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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
up vote 6 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
Glad I could help, and great question by the way! – Zev Chonoles Feb 22 '13 at 12:48
Thanks Zev! This is very interesting! How would one capture the idea of "twists" in this context in terms of mathematical defintions? And how would one go proving that the "twisted" version of $\mathbb{S}^1\times\mathbb{R}^2$ is isomorphic to $\mathbb{S}^1\times\mathbb{R}^2$? I am still in the early stages of learning vector bundles so references to literature would be great! – Steve Feb 22 '13 at 12:54
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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