The lists no covering can have

A count that can say zero ended on a single list of cycle shapes. A covering of the sphere by itself with four sheets, branched over three points, where two points swap the sheets in two pairs and the third cycles three of them and fixes the fourth: the shapes $(2,2)$, $(2,2)$, $(3,1)$. The list passes the Riemann–Hurwitz count, and Frobenius’s character formula returns nought — no covering exists — without saying why.

One failing list is a curiosity. The question it raises is how many there are, and what they have in common. That needs a census: every list of three cycle shapes that passes the count, for every number of sheets that can be searched, each one decided by finding a covering or by proving there is none.

Up to eight sheets there are nineteen: one with four sheets, five with six, thirteen with eight. With three, five or seven sheets there are none. That is the first thing the census shows, and it is the shape of a conjecture forty years old.

A covering is three permutations #

The translation from geometry to algebra is the one the essay on coverings as permutations made. Take a covering of the sphere by a surface with $d$ sheets, branched over three points. Walk once round a loop encircling the first branch point and follow each sheet: the sheets come back permuted, by a permutation $\sigma_1$ — the same lifting of a loop that turns a loop round a circle into a walk along a line in the simplest covering of all, here with finitely many sheets to come back to. The same for the other two points gives $\sigma_2$ and $\sigma_3$. A loop round all three points is a loop round nothing, so

$$\sigma_1 \sigma_2 \sigma_3 = 1,$$

and the covering is connected exactly when the three permutations together can carry any sheet to any other. Conversely, any three permutations with these two properties build a covering, by gluing $d$ copies of the sphere cut along two arcs. The branch data are the cycle shapes of the three permutations, three partitions of $d$.

What a branch point subtracts found the one condition the data obviously must satisfy. A cycle of length $k$ glues $k$ sheets together at one point, losing $k - 1$ points from the count of the cover’s Euler characteristic, so a branch point whose permutation has shape $\lambda$ subtracts $d - \ell(\lambda)$, with $\ell$ the number of cycles. A sphere covering a sphere needs $2 = 2d - \sum (d - \ell(\lambda_i))$, that is

$$\sum_{i=1}^{3} \big(d - \ell(\lambda_i)\big) = 2d - 2.$$

It is Euler’s count of corners, edges and faces applied to a surface built from sheets, and it is the only constraint anyone has found that applies to every list at once. The Hurwitz existence problem asks which data that pass this count are actually realised. For coverings of any surface other than the sphere the answer is known and short: almost every list that passes is realised. Over the sphere it is not known, and lists like $(2,2),(2,2),(3,1)$ are the reason.

The search #

Deciding one list is a finite problem. Conjugating all three permutations by the same relabelling of the sheets changes nothing, so the first permutation can be fixed as any one permutation of its shape — the census takes the shape with the most permutations, so that the fixing saves the most. Then every permutation of the shape with the fewest is tried as $\sigma_3$, the second is forced as $\sigma_2 = (\sigma_1\sigma_3)^{-1}$, and the list is realised the moment some choice gives $\sigma_2$ the right shape and the three a single orbit. If no choice does, the list is not realised, and the failed search is the proof.

Fixing the first permutation is not a shortcut that might miss something. Relabelling the sheets turns a covering into the same covering described differently, and relabelling acts on triples of permutations by conjugating all three at once; every orbit of that action contains triples whose first permutation is any chosen one of its shape. The relabellings that leave the chosen $\sigma_1$ alone — its centraliser — still act on the remaining choices, which is why the search finds each covering several times over, and the relabellings that leave all three permutations alone are the covering’s own symmetries, the deck transformations. None of that matters for existence, which needs only one hit, and it is why the search can stop at the first.

The cost is set by the smaller classes. With eight sheets the largest class has 5,760 permutations and the symmetric group 40,320, so fixing $\sigma_1$ in the largest class and running through the smallest turns a search over pairs — over a billion of them — into at most a few thousand trials per list.

The candidate lists number 1, 6, 11, 39, 85, 244, 562, 1,447, 3,196, 7,918 and 16,944 for three to thirteen sheets, and the failures 0, 1, 0, 5, 0, 13, 6, 34, 0, 91 and 0. The search through eleven sheets takes about a second and a half; twelve took nineteen seconds and thirteen two minutes. Two independent implementations — one listing all of the symmetric group and sorting by shape, one generating each shape’s permutations directly — agree on every count through ten sheets.

