The global rigidity of a framework is not an affine-invariant property

TL;DR

This paper demonstrates that global rigidity of frameworks is not preserved under affine transformations in any Euclidean space, contrasting with the invariance of infinitesimal rigidity under projective transformations.

Contribution

It proves that global rigidity is not an affine-invariant property in Euclidean spaces of any dimension, clarifying a previously less-understood aspect of rigidity theory.

Findings

Global rigidity is not affine-invariant in Euclidean d-space for d ≥ 2.

Infinitesimal rigidity remains invariant under projective transformations.

The result extends known planar cases to higher dimensions.

Abstract

It is well-known that the property of a bar-and-joint framework `to be infinitesimally rigid' is invariant under projective transformations of Eucliean $d$-space for every $d\geqslant 2$. It is less known that the property of a bar-and-joint framework `to be globally rigid' is not invariant even under affine transformations of the Euclidean plane. In this note, we prove of the latter statement for Euclidean $d$-space for every $d\geqslant 2$.

Full text

Keywords: bar-and-joint framework, globally rigid framework, Euclidean space, affine transformation, distance.

Mathematics subject classification (2010): Primary 52C25, Secondary 05C90, 51K05, 92E10.

1. Introduction

A bar-and-joint framework (or a straight-line realization of a graph) in $\mathbb{R}^{d}$, $d\geqslant 2$, is said to be globally rigid in $\mathbb{R}^{d}$ if it is congruent to every other bar-and-joint framework in $\mathbb{R}^{d}$ with the same edge lengths.

The problem of whether a given framework (or all frameworks from a given family) is globally rigid was raised both in mathematics and in its applications. As a purely mathematical problem it was raised, loosely speaking, in distance geometry (see, e. g., [2, 3, 4, 5]), graph theory (see, e. g., [6, 12, 14]), matroid theory (see, e. g., [9])), etc. Since frameworks are a natural model for real-world mechanisms and molecules, this problem appeared also in classical mechanics of mechanisms (see, e. g., [10, 13]), the mechanics of microporous materials (see, e. g., [15]), stereochemistry (see, e. g., [7]), molecular biology (see, e. g., [1, 11]), etc.

A special reason for our interest in globally rigid frameworks is that, for infinitesimally rigid frameworks, which may be treated as an infinitesimal analogue of globally rigid frameworks, a classical theorem reads that the property of a framework ‘to be infinitesimally rigid’ is invariant under projective transformations (see, e. g., [8, 16]). In contrast, Theorem 1, the main result of this note, reads that the property of a framework ‘to be globally rigid’ is not invariant even under affine transformations.

We recall basic definitions and notation from geometric global rigidity of bar-and-joint frameworks (see, e. g., [4]).

Let $d\geqslant 1$ be an integer and $G=(V,E)$ be a graph. Here $V$ and $E$ are the sets of vertices and edges of $G$, respectively.

A *bar-and-joint framework * $G(\mathbf{p})$ in $\mathbb{R}^{d}$ (or a framework, for short) is a graph $G$ and a configuration $\mathbf{p}$ which assigns a point $\mathbf{p}_{i}\in\mathbb{R}^{d}$ to each vertex $i\in V$. For each $i\in V$, $\mathbf{p}_{i}$ is called a joint of $G(\mathbf{p})$ and, for each edge $\{i,j\}\in E$, the staight-line segment with the endpoints $\mathbf{p}_{i}$ and $\mathbf{p}_{j}$ is called a bar.

Note that some authors use the term straight-line realization of a graph $G$ or geometric realization of a graph $G$ instead of the term bar-and-joint framework $G(\mathbf{p})$ (see, e. g., [6, 9]).

We denote the Euclidean distance between $\mathbf{x},\mathbf{y}\in\mathbb{R}^{d}$ by $|\mathbf{x}-\mathbf{y}|$.

Two frameworks $G(\mathbf{p})$ and $G(\mathbf{q})$ are equivalent to each other if $|\mathbf{p}_{i}-\mathbf{p}_{j}|=|\mathbf{q}_{i}-\mathbf{q}_{j}|$ for every $\{i,j\}\in E$ and are congruent to each other if $|\mathbf{p}_{i}-\mathbf{p}_{j}|=|\mathbf{q}_{i}-\mathbf{q}_{j}|$ for all $i,j\in V$.

A framework $G(\mathbf{p})$ is globally rigid in $\mathbb{R}^{d}$ if all frameworks $G(\mathbf{q})$ in $\mathbb{R}^{d}$ which are equivalent to $G(\mathbf{p})$ are congruent to $G(\mathbf{p})$.

