Describe \(H^*(\mathbb{CP}^\infty / \mathbb{CP}^1; \mathbb{Z})\) as a ring with finitely many multiplicative generators. How does this ring compare with \(H^*(S^6 \times \mathbb{HP}^\infty; \mathbb{Z})\)?
Solution
Let \(u\in H^2(\mathbb{CP}^\infty;\mathbb Z)\) be the standard generator. For \[ Q=\mathbb{CP}^\infty/\mathbb{CP}^1 \] one has \[ \widetilde H^*(Q;\mathbb Z) \cong H^*(\mathbb{CP}^\infty,\mathbb{CP}^1;\mathbb Z). \] Since \(\mathbb{CP}^1\) contains exactly the cells in dimensions \(0\) and \(2\), this relative group is infinite cyclic in each even degree \(\ge4\) and zero otherwise. Under the map to \(H^*(\mathbb{CP}^\infty)\), its degree-\(2k\) generator maps to \(u^k\) for \(k\ge2\), and multiplication is inherited from \(\mathbb Z[u]\).
Hence the positive-degree part is the subring generated by \(u^2\) and \(u^3\). Put \[ a=u^2,\qquad b=u^3. \] Then \(|a|=4\), \(|b|=6\), and the only relation is \[ b^2=a^3. \] Thus \[ \boxed{H^*(Q;\mathbb Z)\cong \mathbb Z[a,b]/(b^2-a^3),\qquad |a|=4,\ |b|=6.} \]
For comparison, \[ H^*(S^6\times\mathbb{HP}^\infty;\mathbb Z) \cong \mathbb Z[x,y]/(x^2), \qquad |x|=6,\ |y|=4. \] The two graded groups are isomorphic: both have one copy of \(\mathbb Z\) in degree \(0\) and in every even degree \(\ge4\). But the rings are not isomorphic. In \(H^*(Q)\) the degree-\(6\) generator \(b\) satisfies \[ b^2=a^3\ne0, \] whereas in \(H^*(S^6\times\mathbb{HP}^\infty)\) the degree-\(6\) generator \(x\) has square zero.