7
$\begingroup$

Background: Let $\mathbf{Lat}$ be the 2-category of lattices which can be viewed as a subcategory of the 2-cateogry of posets $\mathbf{Pos}$, that is, objects in $\mathbf{Pos}$ that have all finite products and coproducts (a.k.a meets and joins in lattice-speak); we may or may not require 2-morphisms in $\mathbf{Lat}$ to preserve meets and joins (i.e continuous and cocontinuous).

I am considering (what I call) cellular sheaves valued in lattices which are just functors $F: X \rightarrow \mathbf{Lat}$ where $X$ is the face relation poset of a cell complex. In order to do "sheaf theory" with sheaves valued in $\mathbf{Lat}$, it would be nice to have a notion of a coproduct and product in this category. I think product is fairly clear: just use the product partial order; meets and joins are what you think they would be. As far as a coproduct, I am not sure. If anyone has any suggestions? I have heard of a "free product of lattices" but it is not defined in a language I can understand. Not even clear to me that the "free product" that Gratzer defines is unique.

By "sheaf theory", I mainly mean taking limits and colimits over all the stalks. Another property I would like (or like to know doesn't hold) is the existence of all equalizers and (maybe if coproducts exist) coequalizers which would guarantee the existence of all small (co)limits. That would be lovely.

In general, looking for references on lattice theory, "non-abelian" sheaf theory, or anything about categories and lattice theory.

Thanks in advance!

$\endgroup$
  • $\begingroup$ Do lattices have top and bottom elements for you? $\endgroup$ – Tim Campion Oct 9 '18 at 22:55
  • $\begingroup$ @TimCampion if empty joins and meets count as finite joins and meets, then yes, they should. $\endgroup$ – David Roberts Oct 10 '18 at 0:14
  • 1
    $\begingroup$ I suspect that if you require that maps of lattices are compatible with meets/joins, then the coproduct of lattices might again just be the cartesian product of the underlying lattice, with the left and right injections being defined by sending $x \mapsto (x,\top)$ and $y \mapsto (\top,y)$ respectively. But this might require all joins, not just finite joins. $\endgroup$ – David Roberts Oct 10 '18 at 0:19
  • 1
    $\begingroup$ @TimCampion and I see you have a similar answer here: math.stackexchange.com/a/1195002/3835 $\endgroup$ – David Roberts Oct 10 '18 at 0:21
  • $\begingroup$ @TimCampion Say lattices have bottom elements (not sure I want to restrict to lattices with top elements yet) which I assume is $\perp$ in DavidRoberts's notation. $\endgroup$ – Hans Oct 10 '18 at 21:26
3
$\begingroup$

I'm not sure about sheaf theory, but limits and colimits in categories of lattices are routine to construct. You just have to be clear about your categorical setup. Consider the following categories:

  • $sLat^{\vee}$, the category of posets with $(\vee,\bot)$ and morphisms which preserve these.

  • $Lat$, the category of posets with $(\vee,\wedge,\bot,\top)$ and morphisms which preserve these.

  • $DistLat$ the full subcategory of $Lat$ on the lattices where $\wedge$ distributes over $\vee$.

  • ...

and variants like

  • $Lat^{unbbd}$, the category of posets with $(\vee,\wedge)$ and morphisms which preserve these.

  • $Lat^{\uparrow bdded}$, the category of posets with $(\vee,\wedge,\top)$ and morphisms which preserve these.

  • ...

In each case, we have a variety in the sense of universal algebra, i.e. a category where an object is a set with some $n$-ary functions on it satisfying some universal equations (note that the poset structure can be recovered from $\vee$ or $\wedge$) and morphisms being functions which commute with all the functions. Other examples are categories like groups, rings, etc.

In any such category $\mathcal C$, one has all limits and colimits. They can be constructed in the following way. There is a forgetful functor $U: \mathcal C \to Set$. This functor has a left adjoint $F: Set \to \mathcal C$, which just sends a set to the set of all words in the function symbols, quotiented by the universal relations that hold in objects of $\mathcal C$. For example in the case of groups, $F(X)$ is the free group on the set $X$.

This adjunction $F \dashv U$ yields a monad $UF: Set \to Set$ from which the category $\mathcal C$ can be recovered as the category of algebras. Limits and colimits can be constructed in a standard way. For limits, take the limit of the underlying sets and extend the functions in the obvious way. Colimits are a bit more complicated, but you can read about them here. Filtered colimits, though, are easy -- just take the colimit of the underlying sets and extend the operations in the obvious way (these are what you need to compute stalks). Reflexive coequalizers are likewise computed at the level of the underlying sets. The trickiest class of colimits are coproducts. The coproduct $\amalg_i C_i$ is constructed by taking $F(\amalg_i U(C_i))$ and quotienting by an equivalence relation (for instance, the coproduct of groups is the free amalgam). Alternative descriptions may be available depending on $C$ -- for example see here.

$\endgroup$
  • 3
    $\begingroup$ Even coproducts are not too bad, once the free algebras are at hand. The general construction is a reflexive coequalizer as described in Theorem 2.2 here: ncatlab.org/nlab/show/… On the other hand, the explicit concrete description of free lattices takes some effort (I don't mean as abstract development, but in terms of whether two syntactic descriptions are equivalent as lattice elements), and was originally given by Whitman. See Theorem 6.2 here, for instance: math.hawaii.edu/~jb/math618/os6uh.pdf. $\endgroup$ – Todd Trimble Oct 13 '18 at 17:53
  • 2
    $\begingroup$ Also, for the OP: I think "free products" mean the same thing as coproducts, if group-theoretic terminology is any guide at all. They are certainly well-defined, but the explicit concrete description is about as easy or as hard as understanding free lattices, as hinted by my previous comment -- similar to the situation in group theory. $\endgroup$ – Todd Trimble Oct 13 '18 at 17:56
  • $\begingroup$ @TimCampion Warm thanks for your extremely detailed answer. I will take some time to process what you have said and will return here for a follow-up questions if you would be so kind. $\endgroup$ – Hans Oct 14 '18 at 18:44
  • 1
    $\begingroup$ Also, if a "sheaf" is just a functor to ${\bf Lat}$ as the OP says (most people would call that a "presheaf"), then their limits and colimits are just pointwise. $\endgroup$ – Mike Shulman Jan 3 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.