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I was wondering whether there is an example of a connected, open, smooth manifold $M$, whose tangent bundle $\tau$ is not a sub-bundle of a trivial bundle.

Clearly, for closed $M$ this is impossible, as any bundle over a compact space is the subbundle of a trivial bundle.

Julius J.
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1 Answers1

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Assuming you use the standard definition of "smooth manifold" (i.e. Housdorff and second countable), then the no such example exists.

One way to see this is with the fact that every such manifold admits a finite atlas (see this question for more details). From there, you can use a partition of unity to construct an injective bundle morphism $M\to M\times\mathbb{R}^{mk}$, where $m=\dim(M)$ and $k$ is the number of charts in the finite atlas.

Alternately, you can use the smooth Whitney immersion/embedding theorem to obtain a map $TM\to T\times\mathbb{R}^N$ by pulling back the differential of the immersion/embedding $\iota:T\to\mathbb{R}^N$.

Even if you weaken your definition by switching second countable with paracompact, you still don't have any examples. You'll have to use a very non-standard definition of manifold if you want to construct such an example.

Kajelad
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