Homology computations of $\partial \Delta[n]$ using normalized chain complexes

Question

I have to compute the homology group of the abstract simplicial set $\partial \Delta[n]$ using normalized chain complexes.

I know that the nondegenerate $k$-simplices of $\partial \Delta[n]$ are $\Delta[n]^{\text{nd}}_k$ for $k \neq n$ (which are the strictly increasing maps $[k] \to [n]$) and $\varnothing$ otherwise.

Moreover, I proved that the homology of $\Delta[n]$ is the same as the one of a point, that is, $H_0(\Delta[n]) = \mathbb{Z}$ and it is $0$ otherwise.

However, I don't understand how to put everything together. It seems clear that for $k=n$ the homology of the boundary is $0$, but what happens for $k = n+1, n-1$ in the chain complex?

Of course, I know that I have to get $H_k(\partial \Delta[n])=\mathbb{Z}$ for $k=0, n-1$ and $0$ otherwise, but how can I deduce this from the homology of the abstract simplex $\Delta[n]$?

Thank you very much for your help.

Answer 1

Let $S^{n-1}$ be the complex with $\mathbb{Z}$ in degree $0, n-1$ and $0$ in all other degrees. Let $i_n$ be the unique nondegenerate $n$-simplex of $\Delta^n$. Then the nondegenerate $(n-1)$-simplices of $\partial\Delta^n$ are $d^n_i(i_n), i=0,\ldots, n$ and $c = \sum_{j=0}^n(-1)^jd_j(i_n)$ is a cycle in $N(\partial \Delta^n)$, since it is a boundary in $N(\Delta^n)$. Define a chain map $f\colon S^{n-1}\to N(\partial\Delta^n)$ via $f_{n-1}(k) = kc$ and in degree $0$ $\mathbb{Z}\to N(\partial\Delta^n)_0$ by sending the generator of $\mathbb{Z}$ to the basis element corresponding to the $0$th vertex. There is a retraction $r\colon N(\partial\Delta^n)\to S^{n-1}$. In degree $0$ this maps $N(\partial\Delta^n)_0\to \mathbb{Z}$ the basis element corresponding to the $0$th vertex to $1$ and all other basis elements to $0$. In degree $n-1$ it is given by $r_{n-1}\colon N(\partial\Delta^n)\to \mathbb{Z}$, $$r_{n-1}(d^n_j(i_n)) = \begin{cases} 0 &\text{if }j<n,\\ 1 &\text{if }j=n. \end{cases}$$ I claim that $f\circ r$ is homotopic to the identity. An increasing map $\alpha\colon [k]\to [n]$ corresponds to an increasing sequence $(i_0,\ldots, i_k)$ and this is nondegenerate if all elements are distinct. We define a chain homotopy on basis elements by $$s_k(i_0,\ldots,i_k) = \begin{cases} (-1)^{k+1}(i_0,\ldots, i_k, n) &\text{if } i_k<n,\\ 0 &\text{otherwise}, \end{cases}$$ and we extend this to a linear map $N(\partial\Delta^n)_k\to N(\partial\Delta^n)_{k+1}$. I leave it to you to check that this gives the desired chain homotopy.

One can also use the long exact sequence in homology. Let $\mathbb{Z}[n]$ be the complex with $\mathbb{Z}$ in degree $n$ and $0$ in all other degrees. Then there is a short exact sequence of complexes $$0\to N(\partial\Delta^n)\to N(\Delta^n)\to \mathbb{Z}[n]\to 0,$$ since $\partial\Delta^n$ and $\Delta^n$ have the same nondegenerate $k$-simplices for $k\neq n$, $\partial\Delta^n$ has only degenerate $n$-simplices and $\Delta^n$ has exactly one nondegenerate $n$-simplex. Now if you already know the homology of $N(\Delta^n)$ it is easy to read off the homology of $N(\partial \Delta^n)$ from the long exact sequence in homology.

Answer 2

You just need to write down the basis for each term and the boundary maps. For example, if $n =2$, you have $C_2=0, C_1=\mathbb Z\{[1,2],[2,3],[1,3]\}, C_0 = \mathbb Z\{[1],[2],[3]\}$, and $$\partial_1=\left(\begin{matrix}-1&0&-1\\ 1&-1&0\\ 0&1&1 \end{matrix}\right).$$ Similarly $\partial_0 = \left(\begin{matrix}1&1&1\end{matrix}\right).$ Now just calculate kernels and images. Since you've already done this for te full simplex (not just the boundary) you know that for most cases the kernel of the outgoing operator will equal the image of the incoming one. So the only real work to do is to analyze what happens in the top dimension.