I'm wondering if the next exercise in 'An introduction to homological algebra' by Weibel is correct:

Let $G$ be the profinite group $\widehat{\mathbb{Z}}_p$. Show that $$H^i(G;\mathbb{Z}) = \begin{cases} \widehat{\mathbb{Z}}_p & \text{ if $i$ even}\newline 0 & \text{ if $i$ odd} \end{cases}$$

Is it possible that $\widehat{\mathbb{Z}}_p$ should be replaced by $\mathbb{Z}(p^\infty)$ ? I've found on the website of Weibel that it should be for $H^0(G;\mathbb{Z})=\mathbb{Z}$, $H^2(G;\mathbb{Z})=\mathbb{Z}(p^\infty)$ and for all other $i$ $H^i(G;\mathbb{Z})=0$, but shouldn't that be for odd $i$ and the rest still $\mathbb{Z}(p^\infty)$ ? Since the cohomology groups $H^i(\mathbb{Z}/m\mathbb{Z};\mathbb{Z})$ are $\mathbb{Z}/m\mathbb{Z}$ for $i$ even ?


If you're using continuous cochain cohomology, then things work out as follows. For $H^0$, you get the invariants, all of $\mathbb{Z}$ in this case. For $H^1$, you get the group of continuous homomorpisms from $\mathbb{Z}_p$ to $\mathbb{Z}$. The image of such a homomorphism is necessarily finite (by continuity, compactness of $\mathbb{Z}_p$, and discreteness of $\mathbb{Z}$), but $\mathbb{Z}$ has no non-trivial finite subgroups, so you get zero. The cohomology sequence associated to $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Q}\rightarrow\mathbb{Q}/\mathbb{Z}\rightarrow 0$ shows that $H^2(\mathbb{Z}_p,\mathbb{Z})\cong H^1(\mathbb{Z}_p,\mathbb{Q}/\mathbb{Z})$, which is the group of continuous homomorphisms from $\mathbb{Z}_p$ to $\mathbb{Q}/\mathbb{Z}$, i.e., the Pontryagin dual of $\mathbb{Z}_p$. This is isomorphic to $\mathbb{Q}_p/\mathbb{Z}_p$, or in your notation, $\mathbb{Z}(p^\infty)$. For $H^q(\mathbb{Z}_p,\mathbb{Z})$, $q>2$, all the groups are zero because the strict cohomological dimension of $\mathbb{Z}_p$ is $2$.

So, if Weibel intends the standard continuous cohomology of profinite groups, then what he says is not right. Otherwise I'm afraid I'm not sure.

EDIT: It seems the correction on Weibel's website is correct, again, assuming continuous cochains.

  • $\begingroup$ To prove that $H^2(\mathbb{Z}_p,\mathbb{Z})\cong H^1(\mathbb{Z}_p,\mathbb{Q}/\mathbb{Z})$, don't you use that $H^2(\mathbb{Z}_p,\mathbb{Q}=0)$ (and if so, why is that?) ? Also, can you give another reason instead of the cohomological dimension that $H^q(\mathbb{Z}_p,\mathbb{Z})=0$ ? $\endgroup$ – KevinDL Apr 12 '12 at 7:31
  • $\begingroup$ Yes, I'm using that $H^2(\mathbb{Z}_p,\mathbb{Q})=0$. The reason for this is that the group is $p$-primary (every element has order a power of $p$), while, on the other hand, because $\mathbb{Q}$ is uniquely divisible, multiplication by $p$ is an automorphism on cohomology. The only way both these things can be true is if the $H^2$ is zero. For proving the higher cohomology vanishes, it suffices to prove that $H^q(\mathbb{Z}_p,\mathbb{Q}/\mathbb{Z})$ is zero for $q\geq 2$. I think this can be proved in a fairly elementary way using a limiting argument for the cohomology of finite cyclic groups $\endgroup$ – Keenan Kidwell Apr 12 '12 at 9:27
  • $\begingroup$ as in Proposition 2 of Chapter XIII of Serre's Local Fields, which is for $\hat{\mathbb{Z}}$, but should still work for $\mathbb{Z}_p$, at least for discrete $p$-primary torsion coefficients (one has $H^q(\mathbb{Z}_p,\mathbb{Q}/\mathbb{Z})=H^q(\mathbb{Z}_p,mathbb{Q}_p/\mathbb{Z}_p)$ for $q\geq 1$. See also Exercise 1 of that chapter of Serre. $\endgroup$ – Keenan Kidwell Apr 12 '12 at 9:29
  • $\begingroup$ That should be $H^q(\mathbb{Z}_p,\mathbb{Q}_p/\mathbb{Z}_p)$. $\endgroup$ – Keenan Kidwell Apr 12 '12 at 18:17

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