Rigidity and Symmetry by Robert Connelly Asia Ivić Weiss & Walter Whiteley

Author:Robert Connelly, Asia Ivić Weiss & Walter Whiteley
Language: eng
Format: epub
Publisher: Springer New York, New York, NY

Proof.

We suppose that all realisations of H considered are in standard position with respect to (w, x). For each such realisation (H, q) we suppress the (zero) coordinates of q corresponding to q(w) and the first coordinate of q(x) and consider .

Let

Let be given by (in the corresponding realisation (H, q)) and let f be the restriction of F to S. We can also view the rigidity maps as maps on S. Note that the rigidity map is obtained from f H by adding an extra coordinate corresponding to e 2 i.e. the length of the edge e 2 in the realisation of H + e 2.

We can adapt the proof technique of [12] to show that S is a 1-dimensional manifold diffeomorphic to a circle. For each p ∈ S, [17, Lemma 3.4] gives rank . Thus, for every generic point p ∈ S, we have so p is a regular point of f.

Choose a direction for traversing S and let p 1 be the first point after p 0 we reach when traversing S which satisfies . We will show that . Suppose to the contrary that . Then (C, p 0) is equivalent to (C, p 1).

We first consider the case when C is 3-connected. Then C is globally rigid by [3] so (C, p 0) is congruent to (C, p 1). Since (C, p 0) and (C, p 1) are in standard position , where α is a reflection in one of the two coordinate axes or a rotation of π about the origin. Let a: [0, 1] → S be the smooth path from p 0 to p 1 induced by the diffeomorphism from S to the circle, and let b: [0, 1] → S be obtained by putting b(t) = α(a(t)) for all 0 ≤ t ≤ 1. Then b is a smooth path in S from p 1 to p 0. Furthermore, we claim that a and b do not have the same image in S. For suppose to the contrary that a and b traverse some path P in S in opposite directions. Then by the intermediate value theorem there is some t ∈ [0, 1] with a(t) = b(t). This implies that (H, a(t)) has all vertices on one of the two coordinate axes, which is impossible since (H, p), and hence also (H, a(t)), has 2 | V (H) | − 4 algebraically independent edge-lengths. It follows that a and b trace out two paths that together form the entire manifold S. We can choose with and . Now the intermediate value theorem gives some t between t 1 and t 2 with f(a(t)) = f(p 0). This contradicts the choice of p 1.

We next consider the case when C is not 3-connected. Let be the path in the cleavage unit tree of C with and . (We refer the reader to [15, Section 3] for more details on cleavage unit trees of M-circuits.) Let . If C ′ ≠ C then we can apply induction to C ′ . Hence we may assume that C ′  = C. We will proceed by adapting the proof of the case when C is 3-connected.

Let for 1 ≤ i < m. For each p ∈ S, let ℓ i (p) be the line through for 1 ≤ i < m.

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