A geometric model and proof portfolio

Try

In the upper half-plane, take the complete geodesic x=0 and point P=1+i. The vertical geodesic x=1 passes through P and never meets x=0. So does the semicircle centered at 2 on the real axis with radius √2: P is on it and every point of the semicircle has x≥2−√2>0.

This is an exact demonstration of more than one nonintersecting geodesic through an external point. It is stronger than a viewport showing no crossing, but narrower than a complete verification of every neutral axiom.

Preparation: recall the upper-half-plane metric and complete geodesics, complex-number moduli and the distinction between a hypothesis and a conclusion. Here H means all (x,y) with y positive. Its real boundary is not part of H. We will turn the example into a proof for every external point, then transport the proof to the disk and audit its logical scope.

Predict

Predict whether the Poincaré disk and upper-half-plane pictures are different geometries. The Cayley transform carries their specified metrics into one another, so they represent the same hyperbolic geometry.

Predict whether a projective picture with straight chords makes that geometry Euclidean. Straightness in a coordinate chart is not a metric invariant. The Klein model uses straight chords with a different length and angle structure.

Experiment

Build a comparison that keeps each structure visible.

Geometry	Points and lines	Global incidence/order warning	Triangle metric
Euclidean	R² and affine lines	Ordered neutral plane with Playfair	Sum π
Hyperbolic	H and its geodesics	Multiple parallels; ideal boundary excluded	Sum below π
Spherical	Sphere and great circles	Antipodal joins nonunique; closed lines	Positive excess
Elliptic	Antipodal classes and projective lines	Unique joins but no global neutral line order	Locally spherical

Annotate the angle statements with the exact theorem and triangle-region assumptions used. The table is an index of arguments, not a replacement for them.

Does the complete geodesic meet x = 0?

Predict from the equation before checking. The target line is x = 0, y > 0. The dashed bottom edge is excluded from the geometry.

External pointP = (1, 1)P = (−2, 1) Circle center c on the real axis0123

Circle equation: (x − (1))² + y² = 1. Only its upper semicircle belongs to the model.

Your predictionChoose a conclusionMeets x = 0 at positive heightOnly shares an excluded ideal endpointDisjoint, with no shared ideal endpoint Check geometric claim Reset comparison

Temporary practice only: predictions are not saved or graded as proof. Record the equation and boundary explanation in your portfolio.

Start at P=(1,1), c=1. Predict whether the shared endpoint (0,0) is an intersection in H. Then change c to zero and identify the actual positive intersection height. Finally switch to the point on the left of the target line: which way must the inequality reverse? Check the complete line, not only its visible arc.

Understand

A model-based independence argument has three separate steps. First specify the common axiom system, including the intended order, congruence and continuity requirements. Second verify that both the Euclidean and hyperbolic structures satisfy every one of those axioms. Third demonstrate opposite truth values for the proposed parallel assertion. Soundness then implies that neither assertion follows from the common axioms, relative to the background mathematics supporting the models.

This course has supplied explicit incidence arguments, order calculations, Euclidean isometry classification, construction obstructions and hyperbolic metric mechanisms. Full synthetic representation and model-verification theorems are identified dependencies, not silently replaced by finite calculations. The double-parallel example establishes the relevant metric-model behavior directly, while a complete formal independence theorem requires the full axiom verification.

For transformation comparisons, specify the map and the structure it preserves. Cayley preserves hyperbolic length when the disk metric is carried along. An arbitrary projective map preserves incidence but need not preserve an elliptic or hyperbolic metric. A Euclidean similarity preserves angles and ratios while changing lengths.

