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In Tu's "An Introduction to Manifolds", exercise 16.8, we are supposed to prove that $M$ has a trivial tangent bundle $\Leftrightarrow$ M admits a smooth frame $\{X_1, \dots, X_n\}$ (i.e., smooth vector fields $X_1, \dots, X_n$ such that $\{X_{1_p}, \dots, X_{n_p}\}$ is a basis for $T_pM$ for every $p\in M$).

When I read this, I concluded that $M$ is orientable $\Rightarrow$ M admits a smooth frame $\{X_1, \dots, X_n\}$ $\Rightarrow M$ has trivial tangent bundle. However, I've learned that $\mathbb{S}^2$ is an orientable manifold which has no trivial tangent bundle.

What am I missing here?

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  • $\begingroup$ If $M$ has trivial tangent bundle, then $M$ is orientable. The converse is not true as your example demonstrates. $\endgroup$ – Michael Albanese Nov 29 '16 at 22:19
  • $\begingroup$ How do you read that orientable implies parallelizable(existence of global frame? $\endgroup$ – melomm Nov 29 '16 at 22:29
  • $\begingroup$ @melomm if $M$ is orientable, then there are smooth vector fields which form an oriented base for the tangent spaces, no? Wouldn't that be a smooth frame? $\endgroup$ – rmdmc89 Nov 29 '16 at 22:31
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    $\begingroup$ What definition of orientabiliy you are using? $\endgroup$ – melomm Nov 29 '16 at 22:35
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    $\begingroup$ The @Andreas Cap's answer is the definition of orientability in terms of local vector fields. But I think so you are confuse because for an hypersurface $S$ on $\mathbb{R}^{n}$ the orientability is equivalent to esistence a global nornmal vector field, in particular the normal bundle is trivial. (The same is true if you change $\mathbb{R}^{n}$ by an orientable manifold ) $\endgroup$ – melomm Nov 30 '16 at 13:13
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I am referring to your discussion with @melomm in the comments to the question. The right definition of orientability involving vector fields is the following. You have to specify an orientation of the tangent space $T_xM$ for all $x\in M$, which are compatible in the sense that for each $x\in M$, there is an open neighborhood $U$ of $x\in M$ and there are local vector fields $X_1,\dots X_n\in\mathfrak X(U)$ such that for each $y\in U$ the values $X_1(y),\dots,X_n(y)$ define a positively oriented basis for $T_yM$. Of course, this implies that $TM$ is trivial over $U$, but $TM$ is locally trivial anyway.

To see that this is equivalent to the definition via oriented atlasses roughly goes as follows. It easily follows from the defintions that for a connected local chart $(U_\alpha,\phi_\alpha)$ the choice of an orientation of $T_xM$ for one $x\in U_\alpha$ gives rise to a compatible choice for all $y\in U_\alpha$. If you have an oriented atlas than you can extend this orientation to all points in $M$. Conversely, if you have compatible orientations the charts inducing this given orientation form an oriented atlas.

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  • $\begingroup$ that's a much better explanation then the ones I've found in textbooks, thanks! $\endgroup$ – rmdmc89 Nov 30 '16 at 14:24

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