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I wrote the following argument to prove that $S^1$ is parallelizable, that is, to show that the tangent bundle is trivial. It looks fairly reasonable to me.

Let $\tau=2\pi$.

We define a map $\varphi:S^1\times\Bbb R\to TS^1$ by $\varphi((e^{i\tau\theta}, t))=(e^{i\tau\theta},t\frac{\partial}{\partial x^1}\Big|_{e^{i\tau\theta}})$. This map is clearly a bijection: injectivity follows trivially and surjectivity follows since the tangent space $T_{e^{i\tau\theta}}$ is one-dimensional.

It remains to be shown that it is a diffeomorphism. We will, in fact, show that it is the identity map on suitably chosen coordinates. For any point $(e^{i\tau\theta},t)\in S^1\times\Bbb R$, we can choose the coordinate chart such that $(e^{i\tau(\theta+\varepsilon)},(t+\delta))$, for sufficiently small $\varepsilon$ and $\delta$, is given by $(\varepsilon,\delta)$.

Similarly, for any point $(e^{i\tau\theta},t\frac{\partial}{\partial x^1}\Big|_{e^{i\tau\theta}})\in TS^1$, we can choose the coordinate chart such that $(e^{i\tau(\theta+\varepsilon)},(t+\delta)\frac{\partial}{\partial x^1}\Big|_{e^{i\tau(\theta+\varepsilon)}})$, for sufficiently small $\varepsilon$ and $\delta$, is given by $(\varepsilon,\delta)$. Then, we have that $\widehat\varphi$ (the function with respect to these two coordinates) sends $(\varepsilon,\delta)\mapsto (\varepsilon,\delta)$, as desired.

However, I don't see where this is using the tangent bundle construction in any meaningful way. It seems that this would work just as well for any rank-1 vector bundle over the sphere.

That concerns me, because of course the conclusion is not true: for instance the Möbius strip is (diffeomorphic to) a vector bundle over the sphere. Since it is not orientable, we know that the bundle is nontrivial.

So the question is: Where does this break for a general vector bundle? Or if the argument is simply invalid, can it be fixed without much trouble?

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    $\begingroup$ It is not clear to me what you mean by $t\frac{\partial}{\partial x^1}\Big|_{e^{i\tau\theta}})$ (I would have written $ite^{2\pi\theta}$). But it is clear that your argument depends on writing down a formula (e.g., $ie^{2\pi\theta}$) for a non-zero cross section of the tangent bundle - an option that you don't have for the Möbius bundle. $\endgroup$ – Rob Arthan Jul 4 '16 at 23:08
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    $\begingroup$ Your argument is that given a line bundle $\lambda \to B$, the map $B \times \mathbb{R} \to \lambda$ given by $(b, t) \to t s_b$ for a nowhere-vanishing section $s$ is a homeomorphism, diffeomorphism, etc.. That's a valid argument that a line bundle admitting a nonwhere-vanishing section is trivial (modulo some reasonable assumptions on $B$); it's just that bundles generally do not admit such a section. For example, any section of the $2$-bundle $TS^2$ has to vanish somewhere, since $\chi(S^2)\not = 0$. $\endgroup$ – anomaly Jul 4 '16 at 23:44
  • $\begingroup$ @Rob: I'm following the notation of Lee's Introduction to Smooth Manifolds. I intentionally avoided identifying my tangent vectors with with elements of $\Bbb R^2$ since I am rather inept at using these identifications correctly. $\endgroup$ – Eric Stucky Jul 5 '16 at 8:24
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You're implicitly using the fact that a non-vanishing section (the coordinate vector field $\partial/\partial x_{1}$) exists. The tangent bundle admits such a section, while the non-trivial line bundle doesn't.

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  • $\begingroup$ There's no actual global co-ordinate chart though on $S^1$? Is it the same as for other Lie groups where you push forward with left(right) multiplication? $\endgroup$ – snulty Jul 5 '16 at 9:41
  • $\begingroup$ @snulty: You're perfectly correct that technically, no compact manifold admits a global coordinate vector field. On tori, however, one can reasonably pretend that the Cartesian coordinate frame on the universal cover defines a global coordinate frame, since the Cartesian frame is invariant under deck transformations (which are translations). Since OP had already denoted the section that way, the term "coordinate field" seemed harmless. :) $\endgroup$ – Andrew D. Hwang Jul 5 '16 at 11:33
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    $\begingroup$ Thanks for the reply, what you've said is actually very helpful. I was just looking a the fact that lie groups are parallelizable, and trying to piece together a few things. For a coordinate vector field though, for example on tori, which can be though of as an abelian lie group, can you just pick a chart, find a basis for the tangent space, and then push forward a particular basis vector with left/right multiplication to construct a left/right invariant co-ordinate vector field as above? I'll hopefully understand deck transformations soon! I've a rough idea what they are atm. $\endgroup$ – snulty Jul 5 '16 at 12:55
  • $\begingroup$ @snulty: The full story is a bit involved. First, what you say is correct; in fact, if you pick an arbitrary tangent frame at one point of a torus, the frame extends uniquely to a translation-invariant frame, and the torus is covered by coordinate charts for which the restriction of the frame field is the corresponding coordinate field. On a general Lie group, an arbitrary frame at one point can be extended by left translation to a (smooth) global frame field, but usually the result is not a coordinate frame field. (If it is, the induced left-invariant metric is flat.) $\endgroup$ – Andrew D. Hwang Jul 5 '16 at 21:41

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