All geodesics through an external point

Fix P=(a,b), where b>0 and a≠0, and target line l given by x=0 in H. The geodesic classification gives exactly two forms through P: the vertical line x=a, and semicircles centered at (c,0) with radius r determined by r²=(a−c)²+b². The vertical line is disjoint from l. For any finite center c, intersection with l requires

$$y^2=r^2-c^2=a^2+b^2-2ac.$$

If this number is positive there is exactly one intersection in H, at its positive square root. If it is zero, the only Euclidean intersection is (0,0), excluded from H. If it is negative, there is no real intersection. Thus the semicircle is disjoint from l if and only if 2ac≥a²+b². This covers the whole curve, irrespective of clipping in a diagram.

For a>0, the allowed centers satisfy c≥(a²+b²)/(2a). For a<0, they satisfy c≤(a²+b²)/(2a). Equality gives a shared ideal endpoint, often called a limiting parallel; strict inequality gives no shared ideal endpoint, called ultraparallel. We use “disjoint” for both, avoiding ambiguity about the word parallel. The vertical line shares the ideal endpoint at infinity with l and must also be counted. Infinitely many distinct finite centers satisfy the relevant half-line inequality, so infinitely many disjoint geodesics pass through P.

Worked example: P=(−2,1) gives threshold −5/4. At c=−2, r=1 and the entire arc lies between x=−3 and x=−1. At c=−5/4, r=5/4 and it reaches x=0 only at height zero. At c=−1, r²=2 and setting x=0 gives y²=1: the geodesic intersects l at (0,1). Reusing the right-side inequality without reversing it would reverse the intersecting and strictly disjoint cases; equality still requires a separate boundary check.

Transport the certificate, not just the picture

Let C(z)=(z−i)/(z+i), with z=x+iy in H. The denominator cannot vanish in H. The identity |z+i|²−|z−i|²=4y proves |C(z)|<1. Conversely z=i(1+w)/(1−w) has imaginary part (1−|w|²)/|1−w|²>0 for |w|<1, and direct substitution gives the inverse. Thus C is a bijection between H and the open unit disk.

Its derivative has modulus 2/|z+i|² and 1−|C(z)|²=4y/|z+i|². Consequently the disk length element becomes

$$\frac{2|dC(z)|}{1-|C(z)|^2}=\frac{|dz|}{y}.$$

Lengths of piecewise smooth paths agree. Applying the inverse to competing paths shows that shortest paths correspond, not merely that one displayed arc has the same length. Bijectivity also preserves intersections of sets in both directions. In particular C(iy)=(y−1)/(y+1) maps l onto the open real diameter. The shared ideal endpoint zero maps to −1, which is excluded from the disk. An excluded endpoint never becomes an interior crossing under this change of model.

A precise order lemma and its limit

Here is a bounded repair of the hidden placement step in the neutral exterior-angle argument. In the real Euclidean coordinate model, translate C to zero and let A,B be noncollinear with C. The midpoint of AC is M=A/2. Reflect B through M to E=2M−B=A−B, and take D=−B on the ray beyond C away from B. Then E=A+D. Both coefficients are strictly positive, so ray CE lies strictly inside the convex angle between CA and CD. More explicitly E/2 is the midpoint of AD and lies in its open segment, and ray CE passes through E/2. Linear independence of A,D excludes either boundary ray. This proves the needed placement in this coordinate model and supplies an explicit midpoint. It does not prove midpoint existence or Pasch from an arbitrary synthetic neutral axiom list. Keep that dependency in a formal neutral proof.

Apply

Prepare a proof portfolio with three original components. First prove a construction or impossibility result, listing the initial field and permitted tools. Second compare a triangle in two models using exact formulas and clearly named curvature assumptions. Third audit a synthetic argument for a hidden order, continuity or parallel dependency and repair it.

A written proof needs review of its logical steps; an exact numerical check only verifies its stated arithmetic target. A collection of correct plots does not establish all geometric claims, and report completion does not certify independent proficiency.

A reviewable model-comparison portfolio

Work in three short sessions if useful; these are planning suggestions, not calibrated timings. Retain an initial proof, one changed-case response and a revision explaining each correction. This public portfolio is optional practice within this lesson, not new required progress or an unseen examination. Solutions remain closed until you choose to inspect them.

