Why if $n$ is odd the simplicial homology is $\mathbb Z,$ while if $n$ is even it is $0$?

Question

I am trying to solve this question:

Compute the homology groups of the $\Delta$-complex $X$ obtained from $\Delta^n$ by identifying all faces of the same dimension. Thus $X$ has a single $k$-simplex for each $k \leq n.$

And I found the following solution online:

But I do not understand the difference between the identification in the odd case and i the even case which leads to that if $n$ is odd the simplicial homology is $\mathbb Z,$ while if $n$ is even it is $0,$ could someone explain this to me please?

I read here The space $\Delta^n$ with all faces of the same dimension. but I could not understand the case $\Delta^1,$ why we identify $v_1$ by $- v_2$(maybe this is the reason for the difference between simplicial homology in the even and odd dimension)? Could anyone explain this to me please?

Thanks in advance!

Answer 1

Briefly, you are identifying $v_1$ with $v_2$, no signs involved. The signs are instead built into the formula for the boundary map. (See Hatcher, p. 105, for example.)

In more detail, as long as $k \leq n$, the boundary map $C_k \to C_{k-1}$ has the form $\mathbb{Z} \to \mathbb{Z}$, since there is one $k$-simplex and one $(k-1)$-simplex, and the issue is identifying that map. The boundary map applied to a $k$-simplex is the alternating sum of its faces — the signs are built into the formula for the boundary map. Let's write $\sigma_m$ for the unique $m$-simplex for each $m$, and write $[\sigma_m]$ for the corresponding element of the chain group. Then the boundary map $C_k \to C_{k-1}$ will be $$ [\sigma_k] \mapsto \sum_{i=0}^k (-1)^i [\sigma_{k-1}] = \left(\sum_{i=0}^k (-1)^i \right) [\sigma_{k-1}]. $$ Now evaluate the sum in parentheses: it's either zero or 1 depending on the parity of $k$, so the whole thing is either $[\sigma_{k-1}]$ if $k$ is even, $0$ if $k$ is odd.

Answer 2

The question is Hatcher's exercise 2.1.9. And my answer is almost the same as the accepted answer. It's just that OP asked a new question related: Why the author changed to parity of $k$ in the last paragraph? . So I try to add a maybe more detailed answer. And I don't know if OP uses the definition as Hatcher's or not, but here I am proving via Hatcher's definition anyway.

Recall that the standard $n$-simplex is defined as:

$$ \Delta^{n}=\left\{\left(t_{0}, \cdots, t_{n}\right) \in \mathbb{R}^{n+1} \mid \sum_{i} t_{i}=1 \text { and } t_{i} \geq 0 \text { for all } i\right\} $$

and Hatcher use the description "restriction on a face" to describe a face operation. But I think the way in Lee's Introduction to topological manifolds is much more clear and formal:

For any $i = 0,1,..., n$, define a map $F_{i,n}: \Delta^{n-1} \rightarrow \Delta^n$ (called the $i$-th face map in dimension $n$ by Lee):

$$ F_{i, n}(t_0, t_1, ..., t_{n-1}) = (t_0, t_1, \dots, t_{i-1}, 0, t_i, \dots, t_{n-1}). $$

Where the $0$ is at the $i$-th coordinate, and the following $t$ s are shifted one coordinate right. Hence it's a map from $\Delta^{n-1}$ to $\Delta^n$.

Then back to Hatcher's definition on simplicial homology, for a $\Delta$-complex structure $\{\sigma_\alpha: \Delta^n \rightarrow X\}$ on a space $X$, the boundary homomorphism $\partial_{n}: \Delta_{n}(X) \rightarrow \Delta_{n-1}(X)$ is defined via basis elements as:

$$ \partial_n(\sigma_\alpha) = \sum_{i=0}^{n}(-1)^i \sigma_\alpha \circ F_{i,n}. $$

Recall that for a $\Delta$-complex, $\sigma_\alpha \circ F_{i, n}$ is still in the structure(see Hatcher's book page 103 at the bottom), so the definition is well defined.

Now coming to the spacial case. I think the second description is more useful rather than using the description "by identifying all faces of the same dimension." and in fact it leads to the first description. So let the $\Delta$-complex structure be $\{\sigma_k: \Delta^k \rightarrow X | k = 0,1,..,n \}$.

Recall again that each restriction of the simplex to a face is still in the stucture, we get that:

$$ \sigma_k \circ F_{i, k} = \sigma_{k-1} $$ for all $1 \leq k \leq n$ and $0 \leq i \leq k$.

And then calculate the simplex homology group as the accepted answer:

For any $0 \leq k \leq n$, the elments in $H^\Delta_k(X)$ is represented by cycles in $\Delta_k(X)$. Since $\Delta_k(X)$ has the only basis $\sigma_k$, it's enough to check if it's a cycle or some boundary:

1. if $\sigma_k$ is not a cycle, then $H^\Delta_k(X) = 0$.
2. if $\sigma_k$ is a cycle but not some boundary, then $H^\Delta_k(X) = \Bbb{Z}$.
3. If $\sigma_k$ is some boundary, then $H^\Delta_k(X) = 0$.

Now since for $k = 1,2, .., n$:

$$ \begin{aligned} \partial_k(\sigma_k) & = \sum_{i=0}^{k}(-1)^i \sigma_k \circ F_{i,k} & \text{(by def of $\partial$)} \\ & = \sum_{i=0}^{k}(-1)^i \sigma_{k-1} & \text{(by def of $X$)} \\ & = 0 \text{ if $k$ is odd, } \sigma_{k-1} \text{ if $k$ is even.} \end{aligned} $$

And $\sigma_0$ is always cycle, $\sigma_n$ is not boundary by convention. Hence we get the result of the accepted answer:

For $k=0$, we have $\partial_1(\sigma_1) = 0$. Hence $\sigma_0$ is not a boundary. Hence $H^\Delta_0(X) = \Bbb{Z}$.

For $k=n$, we have $\partial_n(\sigma_n) = 0$ if $n$ is odd, $\partial_n(\sigma_n) = \sigma_{n-1}$ if $n$ is even. Correspondingly, $H^\Delta_n(X) = \Bbb{Z}$ if $n$ is odd, $H^\Delta_n(X) = 0$ if $n$ is even.

For $0 \lt k \lt n$, either $\sigma_k$ is some boundary or not a cycle. In both case we have $H^\Delta_k(X) = 0$.