Twisted Sheaf of Serre identified with Moebius strip Let consider the twisted sheaf $\mathcal{O}_{\mathbb{P}^1 _k}(1)$ over projective line $\mathbb{P}^1 _k$. Can anybody explain to me where the visualisation "twisted" come from. I read that $\mathcal{O}_{\mathbb{P}^1 _k}(1)$ as line bundle can be associated with Moebius strip in some sence which isn't clear to me. So intuitively I suppose the "twist" comes from the geometrical analogy that Moebius strip is a "twisted" cylinder. Presumably.
But I have no idea where the identification of $\mathcal{O}_{\mathbb{P}^1 _k}(1)$ with a line bundle that is isomorphically to Moebius strip comes from. Is there a way to imagine it preferably geometrically?
 A: The key is to look the real points of the bundle $\mathcal O(-1)$. It is a real line bundle over $\Bbb RP^1 \cong S^1$, so it is the cylinder or the Moebius band. Over $\theta \in S^1$ the fiber is $L_{\theta} = \{(x,y) \in \Bbb R^2 : \angle(x,y) = \theta\}$. Under this identification, you can check that $L_{\theta}$ do half a twist between $0$ and $\pi$ and another half-twist between $\pi$ and $2 \pi$ so we get eventually a Moebius band. 
More generally, $\mathcal O(k)$ is the Moebius band if $k$ is odd or the cylinder if $k$ is even. This is related to the fact that real line bundles over $S^1$ are classified by $\Bbb Z/2\Bbb Z$. On the other hand, for each $k$ the bundles $\mathcal O(k)$ are different if you are looking over the complex numbers, because they are classified by their Chern class $c_1(\mathcal O(k)) = k \in \Bbb Z$.
The same happens with $\Sigma_k := P(\mathcal O \oplus \mathcal O(k))$, the famous Hirzebruch surfaces ($k \geq 0$). Over $\Bbb C$ they are all different, and over $\Bbb R$ they are diffeomorphic to the torus $S^1 \times S^1$ if $k$ is even or to the Klein bottle if $k$ is odd. 
For me, the term "twisted" comes from the Veronese embedding, $\mathcal O(k)$ induces a map $\Bbb P^1 \to \Bbb P^k$ and the image is more and more "twisted" by the power of $z$.  
