Why Check a Vertex Separately?
Identifying the edges of a polygon creates a quotient space, but matching edges does not by itself guarantee that every point has a neighborhood like a point of a surface. At the interior of a polygon, a small neighborhood is already a disk. Along an edge, the number of polygon sides meeting there controls the local picture. At a vertex, however, several polygon corners may be identified, and their arrangement matters: they might make one disk neighborhood, or they might meet in separate pieces or branch in a way that is not locally surface-like.
The link records that arrangement without retaining the full polygon. Its vertices represent directions along identified edge germs, and its edges represent polygon corners. For a polygonal quotient, the local question at a vertex becomes a finite graph test.
We use the polygonal, piecewise-linear setting: finitely many polygonal faces are glued along entire edges by endpoint-preserving or endpoint-reversing linear identifications, and the quotient has the quotient topology. A surface with boundary means a space in which every point has a neighborhood piecewise-linearly homeomorphic either to an open disk or to an open half-disk. A point with a disk neighborhood is an interior point; a point with a half-disk neighborhood centered on its straight boundary is a boundary point.
Constructing the Link Graph
An edge germ here means a sufficiently short portion of a polygon edge starting at the relevant endpoint. The endpoint identifications in the definition come from the specified edge gluings, not from a guess based only on which polygon vertices have the same image. Thus the link remembers how the corners are connected around the quotient vertex class.
In a small neighborhood of \(v\), each corner contributes a narrow sector. Its inner tip maps to \(v\), while a short arc across the sector corresponds to the interval for that corner. Gluing the sectors along their sides makes these arcs join exactly as the link endpoints do. Consequently, a sufficiently small neighborhood of \(v\) is a cone on \(\operatorname{Lk}(v)\): it consists of the link’s points at small positive distance from the cone point, together with the cone point itself. This local model explains why the link determines the vertex neighborhood.
Proof. The local sector construction gives a conical neighborhood on the link. The cone on a circle is a disk: identify the circle with the unit circle and send each point at cone radius \(r\) to the point at distance \(r\) from the disk’s center. The cone point maps to the center. The cone on a closed interval is a half-disk up to piecewise-linear homeomorphism: coning the interval gives a triangle, and a triangle is piecewise-linearly homeomorphic to a half-disk, with the cone point on its boundary. These give the two sufficient conditions.
For necessity in the piecewise-linear setting, take a sufficiently small disk or half-disk neighborhood centered at the vertex and cut it by a small arc or segment around that vertex. In a disk, this cross-section is one circle; in a half-disk centered on its boundary, it is one interval. The sectors in the polygon quotient give the same cross-section as its link, with endpoints joined precisely at the identified edge germs. Therefore a disk neighborhood requires a circle link, and a half-disk neighborhood requires an interval link. \(\square\)
For a finite graph, these shapes have simple tests. A graph is a circle exactly when it is connected and every vertex has degree two. A graph is an interval exactly when it is connected, has precisely two vertices of degree one, and all its other vertices have degree two. In the interval case there are no isolated vertices. These criteria include graphs with parallel edges; the degree convention for loop edges is important.
Proof. A circle is connected and has degree two at every point when represented as a finite graph. An interval is connected, has two endpoints of degree one, and has degree two at every other vertex. Conversely, start at any vertex of a finite connected graph in which all degrees are two. Follow edges without reversing direction. Finiteness forces a return to a vertex already visited; degree two then closes a cycle, and connectedness prevents any vertices or edges from lying outside it. The graph is one circle. Now suppose a finite connected graph has exactly two degree-one vertices and every other vertex has degree two. Begin at one degree-one vertex and follow the unique available edge, continuing through each degree-two vertex. The path cannot branch or close into a cycle: a cycle would use both incident edges at each of its vertices and could not be connected to the starting endpoint. Finiteness therefore makes the path end at the other degree-one vertex. Connectedness and the degree conditions leave no additional edges. The graph is an interval. \(\square\)
Worked Link Tests
Worked Example: The Paired Square
Take a square with boundary word \(aba^{-1}b^{-1}\), pairing edges carrying the same letter. Let its successive corners be \(c_0,c_1,c_2,c_3\), starting at the lower-left corner and moving counterclockwise. The pairings identify all four corners into one vertex class. We compute the link at that class.
