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?

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.