Serre's theorem (one of them) states that for a quasi-coherent sheaf $\mathscr F$ on an affine noetherian scheme $H^i(X,\mathscr{F})$ vanish for $i >0$. I used to think that this would imply that on an affine variety there cannot be non-trivial vector bundles, because such a bundle would define a non-trivial cocycle in $\check{H}^1(X, GL_n)$, and this is isomorphic to the derived functor cohomology $H^1(X, GL_n)$ for "good enough" schemes, and the latter vanishes by Serre's theorem.

But here they give examples of non-trivial line bundles on affine varieties.

Where is my reasoning fallacious? Is it because the trivialising opens for the vector bundle can be non-affine?

Does it mean that in general there is no simple way to classify algebraic vector bundles even on an affine variety?


1 Answer 1


$GL_n$ is not a coherent sheaf, because it is not a sheaf of $\mathcal{O}_X$-modules.

  • $\begingroup$ What if I take the sheaf $F$ given by $F(U)=Hom(U,GL_n)$? Oh, I see, since $GL_n$ is not Abelian for $n > 1$, it is not a module. I see that what I've written barely makes sense. Does it make sense at least for n=1, when $GL_1 = \mathcal{O}_X^*$? $\endgroup$ Apr 21, 2011 at 19:00
  • $\begingroup$ For $GL_1$, you have a sheaf of abelian groups, but not an $\mathcal{O}_X$-module. How do you take a section of $GL_1$ and multiply it by a section of $\mathcal{O}_X$ to get another section of $GL_1$? $\endgroup$ Apr 21, 2011 at 20:13
  • $\begingroup$ oh, I am sorry, you are right, of course. I am being dense. $\endgroup$ Apr 21, 2011 at 20:21
  • $\begingroup$ Although it goes a bit out of the scope of the question, I hope you don't mind me asking if there is some characterisation of affine schemes that can only have trivial vector bundles over them? $\endgroup$ Apr 21, 2011 at 20:30
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    $\begingroup$ I don't know, but I imagine this should be really hard. Remember that the Quillen-Suslin theorem -- every algebraic vector bundle on $\mathbb{C}^n$ is trivial -- took 20 years to prove. $\endgroup$ Apr 23, 2011 at 4:32

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