# Homotopy groups $\pi_{n+1}(S^{n})$

In section $$5.1$$ of Hatcher's note about spectral sequences, he starts to compute stable homotopy group $$\pi_{n+k}(S^{n}),k \leq 3$$. Particularly for $$\pi_{n+1}(S^{n})$$, by Freudenthal suspension theorem it's enough to compute $$\pi_{4}(S^{3})$$. A part of Postnikov tower for $$S^{3}$$ is: $$K(\pi_{4}S^{3},4) \to X_{4} \to K(\mathbb{Z},3)$$ We know the cohomology ring of $$K(\mathbb{Z},3)$$ with coefficients in $$\mathbb{Z}/2$$ is polynomial ring with generators $$Sq^{I}\imath_{3}$$ where $$\imath_{3}$$ is generator of $$H^{3}(K(\mathbb{Z},3),\mathbb{Z}/2)$$ and $$I$$ is amissible with excess strictly less than $$3$$. Using Bockstein homomorphism $$\beta = Sq^{1}$$ associates with $$0 \to \mathbb{Z}/2 \to \mathbb{Z}/4 \to \mathbb{Z}/2 \to 0$$ we have: $$Sq^{1}Sq^{2}\imath_{3}=Sq^{3}\imath_{3}=\imath_{3}^{2}$$ $$Sq^{1}(\imath_{3}Sq^{2}\imath_{3})=\imath_{3}Sq^{1}Sq^{2}\imath_{3}=\imath_{3}^{3}$$ $$Sq^{1}Sq^{4}Sq^{2}\imath_{3}=Sq^{5}Sq^{2}\imath_{3}=(Sq^{2}\imath_{3})^{2}$$ $$\textbf{First Question}$$:Look at dimension of those above he concludes that $$\text{Ker}\beta = \text{Im}\beta$$ through dimension $$5 \to 9$$ and $$2$$-torsion in these dimensions consist of elements of order $$2$$? $$\textbf{Second question}$$: The open circles in the diagram (page $$E_{6}$$) are $$\mathbb{Z}$$ cohomology but have been reduced to $$\mathbb{Z}/2$$ classes, the image of coefficient homomorphism $$\mathbb{Z} \to \mathbb{Z}/2$$. Why this induced map is injective on $$\mathbb{Z}/2$$ summands and the image is same as image of Bockstein homomorphism $$\beta = Sq^{1}$$?

$$\textbf{Last question}:$$ To finish, we look at the differential $$d_{6}^{0,5}$$ and conclude if it is not an isomorphism then something survives to $$E_{\infty}$$ and nonzero torsion appears in $$H^{5}(X_{4})$$ or $$H^{6}(X_{4})$$. Why $$H^{6}$$ of $$X_{4}$$ here? I thought it must be $$H^{6}$$ of $$K(\mathbb{Z},3)$$?

• Is your first 'question' actually a question? – Tyrone Nov 25 '18 at 12:25

## 1 Answer

If I recall, Hatcher caclulates the integral cohomlogy of $$K(\mathbb{Z},3)$$ up to degrees 11 or so at some point earlier in the notes. He also tells you at some point that the mod 2 cohomology ring of this space is a free commutative algebra on classes $$sq^I\iota$$ for admissible multi-indexes $$I=(i_1,\dots,i_n)$$ satisfying certain 'excess' conditions. In particular, up to degree $$10$$ we have

$$H^*(K(\mathbb{Z},3);\mathbb{Z}_2)=\mathbb{Z}_2\{\iota,Sq^2\iota,\iota^2, \iota\cdot Sq^2\iota,\iota^3,Sq^4Sq^2\iota,(Sq^2\iota)^2\dots\}$$

where $$|\iota|=3$$. Here $$\iota_3\in H^3(K(\mathbb{Z},3))$$ is the fundamental class, and I have written $$\iota=\rho_2\iota_3\in H^3(\mathbb{Z},3);\mathbb{Z}_2)$$ for its mod 2 reduction. Anyway, $$|Sq^2\iota|=5$$, $$|\iota^2|=6$$, $$|\iota\cdot Sq^2\iota|=8$$, $$|\iota^3|=9$$, $$|Sq^4Sq^2\iota|=9$$ and $$|(Sq^2\iota)^2|=10$$. Now using the Adem relations together with the ring structure we get

\begin{align} Sq^1\iota&=0 \\ Sq^1(Sq^2\iota)&=(Sq^1Sq^2)\iota=Sq^3\iota=\iota^2\\ Sq^1\iota^2&=2(Sq^1\iota)\iota=0\\ Sq^1(\iota\cdot Sq^2\iota)&=(Sq^1\iota)(Sq^2\iota)+\iota(Sq^1Sq^2\iota)=\iota(Sq^3\iota)=\iota^3\\ Sq^1\iota^3&=3(sq^1\iota)\iota=0\\ Sq^1(Sq^4Sq^2\iota)&=Sq^5(Sq^2\iota)=(Sq^2\iota)^2. \end{align}

Thus in these dimensions $$\ker(Sq^1)$$ is generated as a $$\mathbb{Z}_2$$-vector space by $$\{\iota,\iota^2,\iota^3\}$$, whist $$im(Sq^1)$$ is generated by $$\{Sq^1(Sq^2\iota)=\iota^2,Sq^1(\iota\cdot Sq^2\iota)=\iota^3\}$$. We see that these agree in dimension $$5-9$$.

