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It seems to me that you can have a nontrivial subbundle sitting inside a trivial vector bundle. Can anyone please give an example of this which one can visualize? Thanks!

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If I recall correctly, on sufficiently nice base spaces (e.g. compact manifolds), any vector bundle whatsoever can be embedded in a trivial one. –  Zhen Lin Feb 22 '13 at 12:21
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up vote 5 down vote accepted

You're right; a good example is the Möbius bundle on $\mathbb{S}^1$ sitting inside the trivial bundle $\mathbb{S}^1\times\mathbb{R}^2$.

In the image below, you can think of $\mathbb{S}^1$ as being the central circle of the whole apparatus, and the "square" cross-section of each slice as being $\mathbb{R}^2$, which is after all diffeomorphic to $(0,1)^2$.

After some quick modifications to my code from this question, here is a depiction:

          enter image description here

For my own future reference / anyone else, the edited parts of the code are

NewFaces[R_,r_,s_,t_] := {F[R][t,s,r], F[R][t,r,s], F[R][t,-r,-s], F[R][t,-s,-r]}

NewEdges[R_,r_,t_] := {F[R][t,r,r], F[R][t,-r,r], F[R][t,r,-r], F[R][t,-r,-r]}

ThickMobius[R_,r_,u_] := Show[ParametricPlot3D[NewFaces[R, r, s, t], {s, -r, r},
{t, 0, 2 Pi}, PlotStyle -> {Blue, Opacity -> 0.15}, PlotPoints -> {2, 50}, 
Mesh -> None, Boxed -> False, Axes -> None], ParametricPlot3D[Strip[R, r, s, t],
{s, -r, r}, {t, 0, 2 Pi}, Mesh -> None, PlotStyle -> Red, PlotPoints -> 50], 
ParametricPlot3D[NewEdges[R, r, t], {t, 0, 2 Pi}, PlotStyle -> {Darker[Blue], 
Thickness[u]}, PlotPoints -> 30]]
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Thanks for your prompt answer! Just one side question: Intuitively shouldn't $\mathbb{S}^1\times\mathbb{R}^2$ just be the Cartesian product of an annulus with an interval (call it $A\times I$)? When you add twists to it (as in the figure) would that still be $\mathbb{S}^1\times\mathbb{R}^2$? Is it possible to just embed the Mobius bundle into $A\times I$? –  Steve Feb 22 '13 at 12:43
    
Awesome! The new figure just answered my previous comment. Thanks for your answer! However if you add twists to a trivial bundle, as in the pictures in your question, would that sill be trivial? –  Steve Feb 22 '13 at 12:46
    
@Steve: Absolutely, and that's why my earlier picture wasn't ideal for demonstrating what was going on; it is true that the "twisted" version of $\mathbb{S}^1\times\mathbb{R}^2$ is isomorphic (as a vector bundle) to $\mathbb{S}^1\times\mathbb{R}^2$, but this is not clear. I've made a new image that shows $\mathbb{S}^1\times\mathbb{R}^2$ properly. –  Zev Chonoles Feb 22 '13 at 12:47
    
Glad I could help, and great question by the way! –  Zev Chonoles Feb 22 '13 at 12:48
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The twistedness can't be seen in terms of the vector bundle structure, so you must look at something else; for example, you could look at local frames. –  Zhen Lin Feb 22 '13 at 13:49
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