From some notes for a class I am taking (I don't believe they are publicly available), it is stated:

Definition. Let $X$ be a simplicial complex. Define the simplicial chain complex of $(C_{\bullet}, \partial)$ to have:

  1. Chain groups $C_k(X)$ which are $\mathbb{Z}_2$ vector spaces generated on a basis of the $k$-simplices of $X$.

  2. A boundary differential $\partial_k : C_k(X) \to C_{k-1}(X)$ defined by its values taken on the basis $\{ \sigma \in \Delta | \dim(\sigma) = k\}$

$$ \partial_k = \sum_{{\tau \subset \sigma} \\ [\tau / \sigma ]=1} \tau$$

The notes then proceed to show what the homology group of a disk is:

$$H_k(D^n) = \begin{cases} \mathbb{Z}_2 & k = 0 \\ 0 & k \geq 1 \end{cases} $$

I am having trouble understanding the base case:

As a base case, $D^0$ is a point, so we have that $C_0(D^0) = \mathbb{Z}_2$ and all other chain groups are empty; the claim trivially follows.

Here is what I don't understand:

  1. What does $C_0(D^0)$ have to do with the homology group? (Furthermore, why does being a point imply $C_0(D^0) = \mathbb{Z}_2$?) I thought the way you computed this homology group is by:

$$H_0(D^0) = \frac{\ker \partial_{0}}{\text{Im}\partial_{1}}$$

  1. What is $\partial_{0} : C_0(D^0) \to C_{-1}(D^0)$?

I think if I can understand those two things, I can probably arrive to the conclusion that the $0$th homology group is $\mathbb{Z}_2$.


By definition, $C_k(X)$ is a vector space whose basis is the set of $k$-simplices of $X$. When $X=D^0$, $X$ has a single $0$-simplex and no $k$-simplices for $k\neq 0$. This means that $C_0(D^0)$ is one-dimensional (i.e., isomorphic to $\mathbb{Z}_2$) and $C_k(D^0)$ is trivial if $k\neq 0$.

Now we want to compute $H_k(D^0)$. That means we want to look at the maps $\partial_k:C_k(D^0)\to C_{k-1}(D^0)$ and $\partial_{k+1}:C_{k+1}(D^0)\to C_k(D^0)$ and compute the kernel of $\partial_k$ mod the image of $\partial_{k+1}$. Now the definitions of these maps are pretty complicated, but fortunately we don't have to worry about the definitions in this case because our vector spaces are so trivial. If $k>0$, then $C_k(D^0)$ is trivial. Since $\ker\partial_k$ is a subspace of $C_k(D^0)$, $\ker\partial_k$ is also trivial, and since $H_k(D^0)$ is a quotient of $\ker\partial_k$, $H_k(D^0)$ is also trivial. This proves $H_k(D^0)=0$ for $k\neq 0$.

Now let's consider $k=0$. The map $\partial_0:C_0(D^0)\to C_{-1}(D^0)$ has domain $\mathbb{Z}_2$ and codomain $0$ (we usually don't talk about $C_{-1}(D^0)$; it is always just $0$ because there is no such thing as a $(-1)$-simplex). So $\partial_0$ must send everything to $0$, so its kernel is all of $\mathbb{Z}_2$. On the other hand, the map $\partial_1:C_1(D^0)\to C_0(D^0)$ must be $0$, since its domain is $0$. So the image of $\partial_1$ is trivial. Thus $H_0(D^0)$ is $\mathbb{Z}_2$ mod the trivial subspace, or just $\mathbb{Z}_2$.

  • $\begingroup$ When you say $\partial_0 : C_0(D^0) \to C_{-1}(D^0)$ has domain $\mathbb{Z}_2$. Why is the domain not a space isomorphic to $\mathbb{Z}_2$? Otherwise I think I understand everything. Thanks! $\endgroup$ – Dair Mar 29 '17 at 4:30
  • $\begingroup$ Oh, I guess we need only a space up to isomorphism. I guess it doesn't matter. My bad haha. $\endgroup$ – Dair Mar 29 '17 at 4:38
  • $\begingroup$ @ericwolsey Hang on if $C_0(D^0)$ takes one $0-$simplex, then $C_0(D^0) = \{nP\}$ where $P$ is the $0-$simplex, so shoundn't $\{nP\} \approx \mathbb{Z}$? How did you get $\mathbb{Z}_2$? $\endgroup$ – Hawk Dec 2 '18 at 2:37
  • $\begingroup$ @Hawk: All homology in this question is with coefficients in $\mathbb{Z}_2$. $\endgroup$ – Eric Wofsey Dec 2 '18 at 3:01

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