Now recall Hatcher's discussion of the Bockstein spectral sequence, which takes as input $$E^*_1=H^*(K(\mathbb{Z},3);\mathbb{Z}_2)$$ and converges to $$E^*_\infty= H^*(K(\mathbb{Z},3))/(torsion)\otimes\mathbb{Z}_2$$. The differential on the $$E_1$$-page of this spectral sequence is the Bockstein $$\beta$$, which, in the case $$p=2$$ is exactly the Steenrod operators $$Sq^1$$. Moreover, the $$E_r$$-page can be identified with the subgroup $$2^{r-1}\cdot H^*(K(\mathbb{Z},3);\mathbb{Z}_{2^r})$$ of $$H^*(K(\mathbb{Z},3);\mathbb{Z}_{2^r})$$. The previous calculations tell us that $$E_2^*=0$$ for $$*=5,\dots,9$$. Therefore $$2\cdot H^*(K(\mathbb{Z},3);\mathbb{Z}_4)=0$$ for $$*=5,\dots, 9$$ so the 2-torsion of order at most $$2$$.

I'm not sure I quite understand your second question. If you can be a bit clearer I will be happy to fill in any further details later on. I think you need to consider this. The exact coefficient sequence $$0\rightarrow\mathbb{Z}\xrightarrow{\times 2}\mathbb{Z}\rightarrow\mathbb{Z}_2\rightarrow 0$$ induces a long exact sequence for any space $$X$$

$$\dots\rightarrow H^n(X)\xrightarrow{\times 2}H^n(X)\xrightarrow{\rho_2} H^n(X;\mathbb{Z}_2)\xrightarrow{\delta} H^{n+1}(X)\rightarrow\dots$$

where $$\rho_2$$ is the mod 2 reduction and $$\delta$$ is the connecting map of the long exact sequence. Then the Bockstein $$\beta=Sq^1$$ is the composite

$$\beta=Sq^1=\rho_2\circ\delta:H^n(X;\mathbb{Z}_2)\xrightarrow{\delta} H^{n+1}(X)\xrightarrow{\rho_2} H^{n+1}(X;\mathbb{Z}_2)$$

The point is that not only does it hold that $$Sq^1Sq^1=(\rho_2\delta)(\rho_2\delta)=\rho_2(\delta\rho_2)\delta=\rho_2(0)\delta=0$$, but also that $$Sq^1\rho_2=(\rho_2\delta)\rho_2=(0)\rho_2=0$$ and $$\delta Sq^1=\delta(\rho_2\delta)=(\delta\rho_2)\delta=(0)\delta=0$$. Using what you know about the kernel and image of $$Sq^1$$ you should be able to use these equations to decide why certain mod 2 reductions are injective.

Third question: You are studying the Serre Spectral sequences of the fibration $$K(\mathbb{Z}_2,4)\rightarrow X_4\rightarrow K(\mathbb{Z},3)$$, where $$X_4$$ is obtained as the pullback of the path-space fibration of a map $$k_3=Sq^2\iota:K(\mathbb{Z},3)\rightarrow K(\mathbb{Z}_2,5)$$. This spectral sequence takes as input $$E_2^{p,q}=H^p(K(\mathbb{Z},3);H^q(K(\mathbb{Z}_2,4))$$ and converges to $$E_\infty^{p,q}=H^{p+q}(X_4)$$. This is reason that you end up with classes in $$H^6(X_4)$$ rather than $$H^6(K(\mathbb{Z},3)$$. It is the cohomology of the space $$X_4$$ that you are calculating with this spectral sequence.

With regards to the converges in low degrees of this spectral sequence, there are non non-trivial differentials until the $$E_5$$-page. The class $$\iota\in E_4^{3,0}\cong E_2^{3,0}\cong H^3(K\mathbb{Z},3))$$ clearly survives to $$E_\infty$$. This gives a class in $$H^3(K(\mathbb{Z},3;\mathbb{Z}_2)$$, and the edge homomorphisms tell you that this is the class $$p^*\iota_3$$, where $$p:X_4\rightarrow K(\mathbb{Z},3)$$ is the projection.

Moving onto the $$E_5$$-page, since $$d_5(\iota_4)=Sq^2\iota_3$$ by the construction of the fibration, we have $$E_\infty^{0,4}\cong E_6^{0,4}\cong \ker(d_5)=0$$, and since there are no other classes in $$E_5$$ with total degree $$4$$,we find that $$H^4(X_4;\mathbb{Z}_2)\cong \oplus_{p+q=4}E_\infty^{p,q}=0$$. Here we are on the $$E_5$$-page.

Now we must work on the $$E_6$$-page. In total degree $$5$$ there is only one class, namely, $$Sq^1\iota_4\in E_6^{0,5}\cong H^5(K(\mathbb{Z}_2,4);\mathbb{Z}_2)$$. The class $$Sq^2\iota_3\in E_5^{5,0}\cong H^5(K(\mathbb{Z},3);\mathbb{Z}_2)$$ was killed on the $$E_5$$-page so does no appear at $$E_6$$.

Now you should recall that the differentials of the spectral sequence commute with the action of the Steenrod algebra. This gives us $$d_6(Sq^1\iota_4)=Sq^1(d_6\iota_4)=Sq^1(Sq^2\iota_3)=\iota_3^2$$. Therefore $$E_\infty^{0,5}\cong E_7^{0,5}\cong\ker(d_6)=0$$. Since no other classes of total degree 5 survive to $$E_\infty$$ we have $$H^5(X_4;\mathbb{Z}_2)=0$$.

In total degree $$6$$ we must move to the $$E_7$$-page. Tthere is again only a single class, $$Sq^2\iota_4\in E_7^{0,6}\cong H^6(K(\mathbb{Z}_2,4);\mathbb{Z}_2)$$, since $$\iota^3\in E_6^{6,0}\cong H^6(K(\mathbb{Z},3),\mathbb{Z}_2)$$ was killed on the previous page. We have $$d_6(Sq^2\iota_4)=0$$ since $$E_6^{6,0}=0$$.

Thus we see that $$H^6(X_4;\mathbb{Z}_2)\cong E_\infty^{6,0}\cong\mathbb{Z}_2$$, and is generated by a class $$x_6$$ satisfying $$i^*x_6=Sq^2\iota_4$$, where $$i:K(\mathbb{Z}_2,4)\rightarrow X_4$$ is the fibre inclusion. This class $$x_6$$ turns out to be the class that must be killed at the next stage of the Postnikov tower.

Summing up we have $$H^4(X_4;\mathbb{Z}_2)=0=H^5(X_4;\mathbb{Z}_2)$$. The long exact Bockstein sequence tells us that if it is non-trivial then $$H^4(X_4)$$ must be odd torsion. However we know it cannot be odd torsion, so it must be trivial. Similarly for $$H^5(X_4)$$. Thus $$H^4(X_4)= 0= H^5(X_4)$$ (integral coefficients).

Now to sort out what is happening in $$H^6(X_4)$$ we look at the spectral sequence again. We have $$d_8(Sq^3\iota_4)=Sq^3Sq^2\iota_3=Sq^4Sq^1\iota_3=0$$. Hence $$Sq^3\iota_4$$ survives to $$E_\infty$$ and represents a class $$x_7$$ satisfying $$i^*x_7=Sq^3\iota_4$$. Now we have $$i^*(Sq^1x_6)=Sq^1(Sq^2\iota_4)=Sq^3\iota_4=i^*x_7$$, so that $$Sq^1x_6\neq 0$$.

In particular, since $$Sq^1=\rho_2\delta$$, the class $$x_6$$ cannot be the mod 2 reduction of an integral class - or in fact any $$\mathbb{Z}_{2^r}$$, $$r\geq 2$$, class. Therefore $$H^6(X_6)=0$$ and so $$x_6\in H^6(X_6;\mathbb{Z}_2)$$ is indeed the class to kill to move to the next Postnikov stage.

• $\text{Im}Sq^{1}$ is generated by $\left \{ \imath^{2},\imath^{3} \right \} \neq \left \{ \imath,\imath^{2},\imath^{3} \right \}$ and I haven't read Bockstein spectral sequence ( in Hatcher's note is also not yet written ) so I'm so thankful if you could interpret in another way. Lastly, maybe some latex mistakes between $X_{6}$ and $X_{4}$ ? – bangbang1412 Nov 25 '18 at 14:17
• @DavidGeal, I apolgise for the typo. $im(Sq^1)$ is generated by the classes I have written. The class $\iota$ is of degree $3$, and the statement is about the classes in degrees 5 to 9. Like I have written, these are $Sq^1(Sq^2\iota)=\iota^2$ and $Sq^1(\iota\cdot Sq^2\iota)=\iota^3$. – Tyrone Nov 25 '18 at 15:09
• What I have already written about the Bockstein SS should be enough to help you understand Hatcher's calculations a little better. I shall try to write a little more. – Tyrone Nov 25 '18 at 15:14
• @DavidGeal, I've added some details to the bottom of my answer with regards to your last question. – Tyrone Nov 25 '18 at 16:16
• it's even true that $H_{i}(X_{4},\mathbb{Z}) \cong H_{i}(S^{3},\mathbb{Z}) \forall i \leq 5$ since $X_{n}$ is obtained as $S^{3}$ after attached cells of dimension greater than $n+2$. Thanks for your answer, you're very enthusiastic. – bangbang1412 Nov 25 '18 at 16:29