Let $S$ be a graded ring which is finitely generated by $S_1$ as an $S_0$-algebra. Let $X = \text{Proj}(S)$. Let $E$ be a vector bundle over $S$. Is $\oplus_{n \in \mathbb{Z}} H^0(X,E(n))$ a graded projective $S$-module?

The answer should be no. E.g. there exists indecomposable vector bundles of rank $> 1$ over complex projective space of dimension $> 1$. If the graded module (associated to such bundles) were projective then the module must be free (as graded projective modules over polynomials rings over a field are free) and hence the bundle must be a direct sum of line bundles.

My question is this: can someone give more insight into why $\oplus_{n \in \mathbb{Z}} H^0(X,E(n))$ is not projective module when $X$ is a projective variety.

  • $\begingroup$ I'm confused. Take $S=k[x,y]$ and $E=\mathcal{O}_S$, then your is just $S$? $\endgroup$ – Alex Youcis Mar 4 '14 at 8:31
  • $\begingroup$ @AlexYoucis : Yes, in that case our graded module is just $S$. What is the doubt? $\endgroup$ – user132726 Mar 4 '14 at 16:19

Any graded projective module (as in your case, Noetherian etc.) over a graded ring is free and thus your intuition is correct. If $E$ is not a direct sum of line bundles, the corresponding graded module will not be projective. (Even if the vector bundle is direct sum of line bundles, it may not be, unless it is the direct sum of $\mathcal{O}(n)$'s). For a simple example, take the tangent bundle of $\mathbb{P}^2$. Then $S=k[x,y,z]$ and you can easily check that $\oplus H^0(T_{\mathbb{P}^2}(n))$ is just $Se_1\oplus Se_2\oplus Se_3$ modulo $S(xe_1+ye_2+ze_3)$, which is non-projective.

  • $\begingroup$ Thank you for a specific example. $\endgroup$ – User3568 Oct 20 '15 at 15:53

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