A Reliable Order for Surface Identification
A polygon word records how the sides of a polygonal disk are paired. Reading the word is only the beginning: a valid identification problem also requires checking the vertex neighborhoods. Once the quotient is verified to be a closed surface, its orientability and Euler characteristic determine its type by the classification theorem for compact surfaces.
Throughout, each polygon is a disk whose boundary sides are paired in the manner indicated by its word. A letter and its inverse mark two paired sides; equal letters indicate that their boundary directions agree, while inverse letters indicate that the boundary directions are opposite. The quotient has one 2-cell, one 1-cell for each pair of sides, and one 0-cell for each equivalence class of polygon corners.
The link test is important because edge pairings alone do not guarantee a surface. Points in the interior of the polygon have disk neighborhoods, and points in the interior of a paired edge have disk neighborhoods when exactly two edge sides are glued there. At a vertex class, the corner sectors must fit together around the vertex without gaps or branching.
Proof. Choose a sufficiently small neighborhood of a vertex class in the quotient, made from small sectors near its polygon corners. Before the side identifications, each sector is a small cone on an interval: the corner is the cone point, and the interval records directions out of the corner. After gluing, the directions are joined exactly as specified by the link graph. Thus the neighborhood is a cone on that graph. If the graph is a circle, its cone is a disk, so the vertex has a disk neighborhood. If the graph is not a circle, it is not a single circle of directions, and this neighborhood is not a disk neighborhood at the vertex. Since every edge is paired, there are no unpaired boundary points; consequently the quotient is a closed surface precisely when all vertex links are circles. \(\square\)
Worked Identifications
Worked Example: The Two-Sided Word \(aa\)
Take a polygonal disk with two boundary sides and word \(aa\). Name its corners \(v_0,v_1\), and read the boundary cyclically: the first side runs from \(v_0\) to \(v_1\), and the second from \(v_1\) to \(v_0\). The two sides have the same label. Pairing them in the same boundary direction identifies the initial endpoint of the first side with the initial endpoint of the second, and likewise identifies their terminal endpoints. Hence \(v_0\sim v_1\), so there is one vertex class.
There is one paired edge and one polygonal face, so \(V=1\), \(E=1\), and \(F=1\). Therefore
There are two corner intervals at the single vertex class. Following their endpoints through the same-direction side pairing joins the two intervals into one cycle, so the vertex link is a circle. The quotient is therefore a closed surface. The pairing uses the same boundary direction, so the orientation induced by the polygon cannot fit across this edge pair: the surface is nonorientable. The classification theorem says that a connected closed nonorientable surface with Euler characteristic \(1\) is the projective plane. This gluing gives the projective plane.
Check: the single vertex, single edge class, and single face give \(\chi=1\) directly. The two corner sectors form one circular link, rather than two separate links, so there is one disk neighborhood at the identified vertex.
Worked Example: The Torus Word \(aba^{-1}b^{-1}\)
Use a square with corners \(v_0,v_1,v_2,v_3\) in boundary order. Its sides, also in boundary order, are \(a,b,a^{-1},b^{-1}\). Pair the \(a\)-sides first. Because their labels are inverse, their boundary directions are opposite: the initial endpoint of the \(a\)-side is identified with the terminal endpoint of the \(a^{-1}\)-side, giving \(v_0\sim v_3\); the other endpoints give \(v_1\sim v_2\). Pairing \(b\) with \(b^{-1}\) gives \(v_1\sim v_0\) and \(v_2\sim v_3\). Together these relations put all four corners in one class.
There are two edge classes, one for \(a\) and one for \(b\), and one face. Thus \(V=1\), \(E=2\), and \(F=1\), and
For the vertex-link check, the four corner intervals are all in the same link. Tracing their endpoints through the two inverse-letter pairings joins them in one cycle; no interval is left separate. The link is a circle, so the quotient is a closed surface. Both pairs of sides have opposite boundary directions. The polygon orientation therefore fits consistently across both pairs, and the quotient is orientable. The classification theorem identifies a connected closed orientable surface of Euler characteristic \(0\) as the torus.
