18
$\begingroup$

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?

  • 6
    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!2011-05-11
  • 0
    @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 $\mathbb{Z}[x]/(x^n)$!2011-05-11
  • 2
    @Qwirk: (It's not.)2011-05-11
  • 0
    @Rasmus - sorry $\mathbb{Z}[x]/(x^2)$ where $x$ is the generator of $H^n(S^n,\mathbb{Z})$2011-05-11
  • 1
    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.2011-05-11
  • 0
    @Rasmus - that makes sense - I will try some less trivial examples!2011-05-11

1 Answers 1