Changing a Diagram Without Changing Its Surface
A polygon model records a surface by identifying pairs of boundary edges of a polygon. The same surface can be represented by more than one polygon diagram: a cut can expose edges that were formerly joined, and a later re-pairing can join them in a different-looking arrangement. The key test is not whether the pictures look alike, but whether the new identifications reconstruct the same quotient space.
We use the quotient topology throughout: if \(P\) is a polygon and \(\sim\) is the equivalence relation generated by its edge identifications, its surface model is the quotient \(P/{\sim}\). A cut along an arc separates the two banks of that arc. Regluing the banks by the identifications inherited from the original surface restores the original quotient. An edge slide is a particular change of polygon boundary in which one paired edge is moved past another while the pairing data are adjusted consistently.
There are two useful ways to verify a transformation. For a cut and restoration, track which points are identified before and after the operation. For an edge slide in a standard model, track the generators of the identifications: if the new generators produce exactly the same identifications, they describe the same quotient.
Cutting and Restoring a Paired Edge
Begin with a polygon quotient and choose a paired edge in its boundary. Cutting the corresponding seam in the quotient separates it into two boundary banks. Before regluing, points on opposite banks are no longer identified along that seam. To restore the original surface, identify each point on one bank with the point on the other bank that represented the same point before the cut.
Proof. Let \(S=P/{\sim}\) be the original quotient, and let \(S_{\mathrm{cut}}\) denote the cut surface, in which the two banks of the seam are kept separate. The restoration identifies two points of \(S_{\mathrm{cut}}\) exactly when their images in \(S\) were the same point. Thus the map sending each point of \(S_{\mathrm{cut}}\) to its original point in \(S\) is constant precisely on the classes used in the restored quotient. It therefore induces a bijection from the restored quotient to \(S\). The quotient topology on the restored space is defined by those same identifications: a set is open exactly when its preimage in the cut surface is open. This is also the quotient topology that defines \(S\). The induced bijection and its inverse are therefore continuous, so it is a homeomorphism. \(\square\)
The proposition concerns restoring the original point-identifications. If the banks are paired in a different way, the proposition does not apply; the quotient may change. In a polygon diagram, subdivisions and edge slides are useful because they specify how the boundary pieces are to be matched after the cut.
Worked Example: Subdividing a Paired Edge in the Torus Square
Use the square torus with boundary word \(a b a^{-1}b^{-1}\). The symbol \(a^{-1}\) means that the paired edge is traversed in the direction opposite to the \(a\)-edge. Subdivide the \(a\)-edge into two consecutive pieces \(a_1,a_2\), and subdivide its paired edge at the corresponding point.
The boundary traversal of the original \(a^{-1}\)-edge encounters the matching pieces in reverse order, so the new boundary word is \(a_1a_2 b a_2^{-1}a_1^{-1}b^{-1}\). Pair \(a_1\) with \(a_1^{-1}\), \(a_2\) with \(a_2^{-1}\), and \(b\) with \(b^{-1}\). This subdivision adds no new identifications between points of the original square: it only marks the same paired edges with an extra point. Collapsing the subdivision marks recovers the original square quotient.
Check with the cell counts: before subdivision there is one face, two edge classes, and one vertex class, so \(V-E+F=1-2+1=0\). After subdivision there is one face, three edge classes, and two vertex classes: the two new midpoint marks are paired and give one additional vertex class, while the split \(a\)-pair gives one additional edge class. Thus \(V-E+F=2-3+1=0\), as before. This count is a consistency check; the point-identification argument above verifies that the quotient itself is unchanged.
Edge Slides in the Torus
The torus can be represented as \(\mathbb{R}^2/\mathbb{Z}^2\), where points whose difference has integer coordinates are identified. A square fundamental polygon has edge directions \((1,0)\) and \((0,1)\). An edge slide can replace one direction by its sum with the other. The replacement works here because it is an invertible change of lattice basis.
