The Klein Bottle from a Paired Square

Two Different Pairings of a Square

A square can produce different surfaces depending on which boundary points are identified and how the paired sides are matched. For the Klein bottle, one pair is matched in the same direction as the boundary traversal, while the other is matched in the opposite direction. Keeping track of these directions also helps determine how the square’s vertices, edges, and face become cells in the quotient.

Let \(Q=[0,1]\times[0,1]\), with its usual topology as a subspace of \(\mathbb{R}^2\). Name its corners \(P=(0,0)\), \(R=(1,0)\), \(S=(1,1)\), and \(T=(0,1)\), in counterclockwise order. We pair the left and right sides by \((0,t)\sim(1,t)\), and the bottom and top sides by \((s,0)\sim(1-s,1)\), for \(s,t\in[0,1]\). Here \(\sim\) is the equivalence relation generated by these pairings: it includes all pairs required by the identifications and is transitive.

On the right side, counterclockwise traversal goes from \(R\) to \(S\); on the left side, it goes from \(T\) to \(P\). The map \((0,t)\mapsto(1,t)\) sends the left-side traversal from \(T\) to \(P\) to a traversal from \(S\) to \(R\), which is opposite to the boundary direction on the right side. Thus the vertical sides are paired in opposite directions. On the bottom side the boundary runs from \(P\) to \(R\), and on the top it runs from \(S\) to \(T\). The map \((s,0)\mapsto(1-s,1)\) sends \(P\) to \(S\) and \(R\) to \(T\), so it sends the bottom traversal to the top traversal in the same direction.

One way to picture the construction is first to join the vertical sides, making a cylinder. The remaining boundary consists of the top and bottom circles; the second pairing identifies them with their circle parameters reversed. The quotient model, rather than the picture, determines the identifications exactly.

Tracing Points and Vertices

For any point on a paired side, its equivalence class includes the point on the matching side. At the corners, both pairing rules act, so it is important to apply them together rather than count the four corners as separate points.

The quotient has one 0-cell, represented by the common corner class. The two pairs of sides give two 1-cells: one from the vertical sides and one from the horizontal sides. The interior of the square is not identified with any other interior point, so it gives one 2-cell. These cells form a cell decomposition of the quotient. At an interior point of a paired edge, the two half-neighborhoods on either side of the identified edge join to a disk neighborhood. At the vertex, the four corner sectors join in one cycle: the side identifications connect the sectors at \(P\) to \(S\), \(S\) to \(T\), \(T\) to \(R\), and \(R\) to \(P\). Thus the vertex also has a disk neighborhood, as required for this polygon quotient to be a surface.

The Boundary Word and the Quotient Cells

To record the edge pairings, give each edge pair a letter and choose an arrow on one representative edge. Let \(a\) denote the bottom edge directed from \(P\) to \(R\), and let \(b\) denote the right edge directed from \(R\) to \(S\). The bottom pairing sends the arrow \(P\to R\) to \(S\to T\), which is the counterclockwise traversal of the top edge. The top edge therefore contributes \(a\). The vertical pairing sends the arrow \(R\to S\) to \(P\to T\), opposite to the counterclockwise traversal \(T\to P\) of the left edge. The left edge contributes \(b^{-1}\). Reading counterclockwise from \(P\), the boundary word is \(a b a b^{-1}\).

Proof. The four corners are equivalent because \(P\sim R\), \(P\sim S\), and \(R\sim T\); these relations place all four in one class. There are no other corner points, so there is exactly one quotient vertex. Each side is paired with exactly one other side, and the vertical pairings do not identify an edge interior with a horizontal edge interior. Consequently there are exactly two edge classes. No two points in the square interior are paired, so its image is one 2-cell. Applying the definition of Euler characteristic gives \(1-2+1=0\). \(\square\)

A Common Counting Pitfall

A common error is to count the four square corners as four quotient vertices simply because they are drawn in four different places. That would give \(V=4\), \(E=2\), and \(F=1\), and the incorrect calculation \(4-2+1=3\). The correction is to apply the equivalence relation before counting cells: the side pairings put all four corners in one class, so \(V=1\), and the correct value is \(1-2+1=0\).

A different error is to infer the direction of a pairing from how it looks on the page. For this model the bottom-top map \(s\mapsto1-s\) is same-direction because it sends the bottom boundary traversal \(P\to R\) to the top traversal \(S\to T\). The left-right map preserves the vertical coordinate but is opposite-direction relative to the counterclockwise boundary traversals. The coordinate rule and the boundary direction must both be checked.

The quotient-cell calculation is useful because it extracts a numerical invariant directly from the identifications, without treating the square’s drawn corners or edges as distinct when the quotient says otherwise. Here the boundary word records how the two edge classes meet, while the cell count gives the Euler characteristic. These are different pieces of information: the edge directions matter for describing the gluing, and the vertex, edge, and face counts determine the Euler characteristic.