13
$\begingroup$

I was trying to calculate $H^q(K(\mathbb{Z}, 3); \mathbb{Z})$ for some $q$ with the Serre spectral sequence associated to the fibration $K(\mathbb{Z}, 2) \to PK(\mathbb{Z}, 3) \simeq * \to K(\mathbb{Z}, 3)$.

I obtained that: $$ H^q(K(\mathbb{Z}, 3)) = \mathbb{Z}, 0, 0, \mathbb{Z}x, 0, 0, \mathbb{Z}_2x^2, 0, \mathbb{Z}_3y, \mathbb{Z}_2x^3, 0, \mathbb{Z}_3xy, \mathbb{Z}_2x^4\oplus\mathbb{Z}_5w. $$ But in Hatcher's book on spectral sequences, chapter 1, he claims that $$ H^q(K(\mathbb{Z}, 3)) = \mathbb{Z}, 0, 0, \mathbb{Z}x, 0, 0, \mathbb{Z}_2x^2, 0, \mathbb{Z}_3y, \mathbb{Z}_2x^3, \mathbb{Z}_2z, \mathbb{Z}_3xy, \mathbb{Z}_2x^4\oplus\mathbb{Z}_5w. $$

The only difference is in $H^{10}(K(\mathbb{Z}, 3))$, that for me is $0$, while for Hatcher is $\mathbb{Z}_2z$. I cannot understand why this happens.

My reasoning is: $H^{10}(K(\mathbb{Z}, 3))$ is in position $(10, 0)$ and it can be reached by groups in position $(9 - r, r)$. These are non trivial only if $r$ is even and $9-r = 0, 3, 6$. The only possibility is then $r = 6$, i.e., $$ (3, 6) = H^3(K(\mathbb{Z}, 3); H^6(K(\mathbb{Z}, 2))) = H^3(K(\mathbb{Z}, 3); \mathbb{Z}n^3) = \mathbb{Z}n^3x. $$ But my claim is that $(10, 0)$ could not be reached neither by $(3, 6)$ because this dies turning $E_2$. In fact $d_2(n) = x$, so $d_2(n^3x) = 3n^2x^2 = n^2x^2$ and so $d_2: (3, 6) \to (6, 4) = \mathbb{Z}_2n^2x^2$ is an isomorphism (edit: the error is here, it is not an isomorphism, but it has kernel $2\mathbb{Z}n^3x$). Then $(10, 0)$ would survive at $\infty$, which is not possible.

What's wrong with this?

$\endgroup$
  • 2
    $\begingroup$ Typo: the group in dimension 6 is $\mathbb{Z}/2$, not $\mathbb{Z}$. $\endgroup$ – John Palmieri Jan 3 at 23:59
18
$\begingroup$

I do not like naming a cohomology class $n$ because that deserves to be the name of an integer. I will use the name Hatcher does and call the generator of $H^2(K(\Bbb Z, 2); \Bbb Z)$ by the name "$a$".

The map $d_3: E_3^{0, 8} \to E_3^{3,6}$ sends $d_3(a^4) = 4a^3 x$ by the Leibniz rule. The map $d_3: E_3^{3,6} \to E_3^{6,4} \cong \Bbb Z_2 a^3 x^2$ sends $a^3x$ to $a^3x^2$; that is, this map is reduction mod 2 in this basis. (You already calculated that this must be true earlier in the computation.)

The homology group of $$\Bbb Z \xrightarrow{4} \Bbb Z \xrightarrow{\pmod 2} \Bbb Z_2$$ in the middle is $\Bbb Z_2$. Thus $$E_4^{3,6} = E_7^{3,6} = \Bbb Z_2\langle 2a^3 x\rangle,$$ where the angle brackets indicate that the nonzero class came from the element $2a^3 x$ on the $E_3$ page.

Thus there is indeed something left at $E_7$ in this position, so that the differential $d_7: E_7^{3,6} \to E_7^{10,0}$ must be an isomorphism.

It seems what you missed is that $d_3: \Bbb Z = E_3^{0,8} \to E_3^{3,6} = \Bbb Z$ is $4$, not $2$.

$\endgroup$

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.