Two islands are three numbers, and the proof does not survive them

A ring can be drawn whole, and only one way ends with a count and a question. The count is classical: a flat region with two holes carries three numbers that no angle-keeping map can change, and a region with $k$ holes carries $3k - 3$ of them. The question is what those numbers force. For one hole the answer was clean — a single map, the stereographic, and a single price for refusing to cut.

A sea with two islands in it is the two-hole case, and so is a country with two enclaves, a lake district with two lakes, and an ocean basin with two seamounts drawn as coastline. This essay puts one on the page and measures it. The three numbers turn out to be readable in closed form, with no iteration and no solver, for a reason that belongs to the projection rather than to the region. What does not survive is the theorem.

Circles on the sphere arrive as circles on the page #

The standard way to put a many-holed flat region into a canonical form is to map it onto a circle domain — a disc with round holes cut out of it — which Koebe’s theorem says is always possible and which in practice means an iteration, converging, with a tolerance to argue about.

None of that is needed here, and the reason is one line. Stereographic projection carries every circle on the sphere to a circle in the plane. A cap’s edge is a circle on the sphere. So a sphere with three caps taken out of it lands on the page as a domain bounded by three circles already, exactly, at centres and radii that come out of the cap’s own angular radius and its distance from the projection’s centre by a formula rather than a fit.

The formula is worth stating because it covers the awkward case for free. Put the map’s centre at the sea’s own centre, so a place at angular distance θ lands at radius 2 tan(θ/2). An island whose centre is δ away and whose radius is ρ reaches its nearest and farthest points along the ray through its own centre, at θ = δ − ρ and θ = δ + ρ, so its image is the circle whose diameter runs between 2 tan((δ − ρ)/2) and 2 tan((δ + ρ)/2) on that ray. When the island swallows the map’s centre the first of those is negative, the diameter runs across the origin, and the same two expressions still give the right circle.

So the region is a circle domain by construction. What remains is to decide which circle domain, because the same region can be drawn as many of them.

Six parameters in, three left over #

The counting is the whole of why there are three numbers, and doing it is more convincing than citing it.

Every conformal map of the sphere to itself is a Möbius transformation, and those form a group with six real parameters. Any of them turns one circle domain into another, so two circle domains related by one of them are the same region wearing a different picture. To name the region rather than the picture, the six parameters have to be spent.

Three go on the outer edge: sending a given circle to the unit circle is three conditions. That leaves the transformations carrying the unit disc to itself, which are three-parameter — a rotation and a choice of which interior point goes to the centre. Spend two of those on sending the first island’s hyperbolic centre to the origin, so that island becomes a disc centred on nothing, and the last on a rotation putting the second island’s centre on the positive real axis.

Sending a disc inside the disc to one centred at the origin is not the same as sending its Euclidean centre there, and the difference is a quadratic. If the island meets the real axis at p and q, the automorphism parameter a has to send those two to values that sum to zero, which is

$$a^2(p+q) - 2a(1 + pq) + (p+q) = 0,$$

and the root inside the unit disc is the one. There is no iteration in that either.

What is left when the group has been exhausted is the first island’s radius, the second island’s centre and the second island’s radius: three numbers for two holes, which is $3k - 3$ at $k = 2$. The arithmetic also explains why a ring is different. With one hole there is a fourth parameter left unspent — nothing fixes the rotation, because a single centred disc has none to fix — so the leftover is one number rather than zero, which is the modulus the ring essay reads off a band of latitude.

Three pictures, one region. A sea a hundred and three degrees across with islands twenty-one degrees wide is not a picture anybody would confuse with a sea thirty-six degrees across, and every angle-keeping map of one is an angle-keeping map of the other with a Möbius transformation in front of it. No measurement made inside the sea — no angle, no ratio of lengths at a point, nothing an angle-preserving map leaves alone — can tell the three apart. Moving one island by thirty degrees of azimuth moves the second invariant by more than a thousandth, so the triple is not merely invariant; it also separates.

What the one-hole proof actually used #

The ring’s result is a theorem and it deserves to be restated exactly, because the restatement is where it breaks.

Write any angle-keeping map of the region as the stereographic followed by a conformal rearrangement of the flat picture. The logarithm of the rearrangement’s local magnification is a harmonic function, and for a map that leaves the region whole and is one to one, it is the real part of a single-valued analytic function — so its average round every circle is one constant, independent of which circle. On a band of latitude the stereographic’s own scale is constant on each of those circles, and a quantity cannot vary less than its own averages vary. Hence no uncut rival beats the stereographic.

The load-bearing step is the second one: the stereographic’s scale is constant on the curves the averaging runs round. That holds because a band of latitude is unchanged by rotation about the map’s centre. It is a statement about symmetry, and it says nothing about how many holes the region has.

