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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?

user267839
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1 Answers1

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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 $z$.

Nicolas Hemelsoet
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  • So in more elementar geometric language every homogenious $(x : y) \in \Bbb RP^1$ determinates an angle $\theta$. Can I interpret this as the angle which determines the direction of the basis vector $e_{(x : y) }$ of the one dimensional vector space $V_{(x : y)}$ assigned to $(x : y)$ (because of line bundle structure)? Futhermore $e_{(x : y) }$ stays parallel to the Möbius strip along the changing values of $(x : y)$? So the direction/rotation trajectory of $e_{(x : y) }$ while running trought the $(x : y)$s determines that it's a Möbius strip? Or is that a wrong imagination? – user267839 Nov 03 '17 at 14:30
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    Yes exactly. I used that the fiber of $\mathcal O(-1)$ over $(x:y) \in \Bbb RP^1$ is canonically identified with the line $\Bbb R(x,y)$ in $\Bbb R^2$. I don't really understand what does "$e_{x:y}$ stays parallel to the Moebius strip" means, the point is that $\mathcal O(-1)$ precisely form a Moebius strip, exactly as your second question said. You can embedd $\mathcal O(-1)$ into $S^1 \times \Bbb R^2$, and if you delete a point you can even embedd it into $(0,1) \times R^2$, and from this it is easy to see on the picture that you got indeed the Moebius band. – Nicolas Hemelsoet Nov 03 '17 at 15:30
  • By "$e_{(x:y)}$ stays parallel to the Möbius strip along the changing values of $(x:y)$" I meant that the basisvector $e_{(x:y)}$ corresponding to $(x:y)$ stays always parallel to the red line in this picture: http://mathforum.org/mathimages/index.php/Image:Mobius-axes.gif – user267839 Nov 03 '17 at 15:53
  • ...or is $e_{(x:y)}$ the normal vector of the surface? – user267839 Nov 03 '17 at 15:56
  • @KarlPeter : In your picture $e_{(x:y)}$ will be parallel to the red line as you said. (Sorry not to reply before). – Nicolas Hemelsoet Nov 03 '17 at 17:31