What a branch point subtracts

Over every point of the sphere but four, the square root of $(x - 1)(x - 2)(x - 3)(x - 4)$ has two values, and following a small loop round any one of those four points swaps them. So its values form a two-sheeted surface over the sphere — except at the four points, where the two sheets meet in one.

An unbranched covering multiplies the Euler characteristic by the number of sheets, and two sheets over the sphere would have to be a surface of Euler characteristic 4, which does not exist. The square root’s surface does exist. It is a torus, and its Euler characteristic is nought: exactly four less than the multiplication promised, one for each point where two sheets became one.

A branched covering of the sphere with $d$ sheets is a permutation for each branch point, the permutations multiplying to the identity, and its Euler characteristic is $2d$ minus, at each branch point, the number of sheets less the number of that permutation’s cycles. Every point where sheets meet takes away exactly what it merged. And the count is necessary without being sufficient: there are lists of cycles that pass it and describe no covering at all.

Cutting the sphere into a star #

To count a covering’s cells, the base has to be cut into cells that the branching respects. Mark the $k$ branch points on the sphere, pick one more point as a centre, and join the centre to each branch point by a spoke that avoids the others. That is $k + 1$ vertices and $k$ edges, a star; what is left of the sphere when the star is removed is a single disc, one face. The count is $(k + 1) - k + 1 = 2$, as it must be for anything drawn on a sphere.

The loops that matter are small loops round the branch points, each starting at the centre, running out along its spoke, circling the point and returning. Around each one the sheets are permuted, exactly as a loop in a covering permutes them, and that permutation is what the figures print under each point.

The disc supplies the one relation. Its boundary runs round every branch point in turn, and since the disc can be shrunk to a point, going round all of them in order must return every sheet to where it started. The product of the permutations, taken in order round the centre, is the identity — the sphere’s version of the relation a closed surface’s one face imposes on its loops.

One vertex for each cycle #

Lifting the cells is the same procedure as for an unbranched covering, with one change.

Over the centre, which is not a branch point, there are $d$ vertices, one on each sheet. Over each spoke, whose inside meets no branch point, there are $d$ edges, so $dk$ in all. Over the disc there are as many faces as the permutation round its boundary has cycles; that permutation is the identity, so there are $d$ faces, one per sheet.

The change is over a branch point. A small loop round it sends sheet $s$ to sheet $p(s)$, then on to $p(p(s))$, and so round a cycle, and every sheet in that cycle can be reached from every other by circling the point. Above the point itself they cannot be kept apart: a cycle of length $m$ looks, close to the point, like the map $z \mapsto z^m$ close to nought, where $m$ sheets wind round one another and meet in a single point. Over a branch point there is one vertex for each cycle of its permutation. The figures do not read that number off the cycles; they merge every pair of sheets the permutation joins and count what is left, which is the same number found a different way.

Adding up, with $c_i$ the number of cycles at the $i$-th point:

$$\chi = d + \sum_i c_i - dk + d = 2d - \sum_i (d - c_i).$$

That is the Riemann–Hurwitz formula for coverings of the sphere, and a subtraction was all it took. The term $d - c_i$ is exactly how many vertices were lost at the $i$-th point: $d$ sheets would have given $d$ points above it, and they gave $c_i$. Over a base of Euler characteristic $\chi_0$ the same argument, with a different cell structure below, gives $d\,\chi_0$ minus the same losses.

Two sheets need an even number of points #

With two sheets there is only one permutation that branches, the swap $(1\,2)$, and it loses one vertex each time. So $k$ branch points give $\chi = 4 - k$ and genus $(k - 2)/2$.

The product condition decides which $k$ are allowed. The swap done $k$ times is the identity exactly when $k$ is even, so a two-sheeted covering of the sphere branches at an even number of points — never three, never five. The square root of a polynomial with five roots therefore has to branch at infinity as well: a large loop enclosing all five roots swaps the two values five times, which is a swap, so the point at infinity is a branch point too, and the true count is six.

