Triangulated manifold implies properties of simplicial complex?

Question

I've been told that a compact $d$-dimensional manifold can be realized as a finite $d$-dimensional simplicial complex. Since any manifold is locally compact (I think), if it can be realized as a $d$-dimensional simplicial complex, is it somehow locally finite? Perhaps in the sense that every simplex is contained in a finite number of simplices?

Can we then also say this sort of statement for locally euclidean spaces which can be realized as a simplicial complex?

Answer 1

For CW-complexes (which cover simplicial complexes as special cases) local finiteness is equivalent to local compactness.

1. Let $X$ be a locally compact CW-complex. Then each $x \in X$ has a relatively compact open neighborhood $U$. The set $\overline U $ is compact, thus contained in a finite subcomplex $F \subset X$. Hence $U$ intersects only finitely many cells.

2. Let $X$ be a locally finite CW-complex. Then each $x \in X$ has an open neighborhood $V$ which intersects only finitely many cells. Hence $V$ is contained in a finite subcomplex $G \subset X$. This is compact, thus $V$ is relatively compact.

Answer 2

A compact manifold can be covered in finitely many charts (compactness) and we can assume without loss of generality that the domain of each chart is a contractible open set. Does that give you the kind of local finiteness you are looking for?