Covering a surface multiplies its count

A covering of a wedge of circles is nothing but a permutation for each loop: say where each loop sends each sheet, and the whole covering is determined. Any choice of permutations will do, as long as together they reach every sheet, because a wedge of circles has no relations for them to respect.

A closed surface has one. And that single relation is enough to make the covering’s size decide its shape.

A $d$-sheeted covering of the closed surface of genus $g$ is a choice of $2g$ permutations of the sheets whose commutators multiply to the identity, and its Euler characteristic is $d$ times the base’s. So its genus is forced to be $d(g - 1) + 1$, whatever the permutations are, and one surface can cover another only if the second’s Euler characteristic divides the first’s.

One face, and one relation #

Every closed orientable surface of genus $g$ can be built from a single polygon with $4g$ sides by gluing them in pairs according to the word $a_1 b_1 a_1^{-1} b_1^{-1} \cdots a_g b_g a_g^{-1} b_g^{-1}$. After the gluing, all the polygon’s corners become one point, its sides become $2g$ loops through that point, and its interior becomes one face. That is one vertex, $2g$ edges and one face.

The loops generate the surface’s fundamental group, and cutting the surface to find its group gives exactly one relation. Going once round the boundary of the face is a loop that, unlike a loop round a handle, can be pulled across the face and shrunk to a point — so the product of the commutators is trivial.

A covering assigns a permutation to each loop, as it did for a wedge of circles. But now the face has to lift too. Start on some sheet at a corner and walk round the face’s boundary, applying the permutation of each side as it is crossed — forwards for a side read forwards, backwards for a side read backwards. The face lifts to that sheet only if the walk comes back to the sheet it started on. For every sheet to carry a face, the product of the commutators must send every sheet to itself: it must be the identity permutation.

In the first figure the walk can be followed by eye. Starting on sheet 1, the side $a_1$ leads to sheet 2; $b_1$ is the identity, so the walk stays; $a_1^{-1}$, read backwards, leads back to sheet 1; $b_1^{-1}$ and $a_2$ leave it there; $b_2$ leads to sheet 2; $a_2^{-1}$ leaves it; and $b_2^{-1}$ returns it to sheet 1. The route is 1, 2, 2, 1, 1, 1, 2, 2, 1, and it closes. It closes for a reason visible in the choice of permutations: in each commutator one of the two is the identity, and a commutator with the identity in it undoes itself.

That is the condition a graph never imposed. Two permutations of three sheets that do not commute — $(1\,2)$ for $a$ and $(2\,3)$ for $b$ — describe a perfectly good covering of the wedge of two circles, and no covering of the torus at all. Their commutator cycles all three sheets, so a walk round the torus’s square starting on sheet 1 ends on another sheet, and the square has nowhere to go.

On the torus the word is $a b a^{-1} b^{-1}$, and asking it to be the identity is asking that $a$ and $b$ commute. That is a real restriction, and it can be counted. Of the 36 ordered pairs of permutations of three sheets, 18 commute, and 8 of those reach every sheet. Two pairs that differ only by swapping the names of sheets 2 and 3 describe the same covering with the same starting sheet, so the 8 come in pairs: the torus has exactly four connected three-sheeted coverings. The wedge of two circles, where nothing has to commute, has 26 pairs reaching every sheet and so thirteen.

The cells, counted upstairs #

When the condition holds, the cover’s cells can be listed directly. There are $d$ points over the base’s vertex, one for each sheet, so $d$ vertices. There is one edge over each of the base’s $2g$ edges on each sheet, so $2gd$ edges. And there is one face on each sheet, so $d$ faces. The figures build these cells explicitly, walking each face’s boundary through the permutations, rather than quoting the counts.

Two checks turn the list into a surface. Every lifted edge must lie on exactly two faces, once read forwards and once read backwards — which is what makes the result a closed orientable surface with nothing left dangling — and the figure requires it of all $2gd$ edges. And the permutations must reach every sheet from every other, which makes the cover connected.

A third check is silent in the formula and worth making once. Each face has $4g$ corners, so the cover has $4gd$ corners in all, shared among $d$ vertices. In a covering the neighbourhood of every point upstairs is an exact copy of the neighbourhood of the point below it, and around the base’s one vertex all $4g$ corners of the polygon meet. So every vertex upstairs gathers exactly $4g$ corners — $4gd$ corners over $d$ vertices can come out no other way. A vertex that gathered more would be a place where several copies of the base’s vertex had been pressed into one point, and that is exactly what a covering is not allowed to do.

