From Polygon Gluing to a Group Presentation
A polygon description turns a surface into a space built from edges and a disk. The edge identifications determine a graph, and the way the polygon boundary travels around that graph determines the relation in its fundamental group. For a closed nonorientable surface, the boundary word records each generator twice with the same sign. We will derive the resulting product-of-squares presentation.
Here, a closed surface means a compact surface without boundary; “closed” does not mean a closed subset of some ambient space. For \(k\geq 1\), let \(N_k\) denote the standard closed nonorientable surface of genus \(k\), represented by a \(2k\)-sided polygon whose boundary word is \(a_1a_1a_2a_2\cdots a_ka_k\). Each pair of edges bearing the same label is identified according to that boundary-word convention. We use this polygon model as the given description of \(N_k\).
The presentation we seek is \(\langle x_1,\ldots,x_k\mid x_1^2x_2^2\cdots x_k^2=1\rangle\). To justify it, we first identify the graph left by the edge identifications, then read the boundary of the polygon as a loop in that graph.
The Graph in the Polygon Model
Write the polygon’s vertices in cyclic order as \(v_0,v_1,\ldots,v_{2k-1}\), and let its consecutive edges be \(e_1,\ldots,e_{2k}\), where \(e_j\) runs from \(v_{j-1}\) to \(v_j\), with indices taken cyclically. The edges \(e_{2i-1}\) and \(e_{2i}\) are the two occurrences of the label \(a_i\). The standard identification matches their starting endpoints and their ending endpoints. In particular, it identifies \(v_{2i-2}\) with \(v_{2i-1}\) and \(v_{2i-1}\) with \(v_{2i}\). Repeating this for \(i=1,\ldots,k\) identifies every vertex with every other vertex.
Proof. The endpoint identifications just described give \(v_{2i-2}\sim v_{2i-1}\sim v_{2i}\) for each \(i\). As \(i\) runs from \(1\) to \(k\), these relations connect all consecutive vertices around the polygon, including the last vertex and \(v_0\). Thus there is exactly one vertex class. Each pair \(e_{2i-1},e_{2i}\) is identified to one edge, and distinct labels are not identified to one another, so there are exactly \(k\) edge classes. Both endpoints of each edge class are the sole vertex class, making each edge a loop. \(\square\)
A graph consisting of one vertex and \(k\) loop edges is a wedge of \(k\) circles. Choose the orientation of the \(i\)-th loop to agree with the first boundary occurrence of \(a_i\), and call the resulting based loop \(x_i\). With this choice, the second occurrence of \(a_i\) also traverses the loop in the \(x_i\) direction: the paired edges have the same label in the boundary word, not inverse labels. Therefore the polygon boundary loop reads \(x_1x_1x_2x_2\cdots x_kx_k\).
The Relator from the Attaching Map
The polygon interior supplies one 2-cell. Its boundary attaches to the graph just described by the loop read around the polygon. We use the attaching-a-disk form of the Seifert-van Kampen theorem, established earlier in the course: if a connected space \(Y\) is enlarged by attaching a disk along a based loop representing \(R\in\pi_1(Y,y_0)\), then the fundamental group of the enlarged space is \(\pi_1(Y,y_0)\) modulo the normal subgroup generated by \(R\). In particular, if \(Y\) is a wedge of \(k\) circles with loop generators \(x_1,\ldots,x_k\), its fundamental group is the free group on those generators, and attaching the disk imposes the single relator given by its boundary loop.
Proof. By the proposition, the polygon boundary after identification is a connected graph with one vertex and \(k\) loop edges. Choose that vertex as basepoint. The fundamental group of this graph is the free group on the loops \(x_1,\ldots,x_k\). Following the polygon boundary in its given order traverses each \(i\)-th edge twice in the direction \(x_i\), so the attaching loop represents \(x_1x_1x_2x_2\cdots x_kx_k=x_1^2x_2^2\cdots x_k^2\). Attaching the polygon’s disk therefore quotients the free group by the normal subgroup generated by this boundary word, by the attaching-a-disk form of Seifert-van Kampen. This is exactly the stated presentation. \(\square\)
The relator is a product of squares, not a product of commutators. The order of the factors is the order in which their paired edges occur around the polygon. The presentation contains one relator because the polygon model has one 2-cell; it does not assert that every word representing the identity is literally the displayed word, only that the relator and its conjugates and products determine the normal subgroup imposed.
Worked Example: Genus One
For \(N_1\), the polygon has two edges with boundary word \(a_1a_1\). The identifications give one vertex and one loop edge \(x_1\). The attaching loop traverses that edge twice in the same direction, so its word is \(x_1x_1=x_1^2\). Attaching the disk gives \(\pi_1(N_1)\cong\langle x_1\mid x_1^2=1\rangle\).
