From Edge Pairings to Vertex Classes
A polygon quotient is built by identifying paired edges of a polygon, point by point. The endpoints are its vertices, so each edge pairing tells us which vertices are identified. But the final classes are not generally just the pairs that appear in one edge match: identifications can chain through several vertices. The reliable method is to record every endpoint match and then group vertices using transitivity.
We use the boundary-word convention in which the polygon boundary is traversed in order and each edge is labeled by a symbol with a sign. Each symbol occurs on exactly two edges. A positive occurrence is directed along the boundary traversal, and a negative occurrence is directed against it. Paired edges are glued by matching points with the same label-coordinate along their directed arrows. Thus two arrows pointing the same way match start to start and end to end; arrows pointing opposite ways match start to end and end to start.
For example, if \(e_i\) and \(e_j\) have the same signed label, their boundary-traversal directions agree, so the endpoint relations are \(v_i\sim v_j\) and \(v_{i+1}\sim v_{j+1}\). If they have opposite signed labels, the directions disagree, so \(v_i\sim v_{j+1}\) and \(v_{i+1}\sim v_j\). Indices are read cyclically: \(v_n=v_0\). The symbol \(\sim\) here records a relation to be closed under transitivity, not just a list of isolated pairs.
Why the Endpoint Method Gives the Quotient Classes
Proof. Let \(R\) be the relation on polygon vertices given by finite chains of the endpoint identifications. It is an equivalence relation: a vertex is related to itself by a chain of length zero; reversing a chain gives symmetry; and joining two chains gives transitivity. Every endpoint match used in a gluing relates vertices that represent the same quotient point. Equality of quotient points is itself transitive, so vertices related by any finite chain under \(R\) represent the same quotient point.
Conversely, the quotient identifies points only by the paired-edge gluing maps and the equivalence relation those maps generate. If a chain of such identifications starts at a vertex, each step takes an endpoint to its matched endpoint: a vertex is on the polygon boundary, and an edge-pairing map sends endpoints to endpoints. Consequently, a chain that starts at a vertex and ends at a vertex is a chain of the endpoint matches recorded in \(R\). The two vertices are therefore \(R\)-related. This proves both directions. \(\square\)
This proposition justifies the computation as a graph problem. Make one graph vertex for each polygon vertex and draw an undirected link for each endpoint identification. The classes are exactly the connected components of that graph: a path in the graph is a finite chain of identifications, and every such chain is a path. In a hand calculation, a list of relations and a careful transitive grouping do the same job.
Worked Square Pairings
Worked Example: The Square Word \(aba^{-1}b^{-1}\)
Take a square with boundary word \(aba^{-1}b^{-1}\). Its edges in order are \(e_0=a\), \(e_1=b\), \(e_2=a^{-1}\), and \(e_3=b^{-1}\), and its vertices in order are \(v_0,v_1,v_2,v_3\), with \(v_4=v_0\). Find all vertex classes.
Pair \(a\). The occurrences are \(e_0=a\) and \(e_2=a^{-1}\), with opposite directions. The starting endpoint \(v_0\) of \(e_0\) matches the terminal endpoint \(v_3\) of \(e_2\); the terminal endpoint \(v_1\) of \(e_0\) matches the starting endpoint \(v_2\) of \(e_2\). Thus \(v_0\sim v_3\) and \(v_1\sim v_2\).
Pair \(b\). The occurrences are \(e_1=b\) and \(e_3=b^{-1}\), also with opposite directions. The endpoints give \(v_1\sim v_0\) and \(v_2\sim v_3\).
Take the transitive closure. The relations connect all four vertices: \(v_0\sim v_3\), \(v_3\sim v_2\), and \(v_2\sim v_1\). Therefore there is one class, \(\{v_0,v_1,v_2,v_3\}\).
Check: The list of polygon vertices has exactly four entries, and every one appears in the class. The only endpoint relations came from the two paired edge pairs, so there is no unlisted polygon vertex that could form another class.
Worked Example: The Square Word \(aba^{-1}b\)
Now use the square word \(aba^{-1}b\). Again the vertices are \(v_0,v_1,v_2,v_3\), with \(v_4=v_0\). The \(a\)-edges are \(e_0=a\) and \(e_2=a^{-1}\), so their opposite directions give \(v_0\sim v_3\) and \(v_1\sim v_2\).
The \(b\)-edges are \(e_1=b\) and \(e_3=b\). Their directions agree, so the starting endpoints match and the terminal endpoints match: \(v_1\sim v_3\) and \(v_2\sim v_4=v_0\).
Combining the relations gives the chain \(v_0\sim v_3\sim v_1\sim v_2\). Hence the one vertex class is \(\{v_0,v_1,v_2,v_3\}\).
