Why is Klein bottle non-orientable?

Question

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.

Answer 1

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 $\tau$.

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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.

Answer 2

Here are three ways to look at it.

The Klein bottle is the quotient space of the torus by the action of an orientation reversing involution. So it is not orientable.

• One can choose this involution to be rotation by 180 degrees along one generating circle followed by reflection along the second.

The top Z homology of the Klein bottle is zero so it is not orientable.

• One can define orientability for a triangulated manifold to be the existence of a top dimensional Z-cycle. If you triangulate a Klein bottle you will find that there is no two dimensional Z-cycle.

The Klein bottle is covered by the Euclidean plane. The group of covering transformations can be taken to be the standard lattice (points with integer coordinates) of pure translations together with the affine transformation,

T: (x,y) -> (x + 1/2, -y).

T is orientation reversing since its matrix part has determinant -1.

• One checks that this means that the coordinate transition functions are not all orientation preserving.