Can there be arbitrary vertices on the edge of an abstract simplicial complex?

Question

$$P = \{a, b, c\}$$ $$R_t(P) = \{\{a\}, \{b\}, \{c\}, \{a, b\}, \{b, c\}, \{c, a\}, \{a, b, c\}\}$$

For a Rips complex $R_t(P)$, can there be other points(vertices) on the edges({a, b}, {b, c}, {c, a})?

I was wondering this while I was trying to prove the reason Rips complex include "triangles (so called)" when it has all the "edges (so called)". The proof for this I came up with is, "When a Rips complex includes all of the edges, we can consider arbitrary vertices p, q in the edges as well. And the distance between these points is smaller or equal to t. That's why having all of the edges makes the Rips complex include triangles as well."

But I'm not sure if I can choose an arbitrary point on an edge in abstract simplicial complex.

Answer 1

No, you can't.

What exactly is "edge"? Typically this is defined as an abstract simplex of size $2$. Analogously "triangle" is often defined as abstract simplex of size $3$. Here by "abstract simplex" I understand any element of $S$ where $(X,S)$ is an abstract simplicial complex.

And so there are no intermediate points on edge. This is meaningless. You could of course take geometric realization of $(X,S)$ and this one does have intermediate points on edges. But in general the geometric realization does not need to correspond to the original space on which Rips complex was built. In fact the original space may consist finitely many vertices only, and so no intermediate points are there. It would even be difficult to define what "intermediate" means for an arbitrary metric space.

I was wondering this while I was trying to prove the reason Rips complex include "triangles (so called)" when it has all the "edges (so called)".

This has to be properly defined to be sensible. One way is as follows:

Let $(M,S)$ be Rips complex of some (metric) space $M$. Assume that $n>2$ and $\{x_1,\ldots,x_n\}\subseteq M$ is a set of points such that for any $i=1,\ldots,n$ we have $\{x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n\}\in S$. Then $\{x_1,\ldots,x_n\}\in S$.

Proof. Recall that the Rips complex over $M$, for a fixed $\delta>0$, is defined as follows: vertices of the Rips complex is simply the underlying set of $M$ and abstract simplexes are all finite subsets of $M$ with diameter at most $\delta$. So our statement translates to: if $diam(\{x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n\})\leq \delta$ for any $i$, then $diam(\{x_1,\ldots,x_n\})\leq \delta$.

Denote by $d:M\times M\to\mathbb{R}$ the metric of $M$.

Assume that's not the case, i.e. $diam(\{x_1,\ldots,x_n\})> \delta$. Since we deal with a finite set, then there are $x_i,x_j$ such that $diam(\{x_1,\ldots,x_n\})=d(x_i,x_j)$. Since $n>2$ then we can take a point $x_k$ where $k\neq i$ and $k\neq j$. Consider $A=\{x_1,\ldots,x_n\}\backslash\{x_k\}$. Since $x_i,x_j\in A$ and the distance between them was maximal then $diam(A)=d(x_i,x_j)>\delta$. But on the other hand our $A=\{x_1,\ldots,x_{k-1},x_{k+1},\ldots,x_n\}$ is assumed to have $diam(A)\leq\delta$. Contradiction. $\Box$