It is a well known fact that (isomorphism classes of) principal $S^1$-bundles over a base space $B$ are classified by $B$'s second integral cohomology, $H^2(B;\mathbb{Z})$.

There is always the trivial one $B\times S^1$, and for $B=S^3$ for example, these are all. The Hopf bundle is an interesting example when $B=S^2$. The principal bundles over $S^2$ correspond bijectively to $\mathbb{Z}$; much interesting information about them can be extracted using tools such as in Blair[1].

However, over the Klein bottle or $\mathbb{R}P^2$ there is only one such non-trivial bundle (since their second cohomology is $\mathbb{Z}_2$). So in order to classify all of them, one just needs to find out what is the non-trivial bundle.


  1. What are the non-trivial principal circle bundles over the Klein bottle and $\mathbb{R}P^2$ ?
  2. In particular, given a nice base-space with 2nd integral cohomology a finite group (such as $\mathbb{R}P^2$), is there a constructive way to find out what are the non-trivial principal circle bundles over it?

    [1]: D.E. Blair, Riemannian Geometry of Contact and Symplectic Manifolds.

  • 4
    $\begingroup$ If you ignore "principal" then I would say that obstruction theory is constructive: given a CW structure on $B$ and a cellular 2-cycle representing the given cohomology class, you can use that data to construct a corresponding circle bundle. But maybe this procedure is constructive for principal bundles as well? $\endgroup$
    – Lee Mosher
    Aug 18, 2012 at 14:06
  • 2
    $\begingroup$ As Lee mentions, yes there's a constructive way to build such bundles. The construction is in Steenrod's Fiber Bundles book as well as many other bundle books and standard obstruction theory references. $\endgroup$ Aug 18, 2012 at 20:20

4 Answers 4


A third way to think about Anton Petrunin's example is that $S^2 \to {\mathbb R}P^2$ is a ${\mathbb Z}_2$ principal bundle where the action of ${\mathbb Z}_2 = \lbrace +1, -1 \rbrace $ is the obvious action on vectors in ${\mathbb R}^3$. As ${\mathbb Z}_2$ is a subgroup of $U(1)$, the circle group of complex numbers of length one, you can use standard principal bundle constructions to extend this ${\mathbb Z}_2$ principal bundle to a $U(1)$ principal bundle. In the case at hand these constructions just give the quotient of $S^2 \times S^1$ by ${\mathbb Z}_2$ as above.

You can compute the transition functions of this $U(1)$ bundle explicitly with respect to the standard open cover of ${\mathbb R}P^2$ by just computing the same for the ${\mathbb Z}_2$ bundle $S^2 \to {\mathbb R}P^2$ and check that they give you a Cech representative for the non-zero class in $H^2({\mathbb R}P^2, {\mathbb Z}_2) = {\mathbb Z}_2$.

The fact that this $U(1)$ bundle has a reduction to ${\mathbb Z}_2$ also tells us that when we square it we will get a trivial $U(1)$ bundle. Just think of squaring the ${\mathbb Z}_2$ valued transitions functions to get transitions functions for the squared bundle. Obviously they all just take the value $1$.

  • $\begingroup$ I see. I guess that a similar construction would work equally well for the Klein bottle, with it's standard two-to-one cover? $\endgroup$
    – Shlomi A
    Aug 19, 2012 at 13:32
  • $\begingroup$ Yes as Igor Belegradek comments below these are all examples of flat bundles arising by extending the structure group of the universal cover of a space $X$, which is a principal $\pi_1(X)$ bundle, using a homomorphism from $\pi_1(X)$ to $U(1)$ or $SO(2)$. Thanks David! $\endgroup$ Aug 19, 2012 at 14:03
  • $\begingroup$ Minor correction to my comment above. A two-to-one cover of the Klein bottle won't be the universal cover. But it will still be a ${\mathbb Z}_2$ principal bundle and the extension to $U(1)$ construction will work. $\endgroup$ Aug 19, 2012 at 22:20

Take the quotient $\mathbb S^2\times \mathbb S^1$ by the involution $\iota(x,y)=(-x,-y)$.