1. Classify and construct. For P=(2,3) and l:x=0, classify every finite-center geodesic through P and include the vertical case. Exhibit an intersecting, a boundary-only and a strictly disjoint circle. Your reviewer should check the circle equation, necessary and sufficient inequality, equality case and complete enumeration.

Feedback: classification and three witnesses

The threshold is 13/4. Centers c=3, 13/4 and 4 give r²=10, 169/16 and 13 respectively. Subtract c² to obtain intersection-height squares 1, 0 and −3. The first intersects at (0,1); the second shares only an excluded boundary endpoint; the third is strictly disjoint. Every c≥13/4 is disjoint, as is x=2. If your answer used c>13/4, restore the ideal-boundary case. If it listed just these three circles, add the proof that all others have been classified.

2. Transfer between models. Compute C(2+3i). Carry the classification to the open disk without deriving new circle equations. Explain why a length-preserving forward map alone is weaker than the bijective path argument above. Your reviewer should check the inverse domain, length identity, and exclusion of the ideal endpoints.

Feedback: preserve domains and competing paths

(2+2i)/(2+4i)=(3−i)/5, with squared modulus 2/5<1. The target is the open real diameter. Since C is bijective, images of disjoint interior sets are disjoint, and an intersection maps to an intersection. Its inverse carries every competing path back to H, so the infimum of lengths is preserved in both directions. The boundary-only case still has no interior intersection: its shared ideal endpoint maps to −1.

3. Audit a proof, then revise it. A writer says: “SAS makes CE an interior ray; therefore neutral geometry has angle sum π. The sphere has no parallels, so it disproves every neutral parallel theorem.” Identify three unsupported moves. Then use A=(0,4), B=(3,1), C=(0,0) to verify the coordinate placement lemma, listing exactly what it does prove. A reviewer should require explicit missing hypotheses, not just the label “wrong.”

Feedback: separate congruence, order and model hypotheses

SAS proves equality of corresponding angles, not the containment of a ray; order/placement is separate. The exterior-angle inequality does not give the Euclidean sum identity without a further argument or parallel hypothesis. The sphere fails unique joins for antipodal points and the global neutral line-order assumptions, so it is not a countermodel of all those assumptions. Here M=(0,2), E=(−3,3), D=(−3,−1), and E=A+D. The point E/2=(−3/2,3/2) is the midpoint of AD and lies on CE, proving interior placement in the real coordinate model. It establishes neither a synthetic neutral midpoint theorem nor full formal independence.

Changed case for a reviewer: move P to (−1,2), and ask for the complete classification and the image of P in the disk before reopening the solutions. The method must handle the sign change, not just replace numbers. Assess four criteria separately: complete geodesic enumeration; correct boundary reasoning; valid model transport; and honest axiom scope. A missed criterion calls for a targeted rewrite and a fresh point chosen by the reviewer, not an automatic whole-course failure.

Changed-case reference and repair routes

The center threshold is −5/2 and disjoint finite-center geodesics have c≤−5/2; the vertical line x=−1 is also disjoint. The Cayley image is (2+i)/5. If the inequality is reversed, return to the signed 2ac inequality before dividing. If equality is called an interior crossing, revisit the definition y>0. If transport is justified only by a sketch, write the inverse and the path-length identity. If independence is overclaimed, list the unverified neutral axioms explicitly. The independent attempt must be retained before this reference is opened; exposure makes a later repetition practice.

For optional depth, the Warwick geometry outline motivates comparing metric and projective structures, while South Alabama’s MA321 syllabus includes construction and transformation outcomes. This sequence does not claim equivalence to every university geometry course.

Continue

Continue with differential geometry for curvature proofs and Gauss–Bonnet, or with algebraic/projective geometry for deeper conic and incidence structure. Revisit the axiom ledger when a proof seems to depend on the picture. The central habit is to name the objects, assumptions, preserved structure and exact conclusion before deciding what the evidence proves.