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I'd like to know whether it is possible to identify the sphere bundle arising as follow:

Let $\xi \colon BD_{2n}\to BU(1)$ the complex line bundle corresponding to the element $y^2 \in H^2(D_{2n};\Bbb Z) \cong \Bbb Z_2\langle x^2,y^2\rangle$ (we assume $n=0 \pmod{4}$ and $D_{2n}$ is the dihedral group of order $2n$).

Let $q\colon S(\xi) \to BD_{2n}$ be the sphere bundle of $\xi$. It's easy to see that it's a Eilenberg MacLane space $K(G,1)$ where $G$ sits in the following s.e.s. $$0 \to \Bbb Z \to G \to D_{2n} \to 0$$ which arises from the following piece of l.e.s. of a fibration $$0 \to \pi_2(\Bbb CP^{\infty}) \to \pi_1(S(\xi)) \xrightarrow{\pi_1(q)} \pi_1(BD_{2n}) \to 0$$ I'd like to identify such $G$. The Serre exact sequence in homology (integer coefficient) gives us the following exact sequence: enter image description here

Since it's know that $H_2(BD_{2n})\cong \Bbb Z_2$, we have that $H_2(BU(1))\hookrightarrow H_1(S(\xi))$ which means that the exact sequence gives us $$0 \to \Bbb Z \to H_1(S(\xi)) \to \Bbb Z_2 \oplus \Bbb Z_2 \to 0$$

which implies that $H_1(S(\xi))$ cannot be a finite abelian group. In particular we already have that that $G \neq D_{\infty}$ since $H_1(D_{\infty})=ab(D_{\infty})=\Bbb Z_2\oplus \Bbb Z_2$.

From the exact sequence retrieved by the dual Blakers Massey Theorem (see here, in our case the base space is $1$-connected and the map is $1$-connected)

enter image description here

I was able to retrive this informations: $H^1(S(\xi);\Bbb Z)\cong \Bbb Z$, the boundary map $\delta \colon H^1(S(\xi);\Bbb Z) \to H^2(\Bbb C P^{\infty})$ is multiplication by $2$ and $H^2(S(\xi);\Bbb Z)\cong \Bbb Z_2$ since I know explicitly the map $H^2(BU(1))\to H^2(BD_{2n})$.

Is there a way to identify such $G$ completely?

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  • $\begingroup$ What are your x and y? Is this the line bundle corresponding to the representation whose kernel is the subgroup of rotations? I think G will be the infinite dihedral group. The way I would try to show this is by writing the sphere bundle as $ED_{2n} \times_{D_{2n}} S^1$ and to realize this as $E(G/\mathbb{Z}) \times_{G/\mathbb{Z}} B\mathbb{Z}$ which is a model for BG. You just need to check that the two actions on S^1 match. $\endgroup$ – James Cameron Jun 6 '17 at 22:36
  • $\begingroup$ Dear James, $x$ and $y$ comes form the following $H^*(D_{2n};\Bbb Z_2)\cong \Bbb Z_2[ x,y,w]/\langle xy=x^2\rangle$ and $|x|=|y|=1, |w|=2$. Using Bockstein exact sequence then one has that the integral cohomology $H^1$ can be identified as claimed $\endgroup$ – Luigi M Jun 6 '17 at 23:03
  • $\begingroup$ Using hurewicz + the fact that mod $2$ homology is the dual of mod $2$ cohomology, one has that the dual of $x$ is represented by the rotation matrix and the dual of $y$ is represented by the reflection matrix. $\endgroup$ – Luigi M Jun 6 '17 at 23:06
  • $\begingroup$ Dear @JamesCameron maybe I'm missing something, but from the Serre exact sequence it's seems clear that $\Bbb Z$ injects into $H_1(G)$. If $G$ were the infinite dihedral group, then $H_1(G)=\Bbb Z_2\oplus \Bbb Z_2$ $\endgroup$ – Luigi M Jun 7 '17 at 7:13
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    $\begingroup$ you might want to have a look at the Gysin sequence in order to know more about the cohomology of such Sphere bundle $\endgroup$ – Riccardo Jun 8 '17 at 14:38
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Your cohomology class in $H^2(D_{2n},\mathbb{Z})$ comes from a homomorphism $D_{2n} \to S^1$, which factors as $D_{2n} \to \mathbb{Z}/2 \to S^1$, where the map $D_{2n} \to \mathbb{Z}/2$ sends the rotation to $0$ and the reflection to $1$.

