Is it true that every $(n-1)$-dimensional submanifold of a compact (compact without boundary) $n$-dimensional manifold is orientable ?
I think, that the answer is no, but I can’t provide it with an example. I thought about Möbius strip and Klein bottle. But the dimensions are not corresponding (the difference is 2 , not 1)
Is an $(n-1)$-dimensional submanifold of an $n$-dimensional manifold orientable
differential-geometrysubmanifold
Related Solutions
Here is a possible proof (credit to user pappus here):
Recall that a manifold is orientable iff there exists a volume form on it. So in order to show that $K^2$ is non-orientable, it is enough to show that any $2$-form $\omega$ on $K^2$ must vanish at some point.
Let $K^2$ be defined here as the quotient of $\mathbb{R}^2$ by the subgroup of diffeomorphisms of $\mathbb{R}^2$ generated by $\tau : (x,y) \mapsto (x+1, y)$ and $\sigma : (x,y) \mapsto (1-x, y+1)$1.
So let $\omega$ be a $2$-form on $K^2$and $\tilde{\omega}$ be the pull-back to $\mathbb{R}^2$ (by the canonical projection $\mathbb{R}^2 \to K^2$). There exists a smooth function $f :\mathbb{R}^2 \to \mathbb{R}$ such that $\tilde{\omega} = f(x,y) \,\omega_0$, where $\omega_0 = \det = dx \wedge dy$2. Then $\tilde{\omega}$ must be invariant under $\sigma$ (and $\tau$): $\sigma^* \tilde{\omega} = \tilde{\omega}$. You can check that $\sigma^* \omega_0 = -\omega_0$, so $f$ must satisfy $\sigma^* f = -f$, i.e. $f(1-x, y+1) = -f(x,y)$. It follows that $f$ must vanish (by continuity of $f$ and connectedness of $\mathbb{R}^2$), hence $\tilde{\omega}$ must vanish, hence $\omega$ must vanish.
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Edit: for the second part, you can produce the same proof. Replace $\mathbb{R^2}$ with the band $\{1/4 \leqslant x \leqslant 3/4\}$ and forget about $\tau$.
1 This is clearly equivalent to your definition, and is a good way to define $K^2$ together with its smooth structure.
2 This is because the top exterior power of a finite-dimensional vector space is one-dimensional, which implies that any $n$-form on a orientable $n$-manifold is proportional to a given volume form.
I am a little late to the game here, but I wanted to share some pictures, which might help intuition. Paul Frost answers your question, but I thought of this a different way. Take the standard $x,y$-plane in 3-space. This is an unbounded surface. If I were to add a handle to it, it is still unbounded. The only question is whether this handle makes the surface non-orientable or not. In the image below, I have both ways I can attach the handle. These surfaces are topologically a punctured torus and a punctured Klein bottle, but drawn this way, I think it sheds light on what you were thinking about. The orange circle in the bottom is the circle of immersion necessary to put the Klein in 3-space. And while there is obviously nontrivial loops, these are unbounded. Last is an example of a non-orientable surface with boundary, but it is embedded.
Best Answer
If you take $M$ a non-orientable compact $(n-1)$-manifold, then it is a submanifold of $M\times S^1$ which is compact.
For example you can take $M$ to be the Klein bottle.