Let $A:\mathbb{R}^{d}\to\mathbb{R}^{d}$ be an affine transformation and $G(\mathbf{p})$ be a framework in $\mathbb{R}^{d}$. We write $A\mathbf{p}$ for a configuration of the graph $G$ given by the formulas $(A\mathbf{p})_{k}=A(\mathbf{p}_{k})$, $k\in V$. In the sequel, where this cannot cause misunderstanding, we write $A\mathbf{x}$ instead of $A(\mathbf{x})$ for $\mathbf{x}\in\mathbb{R}^{d}$.

The main result of this note is the following

Theorem 1**.**

For every $d\geqslant 2$, there is a framework $G_{d}(\mathbf{p})$ in $\mathbb{R}^{d}$ and an affine transformation $A:\mathbb{R}^{d}\to\mathbb{R}^{d}$ such that $G_{d}(\mathbf{p})$ is not globally rigid in $\mathbb{R}^{d}$ while $G_{d}(A\mathbf{p})$ is globally rigid in $\mathbb{R}^{d}$.

Note that, Theorem 1 is not new for the case $d=2$. In fact, in [4, Example 8.3], a configuration in $\mathbb{R}^{2}$ is constructed that is not globally rigid in $\mathbb{R}^{2}$, while its affine image is globally rigid in $\mathbb{R}^{2}$. That example relies on the properties of coning and stress matrices, specific techniques for the study of global rigidity of bar-and-joint frameworks, developed in [4] and articles mentioned there.

In contrast, the proof of Theorem 1 given in Section 3 below is valid in Euclidean spaces of all dimensions and is straightforward, in particular, it does not involve coning.

2. Preliminary considerations

Let $G^{*}=(V^{*},E^{*})$ denote the graph such that $V^{*}=\{1,2,3,4,5\}$ and E^{*}=\bigl{\{}\{1,2\},\{1,3\},\{1,4\},\{1,5\},\{2,3\},\{2,4\},\{3,5\},\{4,5\}\bigr{\}}. $G^{*}$ is shown in Fig. 1(a). Let $\mathbf{p}^{*}$ denote the configuration of $G^{*}$ in $\mathbb{R}^{2}$ which is given by the following formulas: $\mathbf{p}^{*}_{1}=(0,2)$, $\mathbf{p}^{*}_{2}=(-1/4,1/2)$, $\mathbf{p}^{*}_{3}=(21/20,9/10)$, $\mathbf{p}^{*}_{4}=(-1,0)$, and $\mathbf{p}^{*}_{5}=(1,0)$. $G(\mathbf{p}^{*})$ is shown in Fig. 1(b).

Let $A:\mathbb{R}^{2}\to\mathbb{R}^{2}$ be the affine transformation given by the formula $A(x_{1},x_{2})=(x_{1},{x_{2}}/2)$.

Lemma 1**.**

Let the graph $G^{*}$, the framework $G^{*}(\mathbf{p}^{*})$ and the affine transformation $A$ be as above in this Section. Then $G^{*}(\mathbf{p}^{*})$ is not globally rigid in $\mathbb{R}^{2}$, while $G^{*}(A\mathbf{p}^{*})$ is globally rigid in $\mathbb{R}^{2}$.

Proof.

Let $\mathbf{q}^{*}$ be the configuration of $G^{*}$ in $\mathbb{R}^{2}$ which is given by the following formulas: $\mathbf{q}^{*}_{1}=\mathbf{p}^{*}_{1}=(0,2)$, $\mathbf{q}^{*}_{2}=(-21/20,9/10)$, $\mathbf{q}^{*}_{3}=(1/4,1/2)$, $\mathbf{q}^{*}_{4}=\mathbf{p}^{*}_{4}=(-1,0)$, and $\mathbf{q}^{*}_{5}=\mathbf{p}^{*}_{5}=(1,0)$. $G^{*}(\mathbf{q}^{*})$ is shown in Fig. 1(c).

