# An fundamental exercise in computation in Simplicial Homology

I am studying homology for the first time, and am trying to get a handle on how to do computations in homology (I have not assimilated things fully yet!). In example, I am trying to work through what might, relatively speaking, be a "baby steps" exercise in this vein (at least, I expect that, retrospectively, I will think so). It reads:

Let $K$ be the boundary of a 2-(resp., 3-)simplex. Compute, using only the definition, the homology of $K$.

In the case of the 2-simplex, I felt that as a first step I should start by using notation for the 3 vertices of the 2-simplex and I looked at the image of the corresponding convex hull under the boundary map using the standard summatory formula. But with this computation in hand, I am not sure what I should be doing next. I expect that if I come to grips with the 2-simplex case, the 3-simplex case won't be exponentially more difficult. If anyone visiting has a better handle on computations in simplicial homology, and would be up for walking me through how to approach such computations in the context of this exercise (say in the case of the 2-simplex), I would really appreciate input. Thanks.

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Let me explain this in detail for you.

Firstly, the boundary of a $2$-simplex $\sigma$ is a simplicial complex $K$. An abstract simplicial complex to which the vertex scheme of $K$ is isomorphic is $\{\{v_0\},\{v_1\},\{v_2\},\{v_0,v_1\},\{v_0,v_2\},\{v_1,v_2\}\}$; the geometric realization of this abstract simplicial complex is $K$.

Let us understand the simplicial chain complex $C(K)$. The chain group in dimension $0$, $C_0(K)$, is the free abelian group on the set of vertices of $K$ (by definition), the chain group in dimension $1$, $C_1(K)$, is the free abelian group on the set of (oriented) $1$-simplices of $K$, and the chain groups in other dimensions are trivial (there are no simplices of dimension greater than $1$ or less than $0$ in $K$).

In particular, the chain complex of $K$ looks like $\cdots\to 0\to 0\to C_1(K)\to C_0(K)\to 0\to 0\to\cdots$; we need to compute the kernel and image of the map $C_1(K)\to C_0(K)$, i.e., the set of cycles and boundaries in dimensions $1$ and $0$, respectively. Firstly, your intuition of "cycle" should suggest that an oriented sum of the $1$-simplices of $K$ should be a cycle. Indeed, you can check this by direct computation:

$\partial([v_0,v_1]+[v_1,v_2]+[v_2,v_0])$

$=\partial([v_0,v_1])+\partial([v_1,v_2])+\partial([v_2,v_0])$

$=(v_1-v_0)+(v_2-v_1)+(v_0-v_2)$

$=0$.

I leave the following important step as an exercise:

Exercise 1: Prove that if $n_0[v_0,v_1]+n_1[v_1,v_2]+n_2[v_2,v_0]$ is a cycle, then $n_0=n_1=n_2$. (Hint: directly compute the image of the simplicial $1$-chain under the boundary map $C_1(K)\to C_0(K)$.)

Therefore, the kernel of $C_1(K)\to C_0(K)$ is the cyclic subgroup of $C_1(K)$ generated by (the cycle) $[v_0,v_1]+[v_1,v_2]+[v_2,v_0]$. In particular, the first homology group $H_1(K)=Z_1(K)/B_1(K)=Z_1(K)\cong \mathbb{Z}$; there are no boundaries in dimension $1$!

In the case of $H_0(K)$, it is much easier to compute this group directly than to first compute the groups of cycles and boundaries $Z_0(K)$ and $B_0(K)$, respectively, in dimension $0$. Let me explain. The group $Z_0(K)$ is clearly $C_0(K)$ because $C_{-1}(K)=0$ but $B_0(K)$ is a little more difficult to describe. We can instead observe that the group $H_0(K)$ is generated by the cosets of $v_0,v_1,v_2$ in $Z_0(K)/B_0(K)=C_0(K)/B_0(K)$ because $C_0(K)$ is the free abelian group generated by $v_0,v_1,v_2$. In order to understand $H_0(K)$, we need to understand the relations between $v_0,v_1,v_2$. Let us compute boundaries:

$\partial([v_0,v_1])=v_1-v_0$

$\partial([v_1,v_2])=v_2-v_1$

$\partial([v_2,v_0])=v_0-v_2$.

In other words, $v_0+B_0(K)=v_0+(v_1-v_0)+B_0(K)=v_1+B_0(K)$ and similarly $v_1+B_0(K)=v_2+B_0(K)$; i.e., $v_0,v_1,v_2$ all have the same image in $H_0(K)=C_0(K)/B_0(K)$! Therefore, the group $H_0(K)$ is generated by the image of, say, $v_0$.

Exercise 2: Prove that if $nv_0$ is a boundary, then $n=0$. (Hint: let us define the augmentation map $\epsilon:C_0(K)\to \mathbb{Z}$ by the rule $\epsilon(v_i)=1$ for all $1\leq i\leq 3$. Prove that the composition of $\epsilon$ with the boundary map $\partial:C_1(K)\to C_0(K)$ is zero.)

Finally, we have shown that $H_0(K)$ is the infinite cyclic group generated by $v_0$, that is, $H_0(K)=\mathbb{Z}$!

I leave the case of the simplicial complex $L$ consisting of the proper faces of a $3$-simplex as a (more difficult) exercise.

I hope this helps!

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