Proof. Since the entries of \(M\) are integers, \(M\) maps every integer-coordinate vector to an integer-coordinate vector, so \(M(\mathbb{Z}^2)\subseteq\mathbb{Z}^2\). The inverse matrix is \(M^{-1}=(\det M)^{-1}\operatorname{adj}(M)\). Because \(\det M\) is \(1\) or \(-1\), the entries of \(M^{-1}\) are integers as well. Hence \(M^{-1}(\mathbb{Z}^2)\subseteq\mathbb{Z}^2\), which gives \(M(\mathbb{Z}^2)=\mathbb{Z}^2\). Therefore \(x-y\in\mathbb{Z}^2\) if and only if \(Mx-My\in\mathbb{Z}^2\). The map \(x\mapsto Mx\) consequently gives a well-defined bijection on the quotient, with inverse induced by \(M^{-1}\). Both maps are continuous because they are induced by linear maps on \(\mathbb{R}^2\) and the quotient topology. Thus the induced map is a homeomorphism. \(\square\)
Worked Example: Sliding One Torus Edge Past the Other
Replace the square’s basis directions \(u=(1,0)\), \(v=(0,1)\) by \(u=(1,0)\), \(v'=(1,1)\). The second new direction is obtained by sliding the \(v\)-edge once along the \(u\)-direction. The matrix whose columns are \(u\) and \(v'\) is
Its determinant is \(1\cdot1-1\cdot0=1\). Its inverse is \(\begin{pmatrix}1&-1\\0&1\end{pmatrix}\), also an integer matrix. The new parallelogram has side vectors \(u\) and \(v'\), and its edge identifications give a torus model. By the proposition, the change of basis preserves the quotient torus.
Check the lattice: a vector \(m u+n v'\), for integers \(m,n\), equals \((m+n,n)\), which has integer coordinates. Conversely, an arbitrary integer vector \((r,s)\) equals \((r-s)u+s v'\), with integer coefficients \(r-s\) and \(s\). Thus the new edge directions generate exactly \(\mathbb{Z}^2\), not a proper sublattice.
The determinant condition matters. For instance, replacing \(v=(0,1)\) by \(v''=(0,2)\) gives the matrix \(\begin{pmatrix}1&0\\0&2\end{pmatrix}\), whose determinant is \(2\). The vectors \((1,0)\) and \((0,2)\) generate only the lattice of vectors with even second coordinate. This is not the original lattice, so the preceding homeomorphism argument does not apply. A claimed slide must be checked, not assumed to preserve the surface.
A Klein Bottle Edge Slide
The same bookkeeping works for a nonorientable model. One description of the Klein bottle uses transformations of \(\mathbb{R}^2\): let \(b(x,y)=(x+1,y)\) be a translation and let \(a(x,y)=(-x,y+1)\) be a glide reflection. The group of identifications is generated by \(a\) and \(b\). Direct calculation gives \(a b a^{-1}=b^{-1}\), the relation encoded by the standard Klein bottle boundary word \(a b a^{-1}b\).
Worked Example: Sliding a Klein Bottle Generator
Set \(a'=ab\), which is the glide \(a\) followed by one horizontal translation. The new pair \(a',b\) generates exactly the same group as \(a,b\): it is contained in the original group because \(a'=ab\), and \(a=a'b^{-1}\) shows that the original generators are contained in the new group.
Now verify the boundary relation with the new generator. Since \((a')^{-1}=b^{-1}a^{-1}\),
Thus the new pair still satisfies \(a'b(a')^{-1}b=1\), the Klein bottle boundary relation. The transformation is an edge slide: the glide generator has been multiplied by the translation generator, while the generated identification group remains unchanged.
Check the glide explicitly: applying \(b\) first and then \(a\) gives \(a'(x,y)=a(x+1,y)=(-x-1,y+1)\). This is still a glide reflection, not a translation. The calculation of the generated group verifies that this shifted glide gives the same quotient identifications as before.
Common Pitfall: Pairing the Right Labels in the Wrong Order
A frequent error is to cut a polygon boundary and then pair the new pieces in an arbitrary order because their labels look similar. The subdivision example shows why order matters: when an edge is traversed in reverse, its subdivided pieces occur in reverse order too. Pairing \(a_1\) with the wrong segment can change which boundary points are identified, so the original cut-and-restore proposition no longer proves that the surface is unchanged.
Another error is to treat every integer change of edge directions as an invertible change of torus model. The torus calculation requires an integer matrix with determinant \(1\) or \(-1\); the determinant-\(2\) example fails because the new directions generate only a proper sublattice. In any diagram transformation, list the new pairings or generators explicitly, check that they recover the original identifications, and only then conclude that the quotient is preserved.
Check Your Understanding
For each question, identify the point-identifications or generators that justify your answer.
- When a paired edge is subdivided into three pieces, in what order do the pieces appear along the oppositely traversed edge?
- Why does a determinant-\(-1\) integer matrix still induce a homeomorphism of the square torus?
- Do the vectors \((1,0)\) and \((1,2)\) generate \(\mathbb{Z}^2\)? Check their determinant and explain.
- For the Klein bottle generators above, express \(a\) in terms of \(a'=ab\) and \(b\).
- What additional information is needed before an arbitrary re-pairing of cut banks can be claimed to preserve a polygon quotient?