Glue two triangles together to make the torus \(T = S^1 \times S^1\) as shown on the source sheet. Write down singular chains \(p, a, b, t\) representing generators of its homology groups in dimensions \(0, 1, 1, 2\). Now let \(\alpha, \beta \in H^1\) be the Poincaré duals of \([a], [b] \in H_1\). Show directly from the definition in singular (co)homology that \(\alpha \cup \beta\) is a generator of \(H^2\) and that the cohomology ring of \(T\) is an exterior algebra on two generators of degree \(1\).
Solution
1. Use the standard square model of the torus, triangulated by the diagonal from the lower-left to the upper-right corner. After the boundary identifications there is one vertex \(p\), three oriented \(1\)-simplices \(a,b,c\), and two oriented \(2\)-simplices \(\sigma,\tau\) with\[ \partial\sigma=a+b-c, \qquad \partial\tau=-a-b+c. \]
Take \(a\) to be the horizontal edge, \(b\) the vertical edge, and \(c\) the diagonal. For the lower-right triangle, the oriented boundary is \(a+b-c\). For the upper-left triangle the top and left edges occur with the opposite boundary orientations, giving \(-a-b+c\).
2. Thus\[ [p]\in H_0(T),\qquad [a],[b]\in H_1(T),\qquad [t]:=[\sigma+\tau]\in H_2(T) \]are the standard generators, with\[ H_0(T)\cong\mathbb Z,\quad H_1(T)\cong\mathbb Z[a]\oplus\mathbb Z[b],\quad H_2(T)\cong\mathbb Z[t]. \]
Both \(a\) and \(b\) are loops at the single vertex. Since \(\partial(\sigma+\tau)=0\), \(t=\sigma+\tau\) is a \(2\)-cycle. The standard cellular chain computation for this CW structure gives the displayed homology groups, and these classes generate them.
3. Define \(1\)-cocycles \(\alpha,\beta\) by\[ \alpha(a)=1,\quad \alpha(b)=0,\quad \alpha(c)=1, \]and\[ \beta(a)=0,\quad \beta(b)=1,\quad \beta(c)=1. \]Then \([\alpha],[\beta]\) are the Poincaré-dual basis of \(H^1(T;\mathbb Z)\) corresponding to \([b]\) and \(-[a]\) up to the chosen orientation convention.
The cocycle conditions are \[ \alpha(a+b-c)=1+0-1=0, \qquad \beta(a+b-c)=0+1-1=0, \] and the same for \(\tau\). Their values on the homology basis \([a],[b]\) are respectively \((1,0)\) and \((0,1)\), so they form the dual basis of \(H^1\); identifying this dual basis with Poincaré duals changes only the conventional signs/order determined by the orientation of \(T\).
4. With the Alexander–Whitney definition of the singular cup product,\[ (\alpha\smile\beta)(\sigma) =\alpha([v_0v_1])\,\beta([v_1v_2])=1, \]while\[ (\alpha\smile\beta)(\tau)=0. \]Hence\[ \langle [\alpha]\smile[\beta],[t]\rangle=1. \]
For an oriented singular \(2\)-simplex \([v_0v_1v_2]\), the Alexander–Whitney formula is \[ (\alpha\smile\beta)([v_0v_1v_2]) =\alpha([v_0v_1])\beta([v_1v_2]). \] On \(\sigma\), the first edge is \(a\) and the second is \(b\), giving \(1\cdot1=1\). On \(\tau\), the second edge is the negatively oriented horizontal edge, on which \(\beta\) is zero. Summing over \(t=\sigma+\tau\) gives \(1\).
5. Therefore \([\alpha]\smile[\beta]\) is a generator of \(H^2(T;\mathbb Z)\).
The group \(H^2(T;\mathbb Z)\cong\mathbb Z\) is detected by evaluation on the fundamental class \([t]\). A class evaluating to \(1\) is a generator.
6. The squares \([\alpha]^2\) and \([\beta]^2\) vanish and\[ [\beta]\smile[\alpha]=-[\alpha]\smile[\beta]. \]
Integral cohomology is graded-commutative, so for a degree-\(1\) class \(x\) one has \(x^2=-x^2\), hence \(2x^2=0\). Since \(H^2(T;\mathbb Z)\) is torsion-free, \(x^2=0\). Graded commutativity also gives the displayed anticommutation relation.
7. Consequently\[ \boxed{H^*(T;\mathbb Z)\cong\Lambda_{\mathbb Z}([\alpha],[\beta])} \]with both generators in degree \(1\).
The groups have ranks \(1,2,1\) in degrees \(0,1,2\), respectively; 5 gives the nonzero product generating degree \(2\), while 6 gives exactly the exterior-algebra relations.