Timeline for Holomorphic vector bundles over a Riemann surface does not satisfy $\mathbf{AB2}$ but satisfies $\mathbf{AB1}$
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10 events
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| May 30, 2014 at 13:08 | comment | added | Eric Wofsey | 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. | |
| May 30, 2014 at 12:51 | comment | added | user40276 | @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? | |
| May 30, 2014 at 9:19 | comment | added | Eric Wofsey | 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. | |
| May 30, 2014 at 7:58 | comment | added | Eric Wofsey | 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. | |
| May 30, 2014 at 3:49 | comment | added | user40276 | Kernels are the problem too. The subsheaves are indeed locally free, but the rank is not constant. Think as a vector bundle (not as a sheaf of modules), then kernel will have each fiber equals to the kernel of the linear map induced on the fibers, but the rank of the fibers of the kernel bundle (as vector spaces) will not necessarily be preserved even in a local trivialization. | |
| May 30, 2014 at 0:46 | comment | added | S. Carnahan♦ | @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. | |
| May 29, 2014 at 23:52 | comment | added | user40276 | So, are tou claiming that $\mathbf{AB1}$ is true? I don´t think so. | |
| May 29, 2014 at 23:48 | comment | added | S. Carnahan♦ | @user40276 It seems that you are taking a kernel in the category of coherent sheaves instead of the category of vector bundles - the question is whether the equalizer of any map with the zero map is representable by a vector bundle. The fact that we are working with a Riemann surface is important. | |
| May 29, 2014 at 23:22 | comment | added | user40276 | 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. | |
| May 29, 2014 at 14:50 | history | answered | S. Carnahan♦ | CC BY-SA 3.0 |