Two copies of the polygon, cross-matched

Two sheets over a one-sided surface built a two-sided surface above the projective plane by using a symmetry the icosahedron happens to have: every face has an opposite face, and identifying the opposites leaves the projective plane, with the sphere lying over it two to one. That construction is exact and it is beautiful, and it works because a convenient solid exists. Over the Klein bottle a torus was found the same way, by a symmetry of the torus. What that essay did not have was a construction that needs no convenient symmetry — one that starts from any surface and produces the two-sided surface over it.

The gluing word is enough. Every closed surface is a polygon with its edges glued in pairs, as every surface is a sphere with handles set out, and the word that records the gluing also records which pairs reverse the sense of turning. The cover is two copies of the polygon, glued according to what the word says about each letter.

The recipe, and why it gives a sense of turning #

Take the polygon and a second copy of it. Give the first copy the sense of turning its word is read in, anticlockwise; give the second the opposite sense, which is what drawing it as a mirror image does. Now go through the letters.

A letter used once each way round — $a$ and later $a^{-1}$ — is a gluing that already respects the sense of turning: walking across that edge from inside the polygon brings a traveller back into the polygon with the same sense, as orientation is a sign read off from the determinant of the frame. So glue that letter’s two edges within each copy, the first copy to itself and the second to itself.

A letter used twice the same way round — $a$ and later $a$ again — is a gluing that reverses the sense. A traveller crossing it comes back mirrored. So glue its edges across the copies: the first edge of the first copy to the second edge of the second copy, and the first edge of the second copy to the second edge of the first. The traveller who would have come back mirrored comes back on the other copy instead, where mirrored is correct.

That is the whole construction. And it proves itself two-sided, by a check that is done for every pair of edges in the figures: with the first copy read anticlockwise and the second clockwise, every pair of edges glued in the cover meets with opposite arrows — the condition for a gluing to preserve the sense of turning. A letter glued within a copy meets with opposite arrows because it was used once each way. A letter glued across meets with opposite arrows because it was used twice the same way and one of the two copies is read backwards. The two cases were designed to cancel, and they do.

The projective plane, from a square #

The smallest test is the square $abab$, whose two letters are both used the same way.

The count comes out at four vertices, four edges and two faces: characteristic $2$, one piece, two-sided. By the classification those three facts name a sphere, the same sphere the icosahedron built:

The two constructions look nothing alike — twenty triangles in pairs against two squares glued across — and they give the same surface over the same surface. That is what the classification guarantees, since a two-sided surface with characteristic $2$ has no other name. It is also a reassurance about the recipe: on the one case where the answer was already known by a different route, the word-built cover agrees.

The icosahedron has one advantage the squares lack: the covering map is visible, as a symmetry of a solid. The recipe has the advantage that it never needed one.

The Klein bottle, and a letter used each way #

In $abab^{-1}$ the letter $a$ is used twice the same way and $b$ once each way. So $a$ is glued across the copies and $b$ within them.

The picture can be read as a construction by hand. Glue $b$ within each copy first: each square becomes a cylinder, open at the two ends marked $a$ and $a'$. Then glue the ends across, the first cylinder’s $a$ end to the second’s and the second’s $a'$ end to the first’s. Two cylinders joined end to end in a ring is a torus. The bottle that needs a fourth dimension had the Klein bottle as a cylinder whose ends are joined with a flip; the cover undoes the flip by going round twice.

A two-sided surface covers itself twice #

The recipe does not care whether the surface is one-sided. Given the torus’s word $aba^{-1}b^{-1}$, it runs as before — and finds nothing to glue across.

The cover is two disjoint tori. That is the honest answer, and it is the answer two sheets over a one-sided surface gave in words: a two-sided surface has a double cover of this kind, but it is disconnected, one sheet for each of its two orientations. A surface can be given a sense of turning exactly when the two copies do not join.

