Describe a CW complex structure for the lens space \(L(p, 1)\) and compute \(\pi_1, H_*\) for it.
Solution
1. The lens space \(L(p,1)\) admits a CW structure with one cell in each dimension \(0,1,2,3\).
View \(L(p,1)\) as the quotient of \(S^3\) by the free action of the cyclic group of order \(p\). Equivalently, use the standard lens-space fundamental domain: the quotient has one vertex, one edge generating the cyclic fundamental group, one \(2\)-cell whose attaching map winds \(p\) times around that edge, and one top-dimensional \(3\)-cell.
2. Its fundamental group is\[ \boxed{\pi_1(L(p,1))\cong\mathbb Z/p}. \]
The \(1\)-skeleton is a circle with generator \(a\), and the \(2\)-cell is attached by the degree-\(p\) map, imposing the single relation \(a^p=1\). The \(3\)-cell does not change \(\pi_1\).
3. The cellular chain complex is\[ 0\longrightarrow\mathbb Z\xrightarrow{0}\mathbb Z\xrightarrow{p}\mathbb Z\xrightarrow{0}\mathbb Z\longrightarrow0. \]
There is one cell in each dimension. The \(2\)-cell attaching map has degree \(p\) on the \(1\)-cell, so \(\partial_2\) is multiplication by \(p\). The top boundary is zero because \(L(p,1)\) is a closed orientable \(3\)-manifold, so its top homology is \(\mathbb Z\).
4. Therefore\[ \boxed{H_k(L(p,1);\mathbb Z)\cong \begin{cases} \mathbb Z,&k=0,3,\\ \mathbb Z/p,&k=1,\\ 0,&\text{otherwise}. \end{cases}} \]
Take homology of the chain complex in 3.