A count that can say zero

What a branch point subtracts finishes by exhibiting a failure. Four sheets over three points of a sphere, with two of the points joining the sheets in pairs and the third joining three of them and leaving one alone: the Riemann–Hurwitz count comes out at $2$, the losses add to an even number, and everything the count can see says a sphere covers a sphere this way.

No such covering exists, and the reason is a fact about permutations rather than about surfaces — there is no way to choose the three permutations so that they multiply to the identity.

That raises a question the count asks and does not answer. Which lists of cycle shapes are realised, and how many ways?

The second-last column is the answer, and there is an exact formula for it that involves no permutations at all.

What is being counted #

A branched covering of the sphere with $d$ sheets and $k$ branch points is, by the monodromy reading, a permutation of the $d$ sheets for each branch point, with two conditions: the permutations multiply to the identity, because the disc left when the star is removed can be shrunk to a point; and they act transitively, because the covering surface is connected.

So the existence question is entirely about permutations. Given cycle shapes $C_1, \ldots, C_k$ — partitions of $d$ — is there a list of permutations of those shapes with product the identity, connecting everything?

Two necessary conditions follow immediately and both are visible in the table.

The count. Each branch point loses $d$ minus the number of cycles in its permutation, and the total losses give $\chi = 2d - \sum (d - c_i)$. That has to be the Euler characteristic of a closed orientable surface, so it is even and at most two.

The parity. A permutation with $c$ cycles on $d$ letters is a product of $d - c$ transpositions, so its sign is $(-1)^{d-c}$. The product being the identity forces the signs to multiply to plus, hence the losses to add to an even number — which is the same statement as $\chi$ being even, arrived at from the group instead of from the cells.

The table’s fifth row passes both and has nothing. The two double swaps give losses of $2$ each; the three-cycle-with-a-fixed-point also loses $2$; six in all, even, and $\chi = 8 - 6 = 2$. And the enumeration over all nine candidate pairs of double swaps finds no third permutation of the right shape.

Why that row fails, in one sentence #

The reason can be stated without any search, and it is worth having because it is the only case in this essay where the obstruction is legible.

The permutations of shape $2+2$ in $S_4$ are exactly three: $(1\,2)(3\,4)$, $(1\,3)(2\,4)$ and $(1\,4)(2\,3)$. Those three together with the identity form a group — the Klein four-group, closed under multiplication, which is the smallest group that is not a cycle. So the product of the first two branch points’ permutations is inside that group of four, and the third permutation must be its inverse, which is also inside it.

A three-cycle is not inside it. There is nothing else to check.

The neighbouring list — the same two double swaps with the third point split into two points where a single pair of sheets meets — has twenty-four realisations. So the failure is not about the numbers being tight; it is about one particular set of permutations happening to be closed under multiplication, which no count of losses could possibly detect.

The formula, and where it comes from #

The exact count is Frobenius’s, from 1896, and it computes the number of lists from given classes with product the identity using nothing but the character table of the symmetric group.

$$N \;=\; \frac{|C_1|\cdots|C_k|}{|G|}\;\sum_{\chi} \frac{\chi(C_1)\cdots\chi(C_k)}{\chi(1)^{\,k-2}}$$

The sum runs over the irreducible characters of $G = S_d$, $\chi(C)$ is the character’s value on that class, and $\chi(1)$ is its degree.

The derivation is a page of representation theory and the shape of it is worth stating. The number of ways to write the identity as a product from given classes is a coefficient in the group algebra, class sums multiply to combinations of class sums, and the characters diagonalise that multiplication — so the whole count becomes a sum of products of eigenvalues, one per irreducible representation. The formula is an eigenvalue computation for a multiplication table, and the character table is the change of basis that makes it diagonal. That is the same manoeuvre as diagonalising a matrix to read off what iterating it does, applied to a multiplication with no matrix in sight.

What matters here is that it has nothing to do with surfaces, branch points or coverings. It is a statement about a finite group, and the covering question was translated into one.

The list one shape away from the failing one does exist, and drawing it makes the difference legible: the three permutations there, together with the identity, are closed under multiplication, and the failing list asks a three-cycle to belong to that closed set.

Putting the three four-sheeted rows beside each other is the cleanest statement of what the count does not know.

The two computations, and why both are run #

Every figure here computes the count twice.

