What are the endofunctors on the simplex category? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:30:59Z http://mathoverflow.net/feeds/question/3697 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3697/what-are-the-endofunctors-on-the-simplex-category What are the endofunctors on the simplex category? Florian 2009-11-01T19:31:56Z 2010-06-10T16:48:09Z <p>Is there a 'classification' of the endofunctors F: &Delta; --> &Delta; where &Delta; 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)?</p> <p>For instance there is a shift functor S: &Delta; --> &Delta; 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.</p> http://mathoverflow.net/questions/3697/what-are-the-endofunctors-on-the-simplex-category/3703#3703 Answer by Charles Rezk for What are the endofunctors on the simplex category? Charles Rezk 2009-11-01T20:17:11Z 2009-11-01T22:21:13Z <p>I don't know such a classification, though I'm interested. Another standard endofunctor is op:&Delta; -> &Delta; which is the identity on objects but which relabels morphisms by reversing the sense of the ordering in each set [n]. </p> <p>Thus, if NC is the nerve of a category, op(NC) = NC<sup>op</sup>.</p> <p>Edit: Here is a thought which might lead to a classification.</p> <p>Given an endofunctor F: &Delta;->&Delta;, there is a restriction functor F<sup>*</sup>: S->S, where S=Psh(&Delta;, Set) = simplical sets. This has a left adjoint F<sub>#</sub>, which on representable presheaves is isomorphic to the original functor F. So F is determined by F<sub>#</sub>, which is determined by the value of F<sup>*</sup> on representables.</p> <p>Write K<sub>n</sub> = F<sup>*</sup>&Delta;[n].</p> <p>Since F<sup>*</sup> preserves limits, we know that K<sub>0</sub> = 1 (terminal object.) For all n, there is a monomorphism &Delta;[n] -> &Delta;[1]<sup>n</sup> (n-fold product), and we can use this to regard K<sub>n</sub> as a subobject of (K<sub>1</sub>)<sup>n</sup>.</p> <p>Finally, you can get &Delta;[n], for n>2, as an inverse limit of a diagram involving &Delta;[1], &Delta;[2], and/or products thereof.</p> <p>Thus, the functor F is basically determined once you know the simplicial set K<sub>1</sub> and the subobject K<sub>2</sub> of (K<sub>1</sub>)<sup>2</sup>.</p> <p>(More is true. In the above, what I'm really doing is using the fact that S is the <em>classifying topos</em> for linear orders. In other words, adjoint pairs G: S &lt;==> 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<sup>*</sup>, the object of E with a linear order is K<sub>1</sub>, and the linear order is the "relation" K<sub>2</sub> on K<sub>1</sub>. This fact discussed, for instance, in Mac Lane &amp; Moerdijk, <em>Sheaves in Geometry and Logic</em>.)</p> http://mathoverflow.net/questions/3697/what-are-the-endofunctors-on-the-simplex-category/3704#3704 Answer by Reid Barton for What are the endofunctors on the simplex category? Reid Barton 2009-11-01T20:25:11Z 2009-11-01T21:20:13Z <p>More examples:</p> <ul> <li><p>the functor &Delta; &rarr; &Delta; 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 &Delta;<sup>2</sup> to a complex with four nondegenerate 2-simplices whose geometric realization is homeomorphic to |&Delta;<sup>2</sup>|.</p></li> <li><p>the functor &Delta; &rarr; &Delta; sending S to S<sup>op</sup> ∐ S. Restriction along this functor sends the nerve of a category C to the nerve of the twisted arrow category of C.</p></li> </ul> <p>I guess all the examples I know can be built out of objects of &Delta;, op, and the join &Delta; &times; &Delta; &rarr; &Delta;.</p> http://mathoverflow.net/questions/3697/what-are-the-endofunctors-on-the-simplex-category/27638#27638 Answer by Tom Goodwillie for What are the endofunctors on the simplex category? Tom Goodwillie 2010-06-10T04:26:17Z 2010-06-10T16:48:09Z <p>To carry Charles' train of thought further:</p> <p>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.</p> <p>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. </p> <p>This extra structure (ordering and endpoints) on $\Delta^1$ is inherited by Charles' $K_1=F^*\Delta^1$; it, too, is a simplicial interval.</p> <p>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. </p> <p>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".</p> <p>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.</p> <p>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).</p>