Is a bundle with non-orientable fibers always trivial?

[klackity]

Suppose I have a manifold M. If E is a fiber bundle over M with non-orientable fiber F, is E necessarily trivial?

I'm imagining bundles over S¹. For example, if F is the mobius strip, then consider any section S of E. By local triviality, we can think of this section as a continuous function \gamma_S: [0,2pi) --> F (the mobius strip), but where \gamma_S(0) need not equal \gamma_S(2pi).

But we can continuously deform (with some map we'll call \Phi: [0,1] --> C⁰(S¹, F) ) each such \gamma_S = \Phi(0) into some other continuous function \widehat{\gamma}_S = \Phi(1) such that \widehat{\gamma}_S(0) = \widehat{\gamma}_S(2pi) and the local triviality condition is still satisfied for \Phi(t) for all t in [0,1].

(This is not true if the fibers are some vector space V, because local triviality will be violated if you try to move the fibers close to 2pi -- you would have to reverse orientation, which can't be done. In other words, fix a continuous path in V parametrized over [0,2pi) by \gamma₁. To have local triviality, we need that any other continuous path \gamma₂ in V over (pi,3pi) remains continuous as \Phi messes with \gamma₁. If the fibers near 2pi are considered to have opposite orientation on either side of 2pi in \gamma₁ but the same orientation on either side of 2pi in \gamma₂, this is simply impossible.)

 

[quasar987]

but where \gamma_S(0) need not equal \gamma_S(2pi).

Why do you say that? If \gamma_S(0) is not equal \gamma_S(2pi), then the section is not continuous at 1=e^{i0}\in S^1.

 

[klackity]

Why do you say that? If \gamma_S(0) is not equal \gamma_S(2pi), then the section is not continuous at 1=e^{i0}\in S^1.

The continuity is with respect to a local diffeomorphism, not globally. For all we know, the section itself could even not be continuous anywhere when considered as a subset of S¹ x F under an arbitrary coordinate assignment. We could assign to each point of S¹ a random point in the mobius strip, for example. The fibers are considered disjoint to begin with until you impose a topology.

If we are given the bundle E, however, and the section S we have chosen is continuous with respect to the topology on E, then we can represent S as a continuous "coordinate" section of S¹ x F (with respect to the standard product topology). We do this by beginning with some point in S¹ (for example, pi), and using the fact that locally, E is diffeomorphic to S¹ x F, and extending the coordinate section all the way to the interval (0,2pi). This section of S¹ x F will be locally continuous (which is just continuous!) everywhere on (0,2pi). We can also extend this section to all of S¹ (which we are parameterizing by [0,2pi) ), with the caveat that the section will not (necessarily) be continuous at 0 = 2pi.

If E is a trivial bundle, however, then we CAN extend some coordinate assignments to a section to be continuous on all of S¹.

Basically, E being a bundle implies that the selection of coordinates can always be done in a smooth way locally. But given two local assignments of coordinates, they need not agree value-wise anywhere. They only need to be diffeomorphically related on their intersection.

In this case, one reasonable local diffeomorphism is from the open interval (0,2pi) x F to a subset of the bundle E. (it's still local, even though it's defined on all but one fiber).

By choosing an alternate local diffeomorphism (namely, one from (pi,3pi) x F to the respective subset of the bundle E), we can make that same path continuous at 2pi by applying some diffeomorphism to the section of the original path on (2pi,3pi) (which would just be sliding the mobius strip) and assigning a value for 2pi. On the intersection between the two local diffeomorphisms, we find that they are related. This alternate diffeomorphism cannot be extended to a diffeomorphism of the entire bundle, however, because it has introduced what would be a new discontinuity at the point in the bundle corresponding to 3pi = pi.

 

[quasar987]

Ok, I see. Next, why do you say that the following is true in the case F non orientable (or F the Mobius band say):

But we can continuously deform (with some map we'll call \Phi: [0,1] --> C⁰(S¹, F) ) each such \gamma_S = \Phi(0) into some other continuous function \widehat{\gamma}_S = \Phi(1) such that \widehat{\gamma}_S(0) = \widehat{\gamma}_S(2pi) and the local triviality condition is still satisfied for \Phi(t) for all t in [0,1].

 

[klackity]

Ok, I see. Next, why do you say that the following is true in the case F non orientable (or F the Mobius band say):

We can find a diffeomorphism f between F and F (the mobius strip and itself) such that f(\gamma_S(2pi)) = \gamma_S(0).

For the mobius strip at least, I think any such diffeomorphism will be homotopic to the identity map.

Let H(t,x) be the homotopy between id and f, with H(0,x) = id(x) = x and H(1,x) = f(x).We can now define a map \Phi: [0,1] --> C⁰(S¹, F) by \Phi(t)(s) = H( t*s/2pi , \gamma_S(s) ).

Note that \Phi(0) = \gamma_S, but \Phi(1) is a continuous deformation of \gamma_S where \Phi(1)(0) = \gamma_S(0) but \Phi(1)(2pi) = \gamma_S(0). Moreover, Phi(t) is continuous for each t.

Now... about the local triviality...

 

[klackity]

About the local triviality:

Suppose we have some other assignment \varphi of coordinates (coordinates in S¹ x F) to the same section S such that \varphi is continuous on a neighborhood of 2pi. Then we want our continuous deformation \Phi to not affect the continuity of \varphi at 2pi.

I think I need to think a bit more whether this is true or not...

 

[lavinia]

Suppose I have a manifold M. If E is a fiber bundle over M with non-orientable fiber F, is E necessarily trivial?

I'm imagining bundles over S¹.

let Z/2Z act on the Klein bottle by reflection in the equatorial direction. with this involution construct a bundle over the circle with fiber the Klein bottle. this bundle is not trivial.

a specific construction of this manifold is obtained from its group of covering transformations on R^3. this group is generated by the standard lattice together with the following two transformations.

(x,y,z) -> (x, y+ 1/2, -z) and (x,y,z) -> (x + 1/2, -y,z)

if this bundle were trivial then it would be homeomorphic to the Cartesian product of a circle with a Klein bottle. but this bundle has a different fundamental group.

 

[zhentil]

In practice the only way to make absolutely certain that a fiber bundle is trivial is if every map from the base to the diffeomorphism group of the fiber is nullhomotopic. Why would that be the case here?