Consider a PL $n$-manifold $M$. Define the complexity $c(M)$ of $M$ to be the minimum number of $n$-simplices needed to triangulate $M$.

For $n>2$ getting lower bounds on $c(M)$ is a difficult problem, to say the least, however I wonder if we can approach this using a "local-to-global" type argument, in the following sense:

Suppose we can estimate (in fact, in my case, compute exactly) the complexity of $M$ locally based on the neighbourhoods of the $k$-simplices for $k<n$. For example, consider the vertices of a triangulation $\mathcal{T}$ of $M$. For a vertex $v$, the star of $v$, $\mathrm{St}(M)$ is effectively a triangulation of a regular neighbourhood of $v$. So suppose we have that $K≤c(\mathrm{St(v)})$, for some quantity $K$ (whatever that happens to be), for all vertices $v$ of $\mathcal{T}$. The question then, is can we somehow get the same estimate on the global complexity of $M$ in its entirety, i.e. show $K≤c(M)$.

As someone mentioned in a comment on the first version of this question, one obvious thing to try is to simply sum over all vertices of $\mathcal{T}$, however there is going to be a tremendous amount of overcounting going on here (since e.g. neighbourhoods of vertices could easily overlap). So I would even appreciate if anyone could give me a hand in getting a handle on by how much we are overcounting using this sum approach.

Naively I thought maybe one could also hope to argue something like "Property X holds on a neighbourhood of the vertices (i.e. holds on $\mathrm{St}(v)$, for each vertex $v$ of $\mathcal{T}$), and so extending linearly over the simplices of $\mathcal{T}$, X holds on $\mathcal{T}$" - or something like this...

Happy to clarify anything further if need be.