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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3$\begingroup$ If I recall correctly, on sufficiently nice base spaces (e.g. compact manifolds), any vector bundle whatsoever can be embedded in a trivial one. $\endgroup$– Zhen LinFeb 22, 2013 at 12:21
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2 Answers
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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:
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]]
answered Feb 22, 2013 at 12:19
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1$\begingroup$ 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 $A\times I$? $\endgroup$– SteveFeb 22, 2013 at 12:43
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$\begingroup$ 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? $\endgroup$– SteveFeb 22, 2013 at 12:46
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1$\begingroup$ Glad I could help, and great question by the way! $\endgroup$ Feb 22, 2013 at 12:48
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1$\begingroup$ 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! $\endgroup$– SteveFeb 22, 2013 at 12:54
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2$\begingroup$ 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. $\endgroup$– Zhen LinFeb 22, 2013 at 13:49
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Take any compact manifold $M$ embedded in $\mathbb R^N$, with non-trivial tangent bundle (like $S^2\subset\mathbb R^3$). Then $TM$ sits inside the trivial vector bundle $M\times \mathbb R^N$.
As any compact manifold can be embedded into a suitable $\mathbb R^N$, this means that any tangent bundle of a compact manifold is subbundle of a trivial one. This is even more generally true for any finite-dimensional vector bundle over a compact manifold instead of the tangent bundle.