So the recipe tests orientability as well as building the cover. The copies join up exactly when some letter is used twice the same way, which is the rule orientation is a sign stated for reading orientability off a word, now seen as the condition for a connected cover.

Why the second copy has to be the mirror image #

The mirror image is not decoration. Suppose both copies were given the same sense of turning, anticlockwise. A letter used once each way, glued within each copy, still meets with opposite arrows and preserves the sense. But a letter used twice the same way, glued across, now joins an edge of one anticlockwise copy to an edge of another anticlockwise copy with the same arrows on both — which reverses the sense exactly as it did on the surface below. The cover would inherit the one-sidedness it was built to remove.

Reversing the second copy flips the arrows on every one of its edges, as they are seen from inside it. Across-glued letters, which met with the same arrows, now meet with opposite ones; within-glued letters are unaffected, because both of their edges are on the same copy and both flip. So the two choices — mirror the second copy, and cross exactly the same-way letters — are one choice, made twice. Either one without the other fails.

That also says what a point of the cover is. It is a point of the surface together with a sense of turning at that point: the first copy holds the points with the anticlockwise sense, the second the points with the clockwise sense. Walking along a loop that reverses the sense carries a traveller from one copy to the other, which is the description two sheets over a one-sided surface gave of the cover before it had a way to build one.

Every count doubles, and nothing is pinched #

A double cover must have exactly two points over every point. For the faces and edges that is built in — two copies of the one face, two copies of each edge. For the vertices it is not, because the vertex classes are formed by the gluing, and a careless choice of which letters to cross could merge corners that should have stayed apart, or split a vertex into more than two.

So the figures count. The corners of both copies are pooled by every identification the cover makes, the classes are counted, and the count is checked to be exactly twice the surface’s own. It always is, for the recipe: the vertices of $aabbcc$ are one class on the surface and two in the cover, and those of $abab$ are two and four. Faces double, edges double, vertices double, and so

$$\chi(\text{cover}) = 2V - 2E + 2F = 2\chi.$$

That doubling is the general fact covering a surface multiplies its count proved for covers of any number of sheets. Here it is not only a consequence but a check: a recipe that pinched a vertex would have produced a characteristic that was not twice the surface’s, and it would have been caught.

The whole one-sided list, at once #

With the recipe in hand, the classification of one-sided surfaces translates into a statement about their covers.

The sphere with $k$ cross-caps has characteristic $2 - k$. Its cover is connected, two-sided, and has characteristic $4 - 2k = 2 - 2(k - 1)$. By the classification of two-sided surfaces, a connected two-sided surface is named by its characteristic, and that one is the surface with $k - 1$ handles. So:

• one cross-cap, the projective plane, is covered by the sphere;
• two, the Klein bottle, by the torus;
• three by the surface with two handles, and in general $k$ cross-caps by $k - 1$ handles.

The table checks the first four of these by counting; the argument above proves all of them. The Klein bottle appears twice, from $aabb$ and from $abab^{-1}$, and gets the same cover from both. The cover depends on the surface and not on the word that happened to describe it — as it must, since the reduction of every word driven to a normal form turns one word into the other by cutting and regluing, and each move can be carried out on both copies at once.

Seven covers, and only one of them two-sided #

The recipe crosses the copies on a particular set of letters: the ones used twice the same way. Nothing stops a different choice. Crossing on any other set of letters still gives two copies glued into something — and when the polygon’s corners all become one vertex, the something is always a genuine double cover.

The sphere with three cross-caps has seven connected double covers, one for each non-empty set of its three letters. All seven have characteristic $-2$, because every double cover doubles the count. Six are one-sided — spheres with four cross-caps. One is two-sided, and it is the one the recipe builds.

The reason is short. A connected double cover is decided by which loops swap the copies. It is two-sided exactly when every loop that reverses the sense of turning also swaps the copies — because then a traveller who comes back mirrored always comes back on the other copy, where mirrored is correct — and every loop that preserves the sense does not. Those two conditions pin the swapping loops down completely: they must be exactly the sense-reversing loops. There is only one such cover, and the letters it crosses on are the ones whose loop reverses the sense — the letters used twice the same way.

