Standard Polygon Words for Surface Types

Reading a Standard Polygon Word

A polygon word records how the edges of a polygon are paired. Certain repeated patterns encode handles or crosscaps, so the word can describe how a surface is built as well as how its edges are identified. The patterns are useful even before applying the classification theorem: they let us recognize familiar polygonal constructions and check their cell counts. Here we describe the standard models, not a proof that every compact surface has one of these forms.

Definition: A polygon word is the cyclic list of labels encountered while traversing the boundary of a polygon, where each label occurs exactly twice. The two occurrences mark the paired edges. A letter such as $a$ records traversal in a chosen direction along its labelled edge; $a^{-1}$ records traversal in the opposite direction. The paired edges are identified by matching their labelled directions. The quotient is the space obtained after these edge identifications.

For example, in $aba^{-1}b^{-1}$, the two $a$-edges are paired and the two $b$-edges are paired. The inverse signs say that each pair is traversed in opposite directions around the polygon boundary. In $aa$, the paired edges are both traversed in the same direction around the boundary. This distinction is important: it affects both orientability and the resulting vertex identifications.

We use $F$ for the number of polygonal faces, $E$ for the number of edge classes after gluing, and $V$ for the number of vertex classes. A single polygon gives $F=1$, and a word with $m$ paired labels gives $E=m$. The number $V$ must be found from the endpoint identifications; it is not in general the number of vertices drawn before gluing.

The Two Repeated Patterns

Standard forms: The standard polygon word for the orientable surface with $g\geq 1$ handles is $$ a_1b_1a_1^{-1}b_1^{-1}\;a_2b_2a_2^{-1}b_2^{-1}\;\cdots\;a_gb_ga_g^{-1}b_g^{-1}. $$ The standard polygon word for the nonorientable surface with $k\geq 1$ crosscaps is $$ x_1x_1x_2x_2\cdots x_kx_k. $$ The sphere has a two-edge polygon model with word $aa^{-1}$, which serves as the $g=0$ orientable case.

Each four-letter block $a_i b_i a_i^{-1}b_i^{-1}$ is the standard paired-polygon model of a torus with a disk removed when viewed as a piece for connected sum; the square word $aba^{-1}b^{-1}$ gives the torus itself. Each two-letter block $x_i x_i$ is the standard paired-polygon model of a projective plane. These models follow the edge pairings for the torus and projective plane. The connected sum construction joins such pieces after deleting a disk from each, and its polygon word is obtained by concatenating their blocks.

Proposition: Concatenating standard polygon blocks corresponds to taking connected sums of the corresponding surface models. Thus $g$ commutator blocks give a connected sum of $g$ tori, and $k$ repeated-letter blocks give a connected sum of $k$ projective planes.

Proof. Consider two polygonal surface models with words $u$ and $v$. Choose a small disk in the interior of each polygon, away from all paired edges, and remove its interior. Join the two new boundary circles to form their connected sum. To exhibit a polygon for this joined surface, cut each punctured model along a narrow arc from the new circle to its polygon boundary, using the cuts to open the joined surface into one polygon. Traversing the boundary of this polygon passes through the old edge sequence for the first model and then the old edge sequence for the second. The paired edges within each old sequence are unchanged, so the resulting word is $uv$. Applying this construction repeatedly concatenates the blocks. Since the individual commutator block is the torus model and the individual repeated-letter block is the projective-plane model, the stated connected sums result. $\square$

The sign pattern also records orientability. In a commutator word each label occurs once with exponent $1$ and once with exponent $-1$, so the polygon orientation fits across each paired edge. In a repeated-letter block the pair occurs with the same boundary direction, producing the crosscap model. This is the standard edge-orientation test for these polygon words; it is a test of the given model, not by itself a classification of all surfaces.

Counting Vertices in the Standard Models

The vertex count can be checked directly. Number the successive polygon corners $v_0,v_1,\ldots,v_{n-1}$ cyclically, so edge $i$ goes from $v_{i-1}$ to $v_i$, with indices taken cyclically. If two paired edges are traversed in opposite boundary directions, their initial endpoint on one occurrence is identified with the terminal endpoint on the other, and vice versa. If they are traversed in the same boundary direction, initial endpoints are identified with initial endpoints and terminal with terminal.

Proposition: The standard polygon word for $g\geq1$ handles has one vertex class and $2g$ edge classes. The standard polygon word for $k\geq1$ crosscaps has one vertex class and $k$ edge classes. The sphere word $aa^{-1}$ has two vertex classes and one edge class.

