18
$\begingroup$

Is there an analogy between fibers $ \pi^{-1} ( x ) $ of a vector bundle $ \pi : E \to X $, and the stalk $ \mathcal{F}_x $ of a sheaf $ \mathcal{F} $ défined by : $ \mathcal{F}_x = \displaystyle \lim \mathcal{F} ( U ) $ : the direct limit over all open subsets of $ X $ containing the given point $ x $ ? Thanks a lot.

$\endgroup$
4
  • 2
    $\begingroup$ Please consider accepting some answers to your previous questions. People will be less willing to respond if they think you won't appreciate their answers anyway. $\endgroup$ Dec 19, 2012 at 12:58
  • 1
    $\begingroup$ I don't know how to do it, can you tell me how to do it please ?. Il don't speak and i don't undertand well english, i'm a moroccan men, sorry. $\endgroup$
    – Bryan
    Dec 19, 2012 at 13:04
  • 2
    $\begingroup$ No need to be sorry. Next to each answer, under the arrows for up/downvoting, there is a little check mark. You can just click on the check mark belonging to the answer you like best to accept it. See here for some images explaining the process. $\endgroup$ Dec 19, 2012 at 13:08
  • 1
    $\begingroup$ Ok, i did it. Thank you. :) $\endgroup$
    – Bryan
    Dec 19, 2012 at 13:23

1 Answer 1

17
$\begingroup$

The stalk of a sheaf $\mathscr{F}$ at a point $x$ is naturally isomorphic the fibre of the espace étalé over $x$. However, the espace étalé is in general a very strange space and is very rarely a vector bundle.

For example, let $\mathscr{T}_M$ be the sheaf of sections of the tangent bundle $T M \to M$ of a manifold (or smooth variety, if you prefer). The fibre of $T M \to M$ over a point $x$ of $M$ is automatically a vector space (by definition!) but the stalk of $\mathscr{T}_M$ at $x$ is in general only a module over the local ring $\mathscr{O}_x$, which need not be a field. More explicitly, the stalk $(\mathscr{T}_M)_x$ consists of germs of vector fields at $x$, while the fibre $T_x M$ consists of tangent vectors at $x$.

Addendum. In the algebraic context, we can get something akin to the fibre by taking the stalk $\mathscr{F}_x$ and tensoring it with the residue field $\kappa (x)$. As far as I know this is not something that can be defined as a direct limit. The fibre of a bundle is a limit in the general sense of category theory – it is just the pullback of the bundle along the inclusion of a point – but I don't think it's useful to think of it that way.

$\endgroup$
3
  • $\begingroup$ Thank you very much @Zhen Lin. Can you tell me please, if it's true that the fiber $ E_x = \pi^{-1} ( x ) $ of a vector bundle $ \pi : E \to X $ can be defined as the direct limit of maps $ U \to E_{U} = \pi^{-1} ( U ) $ like the stalk of a sheaf $ \mathcal{F}_x $ ? In this case, what is the direct system which define this direct limit ? Thanks a lot. $\endgroup$
    – Bryan
    Dec 19, 2012 at 13:16
  • $\begingroup$ Thanks ! Can you tell me how the fibre of a bundle is a limit in the general sense of category theory ? Idon"t understand this sentence : it is just the pullback of the bundle along the inclusion of a point. Thanks a lot. $\endgroup$
    – Bryan
    Dec 19, 2012 at 18:22
  • 1
    $\begingroup$ That is precisely the sense I mean. A pullback is a limit. $\endgroup$
    – Zhen Lin
    Dec 19, 2012 at 20:01

You must log in to answer this question.

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