The prime degrees stand out. With 5, 7, 11 and 13 sheets every one of the 11, 85, 3,196 and 16,944 lists that pass the count is realised. Allan Edmonds, Ravi Kulkarni and Robert Stong conjectured in 1984 that this is always so: when the number of sheets is prime, the Riemann–Hurwitz count is the only obstruction. The census confirms it as far as it reaches, which is where computation has always confirmed it; a proof would need a reason, and the reason sketched in the previous essay — that a transitive group of prime degree is too rigid to trap a product — has not been made into one.

Failure is common in count and rare in share #

The failures grow in number with the sheets — 1, 5, 13, 34 and 91 at four, six, eight, ten and twelve — but the candidates grow faster, and the share that fails falls: 17% of the six four-sheet lists, 13% at six sheets, 5.3% at eight, 2.3% at ten and 1.1% at twelve. With nine sheets it is 1.1% already. A list chosen at random among those that pass the count is almost always realised, and the exceptions are a thinning minority whose absolute number nevertheless keeps growing — the shape a conjecture of the form “only these families fail” would have, if the families were known.

Adding a branch point thins them further. A separate run of the same search with four branch points instead of three — a fourth permutation, and the product of all four equal to one — finds no failures at all with four, five or seven sheets, 2 among the 70 lists with six sheets, and 7 among the 773 with eight, 0.9% against 5.3% for three points. Every one of those nine failures contains a pairing of every sheet, the same signature as before. More branch points give more freedom to choose permutations, and each extra permutation is another way round an obstruction; that is why the classical results for coverings of the sphere are strongest when there are many branch points and why three is where the census has to look.

Two pairings make equal halves #

The red rows in the first figure are the clue to most of the failures. A permutation whose cycles are all of length two, with no fixed sheet, pairs every sheet with another — a fixed-point-free involution. Sixteen of the nineteen failures up to eight sheets contain one, and the simplest family contains two.

Draw the eight sheets as points and the two pairings as two perfect matchings, solid and dashed. Every point has exactly one solid and one dashed edge, so the union is a set of cycles alternating solid and dashed. Applying one pairing and then the other moves a point two steps round its alternating cycle, so an alternating cycle of $2m$ points becomes two cycles of length $m$ in the product. If the covering is to be connected, the union must be a single alternating cycle through all $d$ points, and the product then has shape $(d/2, d/2)$ — never anything else.

So the list $([2^k], [2^k], \lambda)$ is realised exactly when $\lambda = (k, k)$. The Riemann–Hurwitz count allows every two-part $\lambda = (a, d - a)$, and all the others fail: $(3,1)$ with four sheets, $(5,1)$ and $(4,2)$ with six, $(7,1)$, $(6,2)$ and $(5,3)$ with eight. The census finds each of these and the alternating cycles explain them — the first failing list of the previous essay is this family’s smallest member, and the “subgroup getting in the way” there is the dihedral group two involutions generate. It is the same reasoning that makes a sliding puzzle’s parity visible in a drawing: a structure forced on every solution, which the shapes asked for cannot have.

What the failures have in common #

The dihedral family is three of the thirteen failures with eight sheets, and four of the thirty-four with ten. The rest need other explanations, but they share a feature.

With ten sheets only 8% of all candidate lists contain a fixed-point-free involution, and 74% of the failures do. Only 20% of candidates contain a shape whose cycles all have the same length — $(5,5)$, $(2,2,2,2,2)$ — and 76% of the failures do. With nine sheets, where no involution can be fixed-point-free, every one of the six failures contains three 3-cycles. Failures crowd onto shapes with equal cycles, because such a permutation commutes with a large group and its products with other permutations are correspondingly constrained, which is the dihedral argument’s mechanism in a more general form.

The figure’s note records the opposite extreme. A list containing a single cycle through every sheet — the branching of a polynomial at infinity — never fails: 9, 29, 51 and 83 such lists at 6, 8, 9 and 10 sheets, all realised. That is a theorem of René Thom from 1965: every list containing a $d$-cycle that passes the count is the branching of some polynomial. Equal cycles make failure likely, and one long cycle makes it impossible.

The failures nothing obvious explains #

Eight of the thirty-four ten-sheet failures contain no shape with equal cycles. Each of them contains $(4,2,2,2)$ or $(3,3,3,1)$ instead — shapes one cycle away from equality — and four of them are made of nothing else: $(4,2,2,2)$ three times, $(3,3,3,1)$ three times, or a mixture of the two. These are the failures that make the problem hard. A census lists them; it does not group them into families, and the families that have been found by hand — by Pervova and Petronio, by Zheng, by others since the 2000s — are each proved by an argument tailored to the shapes involved. Whether a finite list of such families accounts for every failure over the sphere is open.

