# Classification of circle bundles over $\mathbb{RP}^2$

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.

• $BO(2)$ 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. – William Apr 11 at 13:28
• More specifically, $BO(2)$ can be constructed from the universal $\mathbb{Z}/2$ bundle by replacing the fibres with $BSO(2)$. – William Apr 11 at 13:31
• Thanks William! Can we exploit this information to say something about $[\mathbb{RP}^2, BO(2)]$? – Warlock of Firetop Mountain Apr 11 at 14:20
• 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. – William Apr 11 at 15:26
• 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. – William Apr 11 at 15:26

## 1 Answer

(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.