# Join of ordered simplicial complexes

Let $$K_1$$ and $$K_2$$ be (abstract) simplicial complexes on ordered vertex sets $$\{1,...,m_1\}$$ and $$\{1,...,m_2\}$$ respectively.

The join of $$K_1$$ and $$K_2$$ is a simplicial complex $$K$$ with vertex set the disjoint union $$V(K)=V(K_1) \cup V(K_2)$$, with simplices all unions of the form $$\sigma_1 \cup \sigma_2$$, for $$\sigma_1 \in K_1, \sigma_2 \in K_2$$ (where $$\sigma_1$$ and $$\sigma_2$$ can be the empty set).

When working with ordered simplicial complexes (that is, simplicial complexes with ordered vertex sets), if we take the join and want to get an ordered simplicial complex, we would have to pick an ordering on $$V(K)$$. It seems to me that there is no way to make this choice canonically.

Is there a way to define such an operation with a canonical choice of ordering on the vertex set $$V(K)$$? Or, do we have to say that the join of two ordered simplicial complexes is actually a class of ordered simplicial complexes - that class being those simplicial complexes which are the same up to the choice of appropriate ordering on the vertex set $$V(K)$$?

• Do you mean joint Commented Oct 16, 2020 at 11:25
• No I don't, no.
– Matt
Commented Oct 16, 2020 at 12:27
• Well, you could have the vertices of $K_1$ first and $K_2$ after that. That can be fairly canonical depending on your model and will even make the join associative, although non-commutative. A question is why would you insist on working in the category of ordered simplicial complexes. Commented Oct 19, 2020 at 6:56

I'll try to do some discussion on the asked construction. First we have to be clear what properties of ordered simplexes we want to preserve. We will consider at first any partial order on a finite set. We define our category of ordered simplicial complexes as $$SimpCOrd$$ the category of pairs $$(K, \leq)$$ with $$\leq$$ an ordering on $$V(K)$$.
Now most important of all we have to define what an ordered simplicial complex morphism will be as it will be the key component to our construction. Let $$(K_1, \leq_1), (K_2, \leq_2) \in SimpCOrd$$ and $$f : V(K_1) \rightarrow V(K_2)$$, we call $$f$$ an ordered simplicial complex morphism if is a simplicial complex morphism and preserves the order that is $$\forall a, b \in K_1, a \leq_1 \Rightarrow f(a) \leq_2 f(b)$$. That is not a very strong requirement but I would say it is natural in some sense. We seek to prove that $$SimpCOrd$$ admits co-products. If we continue to consider only partial orders, the solution is trivial as the coproduct of the underlying sets $$V(K_1) \sqcup V(K_2)$$ (which is the disjoint union as you pointed) can be endowed a standard order where elements of $$V(K_1)$$ are only comparable with those of $$V(K_1)$$ (the co product of the relation sets underlying $$\leq_1$$ and $$\leq_2$$). Now if we restrict ourselves to (the sub category) the simplicial complexes with total orders, there is no unique construction which brings you there, as simply "gluing" the ordering chain formed by $$V(K_1)$$ on top of $$V(K_2)$$ ($$\forall a \in V(K_1), \forall b \in V(K_2), a > b$$) you get an order compatible with the injection maps of the coproduct but not more. However, the "alterning" order on $$V(K_1) \sqcup V(K_2)$$ defined by taking as smallest element the smallest element of $$V(K_1)$$ then the smallest of $$V(K_2)$$ and et cetera is just as good as the "gluing" one for the purpose of the coproduct, but they produce an asymmetry as the orders defined until there are not the same if we take the coproduct in one way or in another. Now with that intuition let's prove that no coproduct exists in the subcategory of simplicial complexes with total orders. Let $$X$$ be the pseudo "gluing" coproduct with $$K_1$$ "smaller than" $$K_2$$, $$f_1$$ the inclusion of $$K_1$$ in it and $$f_2$$ the inclusion of $$K_2$$. $$Y$$ the pseudo "gluing" coproduct with $$K_2$$ "smaller than" $$K_1$$ with $$g_1$$ and $$g_2$$ the respective inclusions.
Let consider the inclusion maps $$i_{K_1} : K_1 \rightarrow K_1 \sqcup K_2$$ and $$i_{K_2} : K_2 \rightarrow K_1 \sqcup K_2$$, by universal properties we have maps $$\sigma_1, \sigma_2$$ from the co product to the pseudo coproducts :
$$\sigma_1 \circ i_{K_1} (a) < \sigma_1 \circ i_{K_2} (b) \Rightarrow i_{K_1} (a) < i_{K_2} (b)$$. However $$\sigma_2 \circ i_{K_1} (a) > \sigma_2 \circ i_{K_2} (b) \Rightarrow i_{K_1} (a) > i_{K_2} (b)$$ a contradiction.