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In Hatcher's Algebraic Topology section 1.3, Cayley complexes are explained. The book states that we get a Cayley complex out of a Cayley graph by attaching a 2-cell to each loop. There is an example showing the Cayley complex for $\mathbb{Z}\times\mathbb{Z}$ (the fundamental group of the torus). We attach one 2-cell to each loop and we get $\mathbb{R}^{2}$ with vertical and horizontal tiling. I understand this.

The book then says (example 1.47) that the Cayley complex of a cyclic group of order $ n $ is $n$ disks with boundaries identified. I can't for the life of me figure out where the $n$ disks come from. In the Cayley graph, we have one loop $e \to x \to x^2 \to \cdots \to x^n = e$. I guess the relation $x^n = e$ somehow generates $n$ loops, but I don't understand why.

The next example is for $\mathbb{Z}_2*\mathbb{Z}_2$ in which two 2-cells are attached to each loop. I also don't understand why two.

I'm looking for a canonical description of the algorithm to build Cayley complexes, and the application of the algorithm to build Cayley complexes for finite cyclic groups and $\mathbb{ℤ}_2*\mathbb{ℤ}_2$.

Thank you.

PeterM
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This is essentially the same answer as user32240 but I will try to explain it differently. Hatcher's description is a bit sloppy. The correct thing to say is that if $R$ is the set of defining relators for $G$, then each element $r\in R$ labels a loop based at every vertex of the Cayley graph. To each of these based loops, you add a 2-cell.

The reason for this is you want the group $G$ to act freely on the Cayley complex. Now if you have a relator of the form $r=s^n$ where $s$ is not a proper power, then each loop labeled by $s^n$ in the Cayley graph can be read from $n$ different starting vertices and so you need a 2-cell for each one.

So, for example, if $G=\mathbb Z_2$ with presentation $\langle a\mid a^2=1\rangle$, then you want to have 2 2-cells and have $\mathbb Z_2$ permute them so that you have a free action. The two 2-cells come from the loop labelled $a^2$ at 1 and the looped labeled by $a^2$ at $a$. You can think of the 2 2-cells as the northern and southern hemisphere of a sphere.

If you attached only one $2$-cell, you would get a disk. The group $G$ would fix the center of this disk and so the action is not free. Although the projective plane is a disk with antipodal points identified, the quotient map is not a covering map. By attaching 2 disks you get a covering map.

Incidentally, this issue is not handled properly in the book of Lyndon and Schupp if memory recalls. Cohen makes a big point of this in his book and on the necessity of using $n$ disks for relators $r^n$.

Benjamin Steinberg
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  • Thank you for taking the time to write this answer. I think I got it now. I have one question to make sure I understood. We only attach one 2-cell when building the universal cover of $\mathbb{Z}\times\mathbb{Z} = \langle x,y \mid xyx^{-1}y^{-1}\rangle$ because the relator $xyx^{-1}y^{-1}$ is not an $n$-power for $n > 1$. Correct? – PeterM Oct 24 '13 at 22:48
  • You only attach one 2-cell for each square in the grid because the commutator is not a proper power. – Benjamin Steinberg Oct 25 '13 at 00:34
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The Cayley complex of $\mathbb{Z}/n\mathbb{Z}$ contains $n$ vertices and $n$ edges connected in a circle. So if you want to attach a disk, you have $n$ choices because $(123),(231),(312)$ gives three different attaching disks. For example in the $\mathbb{R}\mathbb{P}^{2}$ case, you glue a disk along the relation $(12)$, and another one along the relation $(21)$. The resulting complex is just the universal cover - the sphere.

This is hard to visualize for $n$ large, and as Hatcher commented the resulting complex is not a surface in general. For a picture and more elementary explanation you may check Page 52, where Hatcher constructed $X_{G}$ explicitly. That's why Hatcher say $\widetilde{X}_{G}$ consists of $n$-disks with their boundary circles identified.

The $\mathbb{Z}_{2}*\mathbb{Z}_{2}$ case is similar, but here I think Hatcher is only trying to kill off enough relations so that the resulting space is simply connected. He observed that all the relations (or circles) near a point must pass through either of the two boundary circles, so it is enough to "fill in" four disks to "kill" the 2 circles. If one did this to every vertex then one got his complex.

Bombyx mori
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  • Thank you for your answer. Unfortunately, I still don't understand when to attach 2-cells. For example, $(2\ 1)$ and $(1\ 2)$ are the same permutation, no? So why do we attach two 2-cells? I understand why the sphere is the universal cover of the projective plane, but I don't see how the Cayley complex is built. Another example, in $\mathbb{Z}\times\mathbb{Z}$ each loop consists of four vertices. Why don't we attach four 2-cells? I guess what I'm missing is a description of the general method for deciding where to attach 2-cells. Hatcher's explanation is very brief. – PeterM Oct 16 '13 at 08:45
  • @PeterM: I think there is no need to attach two 2-cells for $\mathbb{R}\mathbb{P}^{2}$. One is enough and the space would be contractible. But what he wants is the universal cover, so he needs to attach two 2-cells. In the case of $\mathbb{Z}\times \mathbb{Z}$, the relation is just $\langle e_{1},e_{2}\rangle:e_{1}e_{2}=e_{2}e_{1}$. So he attach 2-cell to every relation like $e_{1}e_{2}e_{1}^{-1}e_{2}^{-1}$, which would pass four vertices. If you think of the Cayley graph of $Z*Z$, then he has to attach four 2-cells to kill them all. – Bombyx mori Oct 16 '13 at 16:10
  • Another thing to note is you might be confused with $X_{G}$ and $\wildetide{X}{G}$. The first one attach one disk to each relation. The second one being the universal cover has to attach the cardinality of the relation's many disks to cover the original disk. – Bombyx mori Oct 16 '13 at 17:07