1
$\begingroup$

I am trying to compute the mod2 cohomology of the semi direct product $G = S^1 \rtimes \mathbb{Z}/2$ using the extension

$$ 1\rightarrow S^1 \rightarrow G \rightarrow\mathbb{Z}/2\rightarrow 1$$

and the HLS spectral sequence associated to it.

We get then that

$$E_2 = H^*(S^1) \otimes H^*(\mathbb{Z}/2) = \mathbb{F}_2[c,w]$$ where $|c|=2$ and $|w|=1$ are the respective generators.

the second differential depends on the value $d_2(w) = \alpha c$ where $\alpha \in \{0,1\}$.

I am stuck here and I do not know how to compute the value of $\alpha$. I know that if $\alpha = 0$ then $H^*(G) = E_\infty = E_2$ (which is kind of strange cause this is the cohomology of the direct product though).

I really appreciate any help/comment.

$\endgroup$
8
  • $\begingroup$ Well: what is the definition of $dw$? (To do that, you need to get a very comcrete description of what $w$ is, of course) $\endgroup$ Jan 9 '17 at 22:13
  • $\begingroup$ @MarianoSuárez-Álvarez could you be a bit more explicit on your idea please $\endgroup$
    – C. Zhihao
    Jan 10 '17 at 18:39
  • $\begingroup$ My idea is for you to use the definition of the differential of the spectral sequence. To do that you will need a concrete description of the class w. $\endgroup$ Jan 10 '17 at 18:53
  • $\begingroup$ Your $E_2$ term is wrong - $H^*(S^1)$ is a nontrivial $\mathbb{Z}/2$ module with the action $c \mapsto -c$ which induces a nontrivial action in dimensions that are not 0 modulo 4. For instance $E_2^{0,4n+2}=0$. $\endgroup$ Jan 14 '17 at 9:18
  • $\begingroup$ @HariRau-Murthy but is not $c = -c$ in this case? since I am considering the mod2 cohomology. $\endgroup$
    – C. Zhihao
    Jan 17 '17 at 22:15
1
$\begingroup$

Thanks to the comments and edits from the people involved in the OP I got the following; if there is still any mistake please don't hesitate commenting.

To compute the cohomology of $G$, I will use the Lyndon-Hochschild-Serre Spectral Sequence associated to the extension

$$1 \rightarrow S^1 \rightarrow G \rightarrow \mathbb{Z}/2 \rightarrow 1$$

where the $E_2$ page of such spectral sequence is given by

$$E_2^{p,q} \cong H^p(\mathbb{Z}/2; H^q(S^1))$$

where $\mathbb{Z}/2$ acts on $H^q(S^1)$ trivially. (In general the action of $\mathbb{Z}/2 = \{\pm 1\}$ over $H^q(S^1;\mathbb{Z})$ is given by $\pm 1 \cdot c = \pm c$)**.

This implies that the $E_2$ page decomposes into the first quadrant spectral sequence $$E_2 \cong H^*(\mathbb{Z}/2) \otimes H^*(S^1) \cong \mathbb{F}_2[w,c],$$ and the differential $d_2$ depends only on the values over $w$ and $c$ respectively since it is a derivation; indeed, $d_2(w) = 0$ since $w$ lies on the $x$-axis of the spectral sequence, and $d_2(c) = 0$ since $d_2(c) \in E^{2,1}_2 = H^2(\mathbb{Z}/2)\otimes H^1(S^1) = 0$.

Therefore, $d_2 = 0$ implying that the spectral sequence degenerates at page 2 and thus $E_2 \cong E_\infty \cong H^*(G)$.

**Thanks to @Hari Rau-Murthy for the clarification

$\endgroup$
3
  • $\begingroup$ Awesome! I didn't see this coming. But I think the transgression of this spectral sequence, $d_3^{0,2}(c)=w^3$(i.e. its nonzero). $\endgroup$ Feb 5 '17 at 19:01
  • $\begingroup$ (and $d_2$ of a spectral sequence being zero only implies $E_2 =E_3$ and does not imply that $E_2=E_\infty$ - (e.g. for the path fibration $\Omega S^3 \hookrightarrow \{pt\} \to S^3$) $\endgroup$ Feb 5 '17 at 19:08
  • $\begingroup$ Thanks for your comment @HariRau-Murthy. Yeah I have realized that already but I wasn't updated anything here. Actually, there is an easier way to solve the problem: Observe that $G$ is homeomorphic to $O(2)$, therefore $H^*(G) = H^*(BO(2)) = H^*(Gr_2(\mathbb{R}^\infty) \cong \mathbb{F}_2[w_1,w_2]$ where $|w_i| = i$, So the differential $d_3$ here must be zero $\endgroup$
    – C. Zhihao
    Feb 6 '17 at 20:05

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.