Compute the following directly from chain complexes and check using UCT:
\(H_*(\RP^n; \ZZ_2)\)
\(H_*(\RP^n, \ZZ_3)\)
\(H^*(\RP^n, \ZZ_6)\)
Solution
Goal: Compute the following homology and cohomology groups for real projective space \(\mathbb{RP}^n\) directly from the cellular chain complex, and verify each result using the Universal Coefficient Theorem (UCT): 1. \(H_*(\mathbb{RP}^n; \mathbb{Z}_2)\), 2. \(H_*(\mathbb{RP}^n; \mathbb{Z}_3)\), 3. \(H^*(\mathbb{RP}^n; \mathbb{Z}_6)\).
1. Cellular chain complex and integral homology of \(\mathbb{RP}^n\).
1.1. Standard CW structure of \(\mathbb{RP}^n\): One cell \(e^k\) in each dimension \(0 \le k \le n\), so \(C_k(\mathbb{RP}^n; \mathbb{Z}) \cong \mathbb{Z}\) for \(0 \le k \le n\).
1.2. The cellular boundary map \(d_k \colon C_k(\mathbb{RP}^n; \mathbb{Z}) \to C_{k-1}(\mathbb{RP}^n; \mathbb{Z})\) is multiplication by \(1+(-1)^k\):\[d_k = \begin{cases} 0 & k \text{ odd}, \\ 2 & k \text{ even } (k \ge 2). \end{cases}\]
In the standard CW structure, the characteristic map of the \(k\)-cell is the quotient \(D^k\to\mathbb{RP}^k\). To compute the incidence number with the unique \((k-1)\)-cell, collapse the \((k-2)\)-skeleton and compare the two hemispherical lifts of the attaching map to \(S^{k-1}\). Their local degrees are \(1\) and \((-1)^k\), so the cellular boundary coefficient is \(1+(-1)^k\).
1.3. The integral homology \(H_k(\mathbb{RP}^n; \mathbb{Z})\) is:- \(H_0 \cong \mathbb{Z}\),- \(H_k \cong \mathbb{Z}/2\mathbb{Z}\) for odd \(k\) with \(0 < k < n\),- \(H_k \cong 0\) for even \(k\) with \(0 < k < n\),- Top dimension: \(H_n(\mathbb{RP}^n; \mathbb{Z}) \cong \mathbb{Z}\) if \(n\) is odd, and \(0\) if \(n\) is even.
1.4. With \(d_k = 0\) for odd \(k\) and \(d_k = 2\) for even \(k \ge 2\), the chain complex alternates between the zero map and multiplication by \(2\); taking kernels and images gives the groups listed in
1.3.
2. Compute Part 1: \(H_*(\mathbb{RP}^n; \mathbb{Z}_2)\).
2.1. Direct Chain Complex:- \(C_k(\mathbb{RP}^n; \mathbb{Z}_2) = C_k \otimes \mathbb{Z}_2 \cong \mathbb{Z}_2\) for \(0 \le k \le n\).- The boundary maps are \(d_k \otimes \operatorname{id}_{\mathbb{Z}_2}\).- Since \(d_k \in \{0, 2\}\), \(d_k \equiv 0 \pmod 2\) for all \(k\).- Thus all differentials are zero: \(d_k = 0\).- Therefore, \(H_k(\mathbb{RP}^n; \mathbb{Z}_2) \cong \mathbb{Z}_2\) for all \(0 \le k \le n\), and \(0\) for \(k > n\).
2.2. UCT Check:- By UCT for homology: \(H_k(\mathbb{RP}^n; \mathbb{Z}_2) \cong (H_k(\mathbb{RP}^n; \mathbb{Z}) \otimes \mathbb{Z}_2) \oplus \operatorname{Tor}_1(H_{k-1}(\mathbb{RP}^n; \mathbb{Z}), \mathbb{Z}_2)\).- For \(0 < k < n\):- If \(k\) is odd: \(H_k \otimes \mathbb{Z}_2 = (\mathbb{Z}/2) \otimes \mathbb{Z}_2 \cong \mathbb{Z}_2\), and \(\operatorname{Tor}_1(H_{k-1}, \mathbb{Z}_2) = \operatorname{Tor}_1(0, \mathbb{Z}_2) = 0 \implies H_k \cong \mathbb{Z}_2\).- If \(k\) is even: \(H_k \otimes \mathbb{Z}_2 = 0 \otimes \mathbb{Z}_2 = 0\), and \(\operatorname{Tor}_1(H_{k-1}, \mathbb{Z}_2) = \operatorname{Tor}_1(\mathbb{Z}/2, \mathbb{Z}_2) \cong \mathbb{Z}_2 \implies H_k \cong \mathbb{Z}_2\).- Boundary cases \(k=0\) and \(k=n\) also yield \(\mathbb{Z}_2\).
2.3. Both the direct chain computation and the UCT computation give \(H_k(\mathbb{RP}^n; \mathbb{Z}_2) \cong \mathbb{Z}_2\) for \(0 \le k \le n\).