  • $\begingroup$ Are you saying that the Klein bottle is obtained as the orbit space of this quotient under the $S^1$ action which rotates the $S^2$ factor? If so, neat! $\endgroup$
    – Mark Grant
    Aug 18, 2012 at 14:07
  • $\begingroup$ No it is about $\mathbb{R}\mathrm{P}^2$. $\endgroup$ Aug 18, 2012 at 14:29
  • $\begingroup$ The quotient of $S^1$ by involution is $S^1$. The quotient of $S^2\times S^1$ by involution is the trivial bundle over $\mathbb{R} P^2$. $\endgroup$
    – Shlomi A
    Aug 18, 2012 at 15:13
  • 1
    $\begingroup$ @Shlomi, No this way you get a nontrivial bundle over $\mathbb{R}\mathrm{P}^2=\mathbb{S}^2/\mathbb{Z}_2$. Look at its Euler's class. $\endgroup$ Aug 18, 2012 at 16:21

Another way to describe Anton Petrunin's example is to start with the trivial $S^1$-bundle over $\mathbb RP^2$ and to take the fibrewise connect sum with the Hopf fibration $S^3 \to S^2$. By fibrewise connect sum I'm referring to taking the regular 2-manifold connect-sum on the base space, and then match that with fibrewise sum on the level of the bundle maps.

Similarly, the non-trivial $S^1$-bundle over the Klein bottle is the fibrewise sum of the Hopf fibration with the trivial $S^1$-bundle over the Klein bottle.

IMO this perpsective is helpful in seeing why you wouldn't expect any more principal $S^1$-bundles over non-orientable surfaces, since if you take the connect sum with two Hopf fibrations you can slide the connect-sum around a non-orientable loop and turn the Hopf fibration into the opposite Hopf fibration, which allows you to "cancel" them.


Say you have two principal $S^1$-bundles over connected $n$-manifolds $N$ and $M$ respectively. Call the bundles $p_N : E_N \to N$ and $p_2 : E_M \to M$. Let $U_N$ and $U_M$ be open sets in $N$ and $M$ respectively whose closures are diffeomorphic to compact balls. $p_N^{-1}(U_N)$ and $p_M^{-1}(U_M)$ are equivariantly diffeomorphic to $U_N \times S^1$ and $U_M \times S^1$ respectively (with the trivial $S^1$ action). Given a diffeomorphism $f : \partial U_N \to \partial U_M$ the fibrewise connect sum of $p_N$ and $p_M$ with respect to $f$ is the manifold:

$$ (E_N \setminus p_N^{-1}(U_N) \cup E_M \setminus p_M^{-1}(U_M)) / \sim $$

The equivalence relation $\sim$ comes from identifying $\partial p_N^{-1}(U_N)$ with $\partial p_M^{-1}(U_M)$ -- since they are both trivial $S^1$-bundles (with an essentially canonical trivialization since $U_M$ and $U_N$ are contractible), we can identify them in a preferred way. This is the union of two principal $S^1$ bundles over a common principal $S^1$-bundle subspace, so it's a principal $S^1$ bundle. To make it smooth you'll need to adjust the argument slightly using collars.

  • $\begingroup$ Ryan, could you please explain how are things matched up in the total spaces? Or perhaps recommend a reference about it? $\endgroup$
    – Shlomi A
    Aug 19, 2012 at 10:19
  • $\begingroup$ Thank you for your answer Ryan. It has been very helpful. :) $\endgroup$
    – Shlomi A
    Aug 20, 2012 at 7:20

Orientable circle bundle with torsion Euler class have been studied systematically. There are exactly the flat $SO(2)$-bundles, see "A Remark on Torsion Euler Classes of Circle Bundles" by Miyoshi or "Flat circle bundles, pullbacks, and the circle made discrete" by Oprea-Tanré.

It is a standard fact that any flat $G$-bundle over a (connected) finite cell complex $X$ can be written as $(\tilde X\times G)/\pi_1(X)$ where $\tilde X$ is the universal cover and $\pi_1(X)$ acts by deck transformations on the first factor, and via some homomorphism $\pi_1(X)\to G$ on the second factor. Thus all examples look like the one given by Anton.

As a caution I wish to point out that many people also studied flat circle bundles with $G=Diff(S^1)$. The answer there is different, namely one gets the so called Milnor-Wood inequality as a condition on the Euler class.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.