On a body with a hole, north can be up everywhere

Twelve rungs of this ladder change the body. An oblate spheroid, a triaxial one, a body with no sea level, a shape whose radius is not a function of direction. Every one of them is a sphere with the shape pushed about, and being a sphere is doing more work in this collection than any of the shapes are.

Every impossibility here rests on one integer. The Euler characteristic of a sphere is 2, and from that number come the total curvature of 4π, the hairy ball theorem, the two charts an atlas needs and the cut every world map must carry. Change the number and every one of those statements changes with it.

The curvature, both ways, integrating to nothing #

The torus of revolution has the metric $(R + r\cos v)^2\,\mathrm{d}u^2 + r^2\,\mathrm{d}v^2$, with $u$ running the long way round and $v$ the short way. Its Gaussian curvature is

$$K = \frac{\cos v}{r\,(R + r\cos v)},$$

which is positive on the outer half, negative on the inner half, and exactly zero on the two circles at $v = \pm 90°$. A sphere’s curvature is $1/R^2$ everywhere and has one sign; this one has both, and a body carrying both signs is a body on which where the surface curves the other way is not an exception but half the surface.

The integral is the point. Gauss–Bonnet says $\int K\,\mathrm{d}A = 2\pi\chi$, and for a torus $\chi = 0$, so the positive outer half and the negative inner half cancel exactly — at every aspect ratio, for every $R$ and $r$. Total curvature and the scale rule uses the sphere’s $4\pi$ to derive a floor on distortion that no map can beat. On a torus that floor is derived from zero, and it is not a floor at all.

The cancellation is not approximate and it is not a property of the particular tori measured here. Positive curvature is spread over the outer half, where the surface is also widest and its area element largest; negative curvature is concentrated on the inner half, where the area element is smallest. The two effects are exactly reciprocal — $K$ carries $1/(R + r\cos v)$ and $\mathrm{d}A$ carries $(R + r\cos v)$ — so the integrand is $\cos v$ times a constant, and $\cos v$ integrates to nothing over a full turn. The theorem’s answer is visible in the integrand before any arithmetic is done, which is the sort of thing that only happens when a formula is telling the truth.

The field that cannot vanish #

North cannot be up everywhere is the sphere’s version of the hairy ball theorem: any continuous tangent field on a sphere vanishes somewhere, so no map can have north pointing up at every point, and the standard demonstration is the east-pointing field $\partial/\partial\lambda$, whose length is $\cos\varphi$ and which dies at both poles.

Run the same construction on a torus. The east-pointing field is $\partial/\partial u$, its length is $R + r\cos v$, and the smallest that gets is $R - r$.

$R - r$ is positive for any torus, so dividing the field by its own length gives a smooth unit tangent field defined at every point of the surface. East is a direction everywhere. So is north, by the same argument applied to $\partial/\partial v$, whose length is the constant $r$. A map of a torus can have north up at every point of the page, and the reason is not that somebody found a clever projection — it is that the obstruction is absent.

The index sum of any tangent field with isolated zeros is the Euler characteristic, which is 2 on a sphere and 0 on a torus. Two is not zero, so a sphere’s field must vanish somewhere; zero is zero, so a torus’s need not.

Three world maps, and none of them is cut #

No map of the whole sphere is one to one and what a cut buys between them establish that a world map of a sphere must be cut, and price the cutting. A torus needs no cut at all.

The reason is that a torus is a rectangle with its opposite edges glued, so the rectangle is the surface rather than a flattening of it: $u$ runs from $-180°$ to $180°$ and wraps, $v$ runs from $-180°$ to $180°$ and wraps, and no point of the body is missing, doubled or discontinuous.

The conformal one is the surprise. Substituting $\mathrm{d}\sigma = r\,\mathrm{d}v/(R + r\cos v)$ turns the metric into $(R + r\cos v)^2(\mathrm{d}u^2 + \mathrm{d}\sigma^2)$ — a flat metric multiplied by a positive function, which is exactly the definition of isothermal coordinates. So $(u, \sigma)$ is a conformal map of the entire body onto a rectangle, with zero angular deformation at every point, no cut, no pole and nothing omitted. That map does not exist for a sphere at any size, in any aspect, by any construction, and the stereographic projection’s single missing point is the closest a sphere can come.

