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According to a theorem of Serre, the cohomology algebra $H^*(K(\Bbb Z,3); \Bbb Z_2)$ is a polynomial ring on elements $\iota_3, \,\operatorname{Sq}^2(\iota_3), \,\operatorname{Sq}^4\operatorname{Sq}^2(\iota_3), \, \cdots$, where $\iota_3$ is a generator of $H^3(K(\Bbb Z,3); \Bbb Z_2)\cong \Bbb Z_2$.

I would now like to conclude the $\Bbb Z$-cohomology of $K(\Bbb Z,3)$ modulo odd torsion in dimensions $\leq 9$.

Hatcher does this in the proof of theorem 5.39 (chapter 5) using the Bockstein $\beta=\operatorname{Sq}^1$. This is the Bockstein associated to the exact sequence $0\to \Bbb Z_2\to \Bbb Z_4\to \Bbb Z_2\to 0$. Using a few Adem relations, Hatcher computes what $\beta$ does with various elements of $H^*(K(\Bbb Z,3); \Bbb Z_2)$. I have no idea how he uses that to say anything about $\Bbb Z$-cohomology modulo odd torsion.

Here is a screenshot of the relevant part:

enter image description here

iwriteonbananas
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1 Answers1

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I would really appreciate feedback on the following answer, particularly the second part.

Part 1. $H^i(K(\Bbb Z,3);\Bbb Z)$ contains no $\Bbb Z$ summand and it contains no $\Bbb Z_{2^k}$ summand for $k>1$, in the range $i\in\{4,\cdots, 9\}$

The Bockstein chain complex has the following form

$$\underbrace{H^3(K(\Bbb Z,3); \Bbb Z_2)}_{\Bbb Z_2\cdot \iota_3}\to 0\to \underbrace{H^5(K(\Bbb Z,3); \Bbb Z_2)}_{\Bbb Z_2\cdot\operatorname{Sq}^2\iota_3}\stackrel{\beta_5}{\to} \underbrace{H^6(K(\Bbb Z,3); \Bbb Z_2)}_{\Bbb Z_2\cdot\iota_3^2}\to 0\to \underbrace{H^8(K(\Bbb Z,3); \Bbb Z_2)}_{\Bbb Z_2\cdot \iota_3\operatorname{Sq}^2\iota_3}\stackrel{\beta_8}{\to} \underbrace{H^9(K(\Bbb Z,3); \Bbb Z_2)}_{\Bbb Z_2\cdot \operatorname{Sq}^4\operatorname{Sq}^2\iota_3\,\oplus\,\Bbb Z_2\cdot\iota_3^3}\stackrel{\beta_9}{\to}\cdots$$ with Bocksteins $\beta_i=\operatorname{Sq}^1$ as connecting homomorphisms. We can actually compute all of these connecting homomorphisms using Adem relations. $\beta_5$ is an isomorphism, $\beta_8$ is an isomorphism onto the second summand, and $\beta_9$ maps the first factor isomorphically into $H^{10}(K(\Bbb Z,3); \Bbb Z_2)$.

Thus the sequence above is exact and therefore it does not give rise to any cohomology groups. Next we apply proposition 3E.3 (p. 305 in Hatcher), which in the present context tells us that:

  • $H^i(K(\Bbb Z,3); \Bbb Z)$ has no $\Bbb Z$ summand since that would give rise to a $\Bbb Z_2$ summand in the cohomology groups of the Bockstein chain complex.
  • $H^i(K(\Bbb Z,3); \Bbb Z)$ cannot contain a $\Bbb Z_{2^k}$ summand for $k>1$ since that would also give rise to a $\Bbb Z_2$ summand in a Bockstein cohomology group.
  • In $H^i(K(\Bbb Z,3); \Bbb Z)$ there could potentially exist $\Bbb Z_2$ summands only in dimensions $6$ and $9$ because those give rise to $\Bbb Z_2$ summands in both $H^i(K(\Bbb Z,3); \Bbb Z_2)$ and $H^{i-1}(K(\Bbb Z,3); \Bbb Z_2)$.

Part 2. There exist $\Bbb Z_2$ summands in $H^i(K(\Bbb Z,3); \Bbb Z)$ for $i=6$ and $i=9$.

Assume to the contrary that $H^6(K(\Bbb Z,3); \Bbb Z)=0$ (recall that we are ignoring odd torsion). By the universal coefficient theorem this implies that $\operatorname{Ext}\left(H_5(K(\Bbb Z,3);\Bbb Z), \Bbb Z \right)=0$. In particular $H_5(K(\Bbb Z,3);\Bbb Z)$ cannot have any 2-torsion. It also doesn't have a $\Bbb Z$ summand because that would give us a $\Bbb Z$ summand in $H^5$. On the other hand,

$$\begin{eqnarray}\Bbb Z_2\cong H^5(K(\Bbb Z,3); \Bbb Z_2)\cong H_5(K(\Bbb Z,3); \Bbb Z_2)\cong \underbrace{H_5(K(\Bbb Z,3); \Bbb Z)\otimes \Bbb Z_2}_{=0 \text{ since $H_5$ has no $2$-torsion}}\, \oplus\, \underbrace{\operatorname{Tor}\left(H_4(K(\Bbb Z,3); \Bbb Z),\Bbb Z_2\right)}_{0}\end{eqnarray}$$

where the $\operatorname{Tor}$ term vanishes since $H_4$ has no 2-torsion (else we get something nonzero in $H^4(K(\Bbb Z,3); \Bbb Z_2)$.

The case $i=9$ can be dealt with in the same way.

iwriteonbananas
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