- #1
klackity
- 65
- 1
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 [tex]\gamma[/tex]S: [0,2pi) --> F (the mobius strip), but where [tex]\gamma[/tex]S(0) need not equal [tex]\gamma[/tex]S(2pi).
But we can continuously deform (with some map we'll call [tex]\Phi[/tex]: [0,1] --> C0(S1, F) ) each such [tex]\gamma[/tex]S = [tex]\Phi[/tex](0) into some other continuous function [tex]\widehat{\gamma}[/tex]S = [tex]\Phi[/tex](1) such that [tex]\widehat{\gamma}[/tex]S(0) = [tex]\widehat{\gamma}[/tex]S(2pi) and the local triviality condition is still satisfied for [tex]\Phi[/tex](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 [tex]\gamma[/tex]1. To have local triviality, we need that any other continuous path [tex]\gamma[/tex]2 in V over (pi,3pi) remains continuous as [tex]\Phi[/tex] messes with [tex]\gamma[/tex]1. If the fibers near 2pi are considered to have opposite orientation on either side of 2pi in [tex]\gamma[/tex]1 but the same orientation on either side of 2pi in [tex]\gamma[/tex]2, this is simply impossible.)
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 [tex]\gamma[/tex]S: [0,2pi) --> F (the mobius strip), but where [tex]\gamma[/tex]S(0) need not equal [tex]\gamma[/tex]S(2pi).
But we can continuously deform (with some map we'll call [tex]\Phi[/tex]: [0,1] --> C0(S1, F) ) each such [tex]\gamma[/tex]S = [tex]\Phi[/tex](0) into some other continuous function [tex]\widehat{\gamma}[/tex]S = [tex]\Phi[/tex](1) such that [tex]\widehat{\gamma}[/tex]S(0) = [tex]\widehat{\gamma}[/tex]S(2pi) and the local triviality condition is still satisfied for [tex]\Phi[/tex](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 [tex]\gamma[/tex]1. To have local triviality, we need that any other continuous path [tex]\gamma[/tex]2 in V over (pi,3pi) remains continuous as [tex]\Phi[/tex] messes with [tex]\gamma[/tex]1. If the fibers near 2pi are considered to have opposite orientation on either side of 2pi in [tex]\gamma[/tex]1 but the same orientation on either side of 2pi in [tex]\gamma[/tex]2, this is simply impossible.)
Last edited: