The third number a surface needs

Every surface is a sphere with handles is careful about which word in its theorem is doing work, and it names the cost of one of them in a sentence: “A Möbius band is a perfectly good surface, it is one-sided, and it is not on the list — because it has an edge. Adding surfaces with boundary to the classification is possible and it needs a third number, the count of boundary circles.”

Here is the third number, and it is produced by the same counting as the other two.

An unpaired edge is a boundary #

A gluing word pairs the edges of a polygon. The gluing words require every letter to appear exactly twice, and refuses a word where one appears once — “a letter appearing once has nothing to be glued to”.

That refusal is a choice rather than a necessity. An edge with nothing to be glued to is simply an edge of the result: a point on it has a half-disc of surface on one side and nothing on the other, which is exactly what a boundary point is — the thing the Möbius band is famous for having. So allowing letters to appear once widens the class from closed surfaces to surfaces with boundary, and the polygon still describes the result completely.

The disc is the simplest case: a triangle with all three edges unpaired is a triangle, which is a disc. Its characteristic is $3 - 3 + 1 = 1$, it has one boundary circle, and it is two-sided by default since there are no pairings to flip.

Counting the circles #

The unpaired edges are edges of the surface, and they join up: an unpaired edge’s endpoints are corners of the polygon, and those corners may have been identified with others by the paired edges. So the boundary is not a set of separate arcs but a set of closed curves, and counting them means following the identifications.

The procedure is short. Pool the corners as usual — the paired edges force identifications, and a union-find does the pooling. Then walk the unpaired edges, hopping from one to another whenever they share a corner class. Each connected run of unpaired edges is one boundary circle, and the count of runs is the third number.

The annulus is the case worth checking by hand. Take a square with one pair of opposite edges glued straight across and the other pair left alone: $abcb^{-1}$, where $b$ is glued and $a$ and $c$ are free. The $b$ gluing identifies the two ends of $a$ with nothing and carries $a$’s corners to $c$’s… no, it does not: it identifies the corners between $a$ and $b$ with the corners between $b^{-1}$ and $c$ on the other side, and the two free edges end up in separate classes. Two boundary circles, which is what a cylinder has, and $\chi = 0$.

The word that is one-sided and has an edge #

The one-sided bordered surfaces are where the third number does the most work, because the closed one-sided list already uses every characteristic at most two and the bordered ones spread further down.

Take $abab^{-1}c$: the letter $a$ appears twice the same way round, so the surface is one-sided; $b$ appears once each way; $c$ appears once, so there is a boundary. The characteristic comes out at $-1$ and the boundary is one circle. Capping it gives a closed one-sided surface of characteristic nought, which is the Klein bottle.

So the triple $(-1, \text{one-sided}, 1)$ is the Klein bottle with a disc removed — and $(0, \text{one-sided}, 1)$ is the Möbius band, which is the projective plane with a disc removed. The pattern is that every bordered surface is a closed one with discs removed, and the third number says how many. That is the capping identity read backwards and it is the whole classification.

It also explains why the Möbius band is the smallest one-sided bordered surface. Removing a disc from the projective plane, which is the closed one-sided surface of largest characteristic, gives the largest characteristic a one-sided bordered surface can have — and that is nought, the band’s.

Why three numbers are enough #

The reason the list closes up again with one more number is a construction rather than a new theorem.

Take a surface with $b$ boundary circles and glue a disc onto each one. The result is a closed surface, and its characteristic is the original’s plus $b$ — because each disc adds one face and its boundary circle’s edges and vertices were already counted. So

$$\chi_{\text{closed}} = \chi + b.$$

Now the closed surface is named by the two numbers the closed classification supplies, and capping is reversible: cutting $b$ discs out of a closed surface gives the bordered one back. So a bordered surface is determined by its capped version together with $b$ — which is to say by $\chi$, the sidedness, and $b$.

Every triple that can occur occurs exactly once, with the constraints being that $\chi + b$ is the characteristic of a closed surface, so at most two, and of the right parity for the two-sided case. The figures check the capping identity for every word they draw: a triple whose capped characteristic exceeded two would be a word describing nothing, and the figures refuse it.

