Computation of Joins of Simplicial Sets

It turns out that joins of simplicial sets are fairly easy to define, but hard to manage. In lots of cases, we'd like to compute what a join is, does it look like a horn?, a boundary?, etc? and identify it as such, so we can figure out when our morphisms from the join have certain nice properties like being anodyne, having lifting properties, and all of that wonderful stuff.

For example, consider the join, $\Lambda^n_j \star \Delta^m$. The problem that I currently face is, I can't tell what this thing looks like from the definition.

Consider an even simpler case, $\Delta^n \star \partial \Delta^m$. From the definition, we get a very nasty definition of this join, and I'm having trouble applying it and computing the join in terms of nicer simplicial sets.

I ask this, because on p.62 of Higher Topos Theory by Lurie, for example, he states that for some $0 < j \leq n$ $$\Lambda^n_j \star \Delta^m \coprod_{\Lambda^n_j \star \partial \Delta^m} \Delta^n \star \partial \Delta^m$$ and says that we can identify this with the horn $\Lambda^{n+m+1}_j$. Unraveling the definitions seems to make it harder to understand, and I just don't see how this result was achieved. However, my aim here is to understand how the computation was actually carried out, since it is completely omitted.

For convenience, here is the definition of the join of $S$ and $S'$ for each object $J \in \Delta$ $$(S\star S')(J)=\coprod_{J=I\cup I'}S(I) \times S'(I')$$ Where $\forall (i \in I \land i' \in I') i < i'$, which implies that $I$ and $I'$ are disjoint.

EDIT AFTER ANSWER: Both Reid and Greg provided good answers to the question, and I only accepted the one that I did because Greg commented more recently. So for anyone reading this at some point in the future, read both answers, as they are both good.

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By $\delta \Lambda^m$ do you mean $\partial \Delta^m$? –  Reid Barton Dec 1 '09 at 0:04
That I did, ser. Fixed now, hopefully. –  Harry Gindi Dec 1 '09 at 0:11

Since the join of simplicial sets is associative and $\Delta^m = \Delta^0 \star \cdots \star \Delta^0$ ($m+1$ times), we should start by trying to understand things like $\Lambda^n_j \star \Delta^0$, a.k.a. the "final" cone on $\Lambda^n_j$. It's not too hard to see that this is the subcomplex of $\Delta^{n+1}$ consisting of those faces which do not contain the (codimension 2) face $\{0, \ldots, r-1, r+1, \ldots, n\}$. In other words, we are missing the face opposite $r$ and $n+1$, because we were originally missing the face opposite $r$ of $\Delta^n$, as well as the three other faces (including the interior of $\Delta^{n+1}$) it contains. Similarly $\Delta^0 \star \partial \Delta^n$ is the horn $\Lambda^{n+1}_0$ (we are missing the interior of $\Delta^{1,\ldots,n}$ and the cone on it).

In general all the simplicial sets that come up have the form of the subcomplex of $\Delta^N$ consisting of those faces which do not contain a fixed face $\Delta^S$, $S \subset \{0, \ldots, N\}$. Forming the cone (on either side) on such a space results in another such space with $N$ replaced by $N+1$ and $S$ unchanged (as a subset of the vertices of the original $\Delta^N$, which if we formed a cone on the left, means we increment each index in $S$).

After doing these sorts of computations, I expect that $\Lambda^n_j \star \Delta^m$ and $\Delta^n \star \partial \Delta^m$ will be two subcomplexes of $\Delta^{n+m+1}$ each characterized by avoiding faces containing a certain face, and that $\Lambda^n_j \star \partial \Delta^m$ is their intersection and $\Lambda^{n+m+1}_j$ is their union, from which the claim would follow.

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I kinda see what you mean, but what happens when we take the next join in the first thing, since you split it up and all. –  Harry Gindi Dec 1 '09 at 0:38
I elaborated a little on what happens in general. –  Reid Barton Dec 1 '09 at 1:01

It may be helpful to consider the geometric join operation for compact topological spaces or simplicial complexes. If $A$ and $B$ are discrete sets, then their join $A \star B$ is the complete bipartite graph connecting $A$ to $B$. The join in general is the natural generalization of that to topological spaces, or simplicial complexes, or simplicial sets, or (if you like) CW complexes. In fact it always is the complete bipartite graph, but with a topology or a simplicial geometry on the set of line segments. If $A$ and $B$ are simplices, their join is another simplex, whose vertices are the disjoint union of the vertices of $A$ and $B$.

A simplicial set $S$ has a small geometric realization consisting of the non-degenerating simplices glued together. In many early examples, such as for instance horns, the small geometric realization is just a finite simplicial complex with locally ordered vertices. (That is, the vertices of each simplex are compatibly ordered.) For instance, the horn $\Lambda_j^2$ is a V, and the horn $\Lambda_j^3$ is a triangular hat (or horn, hence the name and the clever symbol). (As best I can tell, the subscript referring to the apex of the horn $\Lambda^n_j$ is not intrinsic to it as a simplicial set, but rather comes from its inclusion into the simplex $\Delta^n$.) (Edit: Even though a horn is hollow, the local orderings of its faces induce a total ordering of its vertices when $n \ge 2$, or a partial ordering when $n=1$. The subscript indicates the position of the apex.) The join of $A$ with point $\Delta^0$ is also a cone with base $A$. So as Reid said, the cone $\Delta^0 \star \partial \Delta^n$ over a hollow simplex is a horn. It's also easy to see, by drawing a picture, that the cone over a horn is the next horn. [Edit: It's a good idea to draw the picture of the cone over of a horn, to see that it isn't a horn. :-)]

(I apologize if this geometric discussion is too close to what Reid already said somewhat more algebraically.)

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Just a minor point--the various horns of the n-simplex are nonisomorphic as simplicial sets, since the vertices of a simplex are distinguished (effectively, ordered). This doesn't matter much when you use simplicial sets to model spaces, but it does matter when you use them as quasicategories since the direction of a 1-simplex corresponds to the direction of a morphism in a category. –  Reid Barton Dec 1 '09 at 5:38
No, I just goofed, as the new parenthetical explains. –  Greg Kuperberg Dec 1 '09 at 5:44
Ah, I see. I guess what you wrote is true for n = 1. :) –  Reid Barton Dec 1 '09 at 5:46
No, even then it is only partly true. –  Greg Kuperberg Dec 1 '09 at 5:50
I am not entirely sure what you're asking. Do you mean, how does the geometric description of the join match your coproduct of products formula for $S$ and $S'$? The answer is that each non-degenerate simplex of $S$ and each non-degenerate simplex of $S'$ join together into a simplex of $S \star S'$. This exactly matches your formula. –  Greg Kuperberg Dec 1 '09 at 7:05
If one thinks of joins as most naturally applicable to augmented simplicial sets, then many of the questions asked take a (I hope) clearer aspect. The join of augmented simplicial sets is part of a monoidal category structure that explicitly uses ordinal sums and hence has a directional aspect. The ordinal sum of two finite ordinals works well but $[m]\oplus [n]$ and $[n]\oplus [m]$ although isomorphic are linked in a subtler way. Phil Ehler attacked this in his thesis at Bangor using ideas of Duskin and van Osdol (ultimately derivable from Lawvere). I wrote up Phil's thesis material on this an there are two published papers available on ordinal sum and its relationship with join and also on links with anodyne extensions and quasicategories. ( I can give pointers if it would help.)