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).



Generic infinitesimal rigidity of polyhedra
In short, no, this cannot happen.


To be more specific. A standard definition for ``generic'' is that the coordinates of $P \in \mathbb{R}^{3V}$ form an algebraically independent set. It is a nice exercise proving that: (i) almost all (in the measure theory sense) points in $\mathbb{R}^{3V}$ are generic, and (ii) every generic polyhedra is infinitesimally rigid.


Bob Connelly proved in his paper *Generic global rigidity* (see Proposition 3.3) the following: given a generic $P \in \mathbb{R}^{3V}$ and $Q \in [P]$, if $P$ is infinitesimally rigid then so too is $Q$. (His result is stated in the more general setting of bar-joint frameworks and so is more general.) The result you need now just follows from basic differential geometry.


