Geometric Topology 12

Resolving Boundary Codes

The usual model of the Klein bottle $K$ shows that it is the quotient of a closed square in $\mathbb{R}^2$ by the equivalence relation defined by the boundary code $aba^{-1}b$. The previous explanation (by dividing the square into two triangles and rearranging) shows that $K$ can also be obtained from a square with code $aaxx$. We write $aba^{-1}b \sim aaxx$ to indicate that the two quotients are homeomorphic.

A variant of the proof (focusing on the triangle with sides $a,b$ and adding the other diagonal) goes like this symbolically:

The symbol "$+$" merely unites codes arising from a disjoint union of polygons.

In this section we shall start with a $6$-gon and boundary code $aabcb^{-1}c^{-1}$, and show that we get the same (i.e. a homeomorphic) surface if instead we start with the code $aabbcc$, or the normal form $(c_1c_1)(c_2c_2)(c_3c_3)$ in the notation of the main theorem (v2):

Proposition: $(aa)(bcb^{-1}c^{-1}) \sim (c_1c_1)(c_2c_2)(c_3c_3)$.

Cutting, Flipping and Pasting

We start the proof of the proposition with the boundary code $a \boxed{abc}b^{-1}c^{-1}$ of the lower horizontal rectangle. The black cut (closing the boxed edges $a,b,c$) gives the union of two squares, and the 'sum' of two words:

by first labelling the right-hand square.

By inverting the first summand (corresponding to flipping the square), and starting the second with $a$, this can be combined into:

a code that now represents the vertical rectangle.

Recall that $A \sim B$ means that the two polygons whose pairs of edges are identified by means of the words $A,B$ are homeomorphic surfaces.

Substitution

We have reached the vertical rectangle, with code $x^{-1}c^{-1}b^{-1}x^{-1}b^{-1}c^{-1}$. The first operation is to separate the triangle with two sides $x^{-1}c^{-1}$ top left, flip it over and reattach:

To do this, we have effectively substituted $y=x^{-1}c^{-1}$ so $c^{-1} = xy$. Now we move the triangle bottom right, which amounts to substituting $z = x^{-1}b^{-1}$ (so $b^{-1} = xz$):

This is the code for the parallelogram.

Juxtaposition

We now represent the parallelogram by a smaller rectangle with the same boundary code $xyyxzz$. The diagonal cut splits this into two triangles, but we can skip the cutting and pasting steps by substituting $w = xyy$ (so $x = wy^{-1}y^{-1}$):

Setting $c_1 = w, \ c_2 = y^{-1}, \ c_3 = z$ completes the proof. $\blacksquare$

Both sides in the proposition are juxtapositions of codes corresponding to $P$ and $T$. Soon we shall interpret this as a homeomorphism of connected sums: