Geometric Topology 13

Combinatorial Surfaces

Let $\mathscr{P}$ be a polygon modelling a surface $\mathscr{M} = \hat{\mathscr{P}}$ without boundary. The essence of this model is a boundary code of directed edges, telling us how to identify points of $\partial \mathscr{P}$. Our aim is to transform the code into one of the normal forms:

1. $aa^{-1}$ or $a_1b_1a_1^{-1}b_1^{-1}...a_gb_ga_g^{-1}b_g^{-1}$,
2. $c_1c_1...c_hc_h$.

This will be achieved by a sequence of cutting and pasting operations of the sort we have already seen that lead to homeomorphic surfaces. During this process, we could split a single polygon into one or two smaller ones, each with its own code. However, we would like to carry out these operations without having to draw pictures.

Definition: A combinatorial model of a surface is defined by a collection of letters $a_1,a_2,...$ (representing edges) and one or more words $W_1,W_2,...$ (representing complete boundaries) involving these letters or their inverses (normally at least $3$ per word), so each letter appears exactly twice overall (if $\mathscr{M}$ is to have no boundary).

We call this data $\langle a_1,a_2,... | W_1,W_2,... \rangle$ a presentating of the surface.

Valid Operations

Starting with the presentation $\langle a_1,a_2,...| W_2,W_2,...\rangle$ in which each $a_i$ appears exactly twice on the right, we can recover a surface $\mathscr{M}$ by constructing a regular polygon of unit side for each word, to make it easy to identify the edges in pairs.

Proposition: The following operations on words then result in homeomorphic surfaces:

1. "Rotate": $aB \sim Ba$.
2. "Reflect": $A \sim A^{-1}$
3. "Cut or Paste": $AB \sim Ax^{-1} + xB$, i.e. $\{AB\} \sim \{Ax^{-1},xB\}$
4. "Insert or Fold": $A \sim Axx^{-1}$
5. "Relabelling": $A \sim r$

Moreover:

1. Words should have at least $3$ letters, except that $aa$ and $aa^{-1}$ are words.
2. $A,B,...$ stand for $2$ or more letters (excluding $x$), but are not in general words.
3. We shall explain after a sketch proof of the proposition that the operations can be understood group-theoretically, so inverses are computed in that context. Thus $(a^{-1})^{-1} = a$, and if $A = abc$ then $A^{-1} = c^{-1}b^{-1}a^{-1}$.

Abstract Cutting and Pasting

The first two operations result in homeomorphisms because we merely rotate or reflect the regular polygons. Proving that cut and paste does not affect the quotient follows from the pictorial interpretations earlier.

The two images here explain the insert (left to right) and folding (right to left) operations. Here $A$ is the left boundary consisting of $6$ letters (when the polygon is fully closed up). The edges are then widened to accommodate insertion of $xx^{-1}$, and this process leads to a homeomorphism of the quotients of the two polygons provided identification of the edges of $A$ is adjusted accordingly. $\blacksquare$

As we discovered earlier, all these operations are consistent with simple substitutions in group theory. One can push this further, and interpret the letters as generators of a (typically) infinite group, and each boundary code $W$ as the relation $W=e$ (where $e$ is the identity, and the word describing the sphere).

Notation from Group Theory

Here the proof of the proposition regarding the possible operations is streamlined by replacing all the cutting, rearranging and pasting by substitutions. At each step, on alternate lines, a group of boxed letters is replaced by a new letter, which is also used to replace the second occurrence of the red letter. The symbol $\sim$ means both sides define the same relation (when each is set equal to the identity), which again encodes homeomorphic surfaces:

Exercise: Prove that $abca^{-1}b^{-1}c^{-1} \sim xyx^{-1}y^{-1}$ starting with $\boxed{a\color{red}{b}}$.

Counting Vertices

If the surface $\mathscr{M}$ is the quotient of a $2n$-gon $\mathscr{P}$ with a boundary code, then $\chi = V - n + 1$, where $V$ is the number of equivalence classes of vertices (so $V$ is the number of distinct points of $\mathscr{M} = \hat{\mathscr{P}}$ arising from vertices of $\mathscr{P}$).

