I am reading Hartshorne, and I don't understand something about twisting sheaves. He defines, for $S$ a graded ring, the $n$-th twisting sheaf $O_X(n)$ as $S(n)\tilde{}$. I tried to interpret this in two ways, both having problems:

  1. $S(n)$ is the set of elements having degree $n$. This would be fine for a lot of things I read after, but it is not an $S$ module with multiplication action. Maybe $S_d$ acts as zero multiplication for $d\geq 0$?

  2. $S(n)$ is the set of elements of degree at least $n$. Now this is a module, but then there is a proposition which seems false to me. Let $S=A[x_0,\ldots, x_r]$ for a ring $A$, and let $X=Proj S$. Then $S \simeq \Gamma_* ( O_X) := \bigoplus_n \Gamma(X, O_X(n))$. This seems false because I have too much copies of an element, say, in degree 2.

Thank you and sorry for the confusion!


1 Answer 1


$S(n)$ denotes a shift in grading: $S(n)_m = S_{n+m}$.

Also, a thing I think you missed is that the sections of $\widetilde{M}$ are locally degree zero elements of $M$, not all elements of $M$.

In your example of a polynomial ring, it turns out that this is true globally as well: $\Gamma(X, \mathscr{O}_X(n)) \cong S_n$.

I don't recall in what generality $\Gamma(X, \widetilde{M}(n)) \cong M_n$ (in particular, I don't recall whether its always true even in the case that $M = S$), but I'm pretty sure Hartshorne discusses it in the general vicinity of where he defines the notation $\Gamma_*(X, \mathscr{F})$.

  • $\begingroup$ What about S being isomorphic to $\oplus_n S(n)$? Seems false, isn't it? $\endgroup$
    – frame95
    Commented Apr 24, 2018 at 16:21
  • $\begingroup$ @frame95: I've edited. $\endgroup$
    – user14972
    Commented Apr 24, 2018 at 17:13
  • $\begingroup$ Aaaaaah. So M tilde is defined differently for graded and non-graded modules? I mean, in the non-graded case you have that M tilde is M_f on the open D_f, right? Thank you for the aid! $\endgroup$
    – frame95
    Commented Apr 26, 2018 at 14:32

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