Cartographic Projection
every map is wrong, and by exactly how much
Series

Topology — the series

12 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. How each projection escapes being one to one. Eighteen projections, and the theorem allows no fourth column. A map of the whole sphere either leaves ground undrawn, or draws one place as a curve, or is cut so that one ground curve appears twice — and 11 of the eighteen do more than one of those. The three columns are a solid angle, a count of places and a page length, which is why they are three columns rather than one score: there is no rate at which a hemisphere converts into a pole.

    No map of the whole sphere is one to one

    Eleven essays establish that no map preserves distance, and every step of that argument needs a distance. There is a second impossibility underneath it that needs nothing at all: a sphere is compact and has no boundary, so a continuous map of it into the page cannot also be one to one. Eighteen library projections, eighteen escapes, and not one of them free.

    part 1 · impossibility
  2. Ground north, on a map that has the pole on it. Every arrow points along its own meridian, towards the pole, and the pole is at the centre. Walking once anticlockwise round any loop enclosing it turns the arrow once anticlockwise as well: the index is 1, counted as a winding number with no distance anywhere in the calculation. A field like this cannot be combed flat. There is no way to choose a page direction for north at every point of the neighbourhood without the choice tearing somewhere, and the somewhere is the point in the middle.

    North cannot be up everywhere

    Ground north is a field of arrows on the sphere, and a field of arrows on a sphere must vanish somewhere. The failure is not measured, it is counted: the indices of the zeros sum to two, obtained here as a winding number in seven different charts with no distance anywhere in the calculation, and it is the same two that Gauss–Bonnet gets by integrating curvature.

    part 2 · impossibility
  3. The map that is continuous, and the pair it pays with. The orthographic is defined and continuous at every place on the Earth — it is written in the components of the place itself, with no longitude in it to jump. What it gives up is being one to one, and it gives it up almost everywhere: 47 per cent of the sphere shares its page point with the place directly behind it. Borsuk–Ulam guarantees at least one ANTIPODAL pair among those, and here it is exactly one — the centre and the place on the far side of the world, both at the middle of the picture, found to a residual of 1.5e-14.

    Two opposite places on the same spot

    A projection may be continuous everywhere or one to one everywhere, and the first two rungs price both. What neither says is that the choice is not symmetric: a map that keeps continuity does not lose injectivity somewhere arbitrary. It loses it, always and at minimum, on a pair of places directly opposite each other on the Earth.

    part 3 · impossibility
  4. The world after a map of degree 2. The graticule of the sphere, sent through the square of the stereographic coordinate and then drawn in Mollweide. Every parallel is still a closed curve and every meridian still runs pole to pole, and the whole world has been wrapped round the sphere 2 times: the 2 meridians that used to be 180° apart now lie on top of one another. Nothing has been torn. The degree measured by integrating the area the map sweeps is 2.0004.

    How many times, not whether

    Three rungs of this ladder answer yes or no and have no other kind of answer. The degree is the first quantity here that counts: a continuous map of the sphere to itself covers it a whole number of times, injectivity forces that number to ±1, and the number is recoverable three ways — by counting preimages, by integrating swept area, and from the rate at which the preimages coalesce.

    part 4 · impossibility
  5. The cut makes 20 faces out of 14 regions and needs no more colours. The same partition on a sheet cut at the antimeridian. six of the 14 regions are drawn in two pieces, one against each edge, and they are shown darker. The sheet has 20 faces where the globe had 14 regions, and it needs exactly the same four colours — because the two pieces of a split region between them touch exactly what the region touched, so identifying them gives back the sphere's own graph, edge for edge.

    Four colours, and what a cut cannot do to them

    Every projection removes a set, and a map cut at the antimeridian draws six of its fourteen countries in two pieces — twenty faces where the globe had fourteen regions. The obvious guess is that a map with split countries is the exclave problem and needs a fifth colour. It needs exactly four, and the reason is that the cut adds faces and adds no edges.

    part 5 · impossibility
  6. The whole sphere, in two sheets. Two stereographic charts, one centred on each pole, each carried to 100° from its own centre. Two caps cover a sphere exactly when each reaches 90°, so this is the minimal atlas with a little overlap, and the heavy circle in each is the equator — the one curve both sheets contain. Nothing smaller works: one chart cannot cover the sphere at all, which is what this field's first rung proves. What the counting also fixes is a price: the worst point of any two-chart atlas is at 90° from a centre, where a conformal chart's areal factor is exactly 6 and an equal-area one's angular deformation is 49.07°.

    Two charts are enough, and one is not

    The topological minimum for an atlas of the sphere is two sheets, and the counting fixes a price nobody chose: the worst point of any two-chart atlas is 90° from a chart's centre, where a conformal chart's areal factor is exactly 4 and an equal-area one's angular deformation is 38.94°. A national series has a hundred and twenty thousand sheets, and a hundred and twenty thousand minus two of them are bought by accuracy.

    part 6 · impossibility
  7. What a cut buys. The mean angular deformation of the interrupted sinusoidal against the total length of cut the interruption spends, for lobe counts from one to twenty-four. Goode's interruption — the one actually printed — is the marked point: it spends 100 thousand kilometres and returns 18.0°, where the even-lobed curve returns 9.2° for the same length. It is not on the frontier and it was never trying to be: its cuts are placed to keep continents whole.

