Rigid frameworks with dilation constraints

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Problem: Conjecture 6.1 states: for $1\le k<d$, if a generic framework $(G,p)\subset \mathbb R^d$ is globally $(d,k)$-rigid and $G$ is not complete, then there is

$$\sigma\in \ker DR_k(G,p)^T$$

whose stress matrix satisfies

$$\operatorname{rank}\Omega(\sigma)=|V|-d+k-1 .$$

Here $DR_k(G,p)=[R(G,\tilde p)\ f_{G,1}(p_{d-k+1})\cdots f_{G,1}(p_d)]$, and $\Omega(\sigma)$ is the usual graph Laplacian stress matrix.

Result: The conjecture is false.

Take $d=3,\ k=2$, and let

$$G=K_4-e,\qquad V=\{1,2,3,4\},\quad E=\{12,13,23,14,24\}.$$

So $G$ is two triangles sharing the edge $12$, and is not complete.

Consider the rational framework

$$p^0_1=(0,1,2),\quad p^0_2=(3,5,7),\quad p^0_3=(-2,11,13),\quad p^0_4=(4,17,19).$$

For a $(3,2)$-equivalent framework $q$, write

$$q_i=(X_i,\lambda y_i,\mu z_i),\qquad t=\lambda^2,\ s=\mu^2 .$$

For each triangle $12r$, $r=3,4$, the modified squared edge lengths must be realizable on a line in the $X$-coordinate. The Cayley collinearity equations for $p^0$ are

$$36st-1885s-1480t+3329=0,$$ $$144st-1105s-1168t+2129=0.$$

Their only real solutions are

$$(t,s)=(1,1) \quad\text{and}\quad (t,s)=\left(-\frac{845}{432},\frac{1862}{585}\right).$$

Since $t=\lambda^2\ge0$, $s=\mu^2\ge0$, only $(t,s)=(1,1)$ is admissible.

This property is open under small perturbations of $p^0$. Also, at $p^0$, the matrix $DR_2(G,p^0)$ has full row rank $5$; for example, the minor using columns $x_1,x_2,x_3$ and the two dilation columns has determinant $-168564$. Hence, after shrinking to a nonempty open neighbourhood $U$ of $p^0$, every $p\in U$ still has:

1. the only admissible dilation squares $t,s$ equal to $1,1$;
2. $\operatorname{rank}DR_2(G,p)=5=|E|$.

Choose $p\in U$ with algebraically independent coordinates; such $p$ is generic. For this $p$, any $(3,2)$-equivalent $q$ has $\lambda,\mu=\pm1$. Then the remaining $X$-coordinates form a one-dimensional framework on $K_4-e$ with the same edge lengths as $x$. Since the shared edge $12$ determines vertices $3$ and $4$ from their distances to $1,2$, $X$ is congruent to $x$ on the line. Thus $q$ is obtained from $p$ by an isometry of $\mathbb R^3$. Therefore $(G,p)$ is globally $(3,2)$-rigid.

But $\operatorname{rank}DR_2(G,p)=|E|=5$, so

$$\ker DR_2(G,p)^T=\{0\}.$$

Thus no nonzero stress exists, while the conjectured rank is

$$|V|-d+k-1=4-3+2-1=2.$$

So Conjecture 6.1 fails.

A natural repair would need at least to exclude stressless minimally $(d,k)$-rigid globally rigid examples such as this one, e.g. by adding some redundant-rigidity or nonzero-stress hypothesis.

Citation: No known prior resolution used. Definitions and conjecture are from Dewar–Nixon–Sainsbury, “Rigid frameworks with dilation constraints,” arXiv:2402.14093.