The pattern continues without a gap: two sheets over $2g + 2$ points give genus $g$, so every genus arrives as the square root of some polynomial. The torus in the first figure is the case of four points, and it is why a curve given by $y^2$ equal to a cubic is a torus — three roots and infinity. A surface of each genus can be built by gluing the sides of a polygon; here it is built by letting two sheets meet, and the number of meeting points decides which one arrives.

The parity every list obeys #

The evenness is not special to two sheets. Every permutation has a sign, plus for an even number of swaps and minus for an odd number, and a permutation of $d$ sheets with $c$ cycles has sign $(-1)^{d - c}$, since a cycle of length $m$ is a product of $m - 1$ swaps. The permutations round the centre multiply to the identity, whose sign is plus, so the losses $d - c_i$ always add to an even number.

That is why the Euler characteristic the count produces is always even, as a closed orientable surface’s must be. The figures check that it is, and a list for which it came out odd would already have failed the product condition. Parity is the first thing the count asks of a list of cycles, before anything else about it is tried.

Cycles of three and of four #

When more than two sheets meet at a point, one point can take away more than one vertex.

This is the cube root of $x(x - 1)$. Circling nought multiplies the cube root by a third of a turn; circling one does the same; and the loop round infinity, taken in the direction that makes the three multiply to the identity, does it a third time. The surface is a torus, like the square root’s over four points, reached by a different branching: three points losing two vertices each instead of four points losing one.

The power map is the extreme case. All four sheets meet at each of two points, each point takes away three, and $8 - 6 = 2$: the covering surface is the sphere again. The fourth root of $z$ is four-valued everywhere except at nought and infinity, and the surface it lives on is no more complicated than the sphere it was defined on.

A sphere over a sphere loses twice the sheets, less two #

Setting the cover’s Euler characteristic to 2 in the formula gives the condition for a sphere to cover a sphere: the losses must add to $2d - 2$. The power map spends them at two points. The opposite extreme spends them one at a time, with only two sheets meeting at each point.

A rational function of degree $d$ — a ratio of polynomials — is exactly such a covering of the sphere by the sphere, so its critical points lose $2d - 2$ in total, and a typical one of degree 3 has four, each simple, which is the picture. A polynomial of degree $d$ spends $d - 1$ of them at infinity, where all $d$ values run off together as one cycle, and has $d - 1$ left for its finite critical points: the roots of its derivative, which has degree $d - 1$. For $z^3 - 3z$, the derivative vanishes at $1$ and $-1$, the values there are $-2$ and $2$, and the count is $6 - 1 - 1 - 2 = 2$.

Moving the branch points about does not change any of this. Carrying one point round another changes its permutation to a conjugate of itself, which has the same cycles, so the losses — and with them the surface — depend only on which cycle shapes occur, not on where the points sit or in what order the spokes leave the centre.

Passing the count is not enough #

The last covering has four sheets and three branch points, and at each of them the sheets meet in two separate pairs.

It is the sphere divided by its half-turns about three perpendicular axes. Each axis meets the sphere at two points that its half-turn fixes, so above each of three points of the quotient lie two points, each where two of the four sheets meet. The three permutations and the identity multiply among themselves — any two of the three give the third — which is the symmetry a covering has when it is a quotient by a group.

Now change the third point. Instead of two pairs of sheets meeting, let three sheets meet and the fourth stay apart: a cycle of three and a fixed sheet. That point also loses two vertices, so the count is unchanged — $8 - (2 + 2 + 2) = 2$ — and the losses still add to an even number. Everything the formula and the parity can see says a sphere should cover the sphere this way.

No such covering exists. The first two points must each carry a permutation made of two disjoint swaps, and there are only three of those, which with the identity are closed under multiplication, as the figure’s three are. So the product of the first two is the identity or another permutation of the same shape, and the third would have to undo it — which a cycle of three cannot do. Running through all 72 lists of these shapes finds none that multiplies to the identity; the neighbouring shape, with the three-cycle point split into two points where a single pair of sheets meets, has 24 that do and connect every sheet. The count is necessary. It is not sufficient.

Why the sphere is the hard case #

The same shapes behave differently over a torus, and the difference shows exactly what the sphere lacks.