We can say that $G^{*}(\mathbf{q}^{*})$ is obtained from $G^{*}(\mathbf{p}^{*})$ by replacing the joint $\mathbf{p}^{*}_{2}$ by the joint $\mathbf{q}^{*}_{2}$ and by replacing the joint $\mathbf{p}^{*}_{3}$ by the joint $\mathbf{q}^{*}_{3}$. Note that $\mathbf{q}^{*}_{2}$ is chosen in such a way that it is symmetrical to $\mathbf{p}^{*}_{2}$ with respect to the line $\mathbf{p}^{*}_{1}\mathbf{p}^{*}_{4}$ (here and subsequently $\mathbf{x}\mathbf{y}$ denotes the straight line passing through the points $\mathbf{x},\mathbf{y}\in\mathbb{R}^{2}$). Similarly, $\mathbf{q}^{*}_{3}$ is chosen in such a way that it is symmetrical to $\mathbf{p}^{*}_{3}$ with respect to the line $\mathbf{p}^{*}_{1}\mathbf{p}^{*}_{5}$.

Using the symmetry of some parts of the framework $G^{*}(\mathbf{p}^{*})$, it is easy to see that $|\mathbf{q}^{*}_{2}-\mathbf{q}^{*}_{3}|=|\mathbf{p}^{*}_{2}-\mathbf{p}^{*}_{3}|$. Hence, the frameworks $G^{*}(\mathbf{p}^{*})$ and $G^{*}(\mathbf{q}^{*})$ are equivalent to each other. However, we can arrive at the same conclusion by direct calculating the lengths of all bars of these frameworks.

On the other hand, $G^{*}(\mathbf{p}^{*})$ and $G^{*}(\mathbf{q}^{*})$ are not congruent to each other since $|\mathbf{p}^{*}_{2}-\mathbf{p}^{*}_{5}|\neq|\mathbf{q}^{*}_{2}-\mathbf{q}^{*}_{5}|$. In fact, direct calculations show that $|\mathbf{p}^{*}_{2}-\mathbf{p}^{*}_{5}|=\sqrt{29}/4$ and $|\mathbf{q}^{*}_{2}-\mathbf{q}^{*}_{5}|=\sqrt{2005}/20$.

Hence, $G^{*}(\mathbf{p}^{*})$ is globally rigid in $\mathbb{R}^{2}$.

Let $\widetilde{G}^{*}=(\widetilde{V}^{*},\widetilde{E}^{*})$ denote the graph such that $\widetilde{V}^{*}=V^{*}$ and \widetilde{E}^{*}=E^{*}\setminus\bigl{\{}\{2,3\}\bigr{\}}. $\widetilde{G}^{*}$ is shown in Fig. 2(a).

The configuration $\mathbf{p}^{*}$ of the graph $G^{*}$ and the affine transformation $A:\mathbb{R}^{2}\to\mathbb{R}^{2}$ constructed above in this Section define also the frameworks $\widetilde{G}^{*}(\mathbf{p}^{*})$ and $\widetilde{G}^{*}(A\mathbf{p}^{*})$, which are shown in Fig. 2(b) and Fig. 2(c), respectively.

Every framework in $\mathbb{R}^{2}$, which is equivalent to $\widetilde{G}^{*}(A\mathbf{p}^{*})$, is congruent to one of the frameworks shown in Fig. 3, where $\mathbf{x}_{2}=(-3/4,3/4)$ and $\mathbf{x}_{3}=(11/20,-1/20)$.

Note that the points $\mathbf{x}_{2}$ and $\mathbf{x}_{3}$ are chosen in such a way that $\mathbf{x}_{2}$ and $A\mathbf{p}_{2}^{*}$ are symmetrical to each other with respect to the line $A\mathbf{p}_{1}^{*}A\mathbf{p}_{4}^{*}$, and $\mathbf{x}_{3}$ and $A\mathbf{p}_{3}^{*}$ are symmetrical to each other with respect to the line $A\mathbf{p}_{1}^{*}A\mathbf{p}_{5}^{*}$.

Direct calculations show that $|A\mathbf{p}_{2}^{*}-A\mathbf{p}_{3}^{*}|=\sqrt{173}/10$ (see Fig. 3(a)), $|A\mathbf{p}_{2}^{*}-\mathbf{x}_{3}|=\sqrt{73}/10$ (see Fig. 3(b)), $|\mathbf{x}_{2}-\mathbf{x}_{3}|=\sqrt{233}/10$ (see Fig. 3(c)), and $|\mathbf{x}_{2}-A\mathbf{p}_{3}^{*}|=3\sqrt{37}/10$ (see Fig. 3(d)). Since among these numbers there are no two equal, the framework $G^{*}(A\mathbf{p}^{*})$ is globally rigid in $\mathbb{R}^{2}$. For the convenience of the reader $G^{*}(A\mathbf{p}^{*})$ is shown in Fig. 4(a). ∎

3. Proof of Theorem 1

In the proof of Theorem 1 given below, we mainly use the same ideas that were used in Section 2 in the proof of Lemma 1. The novelty is in details that are needed for transition from the plane to space.