This presentation says every element is represented by a power of \(x_1\), and the relation reduces every power to either \(1\) or \(x_1\): for any integer \(n\), write \(n=2m\) or \(n=2m+1\), so \(x_1^n=(x_1^2)^m=1\) in the even case and \(x_1^n=(x_1^2)^m x_1=x_1\) in the odd case. The two elements are distinct: the map sending \(x_1\) to the nonidentity element of the two-element cyclic group respects \(x_1^2=1\) and distinguishes \(x_1\) from \(1\). Thus this presentation gives the cyclic group of order two.
Check: the polygon boundary is \(x_1x_1\), and imposing that loop as the identity gives exactly \(x_1^2=1\), as required by the general formula with \(k=1\).
Worked Example: Genus Two
For \(N_2\), the boundary word is \(a_1a_1a_2a_2\). The quotient graph has one vertex and two loop edges, denoted \(x_1\) and \(x_2\). Reading the four boundary edges in order gives \(x_1x_1x_2x_2=x_1^2x_2^2\), so \(\pi_1(N_2)\cong\langle x_1,x_2\mid x_1^2x_2^2=1\rangle\).
For comparison with another common form of the Klein bottle group, set \(u=x_1x_2\). Then \(x_1=u x_2^{-1}\), and substitution into the relator gives \[ x_1^2x_2^2=(u x_2^{-1})(u x_2^{-1})x_2x_2 =u x_2^{-1}u x_2. \] Thus the relation is equivalent to \(u x_2^{-1}u x_2=1\), or \(x_2^{-1}u x_2=u^{-1}\): to obtain the latter, multiply the former equation on the left by \(u^{-1}\). The substitution is reversible because \(x_1=u x_2^{-1}\) and \(u=x_1x_2\). So the change of generators does not discard or add a relation.
Check: substituting \(u=x_1x_2\) into \(u x_2^{-1}u x_2\) gives \((x_1x_2)x_2^{-1}(x_1x_2)x_2=x_1x_1x_2x_2\), the original boundary word.
Worked Example: Genus Three
For \(N_3\), there are six polygon edges, paired as \(a_1,a_1\), \(a_2,a_2\), and \(a_3,a_3\). The identifications leave one vertex and three loop edges. Traversing the boundary gives the successive loop letters \(x_1,x_1,x_2,x_2,x_3,x_3\), hence the presentation \(\pi_1(N_3)\cong\langle x_1,x_2,x_3\mid x_1^2x_2^2x_3^2=1\rangle\).
The free-group word has length six before any reductions. No adjacent inverse letters occur: the successive pairs are \(x_1x_1\), \(x_2x_2\), and \(x_3x_3\), and each transition between pairs is \(x_i x_{i+1}\), not \(x_i x_{i+1}^{-1}\). Thus there is no cancellation in the boundary word. Attaching the disk imposes this whole word as one relator, not three separate relations \(x_i^2=1\).
Check: the word has \(2\cdot3=6\) letters, with each of the three edge classes appearing twice, matching the six sides of the polygon.
A Common Sign Error
A frequent mistake is to treat the second occurrence of each label as an inverse merely because it is a different side of the polygon. That changes the attaching map. For genus one, the wrong reading \(x_1x_1^{-1}\) reduces to the identity word, and would produce the presentation \(\langle x_1\mid 1=1\rangle\), a free group on one generator. But the polygon convention here labels the two paired boundary traversals with the same letter and identifies their starting endpoints to each other and their ending endpoints to each other. The correct boundary word is \(x_1x_1=x_1^2\), not \(x_1x_1^{-1}\). The error is not a harmless change of notation: it changes the relator and therefore the group calculation.
It is useful to separate three steps when making a polygon computation: first determine which vertices and edges become identified; next choose an orientation for each resulting loop and read the boundary with signs relative to those orientations; finally apply the disk-attaching theorem to the resulting word. The signs come from the boundary path in the quotient graph, not from a visual guess about whether a side appears “flipped.”
Check Your Understanding
For each question, keep the polygon boundary order and the chosen loop orientations explicit.
- Write the boundary word and fundamental-group presentation for \(N_4\).
- Explain why the identified polygon boundary has one vertex for every \(k\geq1\).
- For \(N_2\), substitute \(u=x_1x_2\) into \(x_1^2x_2^2\) and simplify the word.
- What presentation error results from reading each pair of edges as \(x_i x_i^{-1}\) instead of \(x_i x_i\)?
- Why does the polygon construction impose one relator rather than one separate relator for each paired edge?