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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!

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Perhaps the best results in this direction are given by Klee's Dehn-Somerville equations:

Klee, V., A combinatorial analogue of Poincar\'e's duality theorem, Can. J. Math. 16, 517-531 (1964). ZBL0134.42403.

These give linear relations between the numbers of faces of different dimensions of a combinatorial manifold (and the Euler characteristic in the even dimensional case).

Let $f_i=f_i(M)$ denote the number of $i$-dimensional faces of a combinatorial manifold $M$ with Euler characteristic $\chi=\chi(M)$. If $M$ is orientable and $4$-dimensional, then Theorem 3.2 in the cited paper gives that the vector $$ (\frac12\chi,f_0,f_1,\ldots, f_4) $$ is a linear combination of the rows in the matrix $$ \left(\begin{array}{cccccc} 1 & 2 & & & & \newline & 1 & 3 & 2 & & \newline & & 1 & 4 & 5 & 2 \end{array}\right). $$ From this it can be deduced that $$ f_2 = 4f_1-10f_0 +10\chi. $$

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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 $n\ge m$.

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  • $\begingroup$ This is a nice argument, but a bit heavy notationally. In order to follow it, I found it helpful to translate "$s$ with $\#s=2$" into "edges", etc, and likewise, use more descriptive names for $n,m,Y$ and so on (for example, $Y$ is the set of flags). $\endgroup$ – Victor Protsak May 25 '17 at 13:35

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