Proof.

Let $\widetilde{G}_{d}=(\widetilde{V}_{d},\widetilde{E}_{d})$ denote the graph such that $\widetilde{V}_{d}=\{1,2,3,4,\dots,2^{d-1}+2,2^{d-1}+3\}$ and an unordered pair of non-coincident vertices $i,j$ is an edge of $\widetilde{G}_{d}$ (i. e. $\{i,j\}\in\widetilde{E}_{d}$) if and only if one of the conditions (i)–(iii) is fulfilled:

(i) $i,j\in\{1,4,5,\dots,2^{d-1}+2,2^{d-1}+3\}$ (i. e. the complete graph with the vertices $1,4,5,\dots,2^{d-1}+2,2^{d-1}+3$ is contained in $\widetilde{G}_{d}$);

(ii) the unordered pair $\{i,j\}$ coincides with either $\{1,2\}$ or $\{2,2k\}$, where $2\leqslant k\leqslant 2^{d-2}+1$ (i. e. the vertex 2 is connected by an edge to each of the vertices $1,4,6,\dots,2k,\dots,2^{d-1}+2$);

(iii) the unordered pair $\{i,j\}$ coincides with either $\{1,3\}$, or $\{3,2k+1\}$, where $2\leqslant k\leqslant 2^{d-2}+1$ (i. e. the vertex 3 is connected by an edge to each of the vertices $1,5,7,\dots,2k+1,\dots,2^{d-1}+3$).

Let $G_{d}=(V_{d},E_{d})$ denote the graph such that $V_{d}=\widetilde{V}_{d}$ and E_{d}=\widetilde{E}_{d}\cup\bigl{\{}\{2,3\}\bigr{\}}. $G_{2}$ is shown in Fig. 1(a), $\widetilde{G}_{2}$ is shown in Fig. 2(a), and $G_{3}$ is shown in Fig. 4(b).

The points $(\pm 1,\pm 1,\dots,\pm 1)\in\mathbb{R}^{d-2}$ are the vertices of a cube with edge length 2. We enumerate them in an arbitrary way using the numbers $1\leqslant k\leqslant 2^{d-2}$ and denote them by $w_{k}$. So, $w_{k}=(\pm 1,\pm 1,\dots,\pm 1)$ with the proper selection of plus and minus signs.

Let $\mathbf{p}$ denote the configuration of the graphs $G_{d}$ and $\widetilde{G}_{d}$ in $\mathbb{R}^{d}$ that is given by the following formulas: $\mathbf{p}_{1}=(0,2,0,\dots,0)$, $\mathbf{p}_{2}=(-1/4,1/2,0,\dots,0)$, $\mathbf{p}_{3}=(21/20,9/10,0,\dots,0)$, $\mathbf{p}_{2j}=(-1,0,w_{j-1})$, and $\mathbf{p}_{2j+1}=(1,0,w_{j-1})$, where $2\leqslant j\leqslant 2^{d-2}+1$.

Let $\mathbf{q}$ denote the configuration of the graphs $G_{d}$ and $\widetilde{G}_{d}$ in $\mathbb{R}^{d}$ that is given by the following formulas: $\mathbf{q}_{1}=\mathbf{p}_{1}$, $\mathbf{q}_{2}=(-21/20,9/10,0,\dots,0)$, $\mathbf{q}_{3}=(1/4,1/2,0,\dots,0)$, and $\mathbf{q}_{k}=\mathbf{p}_{k}$, where $k=4,5,\dots,2^{d-1}+2,2^{d-1}+3$.

Note that $\mathbf{q}_{2}$ is chosen in such a way that it is symmetrical to $\mathbf{p}_{2}$ with respect to the hyperplane in $\mathbb{R}^{d}$ passing through the points $\mathbf{p}_{1}$, $\mathbf{p}_{4}$, $\mathbf{p}_{6}$, …, $\mathbf{p}_{2^{d-1}}$, $\mathbf{p}_{2^{d-1}+2}$. Similarly, $\mathbf{q}_{3}$ is chosen in such a way that it is symmetrical to $\mathbf{p}_{3}$ with respect to the hyperplane in $\mathbb{R}^{d}$ passing through the points $\mathbf{p}_{1}$, $\mathbf{p}_{5}$, $\mathbf{p}_{7}$, …, $\mathbf{p}_{2^{d-1}+1}$, $\mathbf{p}_{2^{d-1}+3}$.

