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

  1. WWGD

    WWGD 1,321
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    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 I<R be the unit interval
    [0,1]< R (i.e., [0,1] as a subspace of the Reals.)

    We have these maps:

    1) p:I-->S^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(')

    ( where ' is the class of s under p:[0,1]->S^1 as above, so that the clases are:

    { {0,1}, {x} ) : 0<x<1 }


    Then: *****


    A trivialization for U= S^1-{(1,0)} is

    U itself; Pi^-1(U)=(1,0)xR , is already a

    product space. The identity map gives us a homeo.


    *****

    A trivialization containing the point (1,0) in S^1:

    This seems to work, but seems too easy:

    We consider an open set (open in S^1 ; quotient ) containing (1,0). We parametrize

    S^1 by arclength , "based" at (1,0), (i.e., arclength at (1,0)=0) , and we use , e.g.,

    an arc U from (0.1) , to (0,-1) .

    We want to show that Pi^-1(U) is homeo. to UxR . We have:


    Pi^-1(U) =[0,1.57]xR union [4.71,6.28]xR


    Am I on the right track?.

    Thanks.
     
  2. jcsd
  3. quasar987

    quasar987 4,770
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    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).
     
  4. WWGD

    WWGD 1,321
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    "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.?
     
  5. quasar987

    quasar987 4,770
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    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
     
  6. 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.?
     
  7. quasar987

    quasar987 4,770
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    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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