Which rectangle, exactly #

The rectangle a torus maps onto conformally is not arbitrary. Its aspect ratio is a genuine invariant of the surface — the conformal modulus — and it has a closed form.

$$\tau = \frac{1}{2\pi}\int_0^{2\pi}\frac{r\,\mathrm{d}v}{R + r\cos v} = \frac{r}{\sqrt{R^2 - r^2}}.$$

That the integral has a closed form is a convenience; that the integral is an invariant is the content. Two tori with different $R$ and $r$ but the same $\tau$ are conformally the same surface, and one can be mapped onto the other with no angular error anywhere — which is a much stronger relationship than any two spheres have, since two surfaces with the same curvature are locally isometric and these are globally conformal.

What the three maps share is more interesting than what separates them. On a sphere a world map is a compromise between properties and a compromise about where to put the damage: the cut, the pole, the region left out. Here the second half is gone entirely, so what is left is the trade in its pure form, with nothing else in it. That is what makes the next section’s law as clean as it is.

The one obstruction that survives #

Everything above is a statement that a torus is easier than a sphere. This is where it stops.

A torus of revolution has $K \neq 0$ almost everywhere, so no map is faithful applies to it exactly as written: there is no isometry of the surface onto a plane, no map that gets every distance right, and the two-line argument that conformal plus equal-area forces $a = b = 1$ works word for word. The topology went away and Theorema Egregium did not.

So the same trade-off appears, in the same shape, on a body where nothing else about the sphere’s difficulty remains. That is worth more than a curiosity: it separates the two obstructions that a sphere presents together. The cut, the two charts, the vanishing field and the total-curvature floor are all consequences of $\chi = 2$; the impossibility of a faithful map is a consequence of $K \neq 0$; and a sphere confuses them because it has both.

The law, and the plate carrée #

The trade-off is not merely present; on a torus it is exactly solvable, which it is not on a sphere.

Write $\kappa = (R+r)/(R-r)$, the ratio of the outer radius to the inner. Then, with each map given the page aspect that minimises its own worst anisotropy:

Every entry is exact and every one is a function of $\kappa$ alone. On the wide ring, $\kappa = 2$: the conformal map’s areas run over a factor of four, the equal-area map’s worst angular deformation is 38.94°, and the plain rectangle map — sending $(u, v)$ straight to $(x, y)$ — has areas over a factor of two and a worst angular deformation of 19.76°.

The plate carrée of a torus is the exact geometric mean of the conformal and the equal-area map, in both errors at once. On a sphere the equirectangular projection is a compromise in the loose sense of being neither one thing nor the other, and the site’s own measurements show it is not a good one — it is the worst of ten on scale departure at the highest exponents. Here it is a compromise in the exact sense, sitting at the midpoint of a one-parameter trade, and that is because the torus’s trade has one parameter to sit in the middle of.

There is a second reading of the same table that is worth stating because it is the practical one. Anisotropy and areal range are the two axes this collection scores every map on, and on a torus their product in the right powers is fixed: $A\,S^2 = \kappa^2$, whatever map is chosen, so improving one costs the other at a known exchange rate and no cleverness alters the total. On a sphere no such conservation law exists — the trade is real but the frontier has to be searched for, which is what the whole choosing field does — and the difference is that a torus’s shape is one number and a region on a sphere is not.

The exchange rate is also the thing that makes $\kappa$ readable off a picture. A thin ring has $\kappa = 1.4$ and all three maps are nearly good: 19.19° of angular deformation at worst for the equal-area one, and areas over a factor of 1.96 for the conformal one. A thick torus has $\kappa = 6$ and none of them is: 91.17° and a factor of 36. The body’s difficulty is entirely the ratio of its outer radius to its inner one, and its size, its area and its absolute curvature do not enter.

What was computed, and how #

The curvature is computed twice. Once in closed form from the embedding, and once from the first fundamental form alone — from $E = (R + r\cos v)^2$ and $G = r^2$ and their derivatives, with no reference to the three-dimensional shape at all. The gate requires them to agree, which is measuring curvature from inside run on a body that changes sign.

The total curvature is a quadrature of $K$ against the surface’s own area element $r(R + r\cos v)$, and the gate requires it to be zero rather than merely small: it comes out at the last bit the sum carries, on every torus tested.

The modulus is computed twice as well, by the closed form and by the integral that defines it, agreeing to a part in ten million. That matters because the closed form is the thing the essay claims is the map, and a closed form that had drifted from its integral would be a claim about algebra rather than about the surface.

