Is there a reasonable and coherent explanation of why the Serre Twisting Sheaf has the word "twisting" in its name?
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3I've heard people say that $\mathcal{O}(1)$ corresponds to a vector bundle that "is" the Möbius band, which has a twist. Here's a link to start and here's another relevant post. – Viktor Vaughn Apr 05 '17 at 19:42
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Yeah I found that link earlier, too. – Tanner Strunk Apr 05 '17 at 23:37
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1@Rüdiger: you should delete your comment. – bertram Apr 06 '17 at 13:36
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1Here is basically a duplicate. (I, too, do not see how what university you attend has any relevance to this question.) – Viktor Vaughn Apr 07 '17 at 03:03
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Oh ty! (I'll try and figure out how to mark this question as a duplicate.) – Tanner Strunk Apr 07 '17 at 16:03
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The links in the comments are already answering the question, so let me add some remarks.
The $z \mapsto z^n$ can be considered as a "twist" as you are making $n$ turns of the circle ( $ n \in \Bbb Z$). On $\Bbb P^1 \cong S^2$, the clutching functions of $\mathcal O(1)$ are $z \mapsto z$. This implies that the clutching functions on $\mathcal O(n)$ are $z^n$ if $n$ is positive, and if $n$ is negative they are $z^n = (\overline{z})^{-n}$. This is the "twisting". If $\mathscr F$ is a sheaf over $\Bbb P^n$, you can "twist" this sheaf by taking $\mathscr F(k) := \mathscr F \otimes \mathcal O(k)$. You can do the same with any projective variety. This is very useful as the Serre vanishing theorem say that if $\mathscr F$ is a coherent sheaf then $\mathscr F(k)$ has no cohomology in degree $i > 0$ for $k \gg 0$, and having no cohomology is pretty good for computations !
A good way of understanding this is noticing that $\mathcal O(1)(\Bbb R) \cong M$ where $M$ is the Moebius band. Similarly, a good example of such "twisting" is given by the Hirzebruch surface $\Sigma_n$, given by $ \Sigma_n = \Bbb P_{\Bbb P^1}(\mathcal O(n) \oplus \mathcal O), n \geq 0$. As complex variety they all differents, and as real varieties, $\Sigma_n(\mathbb R) \cong T^2$ if $n$ is even (e.g $n = 0$) and the Klein bottle if $n$ is odd (e.g if $n=1$ you get the blow-up of $\Bbb RP^2$ which is diffeomorphic to the Klein bottle). Again, you can see the twist geometrically since $\Sigma_n$ can be seen as the blow-up of the cone over the Veronese curve, given by $f : \Bbb P(1,1,n) \mapsto \Bbb P^{n + 1}(x,y,z) \mapsto (x^n, x^{n-1}y, \dots, y^n, z)$.
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1Why is it obviously that $\mathcal O(1)(\Bbb R) \cong M$ where $M$ Moebius strip holds? – user267839 Nov 03 '17 at 14:59