By enumeration. Fix the first $k-1$ permutations, and the last is determined — it is the inverse of their product. So the work is the product of the first $k-1$ class sizes, and checking whether the forced permutation has the right shape is immediate. For four sheets and three shapes that is nine candidate pairs; for five sheets it is a few thousand.

By the character sum. The formula above, over the table of the relevant symmetric group.

The two agree in every row, and the agreement is the figure’s point rather than a convenience. A table of characters typed into a file is exactly the kind of object where one wrong entry produces plausible answers, so the tables are also checked against themselves: the rows are required to be orthogonal and normalised, which is a set of $n^2$ conditions that no single wrong entry survives.

The enumeration also produces something the formula cannot: how many of the lists are connected. Frobenius counts all lists with product the identity, including those whose permutations split the sheets into independent groups — and such a list describes not one covering but several disjoint ones. So the count that answers the covering question is the enumeration’s second number, and the formula gives an upper bound for it that happens to be exact whenever there is no way to split.

What the formula is worth #

The state of affairs is unusual enough to state plainly.

There is an exact answer. Given any list of cycle shapes, the number of coverings is a finite sum over a finite table, computable without search.

It can return nought. The failing row’s answer is zero, and it is zero because five terms cancel.

And nobody has a criterion that says zero without doing the sum. That is the Hurwitz existence problem, and it has been open since the 1890s in the case that matters.

The problem splits sharply by the base surface, and the split is the informative part. Over any closed orientable surface other than the sphere, the count and the parity are enough — every list passing both is realised, a theorem of Edmonds, Kulkarni and Stong from 1984. The reason is the one the branched count gives: over a surface with loops of its own, the relation the branch points must satisfy is that their product is a commutator rather than the identity, and every even permutation is a commutator, so parity is the only constraint left.

The sphere has no loops to absorb anything, so its branch points must cancel exactly among themselves, and whether given shapes can do that is a question about the symmetric group’s multiplication rather than about any count.

What is known is a long list of families: many families of exceptions have been found, many families shown to be realised, and a conjecture standing since the 1980s holds that when the number of sheets is prime, passing the count is always enough. It is still a conjecture, and the evidence for it is exactly the kind this essay’s figures produce — exhaustive verification at small sizes.

The counts themselves, and what they are counting twice #

The third column of the table is worth reading for its own sake, because the numbers in it are not arbitrary and they are counted by more than one thing.

Two sheets over four swaps: one covering. Three sheets over three three-cycles: two. Four sheets over two four-cycles and a double swap: six. Five sheets over two five-cycles and a double swap: a hundred and twenty.

Those are Hurwitz numbers, and they count more than coverings. The same numbers count factorisations of a permutation into transpositions, and they count certain labelled maps drawn on surfaces.

The cleanest instance of that is a theorem of Dénes from 1959, and it is checkable by hand at small sizes. The number of ways to write a cycle on $d$ letters as a product of $d-1$ transpositions is $d^{\,d-2}$ — three ways for a three-cycle, sixteen for a four-cycle, a hundred and twenty-five for a five-cycle.

That expression is Cayley’s formula for the number of labelled trees on $d$ vertices, and the coincidence is not one. A minimal factorisation into transpositions is a connected graph on the $d$ letters with $d-1$ edges, which is a tree; the ordering of the factors is what the counting has to account for; and the two problems turn out to be the same problem. So the sixteen trees on four points are sixteen ways of writing a four-cycle as three swaps, and the covering-space reading of that is sixteen ways for a sphere to cover a sphere with simple branching at three points.

So the quantity the character sum evaluates is also a count of trees, also a count of factorisations, and — through the ELSV formula of 1999 — also an intersection number on a space of curves. A number with four descriptions is a number worth having, which makes the absence of a fifth description — one that says when it is zero — the more conspicuous.

Why prime might be special #

The conjecture’s shape is worth explaining, because it makes the failing row look less like an accident.

A list of shapes that fails typically fails because some subgroup gets in the way, as the Klein four-group does above. Subgroups of $S_d$ that can trap a product this way are constrained by how $d$ factors — the four-group arises from $4 = 2 \times 2$, and its three double swaps are exactly the ways to split four things into two pairs.

