# Cohomology of $S^1 \rtimes \mathbb{Z}/2$

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.

• Well: what is the definition of $dw$? (To do that, you need to get a very comcrete description of what $w$ is, of course) Jan 9 '17 at 22:13
• @MarianoSuárez-Álvarez could you be a bit more explicit on your idea please Jan 10 '17 at 18:39
• 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. Jan 10 '17 at 18:53
• 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 $E_2^{0,4n+2}=0$. Jan 14 '17 at 9:18
• @HariRau-Murthy but is not $c = -c$ in this case? since I am considering the mod2 cohomology. Jan 17 '17 at 22:15

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

• 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). Feb 5 '17 at 19:01
• (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 $\Omega S^3 \hookrightarrow \{pt\} \to S^3$) Feb 5 '17 at 19:08
• 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 Feb 6 '17 at 20:05