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 \(k \leq n\).
Solution
Let \(\sigma_k\) denote the unique \(k\)-simplex of \(X\). Then \[ C_k(X)\cong\mathbb Z\langle\sigma_k\rangle \qquad(0\le k\le n). \]
1. The simplicial boundary is\[ \partial_k(\sigma_k) = \left(\sum_{i=0}^k(-1)^i\right)\sigma_{k-1} = \begin{cases} \sigma_{k-1},&k\text{ even},\\ 0,&k\text{ odd}. \end{cases} \]
All \((k-1)\)-faces of \(\sigma_k\) are identified with the unique simplex \(\sigma_{k-1}\). Hence the usual alternating boundary formula collapses to the displayed coefficient sum. This alternating sum is \(1\) when \(k\) is even and \(0\) when \(k\) is odd.
2. Thus the chain complex is\[ 0\longrightarrow\mathbb Z \mathop{\longrightarrow}^{\partial_n} \mathbb Z \mathop{\longrightarrow}^{\partial_{n-1}} \cdots \mathbb Z \mathop{\longrightarrow}^{\partial_1} \mathbb Z \longrightarrow0, \]where \(\partial_k\) is the identity for even \(k\) and zero for odd \(k\).
This is exactly 1 after identifying each \(C_k\) with \(\mathbb Z\).
3. For every \(0<k<n\) one has\[ H_k(X)=0. \]
If \(k\) is even then \(\partial_k=1\), so \(\ker\partial_k=0\). If \(k\) is odd then \(\partial_k=0\), but \(k+1\) is even, so \[ \operatorname{im}\partial_{k+1}=\mathbb Z=C_k. \] Thus the homology vanishes in either case.
4. The top homology is\[ H_n(X)\cong \begin{cases} \mathbb Z,&n\text{ odd},\\ 0,&n\text{ even}. \end{cases} \]
There is no incoming boundary in degree \(n\). Hence \[ H_n(X)=\ker\partial_n. \] By 1, \(\partial_n=0\) for odd \(n\) and is the identity for even \(n\).
5. Therefore\[ \boxed{ H_0(X)\cong\mathbb Z, \qquad H_k(X)=0\ (0<k<n), \qquad H_n(X)\cong \begin{cases} \mathbb Z,&n\text{ odd},\\ 0,&n\text{ even}. \end{cases} } \]
The space is connected, so \(H_0(X)\cong\mathbb Z\). The higher groups are given by <1>3–<1>4.