Cohomology groups need not determine cohomology ring

Statement

Cohomology groups need not determine cohomology ring

It is possible to have two topological spaces $M_{1}$ and $M_{2}$ -- in fact, we can choose both $M_{1}$ and $M_{2}$ to be compact connected orientable manifolds -- such that for every $n$, we have an isomorphism of Cohomology group (?)s:

$H^{n}(M_{1};\mathbb {Z} )\cong H^{n}(M_{2};\mathbb {Z} )$

but the Cohomology ring (?) $H^{*}(M_{1};\mathbb {Z} )$ is not isomorphic (as a graded ring, or even as a ring) to the cohomology ring $H^{*}(M_{2};\mathbb {Z} )$. In particular, $M_{1}$ and $M_{2}$ are not homotopy-equivalent spaces and in fact, since our examples are manifolds, they are not even weak homotopy-equivalent spaces.

Homology groups need not determine cohomology ring

Since homology groups determine cohomology groups, an alternative formulation is that it is possible to have two topological spaces $M_{1}$ and $M_{2}$ -- in fact, we can choose both $M_{1}$ and $M_{2}$ to be compact connected orientable manifolds -- such that for every $n$, we have an isomorphism of Cohomology group (?)s:

$H_{n}(M_{1};\mathbb {Z} )\cong H_{n}(M_{2};\mathbb {Z} )$

but the Cohomology ring (?) $H^{*}(M_{1};\mathbb {Z} )$ is not isomorphic (as a graded ring, or even as a ring) to the cohomology ring $H^{*}(M_{2};\mathbb {Z} )$.

Related facts

• Homology groups determine cohomology groups
• Homology groups and fundamental group need not determine homotopy groups
• Homotopy groups need not determine homology groups

Proof

We can in fact construct an example of three different compact connected orientable manifolds with the same cohomology groups but different cohomology ring structures. Take $M_{1}$ to be the product of two 2-spheres $S^{2}\times S^{2}$, $M_{2}$ to be the connected sum of two complex projective planes with same orientation, and $M_{3}$ to be the connected sum of two complex projective planes with opposite orientation. For all these:

$H^{p}(M_{1})=H^{p}(M_{2})=H^{p}(M_{3})=\left\lbrace {\begin{array}{rl}\mathbb {Z} ,&\qquad p=0,4\\\mathbb {Z} \oplus \mathbb {Z} ,&\qquad p=2\\0&\qquad \operatorname {otherwise} \end{array}}\right.$

The ring structures we get are:

• $H^{*}(M_{1};\mathbb {Z} )\cong \mathbb {Z} [x,y]/(x^{2},y^{2})$ where $x,y$ generate $H^{2}$ as a free abelian group of rank two and $xy$ generates $H^{4}$.
• $H^{*}(M_{2};\mathbb {Z} )\cong \mathbb {Z} [x,y]/(x^{2}-y^{2},x^{3},y^{3})$ where $x,y$ generate $H^{2}$ as a free abelian group of rank two and $x^{2}=y^{2}$ generates $H^{4}$.
• $H^{*}(M_{3};\mathbb {Z} )\cong \mathbb {Z} [x,y]/(x^{2}+y^{2},x^{3},y^{3})$ where $x,y$ generate $H^{2}$ as a free abelian group of rank two and $x^{2}=-y^{2}$ generates $H^{4}$.