
#1
Feb2212, 02:48 AM

P: 115

I don't understand the fibre bundle structure of the Mobius band, described by my textbook.
"The base space ##X## is a circle obtained from a line segment L as indicated in Fig. by identifying its ends. The fibre ##Y## is a line segment. The bundle ##B## is obtained from the product ##L \times Y## by matching the two ends with a twist. [The projection ##L \times Y \rightarrow L## carries over under this matching into a projection ##p: B \rightarrow X##. Any curve as indicated with end points that match provides a crosssection. Any two crosssections must agree on at least one point. There is no natural unique homeomorphism of ##Y_x## with ##Y##. There are two such which differ by the map ##g## of ##Y## on itself obtained by reflecting in its midpoint. The group ##G## is the cyclic group of order 2 generated by ##g##.]" What I don't understand are in brackets. Could someone help me to clarify more clearly? Regards. 



#2
Feb2212, 03:59 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

hi rbwang1225!
p is the projection that just "drops" any curve in B (not necessarily stretching all the way round) onto the circle, X mapping Y_{x} onto the "original" Y (or onto any Y_{x'}) can be done in only two topologically different ways … the "right way up", and the "wrong way up" g is the map which turns Y the wrong way up (so g^{2} is the identity) 



#3
Feb2212, 06:08 AM

Sci Advisor
P: 906





#4
Feb2212, 06:24 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

The Mobius band
Hi Deveno!
which is the way it is defined in the textbook … "There are two such which differ by the map g of Y on itself obtained by reflecting in its midpoint." 



#5
Feb2212, 07:06 AM

Sci Advisor
P: 906

my point about "up" is, there is no such thing on a mobius band. we can't orient it. there's just one way and another way, and we can't even decide on "choosing a convention". i mean if U is a neighborhood of x in X, then p^{1}(U) has "2 sides" but B only has 1 (locally flat, but globally twisted). if i'm thinking about this right, it's like we have a cylinder antipodally identified like the sphere is with the projective plane. each fiber is "both versions" of the line Y, one is reflected about the circle X, but we honestly can't say "which is which". by Aut(Y), i mean of course, the group of homeomorphisms Y→Y. strictly speaking, there ought to be more than just 2 (we could "stretch Y in the middle, and shrink it by half the amount we stretched at each end"), so maybe they ought to be isometries, i dunno. 



#7
Feb2212, 07:52 AM

Sci Advisor
P: 1,716

One can think of the Moebius band as the quotient of the cylinder by the action of Z2 which sends a point (x,t) to the point (x + pi, t) where the cylinder is viewed as the Cartesian product of a circle with the interval, (1,1).
This action shows that the Moebius band is a flat Z2 bundle over the circle. As such, it is defined with two coordinate charts and the transition function, t> t. This action of Z2 is an isometry of the flat cylinder, (the cylinder that is made from a sheet of paper by taping opposite edges together), so the Moebius band is a flat Riemannian surface. Parallel translation of a vector that points along the fiber interval around the central circle returns the vector to its negative. Equivalently, parallel translation of an othonormal frame around the circle returns it to an orthonormal frame with the opposite orientation. This shows that the holonomy group of the flat Moebius band is Z2. When one makes a real Moebius band out of paper, it becomes a flat Riemannian manifold that is realized in space. Since it is still flat i.e. it still has a flat geometry when realized in space, parallel translation of that same orthonormal frame around the central circle still reverses orientation. Now add one more vector perpendicular to the Moebius band to create a 3 vector orthonormal frame. Do the parallel translation again. In space, parallel translation of a 3 vector frame around any curve preserves its orientation. But one of the vectors along the Moebius band is reflected, so the normal vector must also be reflected. Thus there is no idea of up and down normal to the Moebius band in space. The Klein bottle is similar to the Moebius band. It is the quotient of a flat torus by the action of Z2 that rotates the first coordinate circle by 180 degrees and reflects the second coordinate circle along an axis. So the Klein bottle is a flat circle bundle over the circle with structure group Z2. The same parallel translation argument holds and the holonomy group of the flat Klein bottle is Z2. If instead of reflecting about an axis, you multiply the second coordinate circle by 1, you get another flat Z2 bundle over the circle. So here is a question. Is this a trivial Z2bundle? 



#8
Feb2312, 12:17 AM

P: 115





#10
Feb2312, 08:33 AM

Sci Advisor
P: 1,716

BTW: The intermediate value theorem tells you that any two cross sections of the Mobius band have to meet at least one point.




#11
Feb2312, 09:42 AM

Emeritus
Sci Advisor
PF Gold
P: 8,987

Aut(Y) denotes the group of automorphisms on Y. That means that it's the set of all isomorphisms from Y into Y.




#12
Feb2312, 01:26 PM

Sci Advisor
P: 1,168

"There is no natural unique homeomorphism of Yx with Y. There are two such which differ by the map g of Y on itself obtained by reflecting in its midpoint. The group G is the cyclic group of order 2 generated by g.]" This is what Aut(Y) is in this case. 


Register to reply 
Related Discussions  
Nonvanishing section for direct sum of Mobius band  Differential Geometry  1  
Mobius Band as Bundle Over S^1: Trivialization Containing (1,0) in S^1  Differential Geometry  5  
mobius band?  General Math  0  
Valance band to Conduction Band (Valance Electron), Energy Band Gap Levels  Advanced Physics Homework  0  
Fundamental Groups of the Mobius Band  General Math  6 