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I am struggling to calculate homology rings.

Even for a simple space such as the sphere, it is easy to calculate the cohomology, but I find it much harder to find the ring structure. (This link gives the answer for the 2-sphere, and the generalisation to the $n$-sphere is clear)

I have tried having a look at Hatcher's notes on this (specifically examples 3.7-3.9 on pp. 207-209). Specifically in Hatcher's book, he claims that $\varphi_1 \cup \psi_1 = 0$ on all 2-simplcies, except the one with outer edge, $b_1$ which is where I got lost.

I tried having a look at the sphere, which has a very simple cohomology, so I figured the cup product should be easy to calculate. I know that the only two non-zero homology groups are $H^0(S^1,\mathbb{Z}) \simeq \mathbb{Z}$ and $H^n(S^n,\mathbb{Z}) \simeq \mathbb{Z}$. So let 1 be the generator of $H^0$ and $x$ the generator of $H^n$ (do we say 1 in the $H^0$ case, as this is the unit of the ring?). Then we have the cup products $1 \smile 1$, $1 \smile x$, $x \smile 1$,$x \smile x$. I can guess that $1 \smile 1 = 1$, but what about the others? How does one calculate this in general? Obviously there is some something simple I am missing?

Is there another nice book that has some nice examples on calculating cohmology rings?

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    $\begingroup$ The sphere really is a simple case...maybe too simple to get a feel for what's going on. Two observations: (i) the cohomology ring does have a multiplicative identity. You should convince yourself that what you've labelled $1$ is indeed this identity. That gives you everything but $x \cup x$: for this, (ii) remember that the cup product takes $H^k \times H^l \rightarrow H^{k+l}$. And then move on to something like complex projective space: that's one of the relatively small number of cases people will expect you to know! $\endgroup$ May 11, 2011 at 11:13
  • $\begingroup$ @Pete - so $1 \smile x = x$ and $x \smile 1 = x$? And in this case $x \smile x \in H^{2n}=0$, so I guess the ring is determined by $x \smile 1 = x = 1 \smile x$ (I still don't see how that is $\mathbb{Z}[x]/(x^n)$! $\endgroup$
    – Juan S
    May 11, 2011 at 11:22
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    $\begingroup$ @Qwirk: (It's not.) $\endgroup$
    – Rasmus
    May 11, 2011 at 12:00
  • $\begingroup$ @Rasmus - sorry $\mathbb{Z}[x]/(x^2)$ where $x$ is the generator of $H^n(S^n,\mathbb{Z})$ $\endgroup$
    – Juan S
    May 11, 2011 at 12:15
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    $\begingroup$ This way it's correct. Observe that as Abelian groups $\mathbb Z[x]/(x^2)$ is the same as $\mathbb Z\cdot 1\oplus \mathbb Z\cdot x$, and that the multiplication is precisely the one you described. $\endgroup$
    – Rasmus
    May 11, 2011 at 14:18

1 Answer 1

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Maybe I will at least show how the cohomology ring for the sphere works, based on the above. I will leave as community wiki, so it can be tidied up if something is not quite right. If I work out any other spaces, I will try explain them here as well.

Firstly, note that we know that there are only two non-zero cohomology groups $H^0(S^n,\mathbb{Z})\simeq \mathbb{Z}$ and $H^n(S^n,\mathbb{Z})\simeq \mathbb{Z}$. The element of degree 0 must be the unit of the ring (noting that cup product with $H^0$ is a map $H^k(X;\mathbb{Z}) \otimes H^0(X;\mathbb{Z}) \to H^k(X;\mathbb{Z})$). We therefore label the generator of $H^0$ as 1 and $H^n$ as $x$. The relations satisfied are therefore $1 \smile 1 = 1, 1 \smile x = x, x \smile 1 = x, x \smile x = 0$ (as we end in degree $H^{2n}=0$). We know that $$H^*(X;\mathbb{Z}) = \bigoplus_{p \ge 0} H^p(X;\mathbb{Z})$$ and so we have that $$H^*(X;\mathbb{Z}) \simeq \alpha_1 \cdot 1 \oplus \alpha_2 \cdot x, \quad \alpha_1,\alpha_2 \in \mathbb{Z}$$ with the relations as above. This is abstractly isomorphic to the polynomial ring $\mathbb{Z}[x]/(x^2)$, where $x$ is the generator of $H^n(S^n,\mathbb{Z})$

A similar calculation shows that the cohomology ring for $H^*(\mathbb{R} P^2, \mathbb{Z})$ is $\mathbb{Z}[x]/(2x,x^2)$ where $x$ is a generator of $H^2(\mathbb{R} P^2,\mathbb{Z})$

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  • $\begingroup$ Next interesting examples: $H^*(\mathbb R P^2,\mathbb Z/2)$. $\endgroup$
    – Rasmus
    May 12, 2011 at 6:21
  • $\begingroup$ @Rasmus - yes - I am trying to understand this (it seems like this is a very common example, of which the (similar) proof is given in a lot of places) $\endgroup$
    – Juan S
    May 12, 2011 at 7:10
  • $\begingroup$ That's true.\\\ $\endgroup$
    – Rasmus
    May 12, 2011 at 7:11
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    $\begingroup$ How about $H^{\ast}(S^2\times S^2)$? By Kunneth's theorem there is an isomorphism $H^{\ast}(S^2\times S^2,\mathbb{Z})\simeq H^{\ast}(S^2)\otimes H^{\ast}(S^2)=\mathbb{Z}[\alpha]/(\alpha^2) \otimes \mathbb{Z}[\beta]/(\beta^2)$ Is there anything else to say in this case? $\endgroup$
    – user54631
    Aug 12, 2014 at 16:37

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