A split into three is not a statement of topology – the plane and the disk are homeomorphic.
The uniformization theorem says that every simply connected Riemann surface is the sphere, the plane or the disk. A split into three sounds like topology, but the plane and the disk are homeomorphic, and what separates them is not topology. NO NEW MATHEMATICAL THEOREM IS CLAIMED: the uniformization theorem, Liouville’s theorem, the modulus of an annulus, the j-invariant with the modular group, and the type problem are all standard since Riemann. Nor does this paper prove uniformization, since what is offered is a measurement and not a proof.
First, a trap. The map sending z to z over one minus the modulus of z is a bijection from the open disk onto the plane, with a worst round-trip error of 1.2 times ten to the minus thirty-two over sixteen sample points. It keeps the angle between the images of the radial and tangential directions, which cross at a right angle, at exactly ninety degrees at each of the points 0.2, 0.5 and 0.8, indistinguishable in this respect from the holomorphic map sending z to z squared plus z. Preserving one angle is not conformality: a radial map necessarily keeps this particular right angle, since the radial and tangential directions remain orthogonal in the image, so preservation of angles must never be checked on one pair of directions. Tilting the same right angle by forty-five degrees opens it to 102.68038, 126.8699 and 157.38014 degrees, and the ratio of the largest stretch to the smallest over all seven hundred and twenty directions is 1.25, 2.0 and 5.0. The holomorphic map has that ratio equal to one exactly at every point. For the homeomorphism it equals one over one minus the radius, and topology sees none of it.
And no conformal map exists at all. A conformal bijection from the plane onto the disk would be a bounded entire function, and Cauchy’s estimate bounds the derivative at the origin by the bound divided by the radius, for every radius. At radii of ten, a thousand, a million and a million million the bound falls to 0.1, a thousandth, a millionth and a million-millionth, forcing the derivative to vanish, which is Liouville’s theorem. There is no non-constant bounded entire function.
Second, and this is the core. Consider the region between the circle of radius R about the origin and the circle of radius r about a point at distance d. Any two such regions are homeomorphic. The modulus was computed along two independent routes. The first solves for the two points inverse with respect to both circles, builds the Mobius map they define, confirms at four hundred points on each circle that the images really are concentric circles, and only then reads the ratio of radii. The second is the inverse hyperbolic cosine of R squared plus r squared minus d squared over twice R r. In all six cases the two routes agreed to twenty digits, giving 1.203972804, 1.15881036, 0.8670147265, 0.4211448485, 2.161580941 and 0.3399260616, with the spread of the image radii at the level of ten to the minus thirty-one, that is numerical dust. The logarithm of the ratio in the concentric case is replaced by an inverse hyperbolic cosine in the eccentric one, and since the hyperbolic cosine of that logarithm equals the same expression at zero offset, the two are one formula read twice.
Third, the family carrying one number was solved explicitly. Fixing the outer radius at one and solving for the inner radius that gives the modulus 1.203972804, the value of the concentric annulus of inner radius three tenths, offsets of 0.1, 0.2, 0.3 and 0.4 give inner radii 0.296706871877, 0.286870020776, 0.270614346149 and 0.248139402709, and remeasuring through the Mobius route returns the same value. Drawn on paper the five regions look nothing alike, one with its hole in the middle and another with the hole pushed towards the rim, and they are conformally the same region, while the region with inner radius 0.45 at offset 0.5, of modulus 0.3399, is a different one. One real number decides and topology sees none of it. This runs opposite to Paper 152, where one and the same character table left two groups non-isomorphic: there, one invariant and different objects; here, different appearances and one object. In both, the answer changes with the invariant one asked by.
Fourth, on a torus the parameter becomes complex. A torus is the complex plane modulo a lattice, and two are conformally equivalent exactly when their parameters agree modulo the modular group. Building the j-invariant from theta functions, as two hundred and fifty-six times a cube over a product of squares in the modular lambda, so that no normalisation is taken on trust, and checking it against a second route, gives j at i equal to 1728 exactly and the modulus of j at the primitive sixth root equal to 1.5 times ten to the minus eighty-eight. At a parameter of 0.31 plus 1.27 i, both generators of the modular group leave j unchanged to within ten to the minus twenty-seven, so j is a function of the torus and not of the lattice one happened to draw. And since j equals 1728 at i, 287496 at twice i and 153553679.39673 at three times i, these three are the same topological surface and no two of them are conformally equivalent. Uniformization does not end in one shape. It ends in one model together with a point of a moduli space.
Fifth, the sign of the Euler characteristic decides only for closed surfaces. By Gauss-Bonnet the integral of the curvature is two pi times the characteristic, so for genus zero, one, two, three and four the characteristic is two, zero, minus two, minus four and minus six, and its sign picks sphere, plane or disk. On open surfaces it decides nothing: the plane and the open disk both have characteristic one and both have the homotopy type of a point, and yet they fall to different models. Which of the two an open surface is, the type problem, is settled by no topological invariant. The trichotomy of the universal cover and the trichotomy by the sign of the characteristic agree on closed surfaces and part company on open ones, by the same discipline with which Papers 67 and 128 fenced in where four pi equals two pi chi actually holds.
Finally, what the word unique means here. Every automorphism of the disk has the form of a rotation times a Blaschke factor, with two real parameters in the point and one in the angle, three real parameters in all. Reading a composition of two of them off its values alone recovers the point as minus 0.4157923416 plus 0.4781102072 i and the angle as minus 0.5242149765, disagreeing at fifteen sample points by 2.9 times ten to the minus thirty-one. Taking the point to be the base point and the angle zero at a base point of 0.35 minus 0.2 i gives a map sending it to zero with derivative 1.194029851, whose imaginary part is two times ten to the minus sixty-nine, that is real and positive. The uni of uniformization is uniqueness after normalisation: before it a three-parameter family, after it exactly one map, and there is no the map before the conditions are imposed.
On the making of this work: The ideas and content of this work stem from the author’s own considerations. Assistance from an AI (a large language model) was used for structuring, English translation, and checking the algebra. Any remaining errors or misinterpretations are solely the author’s. Feedback and corrections are sincerely appreciated.
⚠️ Scope ── what is not claimed
a census of established standard facts (Layer 2). What is not shown: any new mathematical theorem, none is claimed; a proof of uniformization, none is given, since what is shown is a measurement. What is added: computing the modulus along two independent routes and confirming that they agree, solving explicitly for a family of eccentric annuli that share one modulus, exhibiting in numbers the trap that one right angle can be preserved by a map that is not conformal, and setting side by side the fact that the Euler characteristic decides for closed surfaces and does not for open ones. What remains: uniformization is usually told as the theorem that folds every shape into one. It folds as far as the model, and after that one real number, or one complex number, is left over. Whether that leftover parameter is read as a remainder or as part of the answer is what changes how this theorem is read.
Reference
Yamagishi, Y. (2026). Uniformization Gives Not One Shape but One Model and One Parameter [Paper 167]. Zenodo.
DOI: 10.5281/zenodo.22093172
ORCID: 0009-0006-8587-3450