What are the endofunctors on the simplex category? - MathOverflow

What are the endofunctors on the simplex category?

2009-11-01T19:31:56Z

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.

Answer by Charles Rezk for What are the endofunctors on the simplex category?

2009-11-01T20:17:11Z

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) = NC^op.

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 K_n = 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 K_n 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.)

Answer by Reid Barton for What are the endofunctors on the simplex category?

2009-11-01T20:25:11Z

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 S^op ∐ 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.

Answer by Tom Goodwillie for What are the endofunctors on the simplex category?

2010-06-10T04:26:17Z

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).