    What a cut buys

    Cuts are usually counted and never measured. A cut is a curve on the sphere with a length in kilometres, the shape distortion it removes is a falling function of that length, and the interruption everybody prints spends a hundred thousand kilometres to reach a figure the even-lobed curve reaches with sixty.

    part 7 · impossibility
  8. 50° S to 40° S, drawn conformally without a cut and with one. The band of latitude from 50° S to 40° S, drawn two ways that both keep every angle. On the left it is not cut: the only such map with least scale variation is the stereographic from the south pole, the ring closes up, and its scale varies by a factor of 1.0750 across the band. On the right it is cut along one meridian and opened into a conformal conic with cone constant 0.7080: the ring spans only 254.9° and leaves a gap of 105.1°, and its scale varies by 1.0038. Not cutting costs a factor of 1.071 in scale variation.

    A ring can be drawn whole, and only one way

    A band of latitude has a hole in it, and a region with a hole has a number no angle-keeping map can change. Draw the band without cutting it and that number forces the map: the stereographic from one pole is the least-varying picture there is. One cut along a meridian frees the cone constant, and the scale variation the cut buys back runs from nothing on a polar cap to a factor of 2.13 between the tropics.

    part 8 · impossibility
  9. A sea with two islands in it, and the three circles it already is. A stated sea 60° across with two islands 12° across, their centres 12° from its own, drawn on the globe and drawn again under the stereographic projection centred on the sea. Stereographic projection carries every circle on the sphere to a circle in the plane, so the flat picture is bounded by three exact circles with no fitting and no iteration: the sea's edge at radius 0.5359, and the two islands at radius 0.1060 with their centres 0.2108 from the origin. Normalised, the region's three conformal invariants are 0.2361, 0.6848, 0.1221 — three numbers, which is what a region with two holes carries and no angle-keeping map can change.

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

    A sea with one island at its centre has one conformal invariant and a forced best map. Put two islands in it and there are three invariants, readable in closed form because stereographic projection carries circles to circles — and the forcing is gone. The theorem that picked the map needed the region to be unchanged by rotation, not to have one hole, and a single harmonic term two hundredths in size beats the stereographic by 0.19 per cent the moment an island moves off centre.

    part 9 · impossibility
  10. The same two islands, and the cut that buys most is the central one's. A sea 60° across with a large island A, 12° across, and a small one B, 6° across, on opposite sides of its centre, drawn in the plane the stereographic makes of it. Each dashed line is one way of spending a single cut — from A to the open sea, from B to the open sea, or from A to B — and the solid line is the one that buys most, as a share of what two cuts buy. Left, A 18° from the centre and B 6°: A 9% · B 78% · between 21%. Right, A 6° and B 18°: A 81% · B 12% · between 11%. The better cut is the central island's in both, whichever island is large, and a cut joining the islands buys 21% and 11%.

    One cut belongs to the island that would cost most alone

    A sea with two islands pays a few per cent for being drawn whole by an angle-keeping map, and one cut buys back part of it. Which island to cut is a design choice that leaves the ground exactly where it was, and it is not decided by size: across twenty arrangements of a large island and a small one, the cut worth most goes to the island that would cost the sea most if it were alone in it — in eighteen of twenty, the other two near-ties — which is usually the more central one. A cut joining the two islands buys almost nothing, and the better cut is nearly always the longer.

    part 10 · impossibility
  11. The two islands dearest alone are the wrong pair to cut when they sit on the same side. One of forty three-island seas, 60° across, drawn in the plane the stereographic makes of it; beside each island, what it would cost the sea if it were alone in it — P 0.97%, Q 0.72%, R 1.68%. Left: the two dearest islands, P and R, cut to the open sea, which buys 75% of what three cuts buy. Right: the best pair, Q and R, which buys 97%. The dearest islands are 69° apart as seen from the centre, and their single cuts overlap by 40 points.

    The second cut goes across the sea

    With two islands and one cut, the island that would cost the sea most alone names the cut. With three islands and two cuts, the obvious rule — cut the two dearest — names the best pair in only 23 of 40 seas, and loses up to 23 points of what three cuts would buy. It fails when the two dearest sit on the same side of the sea, where their cuts do each other's work. The best pair keeps the dearest island in 37 of 40; its partner is the island across the centre from it, which is right in 32 of 40 and never costs more than 12 points. A cut joining two islands is never worth a second cut to the sea.

    part 11 · impossibility
  12. With five islands the two dearest are still the wrong pair when they share a bearing. One of 20 five-island seas, 60° across, drawn in the plane the stereographic makes of it; below, each island's price to the sea alone — P 0.73%, Q 1.30%, R 0.36%, S 1.96%, T 2.74%. Left: the two dearest, S, T, cut to the open sea, which buys 77% of what five cuts buy. Right: the best pair, Q, T, which buys 100%; the dearest island and then the one farthest from it in bearing — Q, T — buy 100%. One cut buys 77%, three 100%, four 100%.

    A sea with five islands is seldom worth five cuts

    With three islands in a sea, the second cut went across the sea from the first, and the number of cuts worth spending looked like the number of directions the islands point. With four and five islands that survives in part and breaks in part. Choosing the dearest island and then, each time, the island farthest in bearing from every island already cut names the best pair of cuts in 28 seas of 40, where the two dearest name it in 16 — and it fails at three cuts of five, because the best three cuts often do not contain the best two. The dearest island is no longer reliably among the best pair. And 39 seas of 40 are worth fewer cuts than they have islands: the count follows the widest gap in the islands' bearings more closely than it follows how many islands there are.

    part 12 · impossibility

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