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I'm studying the basics of homology on Nakahara, Geometry, Topology and Physics and I'm trying to work out the (simplicial) homology group of the tetrahedron described by the complex $K=\{0,1,2,3,(01),(02),(03),(12),(13),(23),(012),(013),(023),(123)\}$.

I have already calculated $H_2(K)=\mathbb Z$, but I'm stuck on $H_1(K)$.
I know $H_0(K)=\mathbb Z$, because the tetrahedron is connected, but I don't want to use this result.
I found

$$ Z_0(K)=C_0(K)=\{i0+j1+k2+l3\}\\ Z_1(K)=\{i[(01)+(12)+(20)]+j[(02)+(21)+(13)+(30)]+k[(02)+(23)+(30)]\}\\ Z_2(K)=\{i[(012)-(013)+(023)-(123)]\} \\ B_0(K)=\{(-a-b-c)0+(a-d-e)1+(b+d-f)2+(c+e+f)3\} \\ B_1(K)= \{(i+j)(01)+(k-i)(02)+(-j-k)(03)+(i+l)(12)+(j-l)(13)+(k+l)(23)\}\\ B_2(K)= 0 $$

Typically, in the previous examples, Nakahara tries to construct a surjective homomorphism from the cycles group to what will be discovered to be the homology group.
This works for $H_0(K)$: define $$ f\colon Z_0(K) \to \mathbb Z \qquad\text{such that}\qquad i0+j1+k2+l3 \mapsto i+j+k+l $$ Then, the kernel of $f$ is the set of $0$-chains whose coefficients sum to $0$: this is precisely $B_0(K)$. By the first isomorphism theorem, $Z_0(K)/\ker f\simeq \mathbb Z$.

Now, my question: is it possible to construct a similar argument to conclude that $H_1(K)=0$?
I would say no. Also, I feel that this technique is somewhat tricky... maybe there is a more standard and reliable procedure, like the one described here?

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Andrea Orta
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  • Better to pick an orientation for each, so you don't include $(20)$ and $(02)$, but write $-(02)$ for $(20)$ in $C_1$. But perhaps that is just shorthand here? – Thomas Andrews Jul 25 '13 at 17:39
  • Also, is the bondary of $(01)=(1)-(0)$ or $(0)-(1)$? Seems like you are using $(1)-(0)$, but wanted to check. – Thomas Andrews Jul 25 '13 at 17:49
  • @ThomasAndrews: The boundary of (01) is (1)-(0), right. Also, I wrote (20) only for evidentiating the structure of the generators of $Z_1$. In my calculations, I always used the same orientation ((01), (023) and so on). =) – Andrea Orta Jul 25 '13 at 17:56

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It is much easier to just show that $B_1=Z_1$ directly, I think. When $H_i(X)\not\cong 0$ you often have to come up with some clever argument and isomorphisms to figure it out what form it takes, but when $H_i(X)\cong 0$, you just need to show that every element in $Z_i$ is in $B_i$.

You just need to show that each of your generators for $Z_1$ is in $B_1$, which is easy - the only remotely hard one is $$\begin{align}(02)-(12)+(13)-(03)&=\left(-(01)+(02)-(12)\right) + \left((01)+(13)-(03)\right)\\&=-\partial(012)+\partial(013)\end{align}$$ or maybe some sign adjustment is required there, depending on the definition. (edit: fixed)

These early definitions of homology are rarely used in practice after a while - there are other ways to compute homology. The reason to do these computations is to make yourself familiar with the concepts. It helps you familiarize yourself with abelian groups in general, and it helps you see that there is something intuitively "geometric" in all this abstract-seeming nonsense.

Andrea Orta
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Thomas Andrews
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