14
$\begingroup$

Is there a 'classification' of the endofunctors F: Δ --> Δ where Δ denotes the simplex category with objects [n] and the weakly monotone maps [m] --> [n] as morphisms (Actually, I don't know if 'simplex category' is the right name)?

For instance there is a shift functor S: Δ --> Δ defined by S([n])=[n+1] on objects and S(d): [m+1] --> [n+1] being d on [m] and mapping m+1 to n+1 for a morphism d: [m]-->[n]. Hence for a simplicial set X one gets a path-object XoS.

$\endgroup$
1
  • 3
    $\begingroup$ I think "simplex category" is a good name. Some people call it "the simplicial category" but one also often finds "a simplicial category" meaning "a category enriched over simplicial sets." Another unambiguous, though longer, name would be "the simplicial indexing category." $\endgroup$ Commented Nov 2, 2009 at 6:09

3 Answers 3

17
$\begingroup$

To carry Charles' train of thought further:

By an 'interval' let us mean a finite ordered set with at least two elements; let $Int$ be the category of intervals, where a morphism is a monotone map preserving both endpoints.

The simplicial set $\Delta^1$ can be viewed as a simplicial interval. That is, this functor $\Delta^{op}\rightarrow Set$ factors through the forgetful functor from $Int$ to $Set$. In fact, the resulting functor $\Delta^{op}\rightarrow Int$ is an equivalence of categories.

This extra structure (ordering and endpoints) on $\Delta^1$ is inherited by Charles' $K_1=F^*\Delta^1$; it, too, is a simplicial interval.

There aren't that many things that a simplicial interval can be. Its realization must be a compact polytope with a linear order relation that is closed. That makes it at most one-dimensional, and makes each component of it either a point or a closed interval. Simplicially each of these components can be either a $0$-simplex, or a $1$-simplex with its vertices ordered one way, or a $1$-simplex with its vertices ordered the other way, or two or more $1$-simplices each ordered one way or the other and stuck together end to end.

The three simplest things that a simplicial interval can be are: two points, a forward $\Delta^1$, and a backward $\Delta^1$. These arise as $F^*\Delta^1$ for three examples of functors $F:\Delta\rightarrow \Delta$, the only examples that satisfy $F([0])=[0]$, namely the constant functor $[0]$, the identity, and "op".

It's clear that any functor with $F([0])=[n]$ has the form $F_0\coprod\dots\coprod F_n$ where $F_i[0]=[0]$ for each $i$. This means that the corresponding simplicial interval can be made by sticking together those which correspond to the $F_i$. For example, the 'shift' functor mentioned in the question is $id\coprod [0]$; Reid mentioned $id\coprod id$ and $op\coprod id$. These correspond respectively to: a $1$-simplex with an extra point on the right, two $1$-simplices end to end, and two $1$-simplices end to end one of which is backward. As another example, the constant functor $[n]$ corresponds to $n+1$ copies of (two points) stuck together end to end, or $n+2$ points.

In short, every functor $F:\Delta\rightarrow \Delta$ is a concatenation of one or more copies of $[0]$, id, and op. I can more or less see how to prove this directly (without toposes or ordered compact polyhedra).

$\endgroup$
11
$\begingroup$

I don't know such a classification, though I'm interested. Another standard endofunctor is op:Δ -> Δ which is the identity on objects but which relabels morphisms by reversing the sense of the ordering in each set [n].

Thus, if NC is the nerve of a category, op(NC) = NCop.

Edit: Here is a thought which might lead to a classification.

Given an endofunctor F: Δ->Δ, there is a restriction functor F*: S->S, where S=Psh(Δ, Set) = simplical sets. This has a left adjoint F#, which on representable presheaves is isomorphic to the original functor F. So F is determined by F#, which is determined by the value of F* on representables.

Write Kn = F*Δ[n].

Since F* preserves limits, we know that K0 = 1 (terminal object.) For all n, there is a monomorphism Δ[n] -> Δ[1]n (n-fold product), and we can use this to regard Kn as a subobject of (K1)n.

Finally, you can get Δ[n], for n>2, as an inverse limit of a diagram involving Δ[1], Δ[2], and/or products thereof.

Thus, the functor F is basically determined once you know the simplicial set K1 and the subobject K2 of (K1)2.

(More is true. In the above, what I'm really doing is using the fact that S is the classifying topos for linear orders. In other words, adjoint pairs G: S <==> E: H where E is a topos, and the left adjoint G preserves finite limits, correspond to "objects in E equipped with a linear order". In this case, E=S, and G=F*, the object of E with a linear order is K1, and the linear order is the "relation" K2 on K1. This fact discussed, for instance, in Mac Lane & Moerdijk, Sheaves in Geometry and Logic.)

$\endgroup$
4
$\begingroup$

More examples:

  • the functor Δ → Δ sending a totally ordered set S to S ∐ S, where the elements in the left copy are all less than the elements in the right copy. Restriction along this functor is the "edgewise subdivision", e.g., it sends Δ2 to a complex with four nondegenerate 2-simplices whose geometric realization is homeomorphic to |Δ2|.

  • the functor Δ → Δ sending S to Sop ∐ S. Restriction along this functor sends the nerve of a category C to the nerve of the twisted arrow category of C.

I guess all the examples I know can be built out of objects of Δ, op, and the join Δ × Δ → Δ.

$\endgroup$

You must log in to answer this question.

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