3
$\begingroup$

in Grothendieck's Tohoku paper (page 127), Grothendieck asserts that the category of holomorphic vector bundles $\mathbf{Bund}(X)$ over a fixed Riemann surface $X$ does not satisfy $\mathbf{AB2}$, but satisfies $\mathbf{AB1}$. Explicitly, he claims that $\mathbf{Bund}(X)$ has kernels and cokernels, however the canonical morphism $\overline{u}: \text{CoIm}(u) \longrightarrow \text{Im}(u)$ is not an isomorphism. At first I thought it was easy prove, but now I'm not so sure.

I think it's useful to note that the category of vector bundles is not an abelian category in general (Is the category of vector bundles over a topological space abelian?).

Thanks in advance.

Improve this question
$\endgroup$
7
  • $\begingroup$ So, what's the question? $\endgroup$
    – Sasha
    Commented May 29, 2014 at 13:35
  • $\begingroup$ @Sasha Is this true? Why? $\endgroup$
    – user40276
    Commented May 29, 2014 at 13:35
  • $\begingroup$ Grothendieck used the category of vector bundles, not line bundles. $\endgroup$
    – S. Carnahan ♦
    Commented May 29, 2014 at 13:46
  • $\begingroup$ @S.Carnahan You're right, I've edited. $\endgroup$
    – user40276
    Commented May 29, 2014 at 14:00
  • $\begingroup$ Yes, the category of vector bundles is not abelian. $\endgroup$
    – Sasha
    Commented May 29, 2014 at 14:36

2 Answers 2

Reset to default
5
$\begingroup$

(This should really be a comment on S. Carnahan's answer, but don't have enough reputation.)

All coherent sheaves on a Riemann surface split into the direct sum of a torsion sheaf and a vector bundle. This means we get a pair of adjoint functors, the forgetful functor $f:Bund(X)\rightarrow Coh(X)$ and the "kill-torsion" functor $g:Coh(X)\rightarrow Bund(X)$.

The kernel of a map of bundles $A\rightarrow B$ will be a bundle, and so poses no issue.

On the other hand, the cokernel will be $g(\operatorname{cok}(A\rightarrow B)),$ where by $\operatorname{cok}$ I mean cokernel in $Coh(X).$ The fact that this is the cokernel is essentially just the adjunction statement above.

Note that this argument does not work for $X$ of higher dimension, as then we no longer get this splitting.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks, this is indeed an answer (not just a comment). Do you know where can I find about this splitting. Does this (the splitting) hold for any algebraic curve over an algebraically closed field? Any references? $\endgroup$
    – user40276
    Commented May 30, 2014 at 12:38
  • 3
    $\begingroup$ The splitting basically comes from the theory of Dedekind domains: the function field of an affine curve over an algebraically closed field is a Dedekind domain, and you have such a splitting for any Dedekind domain. The case of a projective curve reduces to the case of an affine curve because the problem is local. $\endgroup$
    – dhy
    Commented May 30, 2014 at 15:28
5
$\begingroup$

Take a nonzero map $\mathcal{O} \to \mathcal{O}(1)$ on $\mathbb{P}^1$. In the category of holomorphic vector bundles, the kernel is the inclusion of the zero vector bundle, and the cokernel is the map to the zero vector bundle. The coimage is then $\mathcal{O}$, while the image is $\mathcal{O}(1)$, and the canonical morphism is the same as before (i.e., not an isomorphism).

Improve this answer
$\endgroup$
9
  • 2
    $\begingroup$ @user40276 My previous comment was flawed. Kernels aren't a problem at all. They are subsheaves of locally free sheaves, hence locally free. For cokernels, you should check by yourself that the canonical torsion-free quotient of the sheaf cokernel is a vector bundle cokernel. $\endgroup$
    – S. Carnahan ♦
    Commented May 30, 2014 at 0:46
  • 1
    $\begingroup$ I think your confusion stems from the fact that the kernel of a map $f:A\to B$ of vector bundles is not computed fiberwise. The dimension of the fiber can jump up at a point $p$, but there is no map of vector bundles $g:K\to A$ such that $fg=0$ and the fiberwise dimension of the image of $g$ jumps up at the point $p$ (the dimension of the image can only jump down). Thus it turns out that the kernel in the category of bundles is just a vector bundle whose rank is the dimension of the generic fiber. $\endgroup$
    – Eric Wofsey
    Commented May 30, 2014 at 7:58
  • 4
    $\begingroup$ From the perspective of coherent sheaves and commutative algebra, the fact that kernels are not computed fiberwise is unsurprising: to take the fiber at a point, you tensor with the residue field at that point. Residue fields are not flat, so this tensoring is not left exact and does not preserve kernels. $\endgroup$
    – Eric Wofsey
    Commented May 30, 2014 at 9:19
  • 1
    $\begingroup$ @EricWofsey Thanks for the clarifications, but, for instance, pick the trivial bundle endomorphism of $\mathbb{R} \times \mathbb{R} $, given by $f(x, u) = (x, xu)$, then the fiber collapses at $0$, so the kernel cannot exist, is there something that I´m missing? $\endgroup$
    – user40276
    Commented May 30, 2014 at 12:51
  • 1
    $\begingroup$ Look to the definition of kernel. The kernel of a map $f:A\to B$ is the terminal object in the category of all maps $g:K\to A$ such that $fg=0$. In your example, the kernel is actually 0, because any map $g$ such that $fg=0$ has to vanish at all points except $0$, and therefore also at $0$ by continuity. The definition doesn't care about what happens on fibers. $\endgroup$
    – Eric Wofsey
    Commented May 30, 2014 at 13:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .