Let $M$ be a 1-skeleton of a triangulation of a sphere with $V$ vertices and $E$ edges.
Definition 1 A polyhedron is a map $M\to \mathbb R^3$ that is affine on edges (and non-degenerate on faces). The space of polyhedra is identified with (apparantly an open subset of) $\mathbb R^{3V}$.
Definition 2 Two polyhedra are called isometric if they induce the same metric on $M$. We will denonote this fact by $[P]=[Q]$
Definition 3 A polyhedron $P$ is called infinitesimally rigid if the kernel (at $P$) of the Jacobian of the natural map $f:\mathbb R^{3V}\to\mathbb R^E$ which sends $P$ to the tuple of square-lenghts of its edges has the minimal possible dimension equal to 6. In other words $P$ has no nontrivial infinitesimal deformations (or more algebraically: Zariski tangent space to the scheme $f^{-1}f(P)$ has dimension $6$ at $P$).
It is a well known result of Gluck that the set of infinitesimally rigid polyhedra $M\to \mathbb R^3$ form an open and dense set in $\mathbb{R}^{3V}$. However it can happen that in the set [P] of polyhedra isometric to a given infinitesimally rigid polyhedron $P$ there is a non-rigid representative. Is this phenomenon non-typical? In other words, is the following true:
For a generic $P\in \mathbb R^{3V}$ (you may chose suitable definition) $[P]$ is 6-dimensional (that is, all polyhedra isometric to $P$ are infinitesimally rigid). To put it differently, after quotienting out by isometries of the euclidean space we have that the moduli space of a generic metric polyhedron is a finite union of points (without nilpotents).