The three maps’ errors are computed with each map’s free page constant set to the value that minimises its worst anisotropy — the closed forms $1/(R^2 - r^2)$ for the equal-area map and $r/\sqrt{R^2 - r^2}$ for the rectangle. That constant is genuinely free in a way no sphere map’s is: stretching the page in one direction leaves an equal-area map equal-area, since every area is multiplied alike, so comparing the three at arbitrary constants would be comparing the constants. This is the same normalisation measure.js applies before ranking any projection, for the same reason.

The assertions require six things separately: that the two routes to $K$ agree; that $K$ takes both signs on every torus; that the total curvature is zero against a sphere’s $4\pi$; that the modulus equals its own integral; that the conformal map’s angular deformation is exactly zero and its areal range exactly $\kappa^2$, and the equal-area map’s areal range exactly one; and that the east field’s length is bounded away from zero on every torus while the sphere’s reaches zero exactly.

What it would take to need this #

The measurement is a control rather than a design, and the honest question is what would have to be true for a cartographer to want it.

The nearest real case is not a body at all but a domain. A map of a river basin is a map of a disc; a map of a shipping network round a continent, or of the habitable band of a ring-shaped structure, is a map of an annulus, and an annulus has $\chi = 0$ too. Every result above about the absence of a cut carries over: an annulus has a conformal map onto a flat annulus with no cut and no singular point, and the modulus is again the one invariant. Two charts are enough, and one is not counts charts for a sphere; the same count for an annulus is one.

So the practical residue is a rule about domains rather than about bodies. Whether a region needs cutting is a question about its topology and not about its size, and a region with a hole in it — a coastal band round an island, a corridor that closes on itself — is a different problem from a region without one, in a way that no amount of care about the projection will address.

Where the model stops #

This is a torus of revolution, not every torus. The abstract flat torus — a rectangle with its edges glued and the ordinary flat metric — has $K \equiv 0$ everywhere and is isometric to a piece of the plane, so on that surface even Theorema Egregium stops obstructing. It cannot be embedded smoothly in three dimensions, which is why no body in any solar system has that metric, and it is the reason the last section is about the embedded torus rather than about tori in general.

No body in the solar system is a torus. A coordinate on another body works through the ones that exist, and every one of them is topologically a sphere; even a body whose rays hit the surface twice is a sphere with a waist. Ring systems are not surfaces, and a contact binary is two spheres joined, which is still a sphere.

So the practical content is subtractive. Nothing here tells anybody how to map a real body. What it does is separate two obstructions this collection has been treating as one, by exhibiting a surface that has the second and not the first — which is what a control is for.

The generalisation #

The rule is that an impossibility has a hypothesis, and the hypothesis is usually the part nobody varies.

Every argument in this collection’s impossibility field begins “on a sphere”. Twelve rungs of this anchor vary what “sphere” means as a shape and none varies what it means as a surface, and the shape turns out to have been the less important half: an oblate spheroid, a triaxial ellipsoid and a body with no sea level all behave like a sphere in every topological respect, and change no impossibility at all. One change of topology changes four of them at once.

That is not an argument for studying tori. It is an argument for knowing which hypothesis each result is using, because two results that look alike on a sphere — a world map must be cut and a world map must distort — turn out to have nothing in common, and a reader who has only ever seen them on a sphere has no way to find that out.

Who found it, and when #

None of the mathematics is new or contested. Gauss–Bonnet dates from the 1840s, Poincaré and Hopf’s index theorem from 1885 and 1926, and the uniformisation of genus-one surfaces to flat tori from Riemann. The conformal modulus of a torus of revolution is a textbook exercise.

What is missing is the join to cartography, and the reason is that cartography maps bodies that exist. The subject’s impossibility results are stated for the sphere because the sphere is what there is, and stating a hypothesis nobody can vary feels like pedantry — until the hypothesis turns out to be carrying results that were attributed to a different one.

The place the distinction does get made is in the theory of Riemann surfaces, where the genus is the first thing anybody asks and the curvature is a metric detail. That subject and this one are looking at the same objects from opposite ends: one begins with the topology and adds a metric, the other begins with a measured shape and never asks about the topology at all, because the answer has been 2 every time.

Where the ladder goes next #

Thirteen rungs have varied the body: its flattening, its axes, its sea level, its rotation, and now its topology. Each has changed which formulae apply. What none has changed is the assumption that the body’s shape is known — every one of them starts from a stated ellipsoid or a stated surface, and a real body’s shape is a series that somebody truncated.