Classify hyperbolic Möbius motions

Try

The map T(z)=3z preserves the upper-half-plane metric but moves every interior point. It fixes the ideal boundary points zero and infinity. The translation T(z)=z+1 also preserves the metric and has no interior fixed point, yet has only one ideal fixed point. These are different motion types.

Represent orientation-preserving hyperbolic motions by real matrices with determinant one, acting as T(z)=(az+b)/(cz+d). Matrices M and −M define the same map, so trace squared, rather than signed trace, is the useful invariant.

Predict

Predict how a motion can have infimum displacement zero without fixing any interior point. For T(z)=z+1, move the starting point higher and higher. A Euclidean horizontal shift of one costs less hyperbolic distance there.

Predict whether T(z)=−1/z reverses orientation because of its minus sign. It is holomorphic on H, has determinant-one representative, and preserves orientation. The sign of a formula is not the orientation criterion.

Experiment

Compare canonical examples.

Map	Fixed points in H	Ideal fixed points	Type
−1/z	i	None	Elliptic
z+1	None	Infinity	Parabolic
3z	None	Zero and infinity	Hyperbolic

Compute the fixed-point equation for each. For the first map, z²=−1 has one solution in H. Applying it twice gives the identity, so this is a nontrivial elliptic motion of order two.

Two anchors leave a reflection choice

Predict whether the distances distinguish z=x+iy from −x+iy. The anchors are i,2i and a+i. All displayed numbers are cosh(distance), using the curvature−1 metric.

Point x-2-1012 Height y0.5124 Third anchor a+i1+i: off the anchor geodesici again: degenerate informationReset anchor recovery

Navy dot: supplied point; red square: reflected candidate. The vertical anchor line is a hyperbolic geodesic. The drawing uses coordinate scales, not Euclidean lengths as hyperbolic distances; formulas below determine the distances.

C1=3.00000, C2=2.25000, C3=1.50000. First two recover y=1.00000 and |x|=2.00000. The off-line third anchor recovers x=2.00000.

Third-anchor cosh distance for the reflected point: 5.50000. The third measurement distinguishes the two candidates.

Repair the claim

Fixing two interior points does not force an isometry to be the identity: reflection in their geodesic can still act. An off-geodesic third point removes that freedom. Three points with redundant/collinear information do not supply the same argument.

Forty bounded analytic settings, no fitted data or saved state. These are consistent exact teaching measurements evaluated numerically, not a solver for noisy or arbitrary distance triples. Retain a prediction and proof in the written practice.

Understand

The fixed-point equation is cz²+(d−a)z−b=0. Its discriminant is (d−a)²+4bc=(a+d)²−4, using ad−bc=1. For a nonidentity map, negative discriminant gives a conjugate pair with one interior fixed point, zero discriminant gives one repeated ideal fixed point, and positive discriminant gives two ideal fixed points. The case c=0 includes infinity and must be interpreted projectively. Identity is excluded from the trace-squared-four parabolic label.

For T(z)=kz with k>1, substitute z=x+iy into the distance formula: cosh d(z,kz)=1+(k−1)²(x²+y²)/(2ky²). This is at least (k+1/k)/2=cosh(log k), with equality exactly when x=0. Its translation length is therefore log k, attained along the imaginary axis.

For T(z)=z+1, cosh d(z,z+1)=1+1/(2y²)>1 at every finite height. The infimum displacement is zero as y tends to infinity, but no point attains it.

Apply

Repair “Every isometry without a fixed point translates along a unique axis.” Parabolic motions have no interior fixed point but no minimum-displacement axis. The hyperbolic and parabolic cases differ exactly in this behavior.

Conjugating a motion by another isometry transports its fixed points and displacement structure. To show the Möbius family exhausts the orientation-preserving isometries, we must start with an arbitrary distance-preserving map, not assume its formula. The following argument uses only the already established distance formula and explicit normalization; it requires no theorem asserting that an arbitrary metric isometry is smooth.