Proof. For one commutator block $aba^{-1}b^{-1}$, number the corners $v_0,\ldots,v_3$, with $v_4=v_0$. Pairing the $a$-edges gives $v_0\sim v_3$ and $v_1\sim v_2$; pairing the $b$-edges gives $v_1\sim v_0$ and $v_2\sim v_3$. These identifications put all four corners in one class. In a word with several commutator blocks, each block’s corners are joined in this way, and consecutive blocks share their junction corner. Therefore all corners are in one class. Each block contributes two distinct labels, giving $2g$ edge classes.

For a block $xx$, the two occurrences are traversed in the same boundary direction. If its three successive corners are $v_0,v_1,v_2$, the paired-edge identifications give $v_0\sim v_1$ and $v_1\sim v_2$. Thus the corners of that block are in one class. Consecutive $xx$-blocks share a junction corner, so the full word has one vertex class and one edge class for each of its $k$ labels. Finally, $aa^{-1}$ has two corners $v_0,v_1$. The opposite-direction pairing identifies each corner only with itself, leaving two vertex classes and one edge class. $\square$

For these polygonal cell structures, the Euler characteristic is $V-E+F$. The counts just proved give $2-2g$ for the orientable standard model with $g\geq1$, $2-k$ for the nonorientable standard model with $k\geq1$, and $2$ for the sphere model. The sphere count comes from two vertex classes, one edge class, and one face.

Worked Polygon Constructions

Worked Example: One Handle

Read the word $aba^{-1}b^{-1}$. There are four edge occurrences and two labels, $a$ and $b$. The $a$-edges form one pair, and the $b$-edges form the other. The signs show that each pair is traversed in opposite directions around the boundary. This is the standard square construction of the torus.

For the vertex count, let the four corners in order be $v_0,v_1,v_2,v_3$. The $a$-pair gives $v_0\sim v_3$ and $v_1\sim v_2$. The $b$-pair gives $v_1\sim v_0$ and $v_2\sim v_3$. Together these imply $v_0\sim v_1\sim v_2\sim v_3$, so $V=1$. There is one face and two edge classes, hence $$ \chi=V-E+F=1-2+1=0. $$ Check: substituting the counts into the formula gives $1-2+1=0$, as required for this polygonal cell structure.

Worked Example: Two Handles

The word $a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1}$ consists of two commutator blocks. Each block contributes the standard handle pattern, and concatenating them gives the connected sum of two torus models. There are four paired labels, so $E=4$, and the vertex-count proposition gives $V=1$. There is one polygonal face, so $F=1$. Therefore $$ \chi=V-E+F=1-4+1=-2. $$ Check: the two blocks contain $2\cdot2=4$ distinct paired labels; using those counts directly gives $1-4+1=-2$.

Worked Example: Three Crosscaps

Consider $x_1x_1x_2x_2x_3x_3$. The three repeated-letter blocks are three projective-plane patterns, so concatenation gives the connected sum of three projective-plane models. Each label contributes one edge class, so $E=3$. The repeated-letter vertex identifications join all corners into one class, giving $V=1$, and the polygon supplies $F=1$. Thus $$ \chi=V-E+F=1-3+1=-1. $$ Check: the word has six occurrences, paired into $6/2=3$ edges; substituting $V=1$, $E=3$, and $F=1$ again gives $1-3+1=-1$.

Worked Example: The Sphere Word

For $aa^{-1}$, the two edges are paired in opposite boundary directions. The two polygon corners are not identified with each other, so $V=2$; there is one edge class and one face. Hence $$ \chi=V-E+F=2-1+1=2. $$ Check: the edge pairing leaves two vertex classes, as in the vertex count above, and $2-1+1=2$.

A Common Misreading and What the Words Do Not Prove

Common pitfall. Counting every edge occurrence as a separate edge gives the wrong Euler characteristic. For the torus word, one might incorrectly count four edges and calculate $1-4+1=-2$. But the occurrences $a$ and $a^{-1}$ are the two sides of one paired edge, and $b$ and $b^{-1}$ are the two sides of another. The quotient therefore has $E=2$, not $E=4$, and its correct count is $1-2+1=0$.

Another important caution is that an Euler characteristic is a check on a polygonal cell structure, not a complete identification rule on its own. The formulas above apply to the displayed standard words and their constructions. Establishing that these models account for every compact surface, and using orientability and Euler characteristic to identify a surface in general, requires the classification theorem.

Check Your Understanding

For each question, distinguish edge occurrences from paired edge classes and use the stated polygon word.

How many paired edge classes occur in the standard word for four handles?
Write the standard polygon word for the connected sum of two projective planes. How many crosscap blocks does it contain?
For $x_1x_1x_2x_2$, determine $V$, $E$, and $F$, then calculate the Euler characteristic.
Explain why the two occurrences of $a$ in $aa^{-1}$ represent one edge class but leave two vertex classes.
In one sentence, explain why Euler characteristic alone is not asserted here to classify every compact surface.