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Does anybody know if orientable, closed $3$-manifolds that are circle bundles over $RP^2$ have been classified? One can determine the isomorphism classes of bundles using obstruction theory, but I am interested in what total spaces can appear. I am not assuming the bundle is principal.

Thank you.

flag	asked Nov 19 2009 at 18:01 F.G. 53●3

I apologize if I'm missing something: how do you get a circle bundle that's not principal? Do you mean "trivial" rather than "principal"? – S. Carnahan♦ Nov 19 2009 at 18:27

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There are nonprincipal, nontrivial circle bundles. For example, the Klein bottle is an S^1 bundle over S^1 which is not principal and nontrivial. To see it's not principal, notice that principal circle bundles over S^1 are classified by homotopy classes of maps from S^1 into CP^\infty, but CP^\infty is 1-connected. Thus, the only principal S^1 bundle over S^1 is the trivial one, and hence the Klein bottle is NOT principal. – Jason DeVito Nov 19 2009 at 19:00

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To summarize Scott vs. Jason: for Scott, "circle bundle" means S^1 is the structure group, while for Jason, "circle bundle" means S^1 is the fiber. – Charles Rezk Dec 2 2009 at 23:51

Danny Calegari · Accepted Answer · 2009-11-19 18:13:53Z UTC

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Such manifolds are examples of Seifert fibered spaces, which have, indeed, been classified. A good reference is Montesinos "Classical Tessellations and Three-Manifolds". Basically, such manifolds (over any nonorientable surface base) are classified by their Euler class, which measures the obstruction to the existence of a section.

answered Nov 19 2009 at 18:13

Danny Calegari
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Other references for the classification of Seifert-fibred spaces are Hatcher's 3-manifolds notes (on his webpage), Jaco's CBMS notes on 3-manifolds, the Orlik book on Seifert-fibred spaces. – Ryan Budney Nov 19 2009 at 19:59

HW · Answer 2 · 2009-11-19 18:57:50Z UTC

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According to Scott's paper "The geometries of 3-manifolds", the only closed manifolds that admit an $S^2\times R$ geometry are the two $S^2$ bundles over $S^1$, $P^2\times S^1$ and $P^3\sharp P^3$. It seems like the last one is the only candidate (unless I'm missing a way in which $S^2\times S^1$ can be viewed as such a bundle).

(How do I get a connect sum symbol in maths mode!?)

EDIT: Used \sharp as Richard suggested.

FURTHER EDIT: As pointed out below, I spoke too soon. Of course such bundles could have spherical geometry too. Scott's paper is a nice reference for that too!

edited Nov 19 2009 at 18:57

answered Nov 19 2009 at 18:17

HW
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Try using \sharp – Richard Kent Nov 19 2009 at 18:22

Hmmmm. I'm not thrilled with the result. – HW Nov 19 2009 at 18:28

Hi F.G: the bundles with nontrivial Euler class all have S^3 geometry (i.e. are quotients of S^3). – Danny Calegari Nov 19 2009 at 18:50

@Henry Wilton. Thanks. I think one needs to consider also quotients of S3. Apparently there are also some quotients of the 3 -sphere that are circle bundles over $RP^2$. For example, I read in a paper that $S3^/Q$, where $Q=A_3/Z_2$ is the quaternion group of order 8, and $S^3/Z_4$ , the lens space $L(4,1)$, are such circle bundles, although there are no references given for this. – F.G. Nov 19 2009 at 18:55

Yes, DC just pointed this out to me. My mistake! – HW Nov 19 2009 at 18:56

ivane · Answer 3 · 2009-12-02 19:48:07Z UTC

I think that they have Seifert fiber space presentation as: $(On,1|(1,b))$.

Or $(On,1|(1,b),(a_1,b_1),...,(a_r,b_r))$, if you allow an orbifold with cone points in $RP^2$.

You can look at the cases by decomposing $RP^2=Mo\cup_{\partial}D$, so the orientable 3-manifold will be the

1) orientable $Q=Mo\tilde{\times}S^1$, the twisted circle bundle over the mobius band, very well known being equivalent to the orientable I-bundle over the Klein bottle, with boundary a torus $T$,

2) and a Dehn-filling in the remaining disk $D$, with a whichever fibered solid torus or tori.

We could say that $(On,1\mid (1,b))=Q\cup_T W(1,b)$, for a fibered $(1,b)$ solid torus $W$

Circle bundles over $RP^2$

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