Two distances recover height and horizontal magnitude

For z=x+iy with y>0, write C1=cosh d(z,i) and C2=cosh d(z,2i). The distance formula gives

$$C_1=\frac{x^2+y^2+1}{2y},\qquad C_2=\frac{x^2+y^2+4}{4y},\qquad y=\frac3{4C_2-2C_1},\qquad x^2=2yC_1-y^2-1.$$

For genuine point data the denominator is 3/y, hence positive. These two distances leave only the choice x or −x. Reflection J(z)=−conjugate(z) preserves the half-plane distance formula and fixes the imaginary axis pointwise, so this ambiguity is real rather than an algebraic accident.

With C3=cosh d(z,1+i), subtracting the first equation gives C3−C1=(1−2x)/(2y), and thus x=(1−2y(C3−C1))/2. Therefore distances to i,2i,1+i determine every point uniquely. If an isometry fixes these three points, it fixes every point. The third anchor lies off the geodesic through the first two; replacing it by another point on that axis does not break reflection symmetry.

Normalize any two images using explicit real Möbius maps

Let f:H→H preserve every pairwise distance. Its images p=f(i) and q=f(2i) are distinct and satisfy cosh d(p,q)=5/4. Write p=a+ib with b>0. The affine map L(z)=(z−a)/b sends p to i and is represented by the determinant-one real matrix with rows (1/√b,−a/√b),(0,√b). It preserves distance by the previous lesson. Set w=L(q), so cosh d(i,w)=5/4.

Use the algebraic Cayley coordinate C(z)=(z−i)/(z+i). For z in H,

$$|C(z)|^2=\frac{x^2+(y-1)^2}{x^2+(y+1)^2}=\frac{\cosh d(i,z)-1}{\cosh d(i,z)+1}.$$

Hence |C(w)|=1/3 and C(2i)=1/3. The real determinant-one maps

$$R_\theta(z)=\frac{(\cos\theta)z+\sin\theta}{-(\sin\theta)z+\cos\theta}$$

fix i. Direct substitution gives C(Rθ(z))=exp(2iθ)C(z). Choose θ so exp(2iθ)C(w)=1/3; this is possible since C(w) is nonzero with the required modulus. The Cayley map is injective, with inverse i(1+ζ)/(1−ζ), so Rθ(w)=2i. Thus M=R_θ∘L is a known real Möbius isometry making g=M∘f fix i and 2i. No boundary extension or assumed differentiability of f was used.

The third image decides identity or reflection globally

Because g fixes the first two anchors, the recovery formulas show g(1+i) is either 1+i or −1+i. In the first case g fixes all three anchors and is the identity by the third-distance formula. In the second case J∘g fixes all three, so J∘g is the identity and g=J. This step rules out choosing different reflection signs independently at different points; pairwise distances to the third anchor enforce one global choice.

Undo normalization: every distance-preserving map is either M⁻¹ or M⁻¹∘J. Conversely, each real determinant-one Möbius map and its composition with J preserves distances and is onto; the inverse matrix gives the Möbius inverse and J is its own inverse. Even if surjectivity was not initially included in the definition of f, the classification establishes it.

A real determinant-one Möbius map has nonzero complex derivative 1/(cz+d)², so its real Jacobian determinant is positive. J(x,y)=(−x,y) has determinant −1. Thus the first family is orientation-preserving and the second orientation-reversing. Matrices M and −M define the same Möbius map. Conversely two determinant-one representatives of the same map differ by this sign: their quotient fixes every z, and comparing coefficients in cz²+(d−a)z−b=0 for all z gives c=b=0 and d=a=±1. Therefore the orientation-preserving group is precisely PSL(2,R), with the conjugation family adding all orientation-reversing isometries.

This exhaustion theorem does not extend the trace-squared elliptic/parabolic/hyperbolic test to the orientation-reversing family. For example J fixes an entire geodesic, while z↦−2 conjugate(z) is a reflection combined with translation along that geodesic and has no interior fixed point. State which family is being classified before using the trace test.

Primary reference: Albert Chang, Isometries of the Hyperbolic Plane, August 2010. The explicit three-distance recovery above supplies an elementary exhaustion proof within the current half-plane metric prerequisites. It does not prove every synthetic neutral-geometry axiom or global quotient theorem.

Continue

Record the determinant normalization, identity exception, interior versus ideal fixed points, and whether infimum displacement is attained. Continue to hyperbolic triangles to derive trigonometric identities from the same model geometry.