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Let $\Sigma = \mathbb {RP}^2$, and consider a non-orientable $O(2)$-principal bundle $\pi:P\to \Sigma$, this means that $w_1(P)\neq 0$, the first Steifel-Whitney class.

Form the associated bundle $E= P\times_{O(2)}\mathbb{S}^1$ with fiber $\mathbb S^1$ using the left action of $O(2)$ on $U(1)= \mathbb S^1$,

$P$ is a non-orientable fiber bundle over a non-orientable surface hence $P$ is an orientable manifold.

Similarly, $E$ is a non-orientable fiber bundle over a non-orientable surface hence $E$ is an orientable manifold.

PROBLEM: $E$ carries also a principal right $\mathbb{S}^1$-action inherited from the right multiplication of $\mathbb S^1$ on itself, this action descends to the quotient and makes $E\to \Sigma$ into a principal $\mathbb{S}^1$-bundle.

This is impossible because the total space of a principal $\mathbb{S}^1$-bundle over a non-orientable surface must be not-orientable.

What am I getting wrong?

EXPLANED AGAIN MORE IN DETAIL.

$E$ is defined as the quotient of $P\times \mathbb S^1$ by the relations $(p, z)\sim (p g, g^{-1} z)$ for any $g\in O(2)$, with projection map given by $[p,z]\in E\mapsto \pi(p) \in \Sigma$.

$E\to \Sigma$ is a non-orientable bundle because $w_1(E) = w_1(P)$ (Stiefel-Whitney class). As I showed here Fiber bundle orientability vs manifold orientability This implies that $E$ as a manifold is orientable.

I claim that $E$ carries a freee right $\mathbb S^1$-action. For $w\in \mathbb S^1$ define $$[p,z]\cdot w = [p,zw] \in E.$$ This action is well defined: $[pg,g^{-1}z]\cdot w = [pg,g^{-1}z w] = [p,zw] \in E.$ The action is free: $[p,z]\cdot w = [p,z]$ implies that $[p,gz]=[p,z]$, thus exists $g\in O(2)$ s.t. $pg=p$ and $g^{-1}zw = z$ since $O(2)$ acts freely on $P$ and $\mathbb S^1$ acts freely on itself $g=1, z=1$.

Denote by $\Sigma' = E/\mathbb S^1$ the orbit space of this free $\mathbb S^1$-action. Note that $E\to \Sigma'$ is an $\mathbb S^1$ principal bundle. I claim that $\Sigma = \Sigma'$. This is because the orbit of $\mathbb S^1$ acting on $E$ coincides with the fiber of $E\to \Sigma$. Indeed the former is equal to $\{[p,zw]\}_{w\in \mathbb S^1}$, and the latter is $$\{[pg,x]\}_{g\in O(2), x \in \mathbb S^1} = \{[p,g^{-1}x]\}_{g\in O(2), x \in \mathbb S^1} = \{[p,x]\}_{x \in \mathbb S^1}$$ (I considered the fiber over $\pi(p)$.

Conclusion: $E\to \Sigma$ is an $\mathbb S^1$-principal bundle hence $E$ is non-orientable in contrast to what we have seen above.

Overflowian
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  • Can you write down $P$ explicitly? – Bob Dobbs Sep 27 '23 at 13:19
  • The issue originates in the totally unsupported claim in the paragraph which says PROBLEM. One can look a mistake in your proof once you you write one, but not if you do not. It is clear that $E\to \Sigma$ cannot possibly be a principal $S^1$ bundle simply for the dimension reason (the fiber of this bundle is 2-dimensional). – Moishe Kohan Sep 27 '23 at 14:49
  • @MoisheKohan I will add more details but the fiber is $S^1$ so it's 1-dimensional recall that $E$ is the associated bundle to the action of $O(2)$ on $S^1$. – Overflowian Sep 27 '23 at 14:53
  • Oh, you are right about the dimension, I was careless. – Moishe Kohan Sep 27 '23 at 14:55
  • @MoisheKohan please have a look now, I have included more details. – Overflowian Sep 27 '23 at 16:51
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    For an example, can't you take $P$ to be the (principal bundle associated to the) tangent bundle? Then $E$ is the unit tangent bundle which is $\mathbb{RP}^3$ and which is orientable, right? – Qiaochu Yuan Sep 27 '23 at 17:04
  • @QiaochuYuan Actually we can make it even simpler going down one dimension, the argument above would also apply to the case $\Sigma = \mathbb S^1$, so that $E$ becomes the Klein bottle and the contraddiction would consist in giving the Klein bottle the structure of an $U(1)$-principal bundle (which is impossible). – Overflowian Sep 27 '23 at 17:19
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    I don't believe the action passes down. $g^{-1}\cdot (zw)\neq (g^{-1}\cdot z)w$ for $g^{-1}$ a reflection (e.g., $z\mapsto \bar{z}$), no? – user10354138 Sep 27 '23 at 19:42
  • @user10354138 Thank you, that's the mistake! I was identifying $\mathbb S^1 \simeq SO(2)$ but the action of $O(2)$ on them is completely different, it doesn't even preserve $SO(2)$. So my mistake was that I was writing $g z w$ as if they were all elements of $ O(2)$ via $S^1\simeq SO(2)\subset O(2)$ but in reality the left and right actions were very different as you noticed. – Overflowian Sep 27 '23 at 20:25

1 Answers1

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I think I owe an answer explaining my mistake to those that upvoted my question.

Given an $O(2)$ principal bundle $P\to \Sigma$, we can form an associated $\mathbb S^1$ bundle $E = P\times_{O(2)}\mathbb S^1$ using the left action of $O(2)$ on $\mathbb S^1$.

Then I wanted to define a right $\mathbb S^1$ action on $E$ by $[p,z]\cdot e^{i\theta} = [p,ze^{i\theta}] $ but this was not well defined. Of course $\mathbb S^1\simeq O(2)$ as groups but the action of $O(2)$ on $\mathbb S^1$ is not the multiplication action of $O(2) \curvearrowright SO(2)$ (indeed the latter is not even an action since does not preserve $SO(2)\subset O(2)$.

Thus is false that: $$[pg,g^{-1} z e^{i\theta}] = [p, ze^{i\theta}] \in E.$$ It would be more correct to distinguish the actions: $$[pg,(g^{-1}\cdot_{O(2)} z)\cdot_{\mathbb S^1} e^{i\theta}]$$ which is not equal in general to $[pg,g^{-1}\cdot_{O(2)} (z\cdot_{\mathbb S^1} e^{i\theta})]$.

Overflowian
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