7
$\begingroup$

Why doesn't the join operation on the category of simplicial sets commute up to unique isomorphism? I mean, aren't products and coproducts commutative up to isomorphism? That leads me to conclude at first glance that the join is commutative, but it's not. Recall, given two simplicial sets $S$ and $S'$, we define the join to be the simplicial set such that for all finite nonempty totally ordered sets $J$, $$(S\star S')(J)=\coprod_{J=I\cup I'}S(I) \times S'(I')$$ Where $\forall (i \in I \land i' \in I') i < i'$, which implies that $I$ and $I'$ are disjoint.

Now the thing is, clearly my conclusion is stupid, because we use the fact that it doesn't commute to distinguish between over quasi-categories and under quasi-categories. Where did I go wrong?

I hope this is up to the standards of MO, but if it's not, I'll delete the topic.

$\endgroup$

3 Answers 3

9
$\begingroup$

Implicit in the index of the coproduct is that you're writing J as an ordered disjoint union of I and I', where I comes first.

EDIT: Some elaboration.

For a simplicial set $T$, let's write $T\_n$ for the "n-simplices", i.e. the value of on the ordered set $\{0,1,...,n\}$; these together with the maps between them determine the functor $T$ completely. (Your formula for the join requires the convention that $T$ takes the empty set to a single point.)

Given $S$ and $S'$, let's determine the 0- and 1-simplices of the join.

First, $(S \star S')\_0$. There are exactly two ways to write $\{0\} = I \cup I'$ in an order preserving way as indexed by the coproduct: either $I'$ is empty and $I$ is everything, or vice versa. Thus $(S \star S')\_0 = S\_0 \cup S'\_0$ accordingly. The zero-simplices of the join are the zero-simplices of the original simplicial set.

Next, the 1-simplicies. Similarly $$ (S \star S')\_1 = S(\{0,1\}) \cup (S(\{0\}) \times S'(\{1\})) \cup S'(\{0,1\})= S\_1 \cup (S\_0 \times S'\_0) \cup S'\_1 $$ There are then 3 types of 1-simplices: the 1-simplices from S, those from S', and for each choice of a point of S and a point of S' there is a new 1-simplex.

The two boundary maps $(S \star S')\_1 \to (S \star S')\_0$ are induced by the inclusions of $\{0\}$ and $\{1\}$ into $\{0,1\}$ (the "back" and "front" boundaries respectively). In particular, on the new 1-simplices $S\_0 \times S'\_0$ the back boundary is the projection to $S\_0$ and the front boundary is projection to $S'\_0$. There is asymmetry here because the only ways we're allowed to decompose $\{0,1\}$ in the coproduct have $I$ (the subset corresponding to $S$) first and $I'$ second. None of the "new" edges start at a vertex of $S'$ and end at a vertex of $S$.

$\endgroup$
2
  • $\begingroup$ Actually, I don't see how this gives us a direction or order. Could you elaborate a little further, preferably not with degenerate simplicial sets? I don't see what you mean by ordered disjoint union. $\endgroup$ Nov 29, 2009 at 19:34
  • $\begingroup$ I added some more that will hopefully clarify. $\endgroup$ Nov 29, 2009 at 22:12
8
$\begingroup$

It might be helpful to work through some simple examples. You probably know that Δn ★ Δk = Δn+k+1. This has to do with the ordinal sum: one way of defining joins is as a restriction of the monoidal structure on augmented simplicial sets, which are contravarient functors from the category Δ+ of all finite ordinals (including the empty ordinal) into sets. The category Δ+ has a monoidal structure given by ordinary addition with ∅ as the unit, and this induces the aforementioned monoidal structure on augmented simplicial sets. The thing we call n when we are talking about simplicial sets is really the ordinal n+1, so the formula above holds because

(n+1) + (k+1) = (n+k+1)+1.

Of course, this example doesn't illustrate the asymmetry you asked about, but this one will:

∂Δn ★ Δ0 = Λn+1[n+1] while Δ0 ★ ∂Δn = Λ0[n+1].

To work out the details, you'll need to understand how the face maps of S★T are defined, as alluded to above. Here's my notation: (S★T)n = SnTn ∪ (∪ j+k = n+1 Sj × Tk ).

The i-th boundary map di : (S★T)n → (S★T)n-1 is defined on Sn and Tn using the i-th boundary map on S and T. Given σ∈Sj and τ∈Tk , we have:

di (σ, τ) = (di σ,τ) if i ≤ j, j ≠ 0.
di (σ, τ) = (σ,di-j-1 τ) if i > j, k ≠ 0.

If j = 0, d0(σ, τ) = τ ∈ Tn-1 ⊂ (S★T)n-1. If k = 0, dn(σ, τ) = σ ∈Sn-1 ⊂ (S★T)n-1 .

Try this out for n = 1 or 2 first, to get a feel for things. While these sorts of computations can be quite annoying, I find they do really help me develop my intuition. Best of luck!

$\endgroup$
2
  • $\begingroup$ I'm not familiar with the notation $\Lambda^k [j]$. I am familiar with $\Lambda^k_j$ being the jth k-horn, but $0\leq j \leq k$ in that case. $\endgroup$ Nov 30, 2009 at 10:25
  • $\begingroup$ Sorry. Both notations are fairly standard. I chose the former because it looked better in html. But when writing $\Lambda^k[j]$, one typically means for $j$ to be the dimension and $0 \leq k \leq j$ to indicate which face is missing. (So this is $\Lambda^j_k$ in your notation.) $\endgroup$ Dec 1, 2009 at 5:19
5
$\begingroup$

As mentioned in the other answers, the join of simplicial sets is closely related to "ordered disjoint union", or "concatenation", of (totally, partially, pre-) ordered sets. You can use this both to get simple examples of its non-commutativity, and to help reconcile that with the intuition that it should be commutative.

Any order $X$ can be seen as the simplicial set whose $n$-simplices are chains $(x_0 \leq \ldots \leq x_n)$ from $X$. That is, there's a full and faithful "nerve" embedding $N: \mathrm{PreOrd} \rightarrow \mathrm{SSet}$. Now if $X$ and $Y$ are orders, seen as their nerves, $X \star Y$ is exactly (the nreve of) their ordered disjoint union.

So e.g. $1 \star \mathbb{N} \not \cong \mathbb{N} \star 1$ is an easy and intuitive example of the non-commutativity.

On the other hand, like you say, looking at the definition, there is an immediate intuition that it should be commutative in some sense, and chasing it down, I think what that intuition is coming from is something like the fact: for any simplicial sets $X$, $Y$,

$(X \star Y)^\mathrm{op} \cong Y^\mathrm{op} \star X^\mathrm{op}.$

So commuting $\star$ distributes over $\mathrm{op}$: so in a sense, the only asymmetry in $\star$ is an asymmetry of variance. This is nice and intuitive for ordered sets, and easily shown for all simplicial sets.

$\endgroup$
1
  • $\begingroup$ Isn't it (X * Y)^op = Y^op * X^op? $\endgroup$ Dec 5, 2009 at 18:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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