Then the Euler characteristic is a subtraction:

$$\chi(\text{cover}) = d - 2gd + d = d\,(2 - 2g) = d\,\chi(\text{base}).$$

The count multiplies by the number of sheets, and since a closed orientable surface of genus $h$ has $\chi = 2 - 2h$, the cover’s genus is $h = d(g - 1) + 1$. For the first figure that is $3 \times 1 + 1 = 4$.

Nothing in the subtraction used the particular permutations. Any three-sheeted covering of the genus-2 surface, with any permutations that pass the condition, has genus 4. The permutations decide which surface of genus 4 sits over the base and how it sits there; the number of sheets alone decides its genus.

The same multiplication, on a graph #

The multiplication is not special to surfaces, and the version for graphs has an essay of its own.

A subgroup of index three in the free group on two letters is free on four, and in general a subgroup of index $d$ in a free group of rank $r$ is free of rank $d(r - 1) + 1$. That is the graph formula $\chi(\text{cover}) = d\,\chi(\text{base})$ in disguise, since a connected graph of Euler characteristic $\chi$ has a free fundamental group of rank $1 - \chi$. The rank formula for free groups and the genus formula for surfaces are one multiplication, applied to a space with no faces and to a space with one.

What the surface adds is the face and its condition. A graph accepts any permutations; a surface accepts only those whose commutators cancel, and the count describes only the coverings the condition lets through.

The same multiplication, measured as area #

There is a second reason the count had to multiply, and it involves no cells at all. The corners of a solid fall short of flat by gaps that always add to two full turns, and the smooth form of that statement is the Gauss–Bonnet theorem: on a closed surface, the total curvature is $2\pi$ times the Euler characteristic.

A covering carries the base’s shape upstairs, sheet by sheet. Give the genus-2 surface a shape of constant curvature $-1$, which it can have; its total curvature is $2\pi \times (-2) = -4\pi$, so its area is $4\pi$. Every small patch of the base has $d$ identical copies upstairs, curved the same way, so the cover also has curvature $-1$ everywhere, and its area is $4\pi d$. Its total curvature is $-4\pi d$, and dividing by $2\pi$ gives its Euler characteristic, $-2d$, without a single cell having been counted.

The multiplication is area multiplying, read through a theorem that turns area into a count. It explains the torus from the other side too: a flat torus has total curvature nought however large it is made, so any cover of it is flat as well, and the only flat closed orientable surface is a torus.

What the multiplication forbids #

The formula is most useful for what it rules out, because it applies before any cover has been constructed.

The sphere has no connected covering with more than one sheet. A $d$-sheeted cover of the sphere would have Euler characteristic $2d$, and no closed surface has an Euler characteristic above 2. The count can only speak about coverings with finitely many sheets, so it does not by itself prove that every loop on the sphere shrinks; but it agrees with that fact, and it reaches the finite half of it without drawing a single loop.

The torus is covered only by tori. Its Euler characteristic is nought, so every cover’s is nought too.

Each covering of the torus by a torus is a choice of smaller lattice inside the grid of whole-number pairs the torus is the quotient of: the loops that close up upstairs. The loops whose lifts close are the subgroup the covering corresponds to, and in the torus’s group — pairs of whole numbers under addition — the subgroups of index $d$ are counted by the sum of the divisors of $d$: three for two sheets, four for three, seven for four, which is what counting commuting pairs gives as well. The four three-sheeted lattices are easy to name. Two are the square stretched three times, along $a$ or along $b$; two are sheared, the figure’s, where $m + n$ is a multiple of three, and its mirror image, where $m - n$ is. And because that group is commutative, every one of them is as symmetric as a covering can be, with one symmetry for each sheet.

For genus 2 and above, the genera that can cover are an arithmetic progression. A surface of genus $h$ covers one of genus $g \ge 2$ with $d$ sheets only if $h - 1 = d(g - 1)$. So a surface of genus 3 covers one of genus 2 only with two sheets, and a surface of genus 2 covers no orientable surface except itself. Every value the formula allows is realised: letting $a_1$ cycle all $d$ sheets and every other generator act trivially satisfies the condition, since each commutator involves an identity, and it reaches every sheet.

