# Contradiction in spectral sequence for $K(\mathbb{Z},3)$

$\newcommand{\Z}{\mathbb{Z}}$ Take the fibration $K(\Z,2) \hookrightarrow * \to K(\Z,3)$. Then $d_3^{0,2}$ is an isomorphism since this is the only way to get rid of $H^2(K(\Z,2))$ and to kill $H^3(K(\Z,3))$. Therefore $i \in \Z[i]=H^*(K(\Z,2))$ is sent under $d^3$ to a generator which has to be a fundamental class $j$ of $H^3(K(\Z,3))$. Therefore $d_3^{0,4}$ is the multiplication by $2$ map: $d_3^{0,4}(i^2)=j \otimes i +(-1)^{2+2} i \otimes j \mapsto (-1)^{2*3} ij +i j=2 ij \in \Z\langle ij \rangle=H^3(K(\Z,3),\Z)=E_3^{3,2}$ . Here $\mapsto$ is the cup product map and $\langle \rangle$ denotes 'module generated by'.

$d_3^{3,2}$ is the zero map since $d_3(j \otimes i)=-j\otimes j \mapsto j^2=0$ since $j$ is of odd degree. Therefore $E_4^{3,2}=\Z/2$ and there is nothing to kill or get rid of it. Therefore $H^5(*) \neq 0$. Contradiction.

What is the error in calculating this spectra sequence.

This calculation was done while doing the spurious calculation in Another way to compute $\pi_4(S_3)$: contradiction in spectral sequence calculation

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• Are you sure the first map you specify is a fibration? Is $\ast$ supposed to be a one point space? – Pedro Tamaroff Dec 12 '16 at 1:29
• $*$ is the path space of $K(\Z,3)$. The notation is used for because it is contractible. You can take the model for $K(\Z,2)$ to be the loopspace of $K(\Z,3)$ – user062295 Dec 12 '16 at 1:31

## 1 Answer

The error is that $j^2\neq 0$. The fact that $j$ has odd degree tells you that $j^2=-j^2$, but this just means $2j^2=0$, not $j^2=0$. In fact, your computation gives a proof that $j^2$ cannot be $0$ and hence is an element of order $2$ in $H^6(K(\mathbb{Z},3);\mathbb{Z})$.

• RIght and $d_3^{6,0}$ must be $0$, and there is nothing to kill or get rid of $E_4^{6,0}$, so $H^6(K(\mathbb{Z},3))=\mathbb{Z}/2$. This was the entry I needed for calculating $\pi_4(S_3)$ with my(canonical) method. – user062295 Dec 12 '16 at 2:11
• I guess I also needed that $H^5(K(\mathbb{Z},3))=0$ but this happens since $i^2$ doesn't even survive to $E_4$., so $H^5$ can't be killed by $d_5$. The calculation continues! – user062295 Dec 12 '16 at 2:21