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$\newcommand{\Z}{\mathbb{Z}}$ I decided that I would try another way of computing $\pi_4(S_3)$. Take the fibration $S_3 \to K(\Z,3)$ with fiber defined to be $X_4$. I want to directly use this fibration.

From the usual way of computing $\pi_4(S_3)$, I know that $H^4(X_4)=0$ and $H^5(X_4)=\Z/2$. (Briefly, you continue(loop) the above fibration to the left to get the fibration $K(\Z,2) \hookrightarrow X_4 \to S_3$ and then you use that the fundamental class of $K(\Z,2)$ is sent to the fundamental class of $S_3$ by $d_3$ in the cohomology Serre Spectral Sequence. Using leibniz' rule, get that $d_3^{0,4}$ is the multiplication by $2$ map. Since $d_3^{3,2}$ is the zero map, $H^5(X_4)=\Z/2$).

Back to the spectral sequence of the fibration $X_4 \hookrightarrow S_3 \to K(\Z, 3)$

Now I will use that I know the cohomology of all three spaces in the fibration: $H^*(K(\Z,3))=\Z[j]$ where $j$ has degree 3. The only possible differential that could get rid of the $\Z/2=E_2^{0,5}$ is $d_6: \Z/2=E_5^{0,5} \to \Z=E_5^{6,0}$. But $d_6$ must be zero since there are no homomorphisms from $\Z/2 \to \Z$ and $E_2^{0,5}=E_5^{0,5}$ survives to $E_\infty$. Therefore $H^6(S_3) \neq 0$. Contradiction.

What is the flaw in my computation of the spectral sequence of $X_4 \hookrightarrow S_3 \to K(\Z, 3)$?

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It is not true that $H^*(K(\mathbb{Z},3))=\mathbb{Z}[j]$ where $j$ is the canonical class in degree $3$. Indeed, since the product must be graded-commutative and $j$ has degree $3$, $j^2=(-1)^{|j||j|}j^2=-j^2$, so $2j^2=0$ (and similarly for all higher powers of $j$). In fact, $j^2\neq 0$ and it generates $H^6(K(\mathbb{Z},3))$, so $H^6(K(\mathbb{Z},3))=\mathbb{Z}/2$, not $\mathbb{Z}$ as you claim, and your differential $d_6$ is nonzero and kills $E^{0,5}_2$.

(In addition to $j^2$ having order $2$, $H^*(K(\mathbb{Z},3))$ also has lots of elements in higher degree which are not in the subring generated by $j$.)

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  • $\begingroup$ Thanks. Now I need to understand the contradiction in my calculation of the spectral sequence for $K(\Z,3)$. $\endgroup$ – user062295 Dec 12 '16 at 0:54
  • $\begingroup$ One step closer to understanding spectral sequences! math.stackexchange.com/questions/2054791/… $\endgroup$ – user062295 Dec 12 '16 at 1:23
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$\newcommand{\Z}{\mathbb{Z}}$ Here is how you compute $\pi_4(S_3)$ directly using your fibration, and using the calculations that you did in Contradiction in spectral sequence for $K(\mathbb{Z},3)$. You showed there that $H^5(K(\Z,3))=0$ and that $H^6(K(\Z,3))=\Z/2$. We have $d_5^{0,4}=0$ and since there is nothing else to kill or get rid of $E_6^{0,4}$, $H^4(X_4)=0$. Since there is nothing to get rid of $E_7^{0,5}$ or kill $E_7^{6,0}$, $d_6^{0,5}$ must be an isomorphism so $H^5(X_4)=\Z/2$

Using the exact sequence $ 0 \to Ext(H^{5}(X_4),\Z) \to H_4(X_4,\Z) \to Hom(H^4(X,\Z),\Z) \to 0 $, we have that $\pi_4(S_3)=H_4(X_4, \Z)=\Z/2$.

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