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I have been solving a lot of questions lately in preparation for an exam which is in about 3weeks. Well, I came across this question but I have no clue. I think I would need a lot of help.

I am asked to compute the homology groups of a tetrahedron (i.e a triangulation of the 2-sphere) indicating all the chain groups and boundary homomorphisms. I am familiar with computing the homology groups of some well-known surfaces like the Möbius strip, the Klein bottle, the torus and the projective plane, but I am not sure how to go about the tetrahedron. Any help would be greatly appreciated.

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Which homology groups? The simplicial ones? Can you write down all of the boundary maps explicitly? –  Qiaochu Yuan Sep 1 '12 at 21:52
    
@QiaochuYuan. Yes, the Simplicial ones. –  smanoos Sep 1 '12 at 21:53
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The tetrahedron is formed from a 3-simplex together with all its faces. This should help you to work out the boundary maps quite easily. (Draw a picture)! Let me know what you think they are and I will verify them. –  Edward Hughes Sep 1 '12 at 22:33
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3 Answers

up vote 7 down vote accepted

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).

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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 $H_1(X) = 0$.

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The simplicial complex for the tetrahedron $T$ is exactly the same as that of the filled tetrahedron $F$ (a.k.a. the $3$-simplex) except in degree $3$. As we know the homology of the $3$-simplex (hopefully!) we need only compute $H_2(T)$.

Now, the $2$-cycles in $T$ at the same as in $F$. Since the group of $2$-boundaries in $F$ is cyclic of rank $1$ (because the $3$-chains form such a group) and $H_2(F)=0$, we see at once that $H_2(T)=\mathbb Z$.

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This reasoning applies yo every situation in which have two simplicial complexes, $T\subset F$ with $F$ differing from $T$ by exactly one simplex of dimension larger than all simplices in $T$. –  Mariano Suárez-Alvarez Sep 29 '13 at 9:59
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