Suppose that $M$ is an $n$-dimensional manifold equipped with a triangulation $T$. Given $n\ge 1$, in order to recover $T$ (up to an isomorphism of simplicial complexes) one needs to know at least the 1-dimensional skeleton of $T$. How large $k=k(n)$ does one need to recover the triangulation $T$ from the $k$-skeleton of $T$?
More formally. Let $X^k$ denote the $k$-skeleton of a simplicial complex $X$. Given $n\ge 1$, define $k(n)$ as the least $k$ such that for any two $n$-manifolds (without boundary) equipped with triangulations $X, Y$, if $X^k$ is isomorphic to $Y^k$ then $X$ is isomorphic to $Y$ (as a simplicial complex).
Question. How much is known about the function $k(n)$?
For instance, is it true that $k(n)\ge 2$ for all $n\ge 3$? (In other words, is it true that for each $n\ge 3$ there exist triangulated $n$-manifolds with isomorphic 1-skeleta of triangulations but non-isomorphic triangulations?) Is it true that $k(n)\ge n/2$ if $n$ is sufficiently large? Etc.
One can ask various other versions of this question. For instance, instead of requiring triangulations $X, Y$ to be isomorphic, only require the corresponding triangulated manifolds to be PL isomorphic. One can also restrict to pairs of homeomorphic $n$-manifolds.
Edit. I just found
Jerome Dancis, Triangulated $n$-manifolds are determined by their $[n/2] + 1$-skeletons. Topology Appl., 18(1):17–26, 1984.
He proves that compact triangulated $n$-manifolds are determined by their $\lceil (n+1)/2 \rceil$-dimensional skeleta. Together with Richard Stanley's reference, this yields that in the setting of compact manifolds $$ k(n)= \lceil (n+1)/2 \rceil, $$ except in the case of surfaces.