Write the four link vertices as \(A,B,C,D\), where \(A\) identifies the germ at the start of the first \(a\)-edge with the corresponding germ at the end of the \(a^{-1}\)-edge; \(B\) identifies the other pair of \(a\)-germs. Similarly, \(C,D\) are the two classes of \(b\)-germs. The corner intervals join the following pairs: \(c_0\) joins \(C\) to \(A\); \(c_1\) joins \(B\) to \(C\); \(c_2\) joins \(D\) to \(B\); and \(c_3\) joins \(A\) to \(D\). Thus the link is the cycle \(A-C-B-D-A\).
It is connected, and each of \(A,B,C,D\) has degree two. The link is a circle, so the quotient vertex has a disk neighborhood. The test concerns the vertex; to conclude that the whole quotient is a surface, one must also check points in face interiors and along edge interiors.
Worked Example: A Boundary Vertex of an Unpaired Polygon
Take a single square with no edge identifications, and consider one of its corners. Only one corner lies in this vertex class, so the link consists of one interval: its two endpoints are the two incident, unpaired edge germs. The link is connected, has two degree-one vertices, and has no other vertices. It is therefore a closed interval.
The cone on this interval is a half-disk neighborhood, so the corner is a boundary point of the polygon. This example also shows why an interval link is not a failure of the surface test: it signals a surface with boundary, rather than an interior point of a surface without boundary.
Worked Example: Two Squares Joined at One Corner
Take two disjoint squares and identify just one corner of the first square with one corner of the second. Do not identify any edges. At the shared vertex class there are two polygon corners. Each contributes one interval to the link, and no edge-germ endpoints are identified with endpoints from the other square. The link is therefore a disjoint union of two intervals.
Each link vertex has degree one, but the graph is disconnected and has four degree-one vertices rather than two. It is neither a circle nor an interval. The local cone has two separate sectors meeting only at its cone point, so the shared vertex has neither a disk nor a half-disk neighborhood. The quotient is not a surface at that point, even though each square separately is a surface with boundary.
Checking Edge Interiors and Avoiding a Pitfall
A link test at a vertex does not replace the edge-interior check. At an interior point of a paired edge, the two polygon sides must meet in the local model to give a disk neighborhood. At an interior point of an unpaired edge, there is just one side, giving a half-disk neighborhood. In the usual polygon construction with each edge either paired with exactly one other edge or left unpaired, these are the two possibilities. Interior points of faces already have disk neighborhoods.
Common pitfall: Checking that every link vertex has degree two is not enough to conclude that the link is a circle. In the two-square example, the two intervals have degree-one endpoints, so it already fails that degree condition. A related mistake is to check degrees but forget connectedness. For instance, a graph made of two disjoint cycle components has degree two at every vertex, but it is not one circle. Its cone consists locally of two disk-like sectors meeting at the cone point, not a single disk neighborhood. The correct test requires both the degree condition and connectedness.
For a closed polygon quotient built by pairing every edge exactly once, the local checks can be organized as follows: face-interior points have disk neighborhoods; paired-edge interior points must be checked for disk neighborhoods; and every vertex class must have a circle link. If unpaired edges are allowed, interval links and half-disk edge neighborhoods are also permitted for a surface with boundary. A failure at even one point is enough to rule out a surface at that point.
Check Your Understanding
For each question, specify whether the proposed link gives an interior point, a boundary point, or a singular point.
- A connected finite link graph has degree two at every vertex. What is its homeomorphism type, and what vertex neighborhood does it give?
- A connected link graph has two degree-one vertices and all remaining vertices have degree two. What local neighborhood does it give?
- Why does a disjoint union of two circles fail the vertex test even though every link vertex has degree two?
- In a quotient with unpaired edges, what neighborhood should an interior point of an unpaired edge have?
- Two polygon corners belong to the same vertex class, but their corner intervals remain unconnected in the link. What feature of a surface neighborhood does this call into question?