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Let $M$ be a $n$-manifold and let $S \subset M$ be a non-orientable $n$-dimensional submanifold possibly with boundary. Under what conditions can I conclude that $M$ is also non-orientable? Is compactness sufficient?

In particular consider the case where $M$ is a surface and $S$ is a Mobius strip.

Blake
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    $M$ is always non orientable if $S$ is non orientable. – Georges Elencwajg Aug 23 '15 at 13:29
  • Oh, right... thank you. Alright what about the case where $M$ is a $m$ dimensional manifold with $m > n$. In particular what if $M$ is an $n+1$ dimensional manifold? – Blake Aug 23 '15 at 13:44
  • In general you can deduce nothing because Whitney's embedding theorem states that every manifold, orientable or not, can be embedded iinto some $\mathbb R^N$, which is of course orientable. – Georges Elencwajg Aug 23 '15 at 13:47
  • Which is why I thought compactness might allow one to deduce something, since it specifically rules out that case. – Blake Aug 23 '15 at 13:52
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    @Blake: if you want to add/modify hypotheses, you should edit your post. In addition, since $\mathbb R^N$ can be seen as a submanifold of $S^N$, assuming the compactness of $M$ doesn't change anything. – Taladris Aug 23 '15 at 13:58
  • Perhaps something along the lines of what you're looking for: codimension 1 submanifolds of $S^n$ are orientable. In general, you want to think about Steifel-Whitney classes of normal bundles. (A submanifold of an orientable manifold is orientable iff its normal bundle is orientable. In particular, if it's codim 1, iff its normal bundle is trivial.) –  Aug 23 '15 at 15:15

2 Answers2

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No conditions needed.

Proof. Suppose that that $M$ is orientable, hence there is non-vanishing $n$-form $\omega$ defined on $M.$ As a result $\omega|_S$ is non-vanishing $n$-form on $S.$ Thus $S$ is orientable. Contradition. So $M$ must be non-orientable.

Other approach.

Here and here you have alternative approach to non-orientability. Namely.

Theorem $\star$. Let $N$ be a smooth manifold. $N$ is non-orientable, if and only if there are two charts $(U_a,\phi_a),(U_b,\phi_b),$ such that $U_a,U_b$ are connected, $U_a\cap U_b\neq\emptyset$ and transformation function $\phi_{ab}$ neither preserves nor reverses the orientation.

We just need to use this theorem $\star$ twice.

Proof. Since $S$ is non-orientable, we get by $\star$ that there are two charts $(U_a,\phi_a),(U_b,\phi_b)$ of $S$ such as in $\star$. $S$ is open, hence $(U_a,\phi_a),(U_b,\phi_b)$ are charts of $M$ as well. Again by $\star$ we get that $M$ is non-orientable.

Remark. This equivalent condition is very convinient in proving that something is non-orientable. Generally you would have to prove that something (non-vanishing form, oriented atlas,...) doesn't exist. Here you just need to indicate two charts with some properties.

Community
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Fallen Apart
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Another approach: you can read non-orientability on loops, via the first Stiefel-Whitney class: $M$ is orientable if and only if $w_1(TM)=0$.

If $$\iota : S\hookrightarrow M$$ is your inclusion map, then $$w_1(TS)=\iota^\star(w_1(TM))$$ so that if $M$ is orientable, then so is $S$.

Concretely, a manifold is non-orientable if and only if there is an embedded loop along which the orientation is reversed $(\star)$. The first Stiefel-Whitney class takes the non-trivial value $1\in \Bbb Z/2\Bbb Z$ on such a loop.

$(\star)$ equivalently, a loop that cannot be lifted into a loop in the $2$-fold orientation covering.

Arnaud Mortier
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