Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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! – Pete L. Clark May 11 '11 at 11:13
@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 $\mathbb{Z}[x]/(x^n)$! – Juan S May 11 '11 at 11:22
@Qwirk: (It's not.) – Rasmus May 11 '11 at 12:00
@Rasmus - sorry $\mathbb{Z}[x]/(x^2)$ where $x$ is the generator of $H^n(S^n,\mathbb{Z})$ – Juan S May 11 '11 at 12:15
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. – Rasmus May 11 '11 at 14:18

Maybe I will at least show how the homology 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 homology 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 $H^2(\mathbb{R} P^2,\mathbb{Z})$

share|cite|improve this answer
Next interesting examples: $H^*(\mathbb R P^2,\mathbb Z/2)$. – Rasmus May 12 '11 at 6:21
@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) – Juan S May 12 '11 at 7:10
That's true.\\\ – Rasmus May 12 '11 at 7:11
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? – user54631 Aug 12 '14 at 16:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.