The Klein Bottle Group from Polygon Gluing

From a Square to a Group Presentation

The Klein bottle can be built by identifying pairs of opposite edges of a square, with one pair glued in the orientation-reversing pattern. The edge labels record the attaching loop of the square’s interior. Seifert–van Kampen turns that loop into a relation, and the resulting relation has a useful interpretation: going around one generator conjugates the other to its inverse.

Use the square convention in which its boundary, read once around, has labels \(a,b,a^{-1},b\). Identify the two edges labelled \(a\) according to their arrows, and do the same for the two edges labelled \(b\). After the identifications, the four corners represent one vertex, and the two pairs of edges give two loop edges. Thus the resulting space has a cell structure with one vertex, two edges \(a,b\), and one two-dimensional cell. The boundary of that cell runs along the edge loop \(aba^{-1}b\).

Here the graph is the two loop edges, and the attaching loop is \(aba^{-1}b\). Applying the stated result gives the presentation below. The notation \(\langle a,b\mid r\rangle\) means the group generated by \(a,b\), subject to the relation \(r=1\); equivalently, it is the free group on \(a,b\) divided by the normal subgroup generated by \(r\).

Proof. The cell structure gives a graph with one vertex and two loop edges, so its fundamental group is free on \(a,b\). The attaching map of the two-cell traverses the boundary word \(aba^{-1}b\). By the disk-attachment form of Seifert–van Kampen, attaching this cell divides the free group by the normal subgroup generated by that word. Therefore the presentation is \(\langle a,b\mid aba^{-1}b=1\rangle\). In any group, \(xy=1\) is equivalent to \(x=y^{-1}\): multiplying \(xy=1\) on the right by \(y^{-1}\) gives \(x=y^{-1}\), and multiplying \(x=y^{-1}\) on the right by \(y\) gives \(xy=1\). Taking \(x=aba^{-1}\) and \(y=b\) gives the equivalent relation \(aba^{-1}=b^{-1}\). \(\square\)

The presentation depends on the chosen labels and boundary-reading direction. Reversing the direction replaces a boundary word by its inverse, which gives the same relation in the normal subgroup used in the disk-attachment result. The presentation above uses the stated convention throughout.

Normal Forms: Why Negative Powers Matter

A useful way to organize words in this group is to put every power of \(b\) before every power of \(a\). To justify doing so, one needs conjugation formulas for both \(a\) and \(a^{-1}\). The second formula is not optional: words in the generators may contain \(a^{-1}\).

Proof. The defining relation gives \(aba^{-1}=b^{-1}\). To obtain the rule for \(a^{-1}\), take inverses of this equality to get \(ab^{-1}a^{-1}=b\). The original equality also gives \(b=a^{-1}b^{-1}a\); taking inverses of that equality yields \(b^{-1}=a^{-1}ba\). Thus both \(a\) and \(a^{-1}\) conjugate \(b\) to \(b^{-1}\). Repeated multiplication, and then taking inverses for negative exponents, gives \(a^n b^k a^{-n}=b^{(-1)^n k}\) for every integer \(n,k\). This includes \(n=0\) or \(k=0\), when the formula is the identity \(b^k=b^k\) or \(1=1\).

Equivalently, \(a^n b^k=b^{(-1)^n k}a^n\). Now read any word from left to right, combining consecutive powers of the same generator. If the part already read is \(b^m a^n\), appending \(b^k\) gives \(b^m a^n b^k=b^{m+(-1)^n k}a^n\); appending \(a^r\) gives \(b^m a^{n+r}\). Starting from the identity, these steps put every finite word into the form \(b^m a^n\). This proves existence of the form, not yet its uniqueness. \(\square\)

For uniqueness, and for a concrete way to calculate, define a set \(M=\mathbb Z\times\mathbb Z\) with operation \[ (m,n)(r,s)=\bigl(m+(-1)^n r,\ n+s\bigr). \] This is a group. For associativity, either parenthesization of three pairs \((m,n),(r,s),(u,v)\) gives first coordinate \(m+(-1)^n r+(-1)^{n+s}u\) and second coordinate \(n+s+v\), since \((-1)^{n+s}=(-1)^n(-1)^s\). The identity is \((0,0)\), and the inverse of \((m,n)\) is \((-(-1)^n m,-n)\), as substitution on either side gives \((0,0)\).

Proof. In \(M\), put \(b=(1,0)\) and \(a=(0,1)\). The inverse of \(a\) is \((0,-1)\). The multiplication rule gives \(ab=(-1,1)\), so \(aba^{-1}=(-1,0)=b^{-1}\). Therefore the defining relation holds, and the presentation defines a homomorphism from the presented group to \(M\) that sends \(a\) and \(b\) to these pairs. Every pair \((m,n)\) equals \(b^m a^n\) in \(M\), so this homomorphism is surjective. By the normal-form existence just proved, every element in the presented group is some \(b^m a^n\). Its image in \(M\) is \((m,n)\), since \((m,0)(0,n)=(m,n)\). If two such forms represent the same element, their images satisfy \((m,n)=(r,s)\), which means \(m=r\) and \(n=s\). Thus the normal form is unique. The homomorphism is also injective: an element in its kernel has a normal form whose image is \((0,0)\), so that form is \(b^0a^0\), the identity. Hence it is an isomorphism. \(\square\)

What the Relation Says—and a Common Pitfall

The group is not abelian: its generators do not commute. The defining relation instead says that conjugating \(b\) by \(a\) changes its direction. The pair model makes the distinction checkable. Using its operation, \(ab=(0,1)(1,0)=(-1,1)\), whereas \(ba=(1,0)(0,1)=(1,1)\). These pairs are different, so \(ab\ne ba\).

The polygon computation illustrates the role of van Kampen: a graph supplies generators, and the attached cell supplies a relator. Here the relator does more than record a loop that becomes trivial. Its equivalent conjugation form and the pair model reveal the structure of the resulting group, while the unique normal form ensures that these computations distinguish elements rather than merely suggest a convenient notation.