When $d$ is prime there are far fewer such structures. A transitive group of prime degree is severely restricted — it is either doubly transitive or a subgroup of the affine group of the prime field, by a classical theorem of Burnside — and neither kind produces the closure that traps the product here.

So the conjecture is that the obstructions are all of the kind exhibited above, and that they need a composite number of sheets to exist. That is a plausible statement about groups and it has resisted proof for forty years, which is the usual situation when a question about all cases has only case-by-case answers.

What happens when the count is not one #

There is a distinction the table’s last two columns make that is worth pulling out, because it is the difference between counting coverings and counting something slightly larger.

A list of permutations that does not act transitively describes a disconnected cover — several sheets forming one surface, the rest forming another, each an honest branched covering with fewer sheets. Frobenius’s formula counts those too, since it only asks about the product. So the two columns can differ, and where they do the extra lists are coverings of smaller degree in disguise.

In the six rows drawn they never differ, which is worth explaining rather than passing over. A list whose permutations split the sheets into two groups has every one of its cycle shapes refining that split, so each shape must be a partition of the group sizes — and every shape in these rows has a part too large for any split to survive. The transitivity is forced by the shapes, in every row, which is why the two columns agree.

They would not agree for a list of, say, four sheets with three double swaps chosen to preserve the split into $\{1,2\}$ and $\{3,4\}$. That is a real possibility and the third row’s six connected coverings are what is left after such lists are discounted; the row’s total happens to be six as well, because none of the nine candidate pairs produces a split-preserving triple with the right shapes.

The general moral is that a formula counting algebraic objects and a question about geometric ones rarely line up exactly, and the gap is usually connectivity. Checking that the gap is empty is a separate computation, and it is the one the enumeration does and the character sum cannot.

A cancellation drawn is still a cancellation #

Every row is verified and nothing is proved. Each count is exact for its own list of shapes, and the general question is about infinitely many lists. Exhaustive verification at four and five sheets is what the figures do, and it is evidence for the conjecture rather than progress on it.

The character table is typed in and checked, not derived. Building it from the Murnaghan–Nakayama rule would be a second computation; instead the tables are required to be orthogonal and normalised, which is a set of conditions a mistyped entry cannot pass. That is a strong check and it is not the same as a derivation.

And no covering is drawn. The branched count draws cells in rows and this page draws numbers, so a reader never sees the six coverings the third row counts. They are six lists of permutations, and what they have in common as surfaces — all six are tori — is read off the Euler characteristic rather than seen.

The cancellation is visible and is not an explanation. The Frobenius figure shows five terms adding to nought, which is exactly what happens, and there is no sense in which the reader learns why from it. That is the essay’s subject rather than a defect of the figure: the formula computes the answer and supplies no reason, and drawing it honestly means drawing an answer with no reason attached.

Still open: a criterion that says zero without doing the sum #

The Hurwitz existence problem over the sphere is open. What is wanted is a rule, checkable from the shapes, that decides realisability without evaluating the sum — and the analogy that makes the absence uncomfortable is that over every other base such a rule exists and is two lines long.

Two smaller questions sit beside it. The prime-degree conjecture is one. The other is about the numbers rather than their vanishing: the counts in the third column are the Hurwitz numbers, and they turn out to have generating functions with a great deal of structure — they count maps between Riemann surfaces, they satisfy recursions, and they are connected to intersection numbers on moduli spaces by the ELSV formula of 1999. So the quantity this essay computes by a character sum has an interpretation as a geometric count on a completely different object, and that interpretation has produced more information about the numbers than about which of them are zero.

A formula that answers and does not explain #

The habit is about what to expect from a closed form.

A formula that returns the right number is usually taken to be the end of a question, and here it is not. Frobenius’s sum is exact, fast, and complete; it decides existence for every list of shapes; and after using it on the failing row one knows that no covering exists and nothing about why. The reason came from somewhere else entirely — three permutations happening to form a group with the identity — and no amount of staring at the character sum produces it.

So a closed form is an answer and a proof is a reason, and the two can be genuinely separate. The same shape appears wherever a determinant or a character sum evaluates something combinatorial: a determinant counts spanning trees exactly and says nothing about any particular tree, and a generating function’s coefficient counts objects it never describes.

The question worth asking of any such formula is what it would take to read the answer off it — and when the answer is a cancellation, usually nothing will.