Given an integer \(m > 1\) and integers \(\ell_1, \cdots, \ell_n\) relatively prime to \(m\), the lens space \(L = L_m(\ell_1, \cdots, \ell_n)\) is the orbit space \(S^{2n-1}/\mathbb{Z}_m\) of the unit sphere \(S^{2n-1} \subset \mathbb{C}^n\) with the action of \(\mathbb{Z}_m\) generated by the rotation \(\rho(z_1, \cdots, z_n) = (e^{2\pi i \ell_1/m} z_1, \cdots, e^{2\pi i \ell_n/m} z_n)\). It has a CW structure with one cell \(e^k\) for each \(k \leq 2n-1\), whose cellular chain complex is \[ 0 \to \mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{m} \mathbb{Z} \xrightarrow{0} \cdots \xrightarrow{0} \mathbb{Z} \xrightarrow{m} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \to 0 \] with boundary maps alternately \(0\) and multiplication by \(m\).
For the lens space \(L_m(\ell_1, \cdots, \ell_n)\) defined in Example 2.43, compute the cohomology groups using the cellular cochain complex and taking coefficients in \(\mathbb{Z}\), \(\mathbb{Q}\), \(\mathbb{Z}_m\), and \(\mathbb{Z}_p\) for \(p\) prime. Verify that the answers agree with those given by the universal coefficient theorem.
Solution
Let \[ L=L_m(\ell_1,\ldots,\ell_n) \] be the \((2n-1)\)-dimensional lens space. Its standard CW structure has one cell in every dimension \(0,1,\ldots,2n-1\), with cellular boundary \[ d_k=\begin{cases} 0,&k\text{ odd},\\ m,&k\text{ even}. \end{cases} \] Hence with coefficients in an abelian group \(G\), the cellular cochain complex has one copy of \(G\) in every degree and \[ \delta^k=\begin{cases} 0,&k\text{ even},\\ \times m,&k\text{ odd}. \end{cases} \]
1. For arbitrary \(G\),\[ H^k(L;G)\cong \begin{cases} G,&k=0,2n-1,\\ G/mG,&0<k<2n-1\text{ and }k\text{ even},\\ G[m],&0<k<2n-1\text{ and }k\text{ odd},\\ 0,&\text{otherwise}, \end{cases} \]where \(G[m]=\ker(m:G\to G)\).
This is immediate from the displayed alternating cochain differential.
2. With \(G=\mathbb Z\),\[ H^k(L;\mathbb Z)\cong \begin{cases} \mathbb Z,&k=0,2n-1,\\ \mathbb Z_m,&0<k<2n-1\text{ even},\\ 0,&\text{otherwise}. \end{cases} \]
Multiplication by \(m\) on \(\mathbb Z\) has zero kernel and cokernel \(\mathbb Z_m\).
3. With \(G=\mathbb Q\),\[ H^k(L;\mathbb Q)\cong \begin{cases} \mathbb Q,&k=0,2n-1,\\ 0,&\text{otherwise}. \end{cases} \]
Multiplication by nonzero \(m\) is an isomorphism of \(\mathbb Q\).
4. With \(G=\mathbb Z_m\),\[ \boxed{H^k(L;\mathbb Z_m)\cong\mathbb Z_m\quad(0\le k\le2n-1).} \]
Multiplication by \(m\) is zero on \(\mathbb Z_m\), so every cellular coboundary vanishes.
5. For a prime \(p\),\[ H^k(L;\mathbb Z_p)\cong \begin{cases} \mathbb Z_p,&0\le k\le2n-1,&p\mid m,\\ \mathbb Z_p,&k=0,2n-1,&p\nmid m,\\ 0,&0<k<2n-1,&p\nmid m. \end{cases} \]
If \(p\mid m\), multiplication by \(m\) is zero on \(\mathbb Z_p\). If \(p\nmid m\), it is an automorphism.
6. These answers agree with the universal coefficient theorem.
The integral homology of the lens space is \[ H_k(L;\mathbb Z)\cong \begin{cases} \mathbb Z,&k=0,2n-1,\\ \mathbb Z_m,&0<k<2n-1\text{ odd},\\ 0,&\text{otherwise}. \end{cases} \] The cohomological universal coefficient sequence \[ 0\to\operatorname{Ext}(H_{k-1}(L),G) \to H^k(L;G) \to\operatorname{Hom}(H_k(L),G)\to0 \] has \[ \operatorname{Hom}(\mathbb Z_m,G)=G[m], \qquad \operatorname{Ext}(\mathbb Z_m,G)=G/mG, \] which reproduces exactly the formula in 1.