So, you have maps $BD_{2n} \to B \mathbb{Z}/2 \to BS^1$, and you want to compute the homotopy fiber of this composition. To do this, first compute the homotopy fiber of the map $B \mathbb{Z}/2 \to BS^1$. This corresponds to the unique nontrivial central extention of $\mathbb{Z}/2$ by $\mathbb{Z}$, i.e. $\mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/2$.

Now, you have a homotopy pullback diagram, where $G$ is the group you are trying to compute:

$$ \require{AMScd} \begin{CD} BG @> >> B \mathbb{Z} \\ @VVV @VVV \\ BD_{2n} @> >> B \mathbb{Z}/2 \end{CD} $$

Taking fundamental groups doesn't preserve pullbacks in general, but it does when everything in sight is the classifying space of a discrete group. So, $G$ is the fiber product $D_{2n} \times_{\mathbb{Z}/2} \mathbb{Z}$, where the map $D_{2n} \to \mathbb{Z}/2$ is the one mentioned earlier.

Algebraically speaking, you have some central extension of $D_{2n}$ by $\mathbb{Z}$, and it is pulled back from a known central extension of $\mathbb{Z}/2$ by $\mathbb{Z}$, so you can use this to figure out what the extension is.

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  • $\begingroup$ Is the map $\Bbb Z/2 \to \Bbb S^1$ you use in your factorisation, representing $Sq^1$ in some sense? $\endgroup$ – Luigi M Jul 3 '17 at 13:26
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    $\begingroup$ If you like you could think of the corresponding map $\mathbb{RP}^{\infty} \to \mathbb{CP}^{\infty}$ as representing $Sq^1(w_1)$. $\endgroup$ – James Cameron Jul 3 '17 at 18:30
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As is described in any standard group cohomology textbook, if $Q$ is a group and $A$ is an abelian group, then elements of $H^2(Q;A)$ correspond to central group extensions $A \rightarrow G \rightarrow Q$.

You are asking for a particular case of this. As you have described your particular element of $H^2(D_{2n};\mathbb Z)$ by its image in $H^2(D_{2n};\mathbb Z/2)$, one might also wish to describe the extension $\mathbb Z/2 \rightarrow \bar G \rightarrow D_{2n}$, the quotient of $G$ by $2\mathbb Z$.

It should be easy to calculate the mod 2 cohomology of $\bar G$. I have just be fooling with this when $n$ is a power of 2. One learns that when $n=4$, $\bar G$ will be group no. 4 of order 16 as described on Simon King and David Green's remarkable group cohomology website http://users.minet.uni-jena.de/cohomology/. When $n=8$, $\bar G$ will be group no. 12 of order 32. I am guessing that when $n=2^k$, $\bar G$ will also be the extension associated to the canonical element in $H^2(D_{2^k}; H_2(D_{2^k};\mathbb Z/2))$, and is a 2-central group of rank 2. (2-central = all elements of order 2 are central.)

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  • $\begingroup$ Whoops: that last sentence is not right when $k=2$: this would make $\bar G = (\mathbb Z/4)^2$. $\endgroup$ – Nicholas Kuhn Jun 8 '17 at 12:03
  • $\begingroup$ Dear Nicholas, I'll meditate on this answer! it sounds promising though and I want the thank you for that $\endgroup$ – Luigi M Jun 8 '17 at 12:50
  • $\begingroup$ I was about to ask why my extension was a central one, but it seems it's the case: mathoverflow.net/questions/48489/… $\endgroup$ – Luigi M Jun 8 '17 at 13:46
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    $\begingroup$ your point of view is interesting, what relations can we draw between $G$ and $\overline{G}$? I mean, what can we say about $G$ if we know the mod 2 cohomology of $\overline{G}$? $\endgroup$ – Riccardo Jun 8 '17 at 14:18
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Here is a solution to a toy version of your question. Namely, I will identify $S(\xi)$ in the special case when $\mathbf n = \infty$.

