Homology groups of a tetrahedron

Solution 1:

I don't know what notation you have been using, but the following could help -just in case you haven't arrived so far.

Name the vertexes of the tetrahedron $[1],[2],[3],[4]$, for instance. Then the set of edges can be denoted by $[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]$ and the one of faces as $[1,2,3],[1,2,4],[1,3,4],[2,3,4]$.

So the corresponding groups of $0,1$ and $2$ chains are the free abelian groups on these generators:

$$ \begin{eqnarray*} C_0 &=& \mathbb{Z}\langle [1],[2],[3],[4]\rangle \\ C_1 &=& \mathbb{Z}\langle [1,2],[1,3],[1,4],[2,3],[2,4],[3,4]\rangle \\ C_2 &=& \mathbb{Z}\langle [1,2,3],[1,2,4],[1,3,4],[2,3,4]\rangle \end{eqnarray*} $$

and the boundary operators

$$ C_2 \stackrel{\partial_2}{\longrightarrow} C_1 \stackrel{\partial_1}{\longrightarrow} C_0 $$

can be computed as follows:

$$ \partial_1 [1,2] = [2]-[1] \ , \qquad \partial_2 [1,2,3] = [2,3]-[1,3]+[1,2] \ , \qquad \text{etc}\dots $$

So you can represent these boundary operators by matrices. For instance,

$$ \partial_1 = \begin{pmatrix} -1 & -1 & -1 & 0 & 0 & 0 \\ 1 & 0 & 0 & -1 & -1 & 0 \\ 0 & 1 & 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 & 1 & 1 \end{pmatrix} $$

and pursue your computations from this point.

For instance, if we denote by $T$ the tetrahedron, then

$$ H_2 (T) = \mathrm{ker}\ \partial_2 $$

and some few elementary column transformations with the $\partial_2$ matrix gives us that

$$ H_2 (T) = \mathbb{Z} \langle [2,3,4] - [1,3,4] + [1,2,4] - [1,2,3] \rangle \cong \mathbb{Z}\ . $$

That is, as we already knew, of course: $H_2(T)$ is $\mathbb{Z}$. But we've got more: an explicit generator for this second homology group. Namely, the 2-cycle $[2,3,4] - [1,3,4] + [1,2,4] - [1,2,3] $, which is, of course, the sum of the faces of the tetrahedron (with a sign).

Solution 2:

Let us finish the computation of $H_1(X)$ using Agustí's notation above:

Now I have row reduced Agustí's matrix above and (if my calculations are correct) we have that the kernel of $\partial_1$ is isomorphic to the free abelian group on three generators:

$$\ker \partial_1 = \Bbb{Z}\bigg\langle \color{red}{[1,2] - [1,3] + [2,3]}, \color{green}{[1,2]-[1,4] + [2,4]}, \color{blue}{[1,3] - [1,4] + [3,4]}\bigg\rangle.$$

Now by direct calculation I find that:

$$\begin{eqnarray*} \partial_2([1,2,3]) &=& \color{red}{[2,3] - [1,3] + [1,2]} \\ \partial_2([1,2,4]) &=& \color{green}{[1,2] + [2,4] - [1,4]}\\ \partial_2([1,3,4]) &=& \color{blue}{[1,3] + [3,4] - [1,4]} \\ \partial_2([2,3,4]) &=& [2,3] + [3,4] - [2,4] .\end{eqnarray*}$$

Now when we quotient out by $\textrm{im} \partial_2$ in calculating $H_1(X)$, what you are basically doing is imposing relations on $\ker \partial_1$. What are those relations? Well, you are simply declaring now that $$\partial_2(\text{all 2-simplices}) = 0.$$

From this you immediately see that upon quotiening, you have sent all the generators to zero so that $H_1(X) = 0$.