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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.

Question:

  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.

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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? –  Lee Mosher Aug 18 '12 at 14:06
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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. –  Ryan Budney Aug 18 '12 at 20:20
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4 Answers

up vote 6 down vote accepted

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 $1$.

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Welcome to MO, Michael! –  David Roberts Aug 19 '12 at 13:25
    
I see. I guess that a similar construction would work equally well for the Klein bottle, with it's standard two-to-one cover? –  Shlomi A Aug 19 '12 at 13:32
    
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! –  Michael Murray Aug 19 '12 at 14:03
    
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. –  Michael Murray Aug 19 '12 at 22:20
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Take the quotient $\mathbb S^2\times \mathbb S^1$ by the involution $\iota(x,y)=(-x,-y)$.

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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! –  Mark Grant Aug 18 '12 at 14:07
    
No it is about $\mathbb{R}\mathrm{P}^2$. –  Anton Petrunin Aug 18 '12 at 14:29
    
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 $\mathbb{R} P^2$. –  Shlomi A Aug 18 '12 at 15:13
    
@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. –  Anton Petrunin Aug 18 '12 at 16:21
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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.

edit:

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.

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Ryan, could you please explain how are things matched up in the total spaces? Or perhaps recommend a reference about it? –  Shlomi A Aug 19 '12 at 10:19
    
Thank you for your answer Ryan. It has been very helpful. :) –  Shlomi A Aug 20 '12 at 7:20
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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.

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