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Stalks and Direct Limits

  1. Jun 3, 2004 #1

    Hurkyl

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    (Hrm. There doesn't seem to be a good category for Algebraic Geometry; maybe Linear/Abstract Algebra?)

    Ok, I'm in the process of teaching myself Algebraic Geometry, and I'm reading ahead a bit to see where things are going. :biggrin:

    In particular, just now I'm reading about presheaves on a topological space, and have come across the definition of a stalk. As quoted from Hartshorne:

    Definition. If [itex]\mathcal{F}[/itex] is a presheaf on X, and if P is a point of X, we define the stalk [itex]\mathcal{F}_p[/itex] of [itex]\mathcal{F}[/itex] at P to be the direct limit of the groups [itex]\mathcal{F}[/itex](U) for all open sets U containing P, via the restriction maps [itex]\rho[/itex].

    (Here, Hartshorne is speaking in particular about presheaves of abelian groups)

    I realize that to understand the most general case of this definition that I really will have to understand direct limits, but I imagine that special cases (such as abelian groups) would be much easier to understand.


    So, I guess at what it should be. :smile: Mainly I'm hoping if someone can tell me if this is an equivalent definition:


    Without losing generality, I can start with some open set U of X containing P.
    The restriction maps [itex]\rho_{UV}[/itex] yield isomorphisms of the groups [itex]\mathcal{F}(U) / \ker \rho_{UV} \leftrightarrow \mathcal{F}(V)[/itex].

    If W and V are open sets such that [itex]P \in W \subseteq V \subseteq U[/itex] we have [itex]\ker \rho_{UV} \subseteq \ker \rho_{UW}[/itex], so the natural choice of such a limit would be to take the ideal I generated by all of the [itex]\ker \rho_{UV}[/itex] with [itex]P \in V \subseteq U[/itex] (V open), and then set
    [itex]\mathcal{F}_P \cong \mathcal{F}(U) / I[/itex]

    Is this reasonable?
     
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  3. Jun 4, 2004 #2

    matt grime

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    You aren't taking a direct limit over U though - you need to let U 'get small', and there is no reason to assume that the sets containing P are nested like that - there will be some cardinality restriction on the underlying set that would imply you can take a direct limit over a filter of complexity 1 (not standard notatation), but generally you need to take them over more complicated filters).

    Given a filter, that is a collection of open sets containing P, the direct limit (colimit) is the universal object in the category of abelian groups satisfying some commutativity relations. In AbGrp this always exists and can be defined using generators and relations.


    Here's an example:
    Z --> Z---> Z--> Z ....

    where the maps are multiplication by p a prime, then the resulting group is Z[[p]] the p-local integers got be adjoining a formal inverse to p. note i may have buggered this one up in translation

    In an abelian category, say a module category, with arbitrary direct sums then the limit of:

    M[1]--->M[2]--->M[3]....

    is the cokernel

    sum M --> sum M ---> limM

    where the injection is given by 1-shift, shift being the maps in the original sequence.


    You;re getting at the right idea.

    The colimit in general is the direct sum of all the groups (which is big, yes) modulo the relation that if two elements become identified on restriction to any other group, then they become indentified in the limit.

    I think that would be the same as yours provided the groups stabliized, but I generally don't think about algebraic geometry cos it makes me feel slightly nauseous.

    I think you'd need to look up the Mittag-Leffler Condition.
     
  4. Jun 4, 2004 #3

    Hurkyl

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    I was hoping to avoid the Category Theoretic approach, becuase I'm quite unfamiliar with it, and I'm already set upon learning one fairly involved subject!

    It dawned on me last night that my construction assumes that all of the restriction maps are onto maps, but the definition of presheaf only requires them to be group morphisms, so I think it works, but only in this nice case.


    Right, the open sets containing P don't have to be nested, but for any open set V containing P, the set [itex]U \cap V[/itex] would be also, so I had hoped that whatever information that is in V but not U is irrelevant. In fact, I'm pretty sure my construction winds up being independant of the choice of U, because if [itex]W \subseteq U[/itex], then things will eventually factor through [itex]\mathcal{F}(W) \cong \mathcal{F}(U) / \ker \rho_{UW}[/itex]


    For the record, the category theoretical definition of the things involved is that, given your topological space X, you form the category [itex]\mathcal{IOP}(X)[/itex] to be the category whose objects are open sets, and the morphisms are inclusion maps.

    Then a presheaf is defined to be a contravariant functor from [itex]\mathcal{IOP}(X)[/itex] to the category of abelian groups (or whatever category you see fit to use).

    Doesn't help me much, but it might help you. :smile:


    I'm finally resigned to the fact I'm gonna have to learn it the right way. Your statement of the colimit seems a lot clearer than the others I've read, so maybe it won't be so bad after all.
     
  5. Jun 4, 2004 #4

    matt grime

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    As is often the case with the high powered general category theoretic approach the definitions often disguise some of the easier realizations. For instance a simple use of colimits are cokernels and pushouts when you're just taking a finite diagram.

    The presheaf functor is disguising a lot of what can be clearly viewed in terms of diagrams.

    for instance the example i gave of modules is properly described as a functor from the well-ordered category of the natural numbers (one object for each natural number and one map from a to b if a<b) but that is just a fancy way of saying you look at that sequence.

    colimits are a lot easier to compute if the maps are eventually all surjections, and that is the mittag-leffler condition, and is what i meant by stabilize.

