I am trying to understand the Kunneth formula in Hatcher's Algebraic Topology. Theorem 3.15 says

The cross product $H^∗(X;R)⊗_RH^∗(Y;R)→H^∗(X×Y;R)$ is an isomorphism of rings if $X$ and $Y$ are CW complexes and $H^k(Y;R)$ is a finitely generated free $R$-module for all $k$.

I want to apply this theorem on computing the cohomology ring of the torus. But from the definition tensor product of rings I get $\Bbb Z[x]/\langle x^2\rangle \otimes \Bbb Z[y]/\langle y^2\rangle=\Bbb Z[x,y]/\langle x^2,y^2\rangle$ which is not the standard result $\mathbb Z[x,y]/\langle x^2,y^2,xy+yx\rangle$.

So where does the relation $xy+yx$ comes from? How do we calculate the tensor product of graded rings here?

  • 1
    $\begingroup$ This may help you. Observe that in the ring $\Bbb{Z}[x,y]$ we have $xy=yx$ so $xy+yx=2xy$. Surely it is not the intention that the generator $x\otimes y$ of the top cohomology $H^2$ would be a $2$-torsion element (which is what your "standard" ring would give). $\endgroup$ – Jyrki Lahtonen Oct 27 '17 at 10:21

I think the full statement is that the Kunneth morphism is an isomorphism of graded rings. You need to use the graded tensor product for the theorem to work, in which case you get the correct answer.

  • $\begingroup$ Yeah (+1). My search-fu is weak today. If you find a better link to the description of the tensor product of graded rings, do add it. $\endgroup$ – Jyrki Lahtonen Oct 27 '17 at 10:26
  • $\begingroup$ Thank you. But I have not seen a clear and full definition of graded rings on google. Could you tell me the definition or reference it? $\endgroup$ – non-abelian group of order 9 Oct 27 '17 at 10:36
  • $\begingroup$ Hatcher on page 212. I'll try to find a better reference later. $\endgroup$ – Tyrone Oct 27 '17 at 14:10
  • $\begingroup$ Dear @Tyrone, it is not clear to me how the grading can influence the underlying ring structure. Is the following correct? To compute the desired product, first you take compute the tensor product of the underlying graded $(R = \mathbb{Z})$-modules so that you get as $0$-th, $1$-st, and $2$-nd graded pieces $\mathbb{Z}$, $\mathbb{Z}^2$, and $\mathbb{Z}$ respectively... then you introduce the multiplicative structure as in Hatcher p. 219, which yields the additional relation $(x \otimes 1)\cdot (1 \otimes y) = -(1 \otimes y) \cdot (x \otimes 1)$...? $\endgroup$ – I I Jun 15 '18 at 9:58
  • $\begingroup$ Yes, that's correct. More generally $a\cdot b=(-1)^{|a||b|}b\cdot a$. $\endgroup$ – Tyrone Jun 15 '18 at 14:07

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