It helps to see how small the obstruction can be. In the four-sheet failure the shapes ask for two pairings, $(2,2)$ each, with product $(3,1)$. Two pairings of four sheets can coincide, in which case their product is the identity, or differ, in which case their union is a single alternating cycle through all four sheets and their product is two 2-cycles. A 3-cycle is simply not on the list of things two pairings of four sheets can multiply to, and that is the whole proof — the character sum of the previous essay was computing this fact without stating it. The eight lists in the figure above have no such one-line reason, as far as anyone knows.

A failing list is a statement that a certain system of equations in a symmetric group has no solution. Most such statements, in most branches of mathematics, are proved by finding an invariant the equations would have to preserve — a parity, a sign, a homomorphism to a smaller group. The alternating-cycle argument is of that kind. For the eight lists in the figure no such invariant is known, and the only proof that they fail is that every candidate was tried.

What a success looks like #

A realised list comes with a witness, and the witness is small enough to check by hand. The seven-sheet covering in the figure has branch shapes $(4,3)$, $(3,3,1)$ and $(2,2,2,1)$, which subtract $5 + 4 + 3 = 12 = 2 \cdot 7 - 2$, so the cover is a sphere. Following the third permutation, then the first, then the second returns every sheet to itself, and the three together reach all seven sheets. Gluing seven copies of the sphere according to those permutations builds the covering.

Coverings branched over exactly three points have a significance well beyond this problem. Alexander Grothendieck noticed in the 1980s that such a covering can be drawn as a graph on the covering surface — the preimage of a segment joining two of the branch points — which he called a dessin d’enfant, a child’s drawing. And by a theorem of Gennadii Belyi from 1979, the surfaces that cover the sphere with three branch points are exactly the algebraic curves that can be written down with coefficients that are algebraic numbers. The seven-sheet witness makes the drawing concrete. Its child’s drawing has a black vertex for each cycle of the first permutation, two of them; a white vertex for each cycle of the second, three; an edge for each sheet, seven; and a face for each cycle of the third, four. Five vertices, seven edges and four faces: $5 - 7 + 4 = 2$, a planar drawing, which is the Riemann–Hurwitz count read as Euler’s formula once more. So every witness the search finds is also a curve defined over the algebraic numbers, and the absolute Galois group of the rationals — the symmetries of all algebraic numbers, the most mysterious group in number theory — acts on the set of these little permutation triples, preserving their cycle shapes. The census is a census of objects that number theorists study for reasons having nothing to do with existence.

What the census does not show #

The census decides lists with exactly three branch points, which is where failures are most common — with more branch points there are more permutations to choose and failures become rarer, though they do not disappear. It decides them up to thirteen sheets and no further; the prime-degree conjecture is confirmed for 5, 7, 11 and 13 and conjectured for every prime, and nothing here bears on the infinitely many primes beyond.

The search also proves less than a theorem would. A failed search shows that a list has no covering; it does not show why, in the sense that would let the reason be applied to other lists. That is the same situation the character formula was in: both are complete decision procedures that return a verdict without an explanation, and the census adds only the observation that the verdicts have structure — equal cycles, involutions, the absence of failures in prime degree and in the presence of a full cycle — which a theory would have to account for. A similar census of closed curves by their crossings tells the same story from the other side: an exhaustive search is the most reliable instrument there is, and the least talkative.

Still open: the prime degrees #

Is every list of shapes that passes the Riemann–Hurwitz count realised when the number of sheets is prime? The conjecture of Edmonds, Kulkarni and Stong has been checked by computer far beyond thirteen sheets and proved for special shapes — for instance when one of the three permutations is a full cycle, by Thom’s theorem, or when the data are close to it — but no proof covers all shapes. A general proof would have to explain the composite-degree failures too, at least in outline, because it would have to show what a prime number of sheets prevents. The deeper question, a criterion that decides realisability from the shapes alone without a search or a character sum, remains open over the sphere; over every other surface it is two lines long.

Search first, and look at what fails #

A covering of the sphere by itself with three branch points is three permutations with product one and a single orbit, and every list of shapes that passes the Riemann–Hurwitz count can be decided by fixing one permutation and trying every candidate for another. Up to thirteen sheets, 150 lists fail — 1, 5, 13, 6, 34 and 91 for four, six, eight, nine, ten and twelve sheets — and none in prime degree. Two pairings of the sheets can only multiply to two equal cycles, which explains one family; failures crowd onto shapes with equal cycles and never involve a full cycle; and eight ten-sheet failures have no explanation except the search itself.