The one-vertex condition matters. In the square $abab$ the corners fall into two vertices, and crossing on $a$ alone pinches one of them: the loop round that vertex reads $a$ and $b$, and a cover that swaps on one and not the other cannot close up there. The check on vertex counts refuses it. That matches the fact that the projective plane has exactly one connected double cover, not three — the sphere — as a covering is a permutation would predict from its loop group of order two.

The swap, and nothing else #

The cover has one symmetry that the construction makes obvious: exchange the two copies. A point of the first copy goes to the same point of the second, and the gluings are carried to gluings, because the recipe treated the two copies identically apart from their sense of turning.

The swap moves every point, since no point lies on both copies. It reverses the sense of turning, since the copies carry opposite senses. And it does nothing else: the only covering symmetries of a connected two-sheeted cover are the identity and the exchange of sheets, the fact the symmetries a cover has of its own counts in general. Dividing the cover by the swap gives back the surface.

So the recipe runs in both directions. From a one-sided surface it builds a two-sided cover with a swap that moves every point and reverses the sense. From a two-sided surface with such a symmetry, identifying each point with its image gives a one-sided surface — the icosahedron’s antipodal map is one example, and every two-sided surface with $k - 1$ handles has one whose quotient is the sphere with $k$ cross-caps. The cover and the symmetry are two descriptions of the same thing.

What the pictures cannot show #

The covering map is not drawn. The two copies sit side by side and the map down to the surface is the one that forgets which copy a point is on; no picture of it is attempted, because the surface below does not fit in three dimensions for any word here except the torus’s.

The cover’s own shape is named, not built. The figure for $aabbcc$ says its cover is the surface with two handles because it is connected, two-sided and has characteristic $-2$. The reduction to the standard two-handled word — a single octagon, $abcda^{-1}b^{-1}c^{-1}d^{-1}$ — can be carried out on the two glued hexagons, but it takes a dozen moves and is not drawn here.

Only closed surfaces. A surface with an edge has an orientation cover too, built by the same recipe with the unpaired edges left unpaired in both copies; a disc sewn to a Möbius band starts from the band, whose cover is an annulus with two edges over the band’s one.

Still open: which branched covers exist #

Every cover on this page is unbranched: over every point of the surface there are exactly two points of the cover, and that is why the characteristic exactly doubles. Allow finitely many points where sheets come together — as the square root does over the origin, where two values of $\sqrt z$ merge into one — and the count changes by a correction for each such point. The Riemann–Hurwitz formula says by how much: a cover of degree $d$ has characteristic $d$ times the base’s, minus, for each branch point, $d$ minus the number of points above it.

That formula is a necessary condition, and the question of when it is also sufficient is the Hurwitz problem: given a base surface, a degree and a list of how the sheets come together over each branch point, does a cover with exactly that data exist? When the base has characteristic zero or less, the answer is known — the formula and one parity condition are enough. When the base is the sphere, they are not. The smallest counterexample has degree four and three branch points, with the sheets meeting as two pairs over the first two and as a triple and a single over the third; the formula is satisfied, predicting a sphere, and no such cover exists. Which data over the sphere are realisable is open in general, despite a century of partial answers. The recipe on this page is the easiest case of all — two sheets, no branching, and one bit per letter deciding everything — and the difficulty begins as soon as a single point is allowed to break that rule.

What the word was carrying #

The habit worth keeping is to read a combinatorial description for everything it records.

A gluing word was introduced to say which edges meet. It turned out to say which surface results, then whether that surface is one-sided, and now what lies above it: the letters used twice the same way are exactly the letters across which the two copies must be matched. The cover was in the word all along, one bit per letter, and building it took no symmetry, no solid and no room — only two copies of the polygon and the discipline of gluing each letter where the word said it belonged.