Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there a good way to define a sheaf over a simplicial set - i.e. as a functor from the diagram of the simplicial set to wherever the sheaf takes its values - in a way that while defined on simplex by simplex corresponds in some natural manner to what a sheaf over the geometric realization of the simplicial set would look like?

Edited to add: I'm actually interested in the question in a concrete fashion rather than an abstract one - I'm trying to figure out whether there might be some interesting interpretation of sheaves over the nerve of a category generated by a network; similar to current work by Robert Ghrist that takes a network and views it as a graph, and thus as a topological space (1-dim simplicial complex), and manages to find useful interpretations of sheaves on this particular space in terms of network analysis.

Hence, what I'm really looking for is an interesting definition for, say, the nerve of the category generated by a finite directed graph, or so...

Edited to add: In off-channels, fpqc has clarified his argument in the answer I've accepted. Specifically, $N(C)$ for a category has inherent direction data that is lost in $|N(C)|$. This messes up attempts to formulate an idea of sheaves over $N(C)$ in a way that stays faithful to the definition of sheaves over $|N(C)|$.

share|improve this question
    
Just to check, do you literally mean "sheaves over the geometric realization" (which is a very big category, and depends on the homeomorphism type of the realization) or something more like locally constant sheaf/local system which is more homotopy-theoretic in nature? (I know I often say simply "sheaf" for the latter concept, so maybe others do too) –  Reid Barton Mar 18 '10 at 4:29
    
I think I might have meant sheaves over the geometric realization, but I'm not certain. Certainly, I don't really want to consider the entire category, but rather I'm interested in finding out ways to deal with concrete sheaves. –  Mikael Vejdemo-Johansson Mar 18 '10 at 4:44
    
Here's the problem that I see: First of all, a sheaf over a simplicial set does not exactly make sense using the regular definition of a sheaf and on a general simplicial set. A sheaf is a contravariant functor into sets. Another problem here is that the definition of a sheaf is relative to the grothendieck topology on the underlying site. Certainly, every presheaf is a sheaf in the chaotic grothendieck topology on a category. In general as well, simplicial sets do not in general have diagrams that describe the whole structure. This is what I meant by "in general, this is not well-defined. –  Harry Gindi Mar 18 '10 at 4:55
    
@Mikael: What I mean is: take any open subset $U$ of the geometric realization (there are uncountably many) and define a sheaf $F$ by $F(S) = *$ if $S \subset U$, $F(S) = \emptyset$ otherwise. Is this an example of the kind of object you are after? –  Reid Barton Mar 18 '10 at 5:05
    
A sheaf is a contravariant functor into sets. Yes, quite, from an appropriate Heyting algebra - most often the Heyting algebra of open subsets of a given topological space. This is why I end up thinking about simplicial sets - using the nerve functor, they're an interesting way to get a topology out of a category; and I'm hoping to find interesting structures by considering sheaves on that topology. And, in order to stay combinatorial, I'm hoping to be able to construct at least a subclass of those sheaves as structures defined as functors from the diagram of the simplicial set. –  Mikael Vejdemo-Johansson Mar 18 '10 at 5:06

2 Answers 2

up vote 8 down vote accepted

Clearly looking at sheaves on the geometric realisation gives something too far removed from the simplicial picture. This is essentially because there are too many sheaves on a simplex have (most of which are unrelated to simplicial ideas). What one could do is to consider such sheaves which are constructible with respect to the skeleton filtration, i.e., are constant on each open simplex. This can be described inductively using Artin gluing. I think it amounts to the following for a simplicial set $F$.

For each simplex $c\in F_n$ we have a set $T_c$, the constant value of the sheaf $T$ on the interior of the simplex corresponding to $c$.

For each surjective map $f\colon [n] \to [m]$ in $\Delta$ the corresponding (degeneracy) map on geometric simplices maps the interior of $\Delta_n$ into (onto in fact) the interior of $\Delta_m$ and hence we have a bijection $T_{f(c)} \to T_c$. These bijections are transitive with respect to compositions of $f$'s.

For each injective map $f\colon [m] \to [n]$ in $\Delta$ the corresponding map on geometric simplices maps $\Delta_m$ onto a closed subset of $\Delta_n$. If $j\colon \Delta^o_n \hookrightarrow \Delta_n$ is the inclusion of the interior we get an adjunction map $T \to j_\ast j^\ast T$ and $j_\ast j^\ast T=T_c$ where $T_c$ also denotes the constanct sheaf with value $T_c$. If $f'\colon \Delta^o_m \hookrightarrow \Delta_n$ is the inclusion of the interior composed with $f$ we can restrict the adjunction map to get a map $T_{f(c)}=f'^\ast T \to f'T_c$ and taking global sections we get an actual map $T_{f(c)} \to T_c$. These maps are transitive with respect compositions of $f$'s.