Over a torus the star is drawn on the square with its sides glued, and the one face’s boundary now runs along the torus’s own loops $a$, $b$, $a^{-1}$ and $b^{-1}$ as well as round the branch points. So the relation becomes: the commutator of the two loops, times the branch points’ permutations, is the identity. The branch points no longer have to cancel among themselves. They only have to multiply to something a commutator can undo.

And a commutator can undo almost anything. A commutator is always an even permutation, since it contains each of two permutations once forwards and once backwards; for three, four and five sheets a search of every pair shows that every even permutation arises as one — all 3, all 12 and all 60 of them — and a classical theorem of Ore says the same for any number of sheets. The parity condition says precisely that the branch points multiply to an even permutation. So over a torus the count and the parity leave nothing further to check. The shapes that failed over the sphere — two points of two pairs, one point of a three-cycle — are realised over the torus: a search finds 2,592 lists of permutations that do it, each a surface of Euler characteristic $4 \times 0 - 6 = -6$, genus 4.

The obstruction that remains over a torus is parity alone. Three sheets over a torus with a single point where two sheets swap is impossible, because a swap is odd and a commutator is not; three sheets with a single point where all three meet is realised, as a surface of genus 2. The sphere has no loops of its own to put into the relation, and so no commutator to absorb what the branch points leave over. Its branch points must cancel one another exactly, and whether given shapes can do that is a question about the permutations themselves, which the count does not see.

Where the subtraction needs its hypotheses #

The branch points are finitely many and isolated. The star needs a spoke to each, and the local picture of $z \mapsto z^m$ needs a small disc round each point containing no other.

The base in the figures is the sphere. For any other closed surface the star is replaced by a cell structure with the branch points among its vertices, and the count becomes $d\,\chi_0$ minus the same losses. The surface a polygonal billiard table unfolds into has its genus computed by a count of exactly that shape, with the corners of the table as the points where copies of it meet.

The cover is connected. The permutations must reach every sheet from every other, and the figures require it. Otherwise the cover falls into pieces, and the formula applies to each piece with its own number of sheets.

The surfaces are orientable. A cycle of length $m$ becoming one point is the local picture of a map between oriented surfaces; the one-sided surfaces have their own version of the count, which the figures do not draw.

Cells in rows, and functions the figures never see #

Each figure draws cells as rows and columns, not the surface they make. The torus, the genus-2 surface and the spheres are named by their Euler characteristics, which leans on the classification of surfaces rather than on anything visible. The local picture of $z \mapsto z^m$, which turns a cycle of length $m$ into one point, is quoted and not drawn.

And the identifications with square roots, cube roots and power maps are arguments in the prose about which permutations those functions produce. The figures receive the permutations and know nothing about the functions; that circling a root of a polynomial swaps its square roots is a fact about complex numbers, not about cells.

Still open: which lists of cycles the sphere allows #

Whether a given list of cycle shapes is realised by some covering — the Hurwitz existence problem — splits sharply by the base. When the base is a 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, building on earlier work of Husemoller. On the sphere they are not enough, as four sheets over three points showed, and no criterion is known that decides every case. Many families of exceptions have been found and many families shown to be realised; a conjecture that has stood since the 1980s holds that when the number of sheets is prime, passing the count is always enough. It is still a conjecture.

Counting realisations is a different question, and it has an exact answer: a formula of Frobenius, written in the characters of the symmetric group, gives how many lists of given shapes multiply to the identity. It says how many, and it can say zero; what nobody has is a rule that says zero without doing the sum.

A formula that says what it lost #

The habit is correcting a multiplicative count by local losses.

The unbranched count was a multiplication, and it broke the moment sheets were allowed to meet. The branched count keeps the multiplication and subtracts, at each exceptional point, the difference between what was expected there and what was found. The subtraction is local — it depends on one point’s permutation and nothing else — while the multiplication is global. The angle defects at the corners of a solid work the same way: a total that would be flat, corrected by what each corner fails to be.

When a count stops holding, measure what it lost where it broke. Often the losses sit at a few points, each can be computed alone, and the corrected count is as exact as the original. And once a count has been corrected that way, the next question is the one this page ends on: whether passing it is enough, or merely required.