The test is the ring essay’s own instrument pointed at three regions. A sea with no island at all is a disc, and the stereographic is unbeatable on it: every disturbance raises the variation, by 2.16 per cent at a disturbance of two hundredths and 10.80 per cent at eight. A ring whose island sits at the sea’s centre behaves the same way, more steeply — 3.63 per cent and 15.31 — and both curves are exactly symmetric in the size of the disturbance, which is the signature of a minimum rather than a slope.

Move that one island twelve degrees off centre and the symmetry goes with it. The curve is no longer even: a disturbance of plus two hundredths costs 2.41 per cent and a disturbance of minus two hundredths saves 0.19 per cent. One harmonic term, chosen without a search, beats the map that eight essays have been treating as forced.

So the theorem belonged to the symmetry. A disc and a centred ring are both invariant under rotation about the map’s centre, and for such a region the rotational average of any rival is admissible too — and a rotationally symmetric harmonic function on a disc is a constant, while on a ring it is a constant plus a multiple of the logarithm of the radius, which an uncut map is not allowed. Either way the average is no better than the stereographic, so nothing is. Off centre there is no rotation to average over, and the argument has nothing to stand on. The failure arrives at one hole, not at two; the ring essay never met it because the only ring it drew was a band of latitude, which is symmetric for a reason about latitude rather than about rings.

A cut is a logarithm #

With the theorem gone, the map has to be searched for, and the search has a shape worth setting out because it turns the word cut into an arithmetic object.

The scale of any angle-keeping map of the region is $\ln k = u_0 + h$, where $u_0 = \ln(1 + |z|^2/4)$ is the stereographic’s own and $h$ is harmonic. Two facts then do all the work.

The first is that $u_0$ is not harmonic. Its Laplacian is $-e^{2u_0}$, which is the Theorema Egregium written in this coordinate: no choice of $h$ makes $\ln k$ constant, which is why no map is faithful and why there is a floor to find rather than a zero to reach.

The second is that not every harmonic $h$ is allowed. A map that leaves the region whole needs $\ln|g'|$ to have a single-valued conjugate, and the harmonic functions that fail that test are exactly the ones carrying $\ln|z - a|$ for $a$ inside a hole — one such term per hole. A cut from an island to the open sea is what admits its logarithm, and nothing else in the basis behaves differently.

That is not an analogy. Run the fit on the band of latitude from 40° to 50° N, where the ring essay has a closed form, and with the logarithm refused it returns 1.075032 against that essay’s stereographic value of 1.075029. Admit the logarithm and it returns 1.003824 against the conformal conic’s 1.003823 — and the coefficient it puts on the logarithm is −0.291991, where the conic’s cone constant is 0.708009. The coefficient is the cone constant less one, to six decimal places. What a standard parallel buys describes choosing that exponent as deciding where the distortion vanishes; it is also, and more usefully here, the one number a single cut releases.

The last cut is where Chebyshev’s criterion becomes reachable. That criterion asks for the scale to be constant on the whole boundary, which with the Laplacian fixed determines the map completely; the solution generally has periods, so it is available only when every period has been released. One cut to each island does that, and the two-cut answer is the criterion’s own map. Fewer cuts, and the criterion is a condition the region cannot meet — which is what the one-hole result found, where the stereographic held its scale constant on each edge separately at two values it could not reconcile.

The dashed line deserves its own sentence. Every one-to-one angle-keeping map of the whole sphere into the plane is a stereographic composed with a similarity — a Möbius map sends exactly one point to infinity, and composing with the stereographic that sends the same point there leaves a map fixing infinity, which is a similarity. So the dashed line is not a lazy comparison; it is the best a cartographer choosing an azimuthal aspect can do, in the sense that the aspect is a free choice sets out. The uncut fit beats it because the region is not the whole sphere, and a region’s own conformal maps are a far larger set than the sphere’s.

What a sea pays for each island #

The price of keeping a region whole does grow with each hole, which is the direction the ring essay guessed at, and it grows by less each time, which it did not.

The reason is visible in the two curves underneath the ratio. The best uncut map gets better as islands are added — 1.0719, 1.0607, 1.0553, 1.0524, 1.0505 — because each island brings a new family of harmonic functions with a pole inside it, and more freedom cannot hurt. The best fully cut map improves faster, because it gains a logarithm as well. The price is the gap between two improving quantities, and it widens because the second improves faster; it flattens because the fifth island’s contribution to either is small next to the first’s.

That gives the practical reading. A sea with a single island pays 1.6 per cent for being drawn whole, which nobody would notice. A sea with four pays 4.0 per cent and would pay about the same with ten. There is no configuration in which the count of holes alone makes the whole picture untenable — which is a different answer from the one the ring gives for a band, where the price runs to a factor of 2.13 between the tropics. Position and size do more than count.

Two islands near the middle of a sea cost most, and two islands against the coast cost nothing worth measuring. An island a degree from the shore is not really a hole: the channel round it is narrow, the region is nearly a disc with a dent, and a map that treats it as one is close to right.

