the projective collineations of the projective plane which leave the unit circle invariant.

Answer 2

A proof in outline of this fact can be found in Greenberg's Euclidean and Non-Euclidean Geometries, page 385, Exercise K-21. Greenberg attributes the result to Jenks.

This theorem is interesting because it shows that, in principle, hyperbolic geometry can be based on axioms that mention only incidence (the "belongs" relation between points and lines), since all other concepts of hyperbolic geometry, including order and congruence, can be defined only in terms of incidence. This sets hyperbolic geometry apart from Euclidean geometry, in which we can't do without order and congruence.

I will write a proof here that is adapted from the one outlined by Greenberg.

Let $G$ be the set of bijective mappings of the open disk onto itself such that the image of a chord is a chord. Then $G$ is easily seen to be a group. (If $f \in G$, then $f^{-1} \in G$, since the chord through $A$ and $B$ is the image of the chord through $f^{-1}(A)$ and $f^{-1}(B)$.)

Each element of $G$ induces a permutation of the set of chords, hence an automorphism of what can be called the "incidence structure" of the disk and its chords. The proof will now largely consist of arguing that various concepts are capable of being defined solely in terms of the incidence relation between points and chords, and must therefore be preserved by elements of $G$.

Let $f \in G$. If two chords have a common endpoint, then so too do their images under $f$. For given any two chords $a$ and $b$, they have a common endpoint if and only if there exists a third chord $c$ such that: no two of $a, b, c$ intersect; and given any point $D \in c$, there is a unique chord $d$ through $D$ that meets neither $a$ nor $b$. (Namely, if $a = AB$ and $b = AB'$, then $c = BB'$, and $d = AD$.)

It follows from the above that the family of all chords with a fixed endpoint $A$ on the circle is carried to the family of all chords with some fixed endpoint $A'$ on the circle. Therefore there exists a unique extension of $f$ to the boundary circle in such a way that any chord $AB$ is carried by $f$ to the chord $f(A) f(B)$.

Let $f \in G$. Let $a$ be a chord, and let $P, Q, R$ be points on $a$. If $Q$ is between $P$ and $R$, then $f(Q)$ is between $f(P)$ and $f(R)$. For $Q$ is between $P$ and $R$ if and only if, for any other two chords $b$ and $c$ passing through $P$ and $R$, respectively, and having a common endpoint, there is exactly one chord passing through $Q$ and not intersecting $b$ or $c$ (namely, the chord through $Q$ and the common endpoint of $b$ and $c$).

Let $f \in G$. Let $a = AB$ and $a' = A'B'$ be chords, and let $X, Y \in a, X', Y' \in a'$. If the cross-ratios $(A, B; X, Y)$ and $(A', B'; X', Y')$ are equal, then so too are the cross-ratios $(f(A), f(B); f(X), f(Y))$ and $(f(A'), f(B'); f(X'), f(Y'))$.

We first consider the special case in which $A = A'$. The general case will follow from this by letting $a'' = AA'$, selecting suitable points on $X'', Y'' \in a''$, and applying the special case to the pairs $a, a'',$ and $a'', a'$. Now we prove the claim. The points $A, X, Y, B, B', X'$ being given, there is a unique point $Y'$ on chord $AB'$ with the property that $(A,B;X,Y) = (A',B';X',Y')$, and it is characterized by the fact that it belongs to a configuration like the one shown in the figure. Projecting centrally first from $B'$ and then from $B$, we have $(A,B;X,Y) = (A'',B'';X'',Y'') = (A',B';X',Y')$.

We define the Klein distance between any points $X$ and $Y$ in the disk as $d(X,Y) = \frac{1}{2} |\log(A,B;X,Y)|$, where $X, Y$ are on chord $A,B$. (This doesn't depend on the order of $A$ and $B$.) Then the result above says that $d(X,Y) = d(X',Y')$ implies $d(f(X),f(Y))$ = $d(f(X'),f(Y'))$.

Now let $f \in G$ be arbitrary. We will show that $f$ is projective. Let $O$ be the origin. Then by composing $f$ if necessary with a rotation, we may assume that $f(O) = (\alpha,0)$, with $0 \leq \alpha < 1$. Now composing with $h_{\alpha}^{-1} \in G$, where $h_{\alpha}(x,y) = ((x+\alpha)/(1 + \alpha x), y\sqrt{1-\alpha^2}/(1 + \alpha x))$, we may assume $f(O) = O$. Composing further with a rotation, we may assume that $f$ leaves the diameter $y = 0$ invariant.

