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 



When can a 3-dimensional triangulation be isometricaly embedded in R^n?
The ratio $E/V$ has no upper bound. Moreover, for every $n\ge 4$ there exists a triangulation of a tetrahedron in $\mathbb R^3$ with $V=n$ and $E=n(n-1)/2$ (so that the edges form a complete graph).


To construct such a triangulation, consider a curve $\gamma:\mathbb R\to\mathbb R^4$ defined by
$$
 \gamma(t) = (t,t^2,t^3,t^4) .
$$
Pick $n$ distinct points $p\_1,\dots,p\_n$ on this curve and let $P$ be their convex hull. It is a 4-dimensional convex polytope. It is easy to see (and well-known) that every segment $[p\_ip\_j]$ is an edge of this polytope. It follows that all faces of $P$ are simplices. So the boundary of $P$ is a triangulated 3-sphere with $n$ vertices and $n(n-1)/2$ edges. Now a standard construction (the stereographic projection from a point lying outside $P$ near one of its faces) turns the face complex of $P$ into a triangulation of one of its faces.