Check: substituting the cell counts gives \(1-2+1=0\). All four corners were shown to be equivalent, so using four vertices instead would contradict the endpoint identifications.
Worked Example: The Klein Bottle Word \(aba^{-1}b\)
Again use square corners \(v_0,v_1,v_2,v_3\) in boundary order, with sides \(a,b,a^{-1},b\). Pairing the inverse-labelled \(a\)-sides gives \(v_0\sim v_3\) and \(v_1\sim v_2\). The \(b\)-sides have the same label, so their boundary directions agree. Pairing their initial endpoints gives \(v_1\sim v_3\), and pairing their terminal endpoints gives \(v_2\sim v_0\). Combining all four relations shows that every corner belongs to one vertex class.
There are two edge classes and one face. Consequently \(V=1\), \(E=2\), and \(F=1\), so
The four corner intervals join, under the endpoint identifications, into one cycle. Thus the vertex link is a circle and the quotient is a closed surface. The \(a\)-pair has opposite boundary directions, but the \(b\)-pair has the same boundary direction. A coherent orientation on the single polygonal face cannot extend across that same-direction pair, so the surface is nonorientable. The classification theorem identifies a connected closed nonorientable surface with Euler characteristic \(0\) as the Klein bottle.
Check: the calculation again gives \(1-2+1=0\), but the orientability check differs from the torus: the \(b\)-pair is not orientation-compatible. Equal Euler characteristics alone do not identify a surface; orientability is also needed.
Why These Checks Determine the Surface
Each quotient above is connected because it is the continuous image of a connected polygonal disk. It is compact because the polygonal disk is compact and the quotient map is continuous. Once the vertex-link test verifies disk neighborhoods at all vertices, and every edge is paired, the quotient is a compact connected surface without boundary.
For a surface built from one polygon with paired sides, the cell count is especially direct: \(F=1\), \(E\) is the number of side-pairs, and \(V\) is the number of vertex classes. The Euler characteristic is therefore \(V-E+F\). The classification theorem for compact connected surfaces then uses two pieces of information: whether the surface is orientable, and its Euler characteristic. In the nonorientable case, the Euler characteristic \(2-k\) determines the number \(k\) of projective-plane summands; in the orientable case, \(2-2g\) determines the genus \(g\).
The order of checks matters. First compute the corner classes, then verify their links. Only after the quotient is established to be a surface should its Euler characteristic and orientability be used to identify it. A cell count by itself cannot repair an invalid vertex neighborhood.
Common Pitfall: Counting Polygon Corners as Vertices
A common incorrect calculation for the torus word \(aba^{-1}b^{-1}\) is to count the square’s four corners as four vertices and conclude that \(\chi=4-2+1=3\). This ignores the endpoint identifications. Pairing the \(a\)-sides gives \(v_0\sim v_3\) and \(v_1\sim v_2\); pairing the \(b\)-sides gives \(v_1\sim v_0\) and \(v_2\sim v_3\). These relations put all four corners in one class, so the corrected calculation is \(\chi=1-2+1=0\), as required for the torus.
A second error is to decide orientability from the number of edge pairs or from Euler characteristic. The torus and Klein bottle examples both have two edge classes and Euler characteristic \(0\), but their boundary-direction patterns differ. The correct orientation check examines each pair: opposite boundary directions allow the face orientation to fit across that pair; a same-direction pair obstructs it in a one-face polygon presentation.
Check Your Understanding
For each problem, track endpoint identifications before using the cell count.
- For the word \(aba^{-1}b\), list the corner equivalences produced by the \(a\)-pair and then by the \(b\)-pair.
- Why does the count \(V=4\), \(E=2\), \(F=1\) fail for the square word \(aba^{-1}b^{-1}\)?
- What feature of a vertex link must be checked to ensure a disk neighborhood at the identified vertex?
- Two one-face polygon gluings each have \(V=1\), \(E=2\), and \(F=1\). What additional information distinguishes the torus from the Klein bottle in the examples above?
- For the word \(aa\), state the numbers of vertex classes, edge classes, and faces, and give the resulting Euler characteristic.