Coverings stack, and the formula stacks with them. The genus-3 surface double-covers the genus-2 surface, the genus-5 surface drawn here double-covers the genus-3 surface, and the two together make a four-sheeted covering of genus 2 by genus 5 — which is what $d(g - 1) + 1$ gives for four sheets over genus 2, since $4 \times 1 + 1 = 5$. The formula composes because it is really a statement about $g - 1$, and $g - 1$ simply multiplies: 1, then 2, then 4.

Where the multiplication needs its hypotheses #

The covering must be unbranched. Every point of the base has exactly $d$ points above it, which is what made the vertex count $d$. If going round some point is allowed to leave the sheets permuted, several sheets meet at a single point above it, the count of points there drops, and the Euler characteristic stops multiplying.

The cover must be connected. Disconnected covers exist for any permutations that fail to reach every sheet; they are several covers side by side, and each piece obeys the formula with its own number of sheets.

Orientability is not needed for the multiplication, only for the drawing. The projective plane has Euler characteristic 1, and it is a disc sewn to a Möbius band; its two-sheeted cover is the sphere, Euler characteristic $2 = 2 \times 1$, and the multiplication holds. A one-sided surface has a different boundary word with no commutators in it, so the figures draw orientable surfaces only.

And the drawing is of cells, not of the glued surface. The octagons in the first figure are drawn apart, with numbers saying how they glue; the surface of genus 4 they make is never drawn, and its genus is read off the count rather than seen.

The surface of genus 4 is counted, never assembled #

The genus of a cover is a statement about a surface nobody in the figures has assembled. What the figures establish is that the cells close up correctly — every face returns to its own sheet, every edge lies on exactly two faces — and that the counts multiply. That a closed orientable surface is determined by its Euler characteristic is the classification theorem, which is quoted.

They also show one covering of each kind and not the number of coverings of each kind. How many $d$-sheeted covers a surface of genus $g$ has is a question about counting solutions of the commutator equation among the permutations of $d$ sheets, and it has an exact answer that the figures do not compute.

For two sheets over the genus-2 surface the count is small enough to make by hand. Permutations of two sheets always commute, so every one of the 16 choices for the four edges passes the condition, and all but the choice of four identities reach both sheets: the genus-2 surface has fifteen connected double covers, every one of genus 3, and nothing in the formula tells them apart. With three sheets, an exhaustive search of the 1,296 choices finds 440 that pass and reach every sheet — 220 coverings, once sheets 2 and 3 may swap names — all of genus 4. For the torus it is the divisor sum above; for higher genus the answer involves every way the permutations of $d$ things can be represented, and the pictures give no hint of it.

Still open: what happens where the boundary does not close #

The condition that the commutators multiply to the identity was needed at exactly one place: the walk round the face. A covering that satisfies it carries a separate copy of the face on every sheet. One that does not cannot: the walk starting on one sheet ends on another, so the copies of the face on those sheets have to be joined edge to edge into a single larger polygon, wound more than once around the base’s face.

That polygon still makes a surface, but its sheets all meet at one point, in the middle of the face — the one point downstairs round which the sheets fail to return. The result is a branched covering, with fewer points above that one than there are sheets, and the multiplication in the formula has to be corrected by exactly how many points were lost. What that correction is, and whether the corrected count is enough to say which branched coverings exist, is the question this page leaves.

A count that multiplies is an obstruction #

The habit is about reading a multiplicative invariant.

The Euler characteristic was introduced as a count that does not care how a surface is cut into cells. Under a covering it does something stronger: it multiplies by the number of sheets. A quantity that multiplies under a construction turns every failure of divisibility into an impossibility, without any search. The sphere has no connected cover of more than one sheet because $2d$ exceeds 2 for every $d$ above one; a genus-2 surface cannot cover a genus-3 surface because $-2$ is not a multiple of $-4$.

The same move works wherever a construction scales a count: the order of a subgroup divides the order of a finite group, the degree of a field extension multiplies along a tower of fields, the number of sheets multiplies along a tower of coverings. Each time, the first question to ask of a proposed construction is whether its counts divide — and a surprising number of proposals fail there, before any of the work of building them has been done.