# Holomorphic vector bundles over a Riemann surface does not satisfy $\mathbf{AB2}$ but satisfies $\mathbf{AB1}$

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?).

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

(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.

• 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? – user40276 May 30 '14 at 12:38
• 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. – dhy May 30 '14 at 15:28

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).

• Thanks for the answer, but I think that Grothendieck was wrong in the assertion that $\mathbf{AB1}$ holds, because the fibers of a kernel don´t have constant rank in general. – user40276 May 29 '14 at 23:22
• @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. – S. Carnahan May 30 '14 at 0:46
• 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. – Eric Wofsey May 30 '14 at 7:58
• 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. – Eric Wofsey May 30 '14 at 9:19
• 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. – Eric Wofsey May 30 '14 at 13:08