In this case, I claim that after suspending once, we have $$ \Sigma S(\xi) \,\, \simeq\,\, \Sigma (S^1 \times \Bbb RP^\infty)\, . $$ Secondly, I claim that there is a cofiber sequence $$ S^1 \to S(\xi) \to S^1_+ \wedge \Bbb RP^\infty $$ showing that the homomorphism $$ H^\ast(S^1_+ \wedge \Bbb RP^\infty) \to H^\ast(S(\xi)) $$ is an isomorphism in degrees $\ge 3$. Furthermore, we can also show that the latter is injective in degrees $\le 3$.

Corollary. If $n = \infty$, then with any coefficients we have $$ H^\ast(S(\xi)) \cong H^\ast(S^1) \oplus H^\ast(\Bbb RP^\infty) \oplus H^{\ast-1}(\Bbb RP^\infty) \, . $$

Here is the argument. In this instance $BD_{\infty} = \Bbb RP^\infty \vee \Bbb RP^\infty$ and the map $y^2: BD_{\infty} \to BU(1)$ is given by $$ \require{AMScd} \begin{CD} \Bbb RP^\infty \vee \Bbb RP^\infty @> (\ast,i) >> \Bbb C P^\infty \end{CD} $$ where $i : \Bbb RP^\infty \to \Bbb CP^\infty$ is the evident map, and $\ast : \Bbb RP^\infty \to \Bbb CP^\infty$ is constant.

The homotopy fiber of this map$^\dagger$ (which gives $S(\xi)$ in this case) is given by the pushout of the diagram $$ \begin{CD} S^1 @< \times 2 << S^1 \times \ast @>\subset >> S^1 \times \Bbb RP^\infty . \end{CD} $$ So there is a homotopy pushout $$ \require{AMScd} \begin{CD} S^1 @>>> S^1 \times \Bbb RP^\infty \\ @V \times 2 VV @VVV \\ S^1 @>>> S(\xi)\, . \end{CD} $$ By taking quotients horizontally, the latter gives the desired cofiber sequence $$ S^1 \to S(\xi) \to S^1_+ \wedge \Bbb RP^\infty \, . $$

We can say a bit more: the map $S^1\to S(\xi)$ is a retract (this can be seen using the universal property of the pushout to construct the retraction). In follows that the map $S(\xi) \to S^1_+ \wedge \Bbb RP^\infty$ is injective on cohomology in all degrees. Furthermore, after suspending once, the cofiber sequence splits, so $$ \Sigma S(\xi) \,\, \simeq\,\, \Sigma( S^1 \vee S^1_+ \wedge \Bbb RP^\infty ) \simeq \Sigma (S^1 \times \Bbb RP^\infty) $$ where I've used the fact that $X\times Y$ splits as $X \vee Y \vee X\wedge Y$ after one suspension.

${}^\dagger$The homotopy fiber of a map of based spaces $(f,g): X \vee Y \to Z$ with $Z$ path connected is given by the homotopy pushout of the diagram $F_f \leftarrow \Omega Z \rightarrow F_g$, where $F_f$ is the homotopy fiber of $f$.

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  • $\begingroup$ Thank you a lot for this answer! I will meditate on this tomorrow, it's pretty late in my time zone now. $\endgroup$ – Luigi M Jun 6 '17 at 23:11
  • $\begingroup$ I have modified my post. I can only claim to give an answer to your question in the "toy case" when $n$ is infinite. $\endgroup$ – John Klein Jun 8 '17 at 13:20
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    $\begingroup$ Dear John, I thank you again. Even thought it's a toy case, there is enough material for me to study. You never know if that might be useful in the future! $\endgroup$ – Luigi M Jun 8 '17 at 13:47
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    $\begingroup$ It seems that this toy case is in fact important, as there seems to be some relation about the group(s) $G$ when for $n=4k$, $k$ varies! $\endgroup$ – user51223 Jun 8 '17 at 22:04
  • $\begingroup$ Dear @user51223 could you elaborate? I'm interested in that! $\endgroup$ – Luigi M Jun 8 '17 at 22:34

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