If $X$ is a projective variety then Hilbert's syzygy theorem says that any coherent sheaf of $\mathcal O_X$ modules has a finite global resolution by locally free modules.

By GAGA, I believe this should imply that over any smooth complex submanifold of $\mathbb CP^n$, a coherent sheaf of modules over the sheaf of holomorphic functions has a finite global resolution by holomorphic vector bundles.

Assuming what I said is correct, is there a direct proof of this in the analytic setting? Also, is the analytic statement true for more general complex manifolds (e.g. Kahler)?

Note: this is a crosspost from math.SE.

  • $\begingroup$ You want to add the word "smooth" in the first sentence. $\endgroup$ – Steven Landsburg Jan 9 '14 at 19:58
  • $\begingroup$ Dear Eric, I added a true link towards your question on Math.SE, I hope you do not mind. $\endgroup$ – agtortorella Jan 9 '14 at 20:01
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    $\begingroup$ The analytic input into the proof of GAGA provides everything which is required, so GAGA isn't needed and the algebraic proof (appropriately formulated) literally works in the analytic setting. Namely, by the analytic ampleness properties of $O(1)$, any coherent analytic sheaf on a closed complex-analytic subspace $X$ of $\mathbf{CP}^n$ has a resolution by vector bundles. So if $X$ is smooth then by Serre's theorem on finiteness of global dimension of regular local rings (applied to stalks of $O_X$) this resolution has vector bundle kernel at the $(\dim X)$-th step. $\endgroup$ – user76758 Jan 9 '14 at 20:22
  • $\begingroup$ @user76758 thanks. maybe you should make your comment an answer. also, do you have good references for these statements? thanks. $\endgroup$ – Eric O. Korman Jan 12 '14 at 4:44
  • $\begingroup$ @Eric: Since I assume you have read the proof of GAGA, and hence must know the analytic properties of $O(1)$ and the basic theory of coherent-analytic sheaves (without which that proof doesn't work), you just use exactly the same proof as in the algebraic case. There is probably no reference because it is literally the same argument. Just talk with anyone in algebraic geometry or number theory on the faculty at your university if you have further questions about it. $\endgroup$ – user76758 Jan 12 '14 at 17:08

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