Consider a triangulation of some bounded region of $R^3$ with a (finite) set of tetrahedra (like in Regge calculus). It can be thought of as a simplicial 3-complex with specified lengths of edges. The other way around - a 3-complex with specified lengths of edges can sometimes be isometrically embedded in $R^3$. Now let's modify some of the lengths in an arbitrary way. Chances are - this triangulation can no longer be embedded in $R^3$. My question is: what is the minimal $n$ such, that our triangulation can always be isometrically embedded in $R^n$? My hypothesis is 6.
It is of course possible to modify the lengths in such a way that there is no embedding in any dimension - this happens if some triangle inequality is broken. I'm only interested in modifications small enough that it doesn't happen. Also: I'm only interested in the generic, non-degenerate case.
If there is a proof that $6V-E\geq 0$ (or $nV-E\geq 0$)* for such triangulations, where $V$ is the number of vertices and $E$ is the number of edges, I would be satisfied with that. It would mean that there is more degrees of freedom than equations, when one tries to embed it in $R^6$ (or $R^n$).
*in reality the equations have $O(n)$ symmetry, so there should be some constant on the r.h.s. other than $0$, but I want to know if such inequality makes sense at all for now