The percentages carry the more interesting result. When the islands are close together the first cut does 86 per cent of the work and the second is nearly free — cutting to one island has already opened the region enough that the other’s logarithm has little left to release. Spread them out and the first cut’s share falls to 36 per cent, meaning the second cut is worth almost twice the first. So a cartographer who can afford one cut and not two has a real decision, and it depends on where the islands are rather than on how many there are: with the islands close, one cut is nearly as good as two, and with them apart, one cut is barely started.

The two invariants behave quite differently as the islands grow, and the difference is the clearest demonstration that the triple is not three copies of one fact. The radii track the islands, as anyone would expect. The separation does not move, because growing both islands symmetrically changes almost nothing about how far apart they are in the canonical picture — and it is the picture the invariants live in, not the sphere.

The price follows neither. It turns over at six degrees, because an island that takes up most of its half of the sea leaves a channel rather than an expanse, and a channel is nearly a ring — which one cut already handles well. So the quantity a cartographer cares about is not monotone in any of the three invariants, and a rule of thumb reading bigger islands are worse is wrong in both tails.

What each number was compared against #

Every figure above rests on one solver — a harmonic function fitted to make the scale’s largest value over its smallest as small as it can be — and nothing in it knows any projection by name. So the checks are cases where the answer is known from somewhere else.

A sea with no island must come back as the stereographic. For a cap of angular radius ρ the least-varying angle-keeping map is the stereographic centred on it, whose scale varies by $\sec^2(\rho/2)$ exactly. On a thirty-degree cap that is 1.071797, and the fit returns 1.071912 — one part in ten thousand, which is the resolution of the grid the extremes are read on rather than an error in the map.

A band of latitude must reproduce a closed form twice. With no logarithm the fit must return the stereographic’s variation across the band and with one it must return the conformal conic’s. On the band from 40° to 50° N those are 1.075029 and 1.003823, and the fit gives 1.075032 and 1.003824.

And the two must differ. A solver that quietly admitted a period would return the cut number both times, so the check that the refusal means anything is that the uncut answer is measurably worse — by 7.1 per cent here, which no rounding produces.

The invariants must move when the region does. Three Möbius images of one region must carry identical triples, and do to within $10^{-8}$. Moving one island by thirty degrees of azimuth must move the triple, and moves the middle number by more than a thousandth.

The extreme values are read on a finite grid, so every spread quoted is a slight under-estimate of the true one at a point between samples. The comparisons are between regions measured the same way, and the closed-form checks bound the error at about one part in ten thousand. The average was a choice of norm is the warning that attaches to every such number: largest over smallest is one summary of a scale field and a mean square is another, and they need not rank two maps the same way.

Where the picture is a model rather than a map #

An island is a circle. This is the assumption the whole closed form rests on, and it is not a small one. A real coastline is not a circle, and a region bounded by three real coastlines is a circle domain only after the iteration this essay avoided. What survives the assumption is the count — three invariants for two holes, whatever their shape — and the mechanism, since the failure of the one-hole proof turns on symmetry rather than on roundness. The numbers are the circles’.

The uncut map is fitted, not proved. With the theorem gone there is no argument that the fitted uncut map is the best there is; it is the best found in a stated basis of fifty-odd harmonic functions. Every uncut price is therefore an upper bound on the true one, and every price of keeping a region whole is a lower bound, since a better uncut map would narrow the gap. The cut answers are on firmer ground: Chebyshev’s condition identifies them.

The basis is a truncation. Adding terms improves the fit and the improvement has to stop mattering before the numbers do, which is the same caution a map with no formula records for the one-hole solver and a condition imposed at points is not a condition records for the sampling under it.

The cut costs nothing here. What a cut buys prices a cut in kilometres of coastline separated, and giving up continuity prices what a reader loses when a map stops being one surface. Nothing above charges for either: a cut is treated as free and only its benefit is measured, so a price of 1.0319 for staying whole is what staying whole buys back, not what it is worth.

Still open: whether three numbers can be aimed at #

The three invariants are now measurable, and every one of them has been treated here as something the region hands over. A cartographer choosing a region does not always have to accept it. A sheet’s edge is a choice — two charts are enough, and one is not is about the smallest number of sheets a sphere needs, and a real series has thousands, each with its own boundary drawn by somebody.

That makes the invariants partly a design variable. Moving a sheet’s edge outward by a degree changes the first number and the third; moving it in a direction rather than uniformly changes the second. Since the price of keeping a region whole turns over in both the islands’ size and their distance from the centre, there should be an edge that puts a given pair of islands at the cheapest triple available — and it need not be the edge a national boundary or a depth contour happens to draw.

Whether such an edge exists for a stated pair of islands, how much of the 3 per cent it recovers, and whether the triple that minimises the price is recognisable as anything — a configuration with a symmetry, or the one that makes two of the three numbers equal — are questions a region taken as given cannot ask.