Now consider the regular octagonal configuration shown below, in which all sixteen segments have equal Klein lengths. It consists of eight adjacent equilateral triangles (for the Klein distance). There is exactly one such configuration, if we require $O$ to be at the origin, $X_0$ on the positive $x$-axis, and $X_2$ in the upper half-disk. (Namely, take $OX_i = \sqrt{2}\sqrt{\sqrt{2} - 1}$, and $(OX_i,OX_{i+1}) = \pi/4$. We have $(J,I;O,X_0) = (A,B;X_0,X_1).)$

Therefore $f$ must leave the entire configuration invariant. After applying reflections, if necessary, through the $x$- or $y$-axes, we may assume that $f(X_0) = X_0$, $f(X_2) = X_2$. Then $f(X_i) = X_i$ for all $i$. The problem is to prove that $f$ is the identity mapping.

Now, as a metric space for the Klein distance, the chord $a = (-1,1) \times \{0\}$ is isometric to $\mathbf{R}$ with its ordinary distance, via the mapping $h \colon (x,0) \mapsto \operatorname{arctanh} x$. The mapping $f$ sends $a$ into itself. When we transport the mapping $f$ to $\mathbf{R}$ via the bijection $h$, it has the following properties: $f$ preserves midpoints, i.e., $f((\alpha + \beta)/2 = (f(\alpha) + f(\beta))/2$; $f$ preserves "betweenness," i.e., if $\alpha < \beta < \gamma$, then either $f(\alpha) < f(\beta) < f(\gamma)$ or $f(\gamma) < f(\beta) < f(\alpha)$; $f(0) = 0$; and $f(\alpha_0) = \alpha_0$ for some $\alpha_0 > 0$ (namely, $\alpha_0 = h(X_0)$). Together, these properties imply that $f$ leaves the $x$-axis pointwise invariant.

Similarly, considering $f(X_2) = X_2$, we see that $f$ leaves the $y$-axis pointwise invariant. It follows that any chord in the disk that intersects both axes is left invariant overall by $f$. But any point in the disk can be written as the intersection of two chords of this kind. Therefore all points are left fixed by $f$.

Addendum The above proof clearly relies on the Archimedean property of the field of coordinates. However, the idea in Sz_Z's answer of extending the mapping to a collineation of the whole projective plane, and then applying the fundamental theorem of projective geometry, allows an extension to the non-Archimedean case. Specifically, if the field of coordinates $F$ is any ordered field in which every positive element has a square root, then every collineation of the open disk extends to a collineation of the projective plane, hence a projective semilinear automorphism. This is the same as a projective mapping, but with an automorphism of $F$ then applied to the coordinates of points.

Given any $f \in G$, we can extend it to a bijection of the projective plane onto itself as follows. If $P$ is on the boundary circle, then define $f(P)$ as in the proof above. If $P$ is outside the circle, then identify chords with the lines on which they lie, and define $f(P)$ as the pole of the chord which is the image under $f$ of the polar of $P$, poles and polars being taken relative to the unit circle.

The problem is now to show that the extension of $f$ still preserves alignment. This boils down to showing that if two chords $a$ and $b$ have the property that one passes through the pole of the other, then their images $a'$ and $b'$ under $f$ have the same property. It turns out that this property of the chords is equivalent to saying that they represent perpendicular lines in the Klein model of hyperbolic geometry.

Thus, we need only show that $f$ preserves orthogonality in the sense of the Klein model. But this is an easy consequence of the fact, proved above, that $f$ preserves the congruence relation between segments. Namely, two lines are perpendicular if and only if they have the property that if $O$ is their point of intersection, $A$ and $A'$ lie on the first line equidistant (with respect to the Klein distance) from $O$, and $B$ and $B'$ lie on the second line equidistant from $O$, then the diagonal segments $AB$, $A'B$, $AB'$, $A'B'$ must be congruent. (This fact can easily be proved in the Klein model by moving the point of intersection to $O$ via a projective mapping that belongs to $G$. There is no loss of generality because the projective mapping preserves the relationship of pole to polar and it preserves congruence. The first line can then further be moved to coincide with the $x$-axis. Now it suffices to show that the $x$- and $y$-axes have the property, but not the $x$-axis and any inclined diameter of the circle.)

Naturally, logarithms can no longer be taken in $F$, but the congruence relation can be defined via the cross-ratio, as above.