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I was trying to prove the following fact: given a non orientable manifold $M$ of dimension $m$, $M$ is always contained in an orientable manifold of dimension $m+1$. I have gotten nothing out of it, so I am asking you. I warn you that my background involves no charachteristic classes, but only basic differential geometry.

Thanks in advance

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  • $\begingroup$ Do you know about the exterior bundle? $\endgroup$ – Aloizio Macedo May 7 '18 at 0:06
  • $\begingroup$ Just curious: how do you do this (concretely) for the real projective plane? $\endgroup$ – Lorenzo May 7 '18 at 0:06
  • $\begingroup$ @AloizioMacedo if you mean the bundle of alternating forms on each tangent space then yes, I do. So, $\Lambda^m(M)$ has the right dimesion, but why is it orientable? $\endgroup$ – Dario RANCATI May 7 '18 at 0:17
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    $\begingroup$ @DarioRANCATI Pick the charts you expect for $\Lambda^m(M)$. You can then prove that a change of charts will have determinant equal to $1$ (intuitively, this is due to the fact that the Jacobian of the associated chart of $M$ will appear for both the base and the exterior product part, in the latter one appearing as an inverse). $\endgroup$ – Aloizio Macedo May 7 '18 at 0:53
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Every manifold has an "orientation double cover" $\tilde M \to M$, whose elements are points of $M$ equipped with an orientation at that point. It comes equipped with a rather tautological orientation. If $M$ is connected, $\tilde M$ is connected iff $M$ is non-orientable.

Given a (reasonable; certainly this includes every manifold, CW complex, ...) space $X$ there is a bijection between isomorphism classes of real line bundles, equipped with a fiberwise metric, and double covers. The forward map is given by passing to the subspace of norm-1 elements; the inverse takes a double cover, trivializes it over an open cover of $X$, and then constructs the real line bundle with the same transition maps $U_i \cap U_j \to \pm 1$.

Now pass from $\tilde M$ to the associated real line bundle $\ell$, considered as a smooth manifold. This has an orientation: the most irritating part of checking this is on the 0 section. Explicitly, $T_{m,0}\ell = T_m M \oplus \ell_m$. Pick a basis $B$ of $T_m M$ and a norm 1 element $o$ of $\ell_m$; say that $(B,o)$ is positively oriented if $o$ says that $B$ is positively oriented.

Explicitly for $\Bbb{RP}^{2n}$, this is the tautological line bundle; its total space can be identified with $\Bbb{RP}^{2n+1} \setminus \{pt\}$, which is indeed orientable.

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Build a line bundle $E\to M$ such that the orientation of the fibre gets reversed along a loop in $M$ if and only if the orientation of $M$ itself is reversed along that loop. Then $M$ is embedded in $E$ as the zero section, $E$ is orientable, and it has the required dimension.

This can be worded in terms of the first Stiefel-Whitney class (if you ever feel like learning about characteristic classes).

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