Describe a cell complex structure on the torus \(T = S^1 \times S^1\) and use this to compute the homology groups of \(T\).
To justify your answer you will need to consider the attaching maps in detail.
Solution
1. The torus has a CW structure with one \(0\)-cell \(e^0\), two \(1\)-cells \(a,b\), and one \(2\)-cell \(e^2\).
Represent \[ T=S^1\times S^1 \] as the quotient of the square \([0,1]^2\) obtained by identifying opposite vertical sides and opposite horizontal sides in the same direction. All four vertices become the single \(0\)-cell \(e^0\). The two pairs of opposite sides become the two \(1\)-cells \(a\) and \(b\), and the interior of the square becomes the single open \(2\)-cell \(e^2\).
2. The \(1\)-skeleton is\[ T^{(1)}=a\vee b\cong S^1\vee S^1, \]and the attaching map of \(e^2\) represents the loop\[ aba^{-1}b^{-1}. \]
Traverse the boundary of the square once counterclockwise from a chosen vertex. The four sides project successively to \[ a, \qquad b, \qquad a^{-1}, \qquad b^{-1}. \] After the four vertices are identified, the attaching map \[ \varphi:S^1=\partial D^2\longrightarrow T^{(1)} \] therefore has word \(aba^{-1}b^{-1}\).
3. The cellular chain complex has the form\[ 0 \longrightarrow \ZZ \xrightarrow{d_2} \ZZ^2 \xrightarrow{d_1} \ZZ \longrightarrow0. \]
By 1, the cellular chain groups are free on one \(2\)-cell, two \(1\)-cells, and one \(0\)-cell, with no cells in any other dimension.
4. The first cellular differential is zero:\[ d_1=0. \]
Each \(1\)-cell has both endpoints attached to the single \(0\)-cell. After orienting either \(1\)-cell, its cellular boundary is therefore \[ e^0-e^0=0. \] Hence \(d_1\) vanishes on both generators \(a\) and \(b\).
5. The coefficient of \(a\) in \(d_2(e^2)\) is zero.
By the cellular boundary formula, this coefficient is the degree of the composite \[ S^1 \xrightarrow{\varphi} a\vee b \longrightarrow (a\vee b)/b \cong a\cong S^1, \] where the second map collapses the \(b\)-circle to the wedge point. By 2, after this collapse the attaching word becomes \[ aa^{-1}. \] Thus the resulting map traverses \(a\) once positively and once negatively, so its degree is \[ 1+(-1)=0. \]
6. The coefficient of \(b\) in \(d_2(e^2)\) is also zero.
Collapse the \(a\)-circle instead. The cellular boundary coefficient of \(b\) is the degree of \[ S^1 \xrightarrow{\varphi} a\vee b \longrightarrow (a\vee b)/a \cong b\cong S^1. \] The attaching word reduces to \[ bb^{-1}, \] whose degree is again \(1-1=0\).
7. Therefore\[ d_2=0. \]
The target \(C_1(T)\) is freely generated by \(a\) and \(b\). Steps 5 and 6 show that both coordinates of \(d_2(e^2)\) in this basis are zero.
8. The integral homology groups of the torus are\[ H_k(T;\ZZ) \cong \begin{cases} \ZZ,&k=0,2,\\ \ZZ^2,&k=1,\\ 0,&\text{otherwise}. \end{cases} \]
By 3, 4, and 7, the cellular chain complex is \[ 0\longrightarrow\ZZ\xrightarrow{0}\ZZ^2\xrightarrow{0}\ZZ\longrightarrow0. \] Taking kernels modulo images gives \[ H_2\cong\ZZ, \qquad H_1\cong\ZZ^2, \qquad H_0\cong\ZZ, \] and all other groups vanish because the corresponding cellular chain groups vanish.