Suppose I have a submanifold $M \subset \mathbb{R}^{n}$, of dimension $n-1$. Apparently it's orientable if and only if there exists a unit normal vector field on $M$. Where a unit normal vector field is a section $\nu$ of the normal bundle $ TM^{\bot} \to M$. So the fibers are all the vectors that are perpendicular to the tangent space of the same base point. With the addition that $\| \nu(p) \| =1$, for all $p$ in $M$.

However, since the codimension is 1, can I not simply identify $T_{P}M^{\bot}$ with $\mathbb{R}$, and consider a smooth section $\nu$ of the bundle $\mathbb{R} \times M \to M$ where every point is mapped to either 1 or -1 by $\nu$, therefore having the section and therefore there exists a unit normal vector field and an orientation. Of ccourse this should be wrong, but what goes wrong in my reasoning and why?


For each $p$ you have that $T_pM^\perp \cong \Bbb R$, indeed. But these isomorphisms need not be natural. In some loose sense, for each $p$ you have an isomorphism $\Phi(p)\colon T_pM^\perp \to \Bbb R$, however the map $$TM^\perp \ni (p,v) \mapsto \Phi(p)(v) \in \Bbb R$$need not even be continuous, since $\Phi(p)$ and $\Phi(q)$ can be completely unrelated for $p \neq q$. One possible way to make a consistent choice of isomorphisms (ensuring good properties of the "coupled" map) is via a nonvanishing normal field defined globally along the manifold.

  • $\begingroup$ ah okay. How would you proof that statement of orientable iff existance of unit normal vector field? $\endgroup$ – AkatsukiMaliki Dec 2 '18 at 11:30
  • $\begingroup$ Probably using the normal field to define a nonvanishing top degree form (via determinant). See Spivak's Calculus on Manifolds, for example. $\endgroup$ – Ivo Terek Dec 2 '18 at 17:21

A dimension one real vector bundle is not always trivial, (consider the tautological bundle over the projective space) therefore, you cannot identify $TM^{\perp}$ to $\mathbb{R}\times M$. The existence of a unit normal vector field is equivalent to saying that $TM^{\perp}$ is trivial.

  • $\begingroup$ this is not my problem. The problem is that It looks as if there always is a smooth section. $\endgroup$ – AkatsukiMaliki Dec 2 '18 at 8:04
  • $\begingroup$ I meant identifying the fibers with R. $\endgroup$ – AkatsukiMaliki Dec 2 '18 at 8:05
  • $\begingroup$ @user3397129 There are always smooth sections, but not smooth sections that are globally non-vanishing! $\endgroup$ – Danu Dec 2 '18 at 10:32

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