The Torus Group from a Square

From a Square to the Torus

A square can encode a surface by identifying its opposite edges. For the torus, the bottom and top edges are identified in the same direction, and the left and right edges are identified in the same direction. The resulting space has one vertex, two loop edges, and one square attached along those loops. Seifert–van Kampen turns the path around the square into a relation between the two generators.

Give \(I=[0,1]\) its usual topology and give \(I\times I\) the product topology. Define an equivalence relation on \(I\times I\) by identifying \((0,t)\) with \((1,t)\), and \((s,0)\) with \((s,1)\), for \(s,t\in I\), together with the identifications forced by transitivity. Let \(T^2\) be the quotient space with the quotient topology: a subset of \(T^2\) is open exactly when its preimage under the quotient map is open in \(I\times I\).

This quotient is the familiar product of two circles. In fact, the map \(I\times I\to S^1\times S^1\) given by \((s,t)\mapsto(e^{2\pi i s},e^{2\pi i t})\) is continuous and is constant on each equivalence class. Its fibers are exactly those classes: equality of the two coordinates on the circle means that the corresponding numbers in \(I\) are equal or are the two endpoints. Thus it induces a continuous bijection \(T^2\to S^1\times S^1\). The square is compact, so its quotient \(T^2\) is compact; \(S^1\times S^1\), with the product topology, is Hausdorff. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. To see why, a closed subset of the compact domain is compact, its image is compact and hence closed in a Hausdorff space, so the bijection is a closed map and its inverse is continuous.

Reading the Boundary Word

Before using the theorem, fix the orientation convention. The boundary of the square is traversed counterclockwise, starting at the lower-left corner. The bottom edge is \(a\). The right edge is \(b\), because it is identified with the left edge in the upward direction. The top edge is \(a^{-1}\): its traversal is from right to left, opposite to the chosen direction of \(a\). The left edge is \(b^{-1}\), since it is traversed downwards. Consequently the boundary loop reads \(aba^{-1}b^{-1}\).

The square’s boundary becomes a loop in the union of the two edge loops. The interior of the square is a disk attached to that union along this boundary loop. The disk-attachment form of Seifert–van Kampen says that attaching a disk to a path-connected space \(X\) along a based loop with class \(w\in\pi_1(X)\) changes the fundamental group to \(\pi_1(X)/\langle\!\langle w\rangle\!\rangle\). Here \(\langle\!\langle w\rangle\!\rangle\) denotes the smallest normal subgroup containing \(w\). This is the attaching-a-disk consequence of Seifert–van Kampen: the disk is simply connected, and its boundary loop becomes null-homotopic after attachment, imposing that relation and all its conjugates.

The edge union is a wedge of two circles, with common point \(v\). Its fundamental group is the free group on the loops \(a,b\), as in the standard wedge-of-circles computation. Applying the disk-attachment result imposes the boundary word as a relation.

Proof. The square quotient consists of the wedge of the two edge loops together with the image of the square’s interior as one attached disk. The attaching loop is \(aba^{-1}b^{-1}\), by the edge-by-edge orientation calculation. The wedge has fundamental group free on \(a,b\). The disk-attachment form of Seifert–van Kampen therefore quotients this free group by the normal subgroup generated by \(aba^{-1}b^{-1}\), which gives exactly the stated presentation. \(\square\)

Why the Generators Commute

The relation can be rewritten as \(ab=ba\). Indeed, from \(aba^{-1}b^{-1}=1\), multiply on the right first by \(b\), then by \(a\), to obtain \(ab=ba\). Conversely, if \(ab=ba\), then \(aba^{-1}b^{-1}=1\). The presentation therefore describes two commuting generators. The next result identifies the presented group precisely, rather than only naming its relation.

Proof. Let \(G=\langle a,b\mid aba^{-1}b^{-1}=1\rangle\). Since the relation gives \(ab=ba\), the assignment \(a\mapsto(1,0)\), \(b\mapsto(0,1)\) defines a homomorphism \(\phi:G\to\mathbb{Z}\times\mathbb{Z}\): the defining relation maps to \((1,0)+(0,1)-(1,0)-(0,1)=(0,0)\). Define \(\psi:\mathbb{Z}\times\mathbb{Z}\to G\) by \(\psi(m,n)=a^m b^n\). This is a homomorphism because \(a\) and \(b\) commute: \(a^m b^n a^r b^s=a^{m+r}b^{n+s}\) for all integers \(m,n,r,s\), including negative exponents. Now \(\phi(\psi(m,n))=(m,n)\), so \(\phi\circ\psi\) is the identity. Also \(\psi(\phi(a))=a\) and \(\psi(\phi(b))=b\); since \(a,b\) generate \(G\), \(\psi\circ\phi\) is the identity on \(G\). Thus \(\phi\) and \(\psi\) are inverse isomorphisms. \(\square\)

A Common Orientation Pitfall

A frequent error is to read the top edge as \(a\), forgetting that the boundary traversal runs from right to left. That incorrect choice would produce the word \(abab\), and would impose a different relation. The correction is not a matter of choosing a more convenient presentation: it follows from the orientation of the actual boundary path.

The calculation illustrates the role of the attaching word: the two loops \(a\) and \(b\) are not independent free generators after the square is filled in. The boundary of that disk supplies a relation, and here that relation says exactly that the two directions commute. The resulting group records the two integer winding counts of a loop on the torus.