TITLE: The number of $2$-simplices and the number of $1$-simplices in a $4$-dimensional simplicial complex
QUESTION [7 upvotes]: Given a $4d$ simplicial complex (a triangulation of $4$-manifold), is there any relation between the number of $2$-simplices (triangles) and the number of $1$-simplices (edges)? Generically, is the number of $2$-simplices always greater than the number of $1$-simplices? Thanks!

REPLY [5 votes]: The answer is yes for pure abstract simplicial complexes. Thus we consider a finite set $X$ and a system of subsets $S$ in $X$, such that $S$ is closed under taking subsets, every $s \in S$ has at most $5$ elements and is contained in at least one $t\in S$ with $5$ elements.
Denote by $m$ the number of $s\in S$ with $2$ elements and by $n$ the number of elements $t\in S$ with $3$ elements. (These are the numbers of one-dimensional and two-dimensional simplices in the abstract simplicial complex $(X,S)$, respectively.)
Let $Y$ be the set  of pairs $(s,t)$, such that $s$,$t\in S$, $\# s=2$, $\# t=3$, and $s$ is contained in $t$.
Then the number of elements in $Y$ equals $3n$. In fact every $t\in S$ with three elements, contains exactly three subsets of cardinality two. 
On the other hand, every $s\in S$ with $\# s=2$, is contained in at least one $v\in S$ with $\# v=5$. Now, there are three different $t$ such that $s\subset t \subset v$ and $\# t =3$. All these $t$ are elements of $S$. Therefore the number of elements in $Y$ is at least $3m$. 
Hence $3n \ge 3m$, which is equivalent to $n\ge m$.