Let \(\Delta^n = [\nu_0, \cdots, \nu_n]\) have its natural \(\Delta\)-complex structure with \(k\) simplices \([\nu_{i_0}, \cdots, \nu_{i_k}]\) for \(i_0 < \cdots < i_k\). Compute the ranks of the simplicial (or cellular) chain groups \(\Delta_i(\Delta^n)\) and the subgroups of cycles and boundaries. [Hint: Pascal’s triangle.] Apply this to show that the \(k\) skeleton of \(\Delta^n\) has homology groups \(\tilde{H}_i((\Delta^n)^k)\) equal to 0 for \(i < k\), and free of rank \(\binom{n}{k+1}\) for \(i = k\).
Solution
For the standard simplex \(\Delta^n\), the group of \(i\)-chains is free on the \((i+1)\)-element subsets of its \(n+1\) vertices. Hence \[ \operatorname{rank} C_i(\Delta^n)=\binom{n+1}{i+1}. \]
1. In the reduced simplicial chain complex of \(\Delta^n\),\[ \operatorname{rank} Z_i = \operatorname{rank} B_i = \binom{n}{i+1} \qquad (0\le i\le n). \]Here \(\binom{n}{n+1}=0\).
The simplex is contractible, so its reduced chain complex is exact. Thus \[ Z_i=B_i \] for all \(i\ge0\). Since \[ 0\to Z_i\to C_i\xrightarrow{\partial_i}B_{i-1}\to0 \] is exact, ranks satisfy \[ \operatorname{rank} Z_i = \binom{n+1}{i+1}-\operatorname{rank} B_{i-1}. \] Starting with \[ \operatorname{rank}B_{-1}=1 \] in the augmented complex and using Pascal’s identity gives inductively \[ \operatorname{rank}Z_i = \binom{n+1}{i+1}-\binom{n}{i} = \binom{n}{i+1}. \] Exactness gives the same formula for \(B_i\).
2. In the ordinary chain complex,\[ \operatorname{rank}Z_0=n+1, \qquad \operatorname{rank}B_0=n, \]and for \(i\ge1\) the formulas of 1 remain valid.
Ordinarily \(\partial_0=0\), so \(Z_0=C_0\) has rank \(n+1\). The augmentation kernel has rank \(n\), and because \(\Delta^n\) is connected this kernel is \(B_0\). For \(i\ge1\), reduced and ordinary chains agree.
Now let \[ K=(\Delta^n)^k \] be the \(k\)-skeleton.
3. For \(i<k\),\[ \widetilde H_i(K)=0. \]
In degrees at most \(k\), the chain groups and boundary maps of \(K\) agree with those of the full simplex. If \(i<k\), then both \(C_i\) and \(C_{i+1}\) are present, so \[ Z_i(K)=Z_i(\Delta^n), \qquad B_i(K)=B_i(\Delta^n). \] Since the full simplex has zero reduced homology, the quotient \(Z_i/B_i\) is zero.
4. In degree \(k\),\[ \widetilde H_k(K)\cong\mathbb Z^{\binom{n}{k+1}}. \]
There are no \((k+1)\)-chains in the \(k\)-skeleton, so \[ B_k(K)=0. \] On the other hand the boundary map out of \(C_k\) is unchanged, hence \[ Z_k(K)=Z_k(\Delta^n). \] By 1 this is free of rank \[ \binom{n}{k+1}. \] Therefore \[ \boxed{ \widetilde H_i((\Delta^n)^k)= \begin{cases} 0,&i<k,\\ \mathbb Z^{\binom{n}{k+1}},&i=k. \end{cases}} \]