Compute the cup product structure in \(H^*(S^2 \times S^8 \sharp S^4 \times S^6; \mathbb{Z})\), and in particular show that the only nontrivial cup products are those dictated by Poincaré duality.
Solution
Let \[ M=(S^2\times S^8)\#(S^4\times S^6) \] with the connected-sum orientation. In degrees strictly between \(0\) and \(10\), reduced cohomology is the direct sum of that of the two summands. Choose generators \[ a\in H^2(M),\quad b\in H^8(M),\quad c\in H^4(M),\quad d\in H^6(M), \] coming respectively from the \(S^2,S^8,S^4,S^6\) factors, and let \[ u\in H^{10}(M)\cong\mathbb Z \] be the orientation class.
The connected-sum collapse maps to the two summands preserve the top class. Hence the product structures on the summands give \[ a\smile b=u, \qquad c\smile d=u \] after choosing the generators compatibly with orientation. Since all four degrees are even, the reversed products are the same.
Products between classes coming from different connected-sum summands vanish. Within each product-of-spheres summand, the square of either positive-dimensional generator is zero. Therefore \[ a^2=c^2=0, \] and every other product of positive-dimensional generators vanishes for either this reason or degree reasons, except \[ \boxed{ab=cd=u.} \] Thus the only nonzero positive-dimensional cup products are exactly the Poincaré-dual pairings.