Let $\tilde h^\bullet$ be a reduced generalized cohomology theory, and let $T^2$ be the torus. For what theories $\tilde h^\bullet$ is $\tilde h^\bullet(T^2)$ known (or easily computable)?

For example:

  1. $H_n(T^2) = \mathbb Z$ for $n=0,2$, $\mathbb Z \oplus \mathbb Z$ for $n=1$, and $0$ otherwise, so from the universal coefficient theorem we can get $\widetilde H^\bullet(T^2;A)$ for arbitrary coefficient $A$.

  2. I believe for complex K-theory we have $\widetilde K^n(T^2) = \mathbb Z \oplus \mathbb Z$ for odd $n$ and $\mathbb Z$ for even $n$. It's not hard to find this on the web.

What are some other examples? I'm specifically looking for non-ordinary theories (e.g. cobordism, cohomotopy, etc.).


1 Answer 1


The torus is actually stably equivalent to $S^1\vee S^1\vee S^2$, so $$\tilde{h}^n(T^2)\cong \tilde{h}^n(S^1\vee S^1\vee S^2)\cong \tilde{h}^{n-1}(S^0)\oplus \tilde{h}^{n-1}(S^0)\oplus \tilde{h}^{n-2}(S^0).$$ So you know $\tilde{h}^*(T^2)$ iff you know the cohomology of spheres. More generally, a product $X\times Y$ is always stably equivalent to $X\vee Y\vee (X\wedge Y)$ (see, for instance, Proposition 4I.1 on page 467 of Hatcher, before which you can also find a simpler argument for the case of a torus).

  • $\begingroup$ Thank you! This is a perfect answer! $\endgroup$
    – user46652
    Feb 27, 2016 at 23:16
  • 1
    $\begingroup$ But note that this argument doesn't determine the cup product if your cohomology theory is multiplicative. $\endgroup$ Feb 28, 2016 at 0:56

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