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I was pretty sure about this result but don't know how to prove it. I will state the question again:

Is any smooth line subbundle (or equivalently smooth 1-dimensional distribution) of the tangent bundle of a manifold is always trivial? Namely, once you have a 1-dim distribution on a manifold, you can have a nowhere vanishing vector field on that.

How to prove?

Thank you in advance!

jingyey
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    Do you have any reason to think it's true? Can you say something about how the question arises (especially, is this homework)? – Andrew D. Hwang Jun 23 '21 at 02:00
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    Nice question! In the book by J. M. Lee, Introduction to Riemannian manifolds (2nd), P44, Theorem2.69 says a manifold admits the existence of Lorentz metric if and only if it admits 1-dim tangent distribution. But we know such Lorectz metric exists if and only if there is a non-vanishing vector field, which you can find in O'Neil, Semi-Riemannian Geometry with Applications to General Relativity page 149. However, I believe the latter result is correct, but I can't see why 1-dim distribution can imply a non-vanishing vector field. – jingyey Jun 23 '21 at 02:18

2 Answers2

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This is not true. Consider the Klein bottle $K$.

As $K$ is a closed manifold with $\chi(K) = 0$, it admits a nowhere-zero vector field. The orthogonal complement of such a vector field is a line subbundle $L$ of $TK$. If $L$ were trivial, then $TK \cong L\oplus\varepsilon^1$ would be trivial, but this is impossible as $K$ is non-orientable.

However, as this example illustrates, if $TM$ admits a line subbundle, then it also admits a trivial line subbundle, i.e. $M$ admits a nowhere-zero vector field. See this answer.

Michael Albanese
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  • Wonderful, Wonderful! – jingyey Jun 28 '21 at 22:35
  • “If $M$ admits a line subbundle, then it also admits a trivial subbundle” - Is this specific to smooth vector bundles, or does it also apply to holomorphic or algebraic ones? – isekaijin Sep 12 '22 at 11:06
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    @pyon: It is specific to smooth manifolds (not smooth bundles, as the statement is not true for those). For example, $T\mathbb{CP}^1$ is a holomorphic line bundle, in particular, it admits a line subbundle (itself). However, $T\mathbb{CP}^1$ does not admit a trivial line subbundle as $T\mathbb{CP}^1$ is not trivial. – Michael Albanese Sep 12 '22 at 11:18
  • Oh, duh... How could I overlook the simplest example. Thanks! – isekaijin Sep 12 '22 at 11:21
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The non-orientable foliation below on the punctured disk shows that a $1$-dimensional distribution does not generally have a continuous nowhere vanishing section.

That doesn't mean the punctured disk has no non-vanishing vector field, of course, it just means that we can't generally select a continuous, non-vanishing vector field from a given distribution.

The field of lines on a Möbius strip is another example, but the punctured disk may be more impressive since its tangent bundle is trivial.

A foliation of the punctured disk not inducing a non-vanishing vector field

Andrew D. Hwang
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  • This is very impressive! I guess my original concerns about this problem should be clear, however, I can't help ask if the existence of 1-dim distribution implies a non-vanishing vector field, without restrictions on that distribution. I guess this is very hard and not always true? – jingyey Jun 23 '21 at 03:04