# Mobius Band as Bundle Over S^1: Trivialization Containing (1,0) in S^1

by WWGD
Tags: band, bundle, mobius, trivialization
 P: 400 Hi: I am trying to show that the Mobius Band M is a bundle over S^1. It seems easy to find trivializations for all points except for the trouble point (1,0). This is what I have so far: Let R be the reals, and IS^1 , a qiotient map, i.e., we give S^1 the quotient topology. 2) q: a quotient map on IxR : identify points (0,y) with (1,-y) . The "identified space" --i.e., the quotient of IxR by q -- will be the top space. (this is the Mob. Band.) 3)The Projection map Pi ,from the top space in 2 down to S^1 is defined by : Pi([s,t])= p([s]') ( where [s]' is the class of s under p:[0,1]->S^1 as above, so that the clases are: { {0,1}, {x} ) : 0
 Sci Advisor HW Helper PF Gold P: 4,768 Pi^-1(U) =[0,1.57]xR union [4.71,6.28]xR How can these numbers be right if the bundle is a quotient of [0,1] x R ?! For simplicity, I suggest you consider U=S^1\(-1,0) so that Pi^1(U)=([0,1]\{½} x R )/~. And now the goal is to remove the twist of the bundle that's "at" the fiber over (1,0).
 P: 400 "Pi^-1(U) =[0,1.57]xR union [4.71,6.28]xR How can these numbers be right if the bundle is a quotient of [0,1] x R ?!" Right. My bad; I did not rescale by 2PI. This should be: Pi^-1(U)=[0,1/4]xR union [3/4,1]xR "For simplicity, I suggest you consider U=S^1\(-1,0) so that Pi^1(U)=([0,1]\{½} x R )/~. And now the goal is to remove the twist of the bundle that's "at" the fiber over (1,0)" O.K, thanks. I tried it, but I don't see too well how to remove the twist: This is what I got: (R^+ is the positive reals, etc.) Pi^-1(U)= [[0,1/2)xR union (1/2,1]xR]/~ = 0xR^+ union (0,1/2)xR union (1/2,1)xR union 1xR^+ union (0,0) i.e., I collapsed 1xR^- with 0xR^+ , and I collapsed 0xR^- with 1xR^+ , and (0,0)~(1,0). How does this help, tho.?. Would you please suggest.?
HW Helper
PF Gold
P: 4,768

## Mobius Band as Bundle Over S^1: Trivialization Containing (1,0) in S^1

With U=S^1\(-1,0), we have Pi^1(U)=([0,1]\{½} x R )/~. Write [t,x] for the equivalence class of (t,x) under ~.

Define a diffeomorphism f:Pi^1(U) --> (½,3/2) x R by setting

f([t,x]) = (t,x) for (t,x) in (½,1] x R,
f([t,x])=(t+1,-x) for (t,x) in [0,½) x R
 P: 662 Quasar 987 wrote: " With U=S^1\(-1,0), we have Pi^1(U)=([0,1]\{½} x R )/~. Write [t,x] for the equivalence class of (t,x) under ~. Define a diffeomorphism f:Pi^1(U) --> (½,3/2) x R by setting f([t,x]) = (t,x) for (t,x) in (½,1] x R, f([t,x])=(t+1,-x) for (t,x) in [0,½) x R " Interesting. I too have been curious about the details of the Mobius Bundle as a non-trivial bundle over the circle. Just a couple of comments, though: It seems like a point like, e.g., [5/3xR] would be sent outside of (1/2,3/2)xR. Also: would you give some insight into how the twisting ( at the point (1,0)) is "solved"? Specifically, It just seems strange that we can have a diffeomorphism , given the twisting at (1,0). Would you comment.?
 Sci Advisor HW Helper PF Gold P: 4,768 Mmh, but 5/3 >1 while points of the bundle are of the form [t,x] with t in [0,1]. The idea of the construction of f is simply this: if you have a Mobius strip made of paper and if you take scissors and cut it "vertically" (this corresponds to removing the slice [{½}xR] in post #4 above), then there is no more "twist" as you can now take the remaining piece of paper and lay it flat on a table to form a rectangle: you have locally trivialized the mobius bundle. This is just what the map f does; it take ([0,1]\{½} x R )/~ and flattens it on the table (R^2).

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