Check: The \(a\)-pair relates \(v_0\) to \(v_3\) and \(v_1\) to \(v_2\); the \(b\)-pair connects \(v_1\) to \(v_3\) and \(v_2\) to \(v_0\). Together these relations include every one of the four vertices in the same class.
Hexagon Pairings and Multiple Classes
Worked Example: The Hexagon Word \(abc\,a^{-1}b^{-1}c^{-1}\)
Label the six edges in order \(e_0=a\), \(e_1=b\), \(e_2=c\), \(e_3=a^{-1}\), \(e_4=b^{-1}\), and \(e_5=c^{-1}\). The vertices are \(v_0,\ldots,v_5\), with \(v_6=v_0\). Compute the classes pair by pair.
Pair \(a\). The edges \(e_0\) and \(e_3\) have opposite directions. Their endpoint relations are \(v_0\sim v_4\) and \(v_1\sim v_3\).
Pair \(b\). The edges \(e_1\) and \(e_4\) also have opposite directions, giving \(v_1\sim v_5\) and \(v_2\sim v_4\).
Pair \(c\). The edges \(e_2\) and \(e_5\) have opposite directions, giving \(v_2\sim v_0\) and \(v_3\sim v_5\).
These relations form two chains: \(v_0\sim v_2\sim v_4\), and \(v_1\sim v_3\sim v_5\). The vertex classes are therefore \(\{v_0,v_2,v_4\}\) and \(\{v_1,v_3,v_5\}\).
Check: The classes are disjoint, and together they list all six vertices: \(v_0,v_1,v_2,v_3,v_4,v_5\), each exactly once. Each displayed relation stays within one of the two classes, so no endpoint match joins them.
Worked Example: A Hexagon with Four Vertex Classes
For a second hexagon, take the word \(abc\,c^{-1}b^{-1}a^{-1}\), again with vertices \(v_0,\ldots,v_5\) and \(v_6=v_0\).
Pair \(a\). The edges \(e_0=a\) and \(e_5=a^{-1}\) have opposite directions. Their endpoint relations are \(v_0\sim v_6=v_0\) and \(v_1\sim v_5\). The first relation adds no new vertex to a class.
Pair \(b\). The edges \(e_1=b\) and \(e_4=b^{-1}\) give \(v_1\sim v_5\) and \(v_2\sim v_4\).
Pair \(c\). The edges \(e_2=c\) and \(e_3=c^{-1}\) give \(v_2\sim v_4\) and \(v_3\sim v_3\). These self-relations do not merge distinct vertices. The resulting classes are \(\{v_0\}\), \(\{v_1,v_5\}\), \(\{v_2,v_4\}\), and \(\{v_3\}\).
Check: These four classes are pairwise disjoint and contain all six vertices. Every endpoint relation listed above is either within one of the classes or a self-relation, so the list is complete.
A Reliable Calculation and a Common Pitfall
For a new pairing, use the following order: write the vertices cyclically; list each paired edge occurrence; decide whether its arrows agree or oppose; record both endpoint matches; then take the transitive closure and count the resulting groups. This order keeps direction decisions separate from the later grouping.
For \(n\) edges, use \(v_0,\ldots,v_{n-1}\) and remember \(v_n=v_0\).
Same signed labels match start to start and end to end; opposite signed labels match start to end and end to start.
Keep merging groups whenever a recorded relation has one endpoint in each group.
Confirm that every polygon vertex occurs exactly once and that each recorded relation has both endpoints in a single final class.
Common pitfall: stopping at the first list of pairs. In the square word \(aba^{-1}b^{-1}\), the \(a\)-pair alone gives \(v_0\sim v_3\) and \(v_1\sim v_2\). A wrong count might call these two classes and stop. But the \(b\)-pair adds \(v_1\sim v_0\) and \(v_2\sim v_3\), which joins the two provisional groups. The corrected calculation gives one class, \(\{v_0,v_1,v_2,v_3\}\). A list of direct matches is not the answer until transitivity has been applied.
The vertex count records which polygon corners become the same quotient point; it is not by itself a description of all local neighborhoods in the quotient. The endpoint calculation has a narrower and useful purpose: it determines the vertex classes from the edge data without relying on a diagram’s appearance.
Check Your Understanding
For each question, write the endpoint relations before deciding the classes.
- In the square word \(aba^{-1}b^{-1}\), which endpoint relations come from the \(a\)-pair, and which come from the \(b\)-pair?
- For a pair of edges with the same signed label, which endpoints match? How does this change for opposite signed labels?
- For the hexagon word \(abc\,a^{-1}b^{-1}c^{-1}\), list the two vertex classes and verify that they partition all six vertices.
- Why can an endpoint identification that adds no new relation between distinct vertices still be recorded correctly?
- Give an example from the worked squares of a provisional pair of vertex groups that merge after another edge pair is processed.