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Let M be an (n-1)-manifold in R^n . Let M(e) be the set of end-points of normal vectors (in both directions) of length e and suppose e is small enough so that M(e) is also an (n-1)-manifold. Show that M(e) is orientable (even if M is not)

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What definition of orientability do you want to use? FYI The manifold $M(e)$ is a special-case construction of what people like to call "the orientation cover" of a manifold. – Ryan Budney Nov 9 '10 at 23:26

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up vote 5 down vote accepted

Each point $p$ of $M(\epsilon)$ comes from a point $q$ in $M$. Consider the vector $X$ field on $M(\epsilon)$ which on $p$ takes the value $\vec{qp}$. Then $N$ is a non-zero normal field on $M(\epsilon)$ (maybe one needs to consider the projection of $N$ onto the normal line to $M(\epsilon)$ at each point, but that projection is surely non-zero and, moreover, to normalize this projection)

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Dear Mariano, What if $M \subset \mathbb{R}^2$ is the union of two circles whose radii differ by $2\varepsilon$? In that case there are points $p \in M(\epsilon)$ which come from two points $p_1,p_2 \in M$. Also, how do you know that the resulting vector field is defined continuously? – user1337 Jun 10 at 16:55
@user1337 I am assuming that $e$ is small enough so that doees not happen. – Mariano Suárez-Alvarez Jun 10 at 22:28
Sorry to bother you again, but can you show that $\epsilon$ can be chosen so that it doesn't happen? (provided $M_\epsilon$ is indeed a manifold, of course). Also, could you please elaborate on why the said projection is non-zero? Thanks – user1337 Jun 13 at 15:15
If the manifold is not compact it may well be the case that there is no such epsilon, as my example in the other question shows. We have to decide what "small enough means" and it should mean "small enough so that the normal bundle embeds correctly"; this can be done with a constant epsilon in the compact case and with a varying epsilon in the general case, as shown there. Now, once we have an epsilon which is small enough in this sense, then the fact that the projection is not zero can be computed in local coordinates. – Mariano Suárez-Alvarez Jun 13 at 21:45

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