    Also the part where you mention U, V and UnV makes it easier, and this is the 'filtered' part - for any two elements in the partial order there is one element greater than them both. ordinary colimits do not have the requirement.

    here are some more examples of colimits in different categories so you can see the idea:

    suppose f:X-->Y and g:X-->Z are two morphisms in an (abelian) module category, then the colimit of the diagram is the corner in the commuting square defining the push out, which is realized as (Y+Z)/~ where a~b if f(a)=g(b).

    take the category of topological spaces and let f=g:T-->D be maps sendig the circle into the boundary of two disks. the push out is then the union of two disks identified at the bounday, ie a 2 sphere (homeomorphically, anyway).

    suppose we take SET, and take an infinite family of sets indexed by N with inculsions from S_1 to S_2 to S_3 etc..., then the colimit is the nested union.


    What you need to imagine is that you've got a graph (directed) that defines some kind of partially ordered set. colim, or direct lim or lim with an arrow underneath pointing to the right is a functor from one category to another. we take some objects in the first category and put them on the vertices of the graph, and assign morphisms along the edges.

    The limit is then an object in the image category, call it L, with maps FROM each of the vertices such that whenever you have a--->b in the diagram the composition of a to L is the same as a to b then b to L WHEN IT EXISTS.

    For ABGRP we have generators and relations, and all we need to do is take a group generated by all the generators of the vertices modulo the fact that all the realations in the smaller groups must glue together in a compatible way.
     
  6. Jun 6, 2004 #5

    Hurkyl

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    Hrm, I wouldn't've thought of colimits of a finite collection of things! That will certainly make it easier to play with simple examples. It may make sense. :smile:
     
  7. Jun 6, 2004 #6

    matt grime

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    Here are some more ideas that you might have seen and have a sort of link to these things:

    as well as colimits there are limits. we will assume for ease that the limits are over the partially ordered set of the natural numbers.


    a colimit firstly requires a sequence in some category of objects which we'll label as 1,2, etc by abuse of notation, so we have objects 1,2,3,.. and maps from 1 to 2, 2 to 3, and so on. the colimit is the object R with maps FROM each of the objects.

    a limit is what you get if you have maps from 2 to 1, 3 to 2, 4 to 3 etc, and a univesal objects with maps TO each of the objects.

    an examply of one of these is the p-adics where the object n is the ring Z/p^nZ, if i remember correctly, and i may well not.

    other examples of inverse limits are pro-p-groups.

    the dual statements for all those finite diagrams are the dual constructions in the categories - pull backs kernels etc.

    interestingly the amalgamated product is one of these so when you're working things out using the seifert-van-kampen theorem you're using one of these constructions.


    you might want to think about ultra-filters and moore-smith convergence if you know anything about C* algebras.

    final example - every infinite dimensional vector space is the direct limit of its finite dimensional subspaces, if the vector space has a countable basis the limit may be taken over the set N, otherwise it may need to be a filter of greater complexity (again not a formal definition but it is adequately suggestive)
     
  8. Jun 26, 2004 #7

    Hurkyl

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    Ok, another question.

    (Assuming X is a T1 space)

    Suppose I choose a point P, and I have the category C of all open sets of X containing P.

    Then, my presheaf functor gives me a subcategory FC of abelian groups.

    The stalk at P, F_p, is given by the direct limit of FC in AbG.

    Now, consider the category A comprised of inclusion maps of all subsets of X. C is a subcategory of A, and it so happens that {P} is the inverse limit of C.

    This is a very nice thing; since F is covariant, I would expect to be able to "extend" F so that it maps inverse limits in A to direct limits in AbG.


    Now, suppose instead of A, I stick to the category IOP(X). In this category, the inverse limit of C is the empty set (right?). The nice correspondence fails here. :frown:


    So what breaks down?
     
  9. Jun 26, 2004 #8

    matt grime

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    Why is the inverse limit of C the empty set in the category IOP? The inverse limit doesn't exist (there is no reason to suppose limits must exist), that is different. But seeing as we only want to take limF, and not Flim that's ok, in fact if Flim existed then we'd not have to do anything to define the stalk at P other than say it is F(P)

    Saying a category has limits is a very strong statement. (equivalent to something like: contains all direct sums/products, kernels/cokernels, pushouts/pullbacks etc, I can't remember it exactly.
     
  10. Jun 26, 2004 #9

    Hurkyl

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    Hrm, my reasoning went like this:

    {} has the property that there is a morphism from {} to each object of C, and that collection of morphisms commutes properly with those of C.


    Furthermore (within IOP), {} and this collection of morphisms are the only object and collection of morphisms with this property, and obviously the morphisms from {} to elements of C factor uniquely through {}.


    Doesn't this satisfy the definition of inverse limit?
     
  11. Jun 26, 2004 #10

    matt grime

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    Is the empty set in IOP? i mean if you want inclusion maps, is the empty function an inclusion? dunno right now, let me ponder the philosophical interpretation. but even so, the fact that limF doesn't equal Flim is a good thing, and without it there'd be nothing interesting going on.
     
  12. Jun 26, 2004 #11

    Hurkyl

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    Well, in any other setting I've worked, the empty function is, indeed, a function, and it vacuously satisfies the requirement to be an inclusion. In Hartshorne, {} is, indeed, in the domain of a presheaf functor. (The definition includes that F({}) is the trivial group) And, of course, {} is an open subset of X.

    I'm just curious why Flim = limF would happen to work in one case but not the other.
     
  13. Jun 27, 2004 #12

    matt grime

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    That's easy enough to explain. Firstly even in the case you cite Flim does not equal limF since F is not a functor that assigns an abgrp to a closed set, ie Flim(P) doesn't actually exist since F assigns abgrps to open subsets. F(P) is a stalk not a sheaf.

    THe second bit I've translated into category theory speak, so the empty set is the initial object in set, so yep, ok , you end up with the trivial group as that is the terminal object in Grp.
     
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