The integer $V$ can be quickly evaluated by carrying out the identifications around $\partial \mathscr{P}$.

Key observation: If the word describing $\partial \mathscr{P}$ is in normal form, and not $aa^{-1}$, then $V=1$. Therefore $\chi = 2-2g$ in case (1), and $\chi = 2-h$ in case (2).

Let's verify this for $W = aba^{-1}b^{-1}cdc^{-1}d^{-1}$. Squiggles (here, connected) indicate that the vertices (points between letters) are equivalent:

(Insert image here page 110)

Note: When $W$ is written out linearly as above, its initial 'vertex' will always be equivalent to its final 'vertex', without having to rely on the fact that these two coalesce on $\mathscr{P}$.

Proof of the Main Theorem (v2)

We know that a connected compact surface $\mathscr{M}$ can be regarded as the quotient $\hat{\mathscr{P}}$ of a polygon $\mathscr{P}$, defined by a word $W$ that identifies points on the boundary $\partial \mathscr{P}$. We shall apply operations to find a sequenceof words:

satisfying various conditions, with $W_4$ in one of the forms (1) or (2) of the theorem.

Set ${\mathbb{A}}_g = (a_1b_1a_1^{-1}b_1^{-1})\cdots (a_g b_g a_g^{-1}b_g^{-1})$ for $g \geq 1$, and $A_0 = aa^{-1}$.

Set $\mathbb{C}_h = (c_1c_1)\cdots (c_h c_h)$ for $h \geq 1$, and $\mathbb{C}_0 = \varnothing$.

Step 1: $W \sim W_1$ where all the vertices defined by $W_1$ map to a unique point of $\mathscr{M}$.

Step 2: $W_1 \sim W_2$ where $W_2 = \mathbb{C}_h B$ for $h \geq 0$, and $B$ has no repeated letters like $b \cdots b$ without an inverse. For this, it is enough to observe that $aCaD \sim xxC^{-1}D$.

Step 3: $W_2 \sim W_3$, where $W_3 = \mathbb{C}_h \mathbb{A}_g$, assuming $B \neq \varnothing$.

Step 4: We can assume that one of $h,g$ is zero. This follows immediately from the proposition which tells us that $\mathbb{C}_1 \mathbb{A}_1 \sim \mathbb{C}_3$.

Steps 1 and 2

For steps 1,2 and 3 we shall resort to the cutting and pasting notation, though we invite the reader to streamline the arguments with group-theoretic substitutions.

We ahve to modify $W$ so that there is only 'one' vertex, i.e. $V = 1$. If this is not the case, $\partial \mathscr{P}$ has two inequivalent vertices $p,q$. We can assume they are adjacent and set $a = \overrightarrow{pq}$. Suppose first that $a$ appears as $a \cdots a$, so we can write $W = aBaC$, with $B \neq \varnothing$ (otherwise $p=q$). Then:

In the first cut, the ends of $x$ are both $p$, so the pasting has eliminated one occurrence of $q$. By induction we can get down to one vertex.

We have incidentally succeeded in converting two separated occurrences of $a$ to a single $xx$, which is what is needed for step 2.

If instead $a$ appears as $a \cdots a^{-1}$ then $W = aBa^{-1}C$ with $B \neq \varnothing$ (otherwise is a sphere, or $aa^{-1}$ can be folded). A trickier argument allows us to eliminate one occcurrence of $q$.

Step 3

Suppose that $W \sim \mathbb{C}_h B$ ($\mathbb{C}B$ for shorthand)m with $B \neq \varnothing$. We need to convert $B$ to normal form. Since $\mathscr{P}$ has a 'unique vertex', we must be able to spot a configuration $a \cdots b \cdots a^{-1} \cdots b^{-1}$ for some $a,b$ (otherwise there would be two inequivalent vertices, one before $a^{-1}$ and one after, there being no edges to link them). So:

for some $a,b,B_i,C_j$ and:

We can subtract another term $a_2 b_2 a_2^{-1} b_2^{-1}$ provided $C_1 B_2$, $B_1 C_2$ are not both empty. $\blacksquare$