Let \(\Sigma\) be a closed orientable surface of genus \(g\). Compute \(H_i(S^1 \times \Sigma; \ZZ)\) for \(i = 0, 1, 2, 3\).
Solution
1. The integral homology of the closed orientable genus-\(g\) surface is\[ H_i(\Sigma;\ZZ) \cong \begin{cases} \ZZ, & i=0,\\ \ZZ^{2g}, & i=1,\\ \ZZ, & i=2,\\ 0, & \text{otherwise}. \end{cases} \]
Use the standard CW structure on \(\Sigma\) with one \(0\)-cell, \(2g\) \(1\)-cells \[ a_1,b_1,\dots,a_g,b_g, \] and one \(2\)-cell attached by the word \[ [a_1,b_1]\cdots[a_g,b_g]. \] The cellular differential \[ d_1:C_1\to C_0 \] is zero because every \(1\)-cell is a loop at the unique vertex. The differential \[ d_2:C_2\to C_1 \] is also zero because the total exponent of every \(a_i\) and \(b_i\) in the attaching word is zero. Thus the cellular chain complex is \[ 0\longrightarrow\ZZ \xrightarrow{0} \ZZ^{2g} \xrightarrow{0} \ZZ \longrightarrow0, \] which gives the displayed homology groups.
2. The homology of \(S^1\) is\[ H_i(S^1;\ZZ) \cong \begin{cases} \ZZ, & i=0,1,\\ 0, & \text{otherwise}. \end{cases} \]
Use the CW structure on \(S^1\) with one \(0\)-cell and one \(1\)-cell. Its cellular differential is zero.
3. The integral Künneth theorem gives\[ H_n(S^1\times\Sigma;\ZZ) \cong \bigoplus_{p+q=n} H_p(S^1;\ZZ)\otimes H_q(\Sigma;\ZZ). \]
In the integral Künneth short exact sequence, the remaining terms are \[ \operatorname{Tor}_1^{\ZZ} \bigl(H_p(S^1;\ZZ),H_q(\Sigma;\ZZ)\bigr). \] By 1 and 2, every nonzero homology group of either factor is free abelian. Hence all Tor terms vanish, leaving the displayed direct sum of tensor products.
4. In degree \(0\),\[ \boxed{H_0(S^1\times\Sigma;\ZZ)\cong\ZZ.} \]
By 3, \[ H_0(S^1\times\Sigma) \cong H_0(S^1)\otimes H_0(\Sigma) \cong \ZZ\otimes\ZZ \cong\ZZ. \]
5. In degree \(1\),\[ \boxed{H_1(S^1\times\Sigma;\ZZ)\cong\ZZ^{2g+1}.} \]
The Künneth decomposition is \[ \begin{aligned} H_1(S^1\times\Sigma) &\cong \bigl(H_1(S^1)\otimes H_0(\Sigma)\bigr) \oplus \bigl(H_0(S^1)\otimes H_1(\Sigma)\bigr)\\ &\cong \ZZ\oplus\ZZ^{2g} \cong\ZZ^{2g+1}. \end{aligned} \]
6. In degree \(2\),\[ \boxed{H_2(S^1\times\Sigma;\ZZ)\cong\ZZ^{2g+1}.} \]
Since \(H_2(S^1)=0\), \[ \begin{aligned} H_2(S^1\times\Sigma) &\cong \bigl(H_1(S^1)\otimes H_1(\Sigma)\bigr) \oplus \bigl(H_0(S^1)\otimes H_2(\Sigma)\bigr)\\ &\cong \ZZ^{2g}\oplus\ZZ \cong\ZZ^{2g+1}. \end{aligned} \]
7. In degree \(3\),\[ \boxed{H_3(S^1\times\Sigma;\ZZ)\cong\ZZ.} \]
The only nonzero tensor product with total degree \(3\) is \[ H_1(S^1)\otimes H_2(\Sigma) \cong \ZZ\otimes\ZZ \cong\ZZ. \]
8. Therefore\[ \boxed{ H_i(S^1\times\Sigma;\ZZ) \cong \begin{cases} \ZZ, & i=0,3,\\ \ZZ^{2g+1}, & i=1,2,\\ 0, & i>3. \end{cases}} \]
Steps <1>4–<1>7 give the requested degrees. There are no tensor or Tor contributions above degree \(3\) because the factors have homology only in degrees at most \(1\) and \(2\), respectively.