3. Compute Part 2: \(H_*(\mathbb{RP}^n; \mathbb{Z}_3)\).
3.1. Direct Chain Complex:- \(C_k(\mathbb{RP}^n; \mathbb{Z}_3) \cong \mathbb{Z}_3\) for \(0 \le k \le n\).- On \(\mathbb{Z}_3\), multiplication by \(2\) is an isomorphism (since \(\gcd(2, 3) = 1\), \(2^{-1} \equiv 2 \pmod 3\)).- Thus \(d_k = 0\) for \(k\) odd, and \(d_k \colon \mathbb{Z}_3 \xrightarrow{\cong} \mathbb{Z}_3\) is an isomorphism for \(k\) even (\(k \ge 2\)).- Therefore, the homology vanishes in all intermediate dimensions:- \(H_0(\mathbb{RP}^n; \mathbb{Z}_3) \cong \mathbb{Z}_3\),- \(H_k(\mathbb{RP}^n; \mathbb{Z}_3) = 0\) for \(0 < k < n\),- Top dimension: \(H_n(\mathbb{RP}^n; \mathbb{Z}_3) \cong \mathbb{Z}_3\) if \(n\) is odd, and \(0\) if \(n\) is even.
3.2. UCT Check:- \(\mathbb{Z}/2 \otimes \mathbb{Z}_3 = 0\) and \(\operatorname{Tor}_1(\mathbb{Z}/2, \mathbb{Z}_3) = 0\) since \(\gcd(2, 3) = 1\).- Thus all 2-torsion vanishes under \(\otimes \mathbb{Z}_3\), giving identical results.
3.3. The direct chain computation and the UCT computation agree: both give \(H_0 \cong \mathbb{Z}_3\), \(H_k = 0\) for \(0 < k < n\), and \(H_n \cong \mathbb{Z}_3\) for odd \(n\) (and \(0\) for even \(n\)).
4. Compute Part 3: \(H^*(\mathbb{RP}^n; \mathbb{Z}_6)\).
4.1. Direct Cochain Complex:- \(C^k(\mathbb{RP}^n; \mathbb{Z}_6) = \operatorname{Hom}(C_k, \mathbb{Z}_6) \cong \mathbb{Z}_6\) for \(0 \le k \le n\).- The coboundary \(\delta^k \colon C^k \to C^{k+1}\) is dual to \(d_{k+1}\), so for \(0 \le k < n\):\[\delta^k = d_{k+1}^* = \begin{cases} 0 & k \text{ even}, \\ 2 & k \text{ odd}. \end{cases}\]and \(\delta^n = 0\) because \(C^{n+1} = 0\).- For \(k = 0\): \(\delta^0 = 0\), so \(H^0 = \ker(\delta^0) / \operatorname{im}(\delta^{-1}) = \mathbb{Z}_6 / 0 \cong \mathbb{Z}_6\).- For \(0 < k < n\):- If \(k\) is even: \(\delta^k = 0\) and \(\delta^{k-1} = 2\), so \(H^k = \mathbb{Z}_6 / 2\mathbb{Z}_6 \cong \mathbb{Z}_2\).- If \(k\) is odd: \(\delta^k = 2\) and \(\delta^{k-1} = 0\), so \(H^k = \ker(\times 2) / 0 = \{0, 3\} \cong \mathbb{Z}_2\).- For \(k = n\): \(\delta^n = 0\), and \(\delta^{n-1} = d_n^*\):- If \(n\) is odd: \(d_n = 0\), so \(H^n = \mathbb{Z}_6 / 0 \cong \mathbb{Z}_6\).- If \(n\) is even: \(d_n = 2\), so \(H^n = \mathbb{Z}_6 / 2\mathbb{Z}_6 \cong \mathbb{Z}_2\).
4.2. UCT Check for Cohomology:- UCT: \(H^k(X; G) \cong \operatorname{Hom}(H_k(X), G) \oplus \operatorname{Ext}^1(H_{k-1}(X), G)\).- Note \(\operatorname{Hom}(\mathbb{Z}, \mathbb{Z}_6) \cong \mathbb{Z}_6\), \(\operatorname{Hom}(\mathbb{Z}/2, \mathbb{Z}_6) \cong \mathbb{Z}_2\), \(\operatorname{Ext}^1(\mathbb{Z}/2, \mathbb{Z}_6) \cong \mathbb{Z}_2\), \(\operatorname{Ext}^1(\mathbb{Z}, \mathbb{Z}_6) = 0\).- Using \(H_*(X; \mathbb{Z})\) from 1 gives:- \(H^0 \cong \operatorname{Hom}(\mathbb{Z}, \mathbb{Z}_6) \oplus 0 \cong \mathbb{Z}_6\).- \(0 < k < n\): \(H^k \cong \mathbb{Z}_2\) for all \(k\).- \(k = n\): \(H^n \cong \begin{cases} \mathbb{Z}_6 & n \text{ odd}, \\ \mathbb{Z}_2 & n \text{ even}. \end{cases}\)
4.3. The direct cochain computation and the UCT computation agree in every degree, so both methods give the same groups.
5. Q.E.D.
5.1. <1>1–<1>4 directly compute and cross-verify all three cases.