2
$\begingroup$

$\newcommand{\Z}{\mathbb{Z}}$ 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

.

$\endgroup$
  • $\begingroup$ Are you sure the first map you specify is a fibration? Is $\ast$ supposed to be a one point space? $\endgroup$ – Pedro Tamaroff Dec 12 '16 at 1:29
  • $\begingroup$ $*$ 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 $K(\Z,3)$ $\endgroup$ – user062295 Dec 12 '16 at 1:31
2
$\begingroup$

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 $H^6(K(\mathbb{Z},3);\mathbb{Z})$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user062295 Dec 12 '16 at 2:11
  • $\begingroup$ 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! $\endgroup$ – user062295 Dec 12 '16 at 2:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.