The cohomology ring of an even dimensional sphere is $\mathbb{Z}[\alpha]/(\alpha^2)$ where the degree of $\alpha$ is the dimension of the sphere. From the Künneth formula, the cohomology ring of a product of even dimensional spheres is $\mathbb{Z}[\alpha]/(\alpha^2) \otimes \mathbb{Z}[\beta]/(\beta^2) \simeq \mathbb{Z}[\alpha,\beta]/(\alpha^2,\beta^2)$.
I want to focus on the isomorphism between the above tensor product of polynomial rings with the two variable polynomial ring.
In particular the case of an odd dimensional sphere, according to Hatcher, the cohomology ring of an odd dimensional sphere is the exterior algebra $\Lambda[\alpha]$ where the degree of $\alpha$ is odd. For an exterior algebra of more than one variable, the generators anti-commute which, from my understanding, is the real distinction between an exterior algebra and the polynomial ring $\mathbb{Z}[\alpha]/(\alpha^2)$. In the case of a one variable exterior algebra, it seems there is no difference between the polynomial ring and the exterior algebra.
If, for example, one wishes to compute the polynomial ring of, say, $S^3 \times S^2$, we have by Künneth that the polynomial ring is $\Lambda[\alpha] \otimes \mathbb{Z}[\beta]/(\beta^2)$ where the degrees of $\alpha$ and $\beta$ are 3 and 2, respectively.
My question is, can we say that the above graded tensor product is isomorphic to $\mathbb{Z}[\alpha,\beta]/(\alpha^2,\beta^2)$?
I believe we can, because the cup product multiplication satisfies $\alpha\beta = (-1)^{|\alpha||\beta|}\beta\alpha$ and, because the degree of $\beta$ is even, this means that $\alpha$ and $\beta$ commute.
Is this logic correct?
Best Answer
Let's consider $S^{2n}\times S^{2m+1}$, where $m,n\in\Bbb Z^+$, then $H^*(S^{2n}\times S^{2m+1})\cong \Lambda_{\Bbb Z}[\alpha]\otimes \Bbb Z[\beta]/(\beta^2)$, where $|\alpha|=2m+1,|\beta|=2n$.