I am trying to understand isomorphism classes of bundles $$\mathbb{S}^1\hookrightarrow E\to \mathbb{RP}^2.$$ These are classified by homotopy classes $[\mathbb{RP}^2,BAut(\mathbb{S}^1)]$.

ATTEMPT 1. $Aut(\mathbb{S}^1)$ are the diffeomorphisms $\mathbb{S}^1\to \mathbb{S}^1$, and it is homotopically equivalent to $O(2).$ Therefore the universal classifying space $BAut(\mathbb{S}^1)$ is homotopically equivalent to $BO(2)$. Thus we are interested in homotopy classes in$[\mathbb{RP}^2,BO(2)]$.

Now topologically $O(2) \simeq SO(2) \sqcup SO(2)\simeq \mathbb{S}^1 \sqcup \mathbb{S}^1,$ (it has two connected component that are diffeomorphic to $\mathbb{S}^1$). Therefore $$BO(2)\simeq B\mathbb{S}^1\sqcup B\mathbb{S}^1 \simeq \mathbb{CP}^\infty \sqcup \mathbb{CP}^\infty.$$

Consequently $$[\mathbb{RP}^2,BO(2)] \simeq [\mathbb{RP}^2,\mathbb{CP}^\infty \sqcup \mathbb{CP}^\infty] \simeq [\mathbb{RP}^2,\mathbb{CP}^\infty]\times[\mathbb{RP}^2,\mathbb{CP}^\infty]$$ Using that $\mathbb{CP}^\infty$ is a $K(\mathbb{Z},2)$ we get that $[\mathbb{RP}^2,\mathbb{CP}^\infty]\simeq H^2(\mathbb{RP}^2,\mathbb{Z})\simeq \mathbb{Z}/2\mathbb{Z}$. This would imply that there are $4$ isomorphism classes of circle bundles over $\mathbb{RP}^2$.

Question 1: which characteristic classes give this classification?

Question 2: what are these 4 manifolds $E$?

ATTEMPT 2. This is inspired by Classification of $O(2)$-bundles in terms of characteristic classes. The first Steifel-Whitney class $w_1 \in H^1(\mathbb{RP}^2,\mathbb{Z}/2\mathbb{Z})$ is $0$ iff the bundle is orientable. In this case we have a reduction of the structure group to $SO(2)$ and $SO(2)$ principal bundles are classified by the Euler class $e\in H^2(\mathbb{RP}^2,\mathbb{Z})\simeq [\mathbb{RP}^2, \mathbb{CP}^\infty]\simeq \mathbb{Z}/2\mathbb{Z}.$ Therefore we have $2$ orientable bundles and an unknown number of non-orientable ones when $w_1 = 1\in \mathbb{Z}/2\mathbb{Z} \simeq H^1(\mathbb{RP}^2,\mathbb{Z}/2\mathbb{Z})$.

Question 3: how to find out how many non-orientable bundles we have?

Please tell me if I am correct and if you know, the answer to the above questions.

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    $\begingroup$ $BO(2)$ is not $\mathbb{C}P^\infty\sqcup \mathbb{C}P^\infty$, it is a bundle over $B\mathbb{Z}/2\simeq \mathbb{R}P^\infty$ with fibre $BSO(2)\simeq \mathbb{C}P^\infty$. In fact every classifying space is connected. $\endgroup$ – William Apr 11 at 13:28
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    $\begingroup$ More specifically, $BO(2)$ can be constructed from the universal $\mathbb{Z}/2$ bundle by replacing the fibres with $BSO(2)$. $\endgroup$ – William Apr 11 at 13:31
  • $\begingroup$ Thanks William! Can we exploit this information to say something about $[\mathbb{RP}^2, BO(2)]$? $\endgroup$ – Warlock of Firetop Mountain Apr 11 at 14:20
  • $\begingroup$ If we are looking at pointed homotopy classes then there is a Puppe sequence $ \dots \to [\mathbb{R}P^2,\Omega B\mathbb{Z}/2]^0 \to [\mathbb{R}P^2,BSO(2)]^0 \to [\mathbb{R}P^2, BO(2)]^0 \to [\mathbb{R}P^2,B\mathbb{Z}/2]^0 $, which gives us some information about the oriented case which we already know, and pointed homotopy classes aren't quite the right thing to consider here I think. $\endgroup$ – William Apr 11 at 15:26
  • $\begingroup$ So far I proven that there are 2 iso classes of orientable bundles (distinguished by Euler class), and I've found two non-orientable bundles which are distinguished by $w_2$, but I haven't been able to determine yet if that's all of them. I'll put up my partial answer because it might still be helpful to you. $\endgroup$ – William Apr 11 at 15:26

(This is a partial answer, I haven't completely solved the case of non-orientable bundles yet.)

The short exact sequence $SO(2) \to O(2) \to \mathbb{Z}/2$ induces a fibration

$$\mathbb{C}P^\infty \to BO(2) \to \mathbb{R}P^\infty $$

Then $\pi_2 BO(2) \cong \mathbb{Z}$ and $\pi_1 BO(2)\cong \mathbb{Z}/2$ so unfortunately it's not an Eilenberg-MacLane space.

First I'll deal with the orientable bundles. A map $X \to BO(2)$ can be homotoped into the fibre $BSO(2)$ iff the composition $X\to BO(2) \to B\mathbb{Z}/2$ is null-homotopic (by the homotopy lifting property), which happens iff $w_1\in H^1(BO(2);\mathbb{Z}/2)$ pulls back to $0$. As you pointed out, isomorphism classes of $SO(2)$ bundles over $\mathbb{R}P^2$ are given by $H^2(\mathbb{R}P^2;\mathbb{Z})\cong \mathbb{Z}/2$ and in fact are determined by their Euler class. We have to ask wether the two bundles $E$ and $E'$ can be isomorphic via a map which reverses orientation (since we are interested in $O(2)$ bundles rather than $SO(2)$) but if they were then $e(E) = - e(E') = e(E')$ so they would have to be orientedly isomorphic anyway.

Therefore there are two isomorphism classes of $O(2)$ bundles over $\mathbb{R}P^2$ which are orientable. One is the trivial bundle ($e = 0$), and for the other consider the map $\mathbb{R}P^2 \to \mathbb{C}P^1$ which collapses the $1$-skeleton, and then the pull-back of the tautological complex line bundle will have $e=1$. Since the morphism $H^2(\mathbb{R}P^2;\mathbb{Z}) \to H^2(\mathbb{R}P^2;\mathbb{Z}/2)$ induced by $\mathbb{Z} \to \mathbb{Z}/2$ is an isomorphism, the oriented bundles are also determined by $w_2$.

This leaves bundles with $w_1\neq 0$. One example is $T\mathbb{R}P^2$, which has $w_1\neq 0 \neq w_2$. If $\gamma_1$ is the tautological real line bundle then there is also the $O(2)$ bundle $\gamma_1\oplus \mathbb{R}$, which has $w_1\neq 0$ but $w_2 = 0$. So we have at least two isomorphism classes of non-orientable bundles, but there is still the possibility that there are more non-isomorphic bundles which are not distinguished by Stiefel-Whitney classes, and I haven't figured that part out yet.


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