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The Mobius band

by rbwang1225
Tags: fiber bundle, mobius strip
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rbwang1225
#1
Feb22-12, 02:48 AM
P: 116
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 cross-section. Any two cross-sections 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.
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tiny-tim
#2
Feb22-12, 03:59 AM
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hi rbwang1225!
Quote Quote by rbwang1225 View Post
[The projection ##L \times Y \rightarrow L## carries over under this matching into a projection ##p: B \rightarrow X##.
L is the flat unjoined-up base-line, X is the circle base-line.

p is the projection that just "drops" any curve in B (not necessarily stretching all the way round) onto the circle, X
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.
Yx is the vertical line above the point x on the circle X

mapping Yx onto the "original" Y (or onto any Yx') 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 g2 is the identity)
The group ##G## is the cyclic group of order 2 generated by ##g##.]
G is the group {I,g} with g2 = I
Deveno
#3
Feb22-12, 06:08 AM
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Quote Quote by tiny-tim View Post
hi rbwang1225!


L is the flat unjoined-up base-line, X is the circle base-line.

p is the projection that just "drops" any curve in B (not necessarily stretching all the way round) onto the circle, X


Yx is the vertical line above the point x on the circle X

mapping Yx onto the "original" Y (or onto any Yx') can be done in only two topologically different ways the "right way up", and the "wrong way up"
i don't understand this. as i see it, we have no way of telling which way is "up", we just have two ways to map Yx to Y. "g" isn't one of them, it transforms one homemorphism into the other.

g is the map which turns Y the wrong way up (so g2 is the identity)


G is the group {I,g} with g2 = I
does G = Aut(Y)?

tiny-tim
#4
Feb22-12, 06:24 AM
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The Mobius band

Hi Deveno!
Quote Quote by Deveno View Post
i don't understand this. as i see it, we have no way of telling which way is "up", we just have two ways to map Yx to Y.
"up" is whatever we choose to call it
"g" isn't one of them, it transforms one homemorphism into the other.
i didn't say g was from Yx to Y, I said it was from Y to Y,

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."
does G = Aut(Y)?
pass
Deveno
#5
Feb22-12, 07:06 AM
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Quote Quote by tiny-tim View Post
Hi Deveno!


"up" is whatever we choose to call it


i didn't say g was from Yx to Y, I said it was from Y to Y,

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."


pass
gotcha, g reflects Y.

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.
tiny-tim
#6
Feb22-12, 07:33 AM
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Quote Quote by Deveno View Post
my point about "up" is, there is no such thing on a mobius band
ah, but my "up" was on Y
lavinia
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Feb22-12, 07:52 AM
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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 Z2-bundle?
rbwang1225
#8
Feb23-12, 12:17 AM
P: 116
Quote Quote by Deveno View Post
does G = Aut(Y)?
Actually, I don't understand this notation "Aut(Y)"
rbwang1225
#9
Feb23-12, 01:02 AM
P: 116
Sorry for forgetting to put the figure.
Click image for larger version

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lavinia
#10
Feb23-12, 08:33 AM
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BTW: The intermediate value theorem tells you that any two cross sections of the Mobius band have to meet at least one point.
Fredrik
#11
Feb23-12, 09:42 AM
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Aut(Y) denotes the group of automorphisms on Y. That means that it's the set of all isomorphisms from Y into Y.
Bacle2
#12
Feb23-12, 01:26 PM
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Quote Quote by rbwang1225 View Post
Actually, I don't understand this notation "Aut(Y)"
This answers your own post:

"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.


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