The Möbius band, at last on a list #

The word $aab$ is a triangle with two edges glued the same way round and one left free. That is a Möbius band: the flip is what makes it one-sided, the free edge is its single boundary circle, and $\chi = 1 - 2 + 1 = 0$.

So the triple is $(0, \text{one-sided}, 1)$, and nothing else has it. Capping its single circle with a disc gives a closed one-sided surface of characteristic one, which is the projective plane — and that is exactly the construction a disc sewn to a Möbius band is about, arriving here as an instance of the capping identity rather than as a separate argument.

The annulus is $(0, \text{two-sided}, 2)$ and caps to a sphere. The disc is $(1, \text{two-sided}, 1)$ and caps to a sphere too — two different bordered surfaces capping to the same closed one, distinguished by $b$, which is the clearest demonstration that the third number is carrying information the other two do not.

The piece everything is built from #

There is one bordered surface worth naming on its own, because every other one is made of copies of it.

A pair of pants is a sphere with three discs removed: $\chi = -1$, two-sided, three boundary circles. Glue two pairs of pants along two of their circles and the result is a two-holed torus with two circles left; glue along all three and it is a closed genus-two surface. Every closed two-sided surface of genus at least two decomposes into pairs of pants, and the number needed is $2g - 2$.

That decomposition is the bordered case earning its keep. The closed classification says what the surfaces are; the pair-of-pants decomposition says how to build them from one piece, and the pieces are bordered surfaces. A classification of the closed case alone has no building blocks in it, because a closed surface cannot be cut without producing boundary.

The arithmetic checks. Each pair of pants contributes $-1$ to the characteristic, and gluing two circles together adds nothing — the circle’s edges and vertices were counted twice and now once, and no face is added or removed. So $2g-2$ pairs of pants give $\chi = -(2g-2) = 2 - 2g$, which is the genus-$g$ surface’s.

Gluing two bordered surfaces, and the arithmetic of it #

The reason the bordered case is worth having is that bordered surfaces can be glued, and the arithmetic of gluing is simple enough to do in the margin.

Glue two surfaces along one boundary circle each. The characteristics add and nothing is lost: the circle’s vertices and edges were counted once in each piece and are counted once in the result, so $\chi = \chi_1 + \chi_2$. The boundary count is $b_1 + b_2 - 2$, since two circles were used up. And the result is two-sided exactly when both pieces are.

That single rule reproduces everything the closed classification says about connected sums. A connected sum is two closed surfaces each with a disc removed, glued along the resulting circles: $\chi = (\chi_1 - 1) + (\chi_2 - 1) = \chi_1 + \chi_2 - 2$, which is the formula the closed case derives by a face count. The bordered arithmetic contains the closed arithmetic as a special case, and it is the more natural place for the formula to live.

It also makes Dyck’s identity a computation. A torus with a disc removed has $\chi = -1$ and one circle; a projective plane with a disc removed is the Möbius band, $\chi = 0$ and one circle. Gluing them gives $\chi = -1$, one-sided, no boundary. Three projective planes with discs removed and glued in a chain give $\chi = 0 + 0 + 0 - 2 - 2 \ldots$ — the chain needs care with which circles are used, and doing it properly gives $-1$ again, one-sided. Same triple, same surface, and the identity is arithmetic.

What the third number does not do #

It does not make the boundary circles distinguishable. The classification counts them and says nothing about which is which, so a surface with three circles has an automorphism permuting them and the invariants cannot tell them apart. For many purposes — decomposing a surface, gluing pieces — which circle is which matters, and that is extra data the triple does not carry.

It does not extend to a boundary that is not a circle. Every boundary component of a surface is a circle, which is a theorem and is why counting them suffices. In three dimensions a boundary component is a surface, so the analogous count is a list of surfaces, and the classification acquires the whole two-dimensional problem inside it.

And it does not handle a surface that is not compact. The plane is a two-sided surface with no boundary and it is not a sphere; a disc with its boundary circle removed is an open disc and has no boundary either. That classification records that compactness is load-bearing, and adding boundary does not weaken that: an open surface has no boundary circles to count and is outside the theorem by the same clause.

What the pictures cannot show #

The polygons are drawn with the unpaired edges heavy, and what happens when the paired ones are glued is not drawn — the boundary circles are computed from the corner classes rather than seen. A reader wanting to see the annulus’s two circles has to carry out the gluing mentally, which is the standing limitation of every gluing diagram.

The circles are counted by a walk over the unpaired edges, hopping between any two sharing a corner class. That is a computation on the word, it is exact, and it is the only access the figures have to a number that is a property of the finished surface.

And the capping is checked rather than drawn. That gluing a disc onto each circle gives a closed surface whose characteristic is the original’s plus the count is an identity about faces, edges and vertices; the figures check it on every word and no picture contains it.

The four words, and what each turns out to be #

The hero figure’s four words are worth naming, because a reader who works out what each one is has done the classification by hand.

$abc$ is a triangle with nothing glued, which is a disc: $\chi = 1$, two-sided, one circle. $abcb^{-1}$ glues one pair straight across and is an annulus: $\chi = 0$, two-sided, two circles. $aab$ glues one pair with a flip and is a Möbius band: $\chi = 0$, one-sided, one circle. And $abab^{-1}c$ is a Klein bottle with a disc out: $\chi = -1$, one-sided, one circle.

Four words, four triples, and no two the same — which the figure requires, so a set of words that collapsed onto one triple would refuse to draw rather than presenting four panels saying the same thing. That requirement is the one worth having in a figure of this kind: a table of examples is only informative if the examples differ in the quantity the table is about, and checking it is a line.

The annulus and the Möbius band are the pair to notice. Both have characteristic nought and one is two-sided with two circles while the other is one-sided with one, so neither of the two closed-case numbers separates them and the third does half the work. Cutting each along its middle is the classical way to tell them apart by hand, and the invariants do it without cutting anything.

Still open: which gluings give which triples #

The classification is complete and the enumeration is not obvious. Given a polygon with $2n$ edges and a rule pairing some of them, which triples arise, and how many words give each? The closed case’s answer is known and is a question about the genus distribution of polygon gluings; the bordered case adds a parameter and the counting is harder.

What is known is the extremes: a word pairing nothing gives a disc, and a word pairing everything gives a closed surface. In between the count of words giving each triple is studied under the name of map enumeration, the generating functions are known in some cases, and the asymptotic shape of the distribution is a live subject.

The question with the clearest statement concerns the decomposition rather than the classification. A closed surface of genus $g$ decomposes into $2g-2$ pairs of pants in many ways, and how many ways — up to the surface’s own symmetries — is not known in closed form for general $g$. That count is what a great deal of two-dimensional geometry is organised around, and it is a question about bordered surfaces rather than closed ones.

Six colours, and where the band’s edge enters #

One consequence of the third number is worth having because it is a statement about the surface a reader can check and not about its invariants.

A map drawn on the Möbius band can need six colours and never seven, which is the band’s own colouring number — and the corresponding number for the projective plane is six as well. That coincidence is not one: capping the band’s boundary with a disc gives the projective plane, and the disc can be coloured to match whatever the band did, so the two numbers agree.

The general statement is that the colouring number depends on the characteristic alone, through a formula, and the boundary count drops out — because capping raises the characteristic by the number of circles and the capped surface’s number is what the bound is about. So the third number matters for what a surface is and not for this particular question, which is worth knowing: not every property of a bordered surface needs all three.

What does need all three is anything about cutting. How many disjoint curves can be drawn on a surface without disconnecting it, how it decomposes, what can be glued to it — each depends on the boundary count, and the pair-of-pants decomposition is the standing example. The rule of thumb is that intrinsic questions see the characteristic and constructive ones see all three.

What allowing an edge bought #

The closed classification is tidier than the bordered one and the bordered one is more useful, and the reason is that cutting is how surfaces are studied.

A theorem about closed objects cannot describe how they are built, because building means gluing pieces and pieces have edges. So the extension is not a generalisation for its own sake; it is what makes the pair-of-pants decomposition sayable, and that decomposition is the standard way to reason about a surface of high genus — the same surfaces the normal forms name and do not build.

What it cost is one number, obtained from the same word by the same kind of count. That is a cheap extension by the standards of the subject, and it is cheap for a reason worth stating: every boundary component of a surface is a circle, so counting them loses nothing. One dimension up, that sentence is false, and the corresponding extension is not cheap at all.