As in the proof of Lemma 1, a direct verification shows that the frameworks $G_{d}(\mathbf{p})$ and $G_{d}(\mathbf{q})$ are equivalent to each other, but not congruent. Hence, $G_{d}(\mathbf{p})$ is not globally rigid.

Let $A:\mathbb{R}^{d}\to\mathbb{R}^{d}$ be the affine transformation given by the formula $A(x_{1},x_{2},x_{3},\dots,x_{d})=(x_{1},{x_{2}}/2,x_{3},\dots,x_{d})$ (i. e. $A$ is the contraction with the factor of 2 along the second axis of $\mathbb{R}^{d}$).

Let $\mathbf{r}$ denote the configuration of the graphs $G_{d}$ and $\widetilde{G}_{d}$ in $\mathbb{R}^{d}$ that is given by the following formulas: $\mathbf{r}_{1}=A\mathbf{p}_{1}$, $\mathbf{r}_{2}=A\mathbf{p}_{2}$, $\mathbf{r}_{3}=(11/2,-1/20,0,\dots,0)$, and $\mathbf{r}_{k}=A\mathbf{p}_{k}$, where $k=4,5,\dots,2^{d-1}+2,2^{d-1}+3$. Note that $\mathbf{r}_{3}$ is symmetrical to $A\mathbf{p}_{3}$ with respect to the hyperplane in $\mathbb{R}^{d}$ passing through the points $A\mathbf{p}_{1}$, $A\mathbf{p}_{5}$, $A\mathbf{p}_{7}$, …, $A\mathbf{p}_{2^{d-1}+1}$, $A\mathbf{p}_{2^{d-1}+3}$.

Let $\mathbf{s}$ denote the configuration of the graphs $G_{d}$ and $\widetilde{G}_{d}$ in $\mathbb{R}^{d}$ that is given by the following formulas: $\mathbf{s}_{1}=A\mathbf{p}_{1}$, $\mathbf{s}_{2}=(-3/4,3/4,0,\dots,0)$, $\mathbf{s}_{3}=\mathbf{r}_{3}$, and $\mathbf{s}_{k}=A\mathbf{p}_{k}$, where $k=4,5,\dots,2^{d-1}+2,2^{d-1}+3$. Note that $\mathbf{s}_{2}$ is symmetrical to $A\mathbf{p}_{2}$ with respect to the hyperplane in $\mathbb{R}^{d}$, passing through the points $A\mathbf{p}_{1}$, $A\mathbf{p}_{4}$, $A\mathbf{p}_{6}$, …, $A\mathbf{p}_{2^{d-1}}$, $A\mathbf{p}_{2^{d-1}+2}$.

Finally, let $\mathbf{t}$ denote the configuration of the graphs $G_{d}$ and $\widetilde{G}_{d}$ in $\mathbb{R}^{d}$ that is given by the following formulas: $\mathbf{t}_{1}=A\mathbf{p}_{1}$, $\mathbf{t}_{2}=\mathbf{s}_{2}$, $\mathbf{t}_{3}=\mathbf{r}_{3}$, and $\mathbf{t}_{k}=A\mathbf{p}_{k}$, where $k=4,5,\dots,2^{d-1}+2,2^{d-1}+3$.

As in the proof of Lemma 1, we make sure that any framework equivalent to $\widetilde{G}_{d}(A\mathbf{p})$ in $\mathbb{R}^{d}$ is congruent to one of the frameworks $\widetilde{G}_{d}(A\mathbf{p})$, $\widetilde{G}_{d}(\mathbf{r})$, $\widetilde{G}_{d}(\mathbf{s})$, or $\widetilde{G}_{d}(\mathbf{t})$. Direct calculations show that $|A\mathbf{p}_{2}-A\mathbf{p}_{3}|=\sqrt{173}/10$, $|\mathbf{r}_{2}-\mathbf{r}_{3}|=\sqrt{73}/10$, $|\mathbf{s}_{2}-\mathbf{s}_{3}|=\sqrt{233}/10$, and $|\mathbf{t}_{2}-\mathbf{t}_{3}|=3\sqrt{37}/10$. Since among these numbers there are no two equal, none of the frameworks $G_{d}(A\mathbf{p})$, $G_{d}(\mathbf{r})$, $G_{d}(\mathbf{s})$, and $G_{d}(\mathbf{t})$ are equivalent to each other. Consequently, $G_{d}(A\mathbf{p})$ is globally rigid. ∎

4. Acknowledgements

The author is indebted to Professor Steven Gortler for bringing his attention to Example 8.3 in [4].