I am doing the homework of differential geometry and encounter this problem:
The Klein bottle $K^2$ is defined to be the identification space
$$[0, 1] \times [0, 1]/{\sim}, \text{ where the identification is }
(x, 0)\sim (1-x, 1) \text{ and } (0, y) \sim (1, y).$$i) Prove that $K^2$ is non-orientable.
ii) For a submanifold $M$ on $K^2$ defined by $1/4\leq x\leq3/4, 0\leq y\leq1,$ is $M$ orientatble?
I already know that I need to check the orientation of the tangent space along some curve. But I am not quite sure how I can write down the proof in a formal way.
Could anyone help? Any comment? Thanks!
Edit: I have done the first part. Can anyone help me with the second? I have a feeling that the submanifold $M$ is isomorphic to the Möbius strip. But I don't have good reasons.
Edit: Yes, $M$ is indeed the Möbius strip.
Best Answer
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.
$$ $$
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.