We have a compatibility between maps coming from surjections and injections. Unless something very funny is going on this compatibility should be that we wind up with a function on the comma category $\Delta/F$ which takes surjections $[n] \to [m]$ to isomorphisms.

There is the stronger condition on the sheaf $F$, namely that it is constant on each star of each simplex. This means on the one hand that it is locally constant on the geometric realisation, on the other hand that $T_{f(c)} \to T_c$ is always an isomorpism.

[Added] Some comments intended to give some kind of relation with the answer provided by fpqc. My suggested answer is not homotopy invariant in the sense that a weak (or even homotopy) equivalence of simplicial sets does not induce an equivalence on the category of sheaves. This is so however if one, as per above, adds the condition that all the maps $T_{f(c)} \to T_c$ are isomorphisms. However, that condition is not so good as many maps that are not weak equivalences induces category equivalences (it is enough that the map induce isomorphisms on $\pi_0$ and $\pi_1$). This is a well-known phenomenon and has to do with the fact the $T_c$ are just sets. One could go further and assume that the $T_c$ are topological spaces and the maps $T_{f(c)} \to T_c$ continuous. Of course adding the condition that these maps be homeomorphisms shouldn't be right thing to do, instead one should demand that they be homotopy (or weak) equivalences. Again, this shouldn't be quite it because of the transitivity conditions. We should not have that the composite $T_{g(f(c))} \to T_{f(c)} \to T_c$ should be equal to $T_{g(f(c))} \to T_c$ but rather homotopic to it. Once we have opened that can of worms we should impose higher homotopies between repeated composites. This can no doubt be (has been) done but there seems to be an easier way out. In the first step away from set-valued $T_c$ we have the possibility of they being instead categories. In that case the higher homotopy conditions is that we should have a pseudofunctor $\Delta/F \to \mathcal{C}\mathrm{at}$. Even they are somewhat unpleasant and it is much better to pass to the associated fibred category $\mathcal{T} \to \Delta/F$. In the general case, and admitting that $\Delta/F$ is essentially the same things as $F$ itself, we should therefore look at (Serre) fibrations $X \to |F|$ or if we want to stay completely simplicial, Kan fibrations $X \to F$. This gives another notion of (very flabby) sheaf which now should be homotopy invariant (though that should probably be in the sense of homotopy equivalence of $\Delta$-enriched categories).

share|improve this answer
    
Very interesting. +1 for you! –  Harry Gindi Mar 18 '10 at 11:21
    
This is essentially what HTT is about, by the way. This notion (in the additional comments) is covered in depth in chapter 2. –  Harry Gindi Mar 18 '10 at 14:12

In general, this is not well-defined (you could look at the completion of the category by homotopy colimits, I guess, but for some reason I feel like this isn't very useful). A sheaf on the geometric realization is equivalent to a covering space. The geometric realization throws out too much data for this sort of thing to be very useful. You should really read HTT by Lurie, because this is what the book is about ([generalizations of] sheaf toposes on infinty-categories (which are special simplicial sets). You usually want to look at sheaves of Kan complexes, which are a higher-categorical equivalnt of categories fibered in groupoids.

Technically you can view any simplicial set as the $\omega$-nerve of a strict $\omega$-category, so in that sense, you could look at strict $\omega$-functors, but these are of very little interest in general.

I answered a similar question to this a while back.

Specifically the question was asking about the result that the category covering spaces on BC:=|N(C)| is canonically isomorphic to presheaves on C.

share|improve this answer
    
I don't really get what's going on here, but is there some sort of error that I made, or do you dislike the tone, or what? –  Harry Gindi Mar 18 '10 at 5:54
    
Someone else voted it down - I don't know who. I accepted it, since, coupled with the off-channel discussions we had, it actually pointed out the parts that were important to figure out how my question was not quite well-defined enough to be sensible. –  Mikael Vejdemo-Johansson Mar 18 '10 at 6:13
    
Oh, I wasn't talking to you. I was talking to whoever did it. =) –  Harry Gindi Mar 18 '10 at 10:28
3  
I voted it down for obfuscation. :P –  Marty Mar 18 '10 at 16:26
1  
+1 on both of your houses. –  Harry Gindi Mar 18 '10 at 18:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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