Homology sphere

A homology sphere is a (closed connected oriented) $n$-mainfolds with the same homology as the $n$-sphere. Poincaré first conjectured that any $n$-manifolds homologous (i.e. having the same homology) to the $n$-sphere should be homeomorphic to the $n$-sphere, then later he found a counterexample, which led him to a modified conjecture (which is the Poincaré’s conjecture).

The counterexample he found is very interesting, one of the way to construct it is the folowing:

  1. take a dodecahedron
  2. identify the pairs of opposite faces by rotating each of them by $\dfrac{\pi}{5}$

This will identify $4$ vertices into $1$, $3$ edges into $1$, and $2$ faces into $1$. The resulting manifold will be a CW-complex with $5$ $0$-cells, $10$ $1$-cells and $6$ $2$-cells. And it can be shown that it has the same homology as $\mathbb{S}^3$, but I have never seen anyone worked it out…so I decided to do it by myself with nice diagrams, since it is quite hard to keep track of everything while computing, and I thought that it will be beautiful to draw them out 😃

Clumsy (but colorful) computation

Below is a figure showing the identified vertices (with numbers) and edges (with colors) of the dodecahedron aftering taking quotient:

We will calculates its singular homology by CW-homology. To do so, we need to know the how the generators (the faces, edges, vertices) are mapped by the differential. Let’s do the differential on each faces first. There are 6 of them, I will mark them out below:

Face A

Face B

Face C

Face D

Face E

Face F

And here is the edges with their color code:

So

$d(A) = i+a+b+c+d$,

$d(B)=-a+e-c+f+g$,

$d(C)=-b-g-d-e+h$,

$d(D)=-c-h-i+g+j$,

$d(E)=-d-j+a+h+f$,

$d(F)=-i-f-b+j-e$

The differential of edges should be clear.

So we have the following cellular chain:

$\mathbb{Z}^5 \xleftarrow{A} \mathbb{Z}^{10} \xleftarrow{B} \mathbb{Z}^6 \xleftarrow{0} \mathbb{Z}$

where

$A = \begin{pmatrix} 0 & 0 & 0 & -1 & 0 & -1 & 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 & -1 & 0 \\ -1 & 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 1 \\ 0 & -1 & 1 & 0 & 0 & 1 & 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1 & -1 & 0 & 0 & 0 & 0 & -1 \end{pmatrix}$

$B = \begin{pmatrix} 1 & -1 & 0 & 0 & 1 & 0\\ 1 & 0 & -1 & 0 & 0 & -1\\ 1 & -1 & 0 & -1 & 0 & 0\\ 1 & 0 & -1 & 0 & -1 & 0\\ 0 & 1 & -1 & 0 & 0 & -1\\ 0 & 1 & 0 & 0 & 1 & -1\\ 0 & 1 & -1 & 1 & 0 & 0\\ 0 & 0 & 1 & -1 & 1 & 0\\ 1 & 0 & 0 & -1 & 0 & -1\\ 0 & 0 & 0 & 1 & -1 & 1 \end{pmatrix}$

(God I hate huge matrices)

So

$\texttt{ker} B = 0$, $\texttt{ker} A \cong \mathbb{Z}^6$ and we get $\tilde{H}_n (X) = \mathbb{Z}$ for $n = 3$ and $0$ otherwise, the same as $\mathbb{S}^3$.

It’s been a toil… maybe this is something like diagram chasing - not difficult, but better to do it in private 😄