Notation for set of $n$-simplices in a simplicial complex? Let $K$ be a simplicial complex. Is there a commonly used notation for the set of all $n$-simplices of $K$?
$K_n$ looks natural, but is $K_n$ already meaning something else?
I learn of the notation $sk_n$ meaning the $n$-skeleton of $K$. So basically, I am finding a neat notation for $sk_n-sk_{n-1}$.
Thanks.
 A: The standard and universal notation is, indeed, $K_n$.
A: I don't know if there's anything standard, but you could use $${K \choose n}.$$
My reasoning is this.
If $X$ is a set, then it make sense to define that $${X \choose n} = \{A \subseteq X : |A|=n\}.$$
That way, we have:
$$\left|{X \choose n}\right| = {|X| \choose n}$$ where the RHS is the usual binomial coefficient of natural numbers.
Now by an abstract simplicial complex, lets mean an ordered pair $(X,E)$ where $X$ is a set and $E$ is a collection of subsets of $X$. Usually the empty subset is excluded from consideration, but lets allow it to be included. By the way, in graph theory these are called hypergraphs. (The definition of hypergraphs also typically excludes $\emptyset$ just like in algebraic topology. Again, for the purposes of this question lets allow it to be an element of $E$.)
It makes sense to define that if $K = (X,E)$ is a hypergraph (abstract simplicial complex, whatever), then $${K \choose n} = \{A \in E : |A| = n\}.$$
The reason is this. Each set $X$ naturally associated with two corresponding hypergraphs, namely $(X,\{\})$ and $(X,\{A \subseteq X\}).$ The latter is a bit more useful/natural, so it's convenient to identify $X$ with $(X,\{A \subseteq X\}).$ In light of this identification, we observe that $${X \choose n} = {(X,\{A \subseteq X\}) \choose n} = \{B \in \{A \subseteq X\} : |B| = n\} = \{B \subseteq X : |B| = n\},$$ which agrees with our previous definition.
Remark. You could upgrade the notation to include $k$-skeletons as well. Whenever $p$ is a predicate on the natural numbers, define
$${K \choose p} = \{A \in K : p(|A|)\}.$$
Then $${K \choose \Box = n} = {K \choose n}, \qquad {K \choose \Box \leq n} = sk_n(K).$$
