How to see line bundle on $\mathbb P^1$ intuitively?

Question

We know the tautological line bundle $\mathcal O(-1)$ on $\mathbb P^1$ can be regarded as the Mobius strip (consider the total space, with base field $\mathbb R$). So, is there similar description for any $\mathcal O(k)$ (its total space is a surface, so we should be able to see it somehow)?

Answer 1

If you look at real points of the bundle, then topologically you have either the Moebius band or the cylinder (it is a $\Bbb R$ bundle over $S^1$). You can show it by computing the Whitney-Steifel class of the bundle, but in a more down-to-earth way, express any bundle $\mathcal O(k)$ with $k \leq 0$ as a tensor product of $\mathcal O(-1)$. You should be able to convince yourself that tensoring $k$ times $\mathcal O(-1)$ gives a bundle with $k$ twist. For $k \geq 0$, a Riemannian metric will give you an isomorphism of manifolds $\mathcal O(k) \cong \mathcal O(-k)$.

On a complex viewpoint you won't be able to say anything, because taking the famous Hirzebruch surface $\Sigma_n := \Bbb P_{\mathbb P^1}(\mathcal O \oplus \mathcal O(n))$ gives non-isomorphic complex surfaces for $n \geq 0$. Also, the real points again give either the Klein bottle or the torus.