4.4 Continuous rigidity for countable frameworks
4.4.1 Continuous rigidity for countable generic frameworks in generic
We would naively hope that Theorem 2.1.5 would extend to infinite independent frameworks. Unfortunately, we can construct frameworks in normed planes with open sets of smooth points that are continuously rigid, independent - even generic if the normed plane is generic - but infinitesimally flexible, see Figure 4.2.
If continuous rigidity does not imply infinitesimal rigidity, then does infinitesimal rigidity imply continuous rigidity? We have more luck with this direction, especially for sequentially infinitesimally rigid frameworks, although there are still many open questions.
For the following we define a tower to beconstant and/orregular if each framework
Theorem 4.4.1. Let (G, p) be a framework in a normed space X. Suppose (G, p)
contains a vertex-complete sequentially infinitesimally rigid tower that is constant, then (G, p) is continuously rigid.
Proof. Suppose α:= (αv)v∈V(G) with αv : (−δ, δ)→X is a finite flex of (G, p) and let
((Gn, pn))
n∈N be a vertex-complete sequentially infinitesimally rigid tower of (G, p). We
note thatα|V(Gn)must also be a finite flex of (Gn, pn) for eachn ∈N. By Theorem 2.1.5, α|V(Gn) is trivial for eachn ∈N, thus for allt∈(−ϵ, ϵ) and alln∈N,α(t)|V(Gn)∈ Opn,
i.e. there exists for each t ∈ (−δ, δ) and n ∈ N some isometry hn
t ∈Isom(X) so that
hn
t.pn =α(t)|V(Gn).
Define for each t ∈ (−ϵ, ϵ) and n ∈ N the set Htn of such isometries g where g.pn=α(t)|V(Gn), then we note Htn =hnt Stabp. By Corollary 1.2.11, Htn is compact.
We note that for n ≤ m, Hm
t ⊆ Htn, and so as Htn ̸= ∅ for all n ∈ N, there exists
ht∈ ∩n∈NHtn. We now note ht.p=α(t), thus α is trivial as required.
Corollary 4.4.2. Suppose (G, p) is sequentially infinitesimally rigid in a normed space X with an open set of smooth points, then (G, p) is continuously rigid.
Proof. By Proposition 2.1.1, every regular framework will be constant. Since (G, p) is
sequentially infinitesimally rigid it contains a sequentially infinitesimally rigid tower. As this tower will be regular (and so constant), the result follows from Theorem 4.4.1.
Corollary 4.4.3. Suppose (G, p) is infinitesimally rigid and generic in a generic normed
plane X, then (G, p) is continuously rigid.
Proof. By Theorem 4.3.14, (G, p) is sequentially rigid. The result now follows from
Corollary 4.4.2.
4.4 Continuous rigidity for countable frameworks 175
Let (G, p) be a framework in a normed space X and αbe a finite flex of (G, p). We
define α to beproper if there existsϵ >0 and v, w∈V(G) so that
∥αv(t)−αw(t)∥ ̸=∥pv−pw∥
for all t∈(−ϵ,0)∪(0, ϵ). We note that all proper finite flexes are non-trivial. If (G, p)
has no proper finite flexes then (G, p) is weakly continuously rigid. It is immediate
that continuous rigidity implies weak continuous rigidity, and for finite frameworks, weak continuous rigidity and continuous rigidity are equivalent. We have the following conjecture.
Conjecture 4.4.4. A framework is weakly continuously rigid if and only if it is
continuously rigid, or equivalently, a framework has a proper finite flex if it has a non-trivial finite flex.
It is possible we would have to restrict to certain categories of frameworks (e.g. generic frameworks, periodic frameworks) for the above conjecture. So far the conjecture holds for all known examples.
Lemma 4.4.5. Suppose (G, p) is a finite generic framework in a normed spaceX with
an open set of smooth points, then there exists an open neighbourhood U of p such
that
fG−1[fG(p)]∩U =f⟨−G1⟩[f⟨G⟩(p)]∩U.
Proof. As (G, p) is generic then (G, p) and (⟨G⟩, p) are regular, thus by Proposition
2.1.1, both are constant. By Lemma 4.1.19, F(⟨G⟩, p) = F(G, p), thus it follows
from Lemma 2.1.4 that there exists an open neighbourhoods U′ of p such that both fG−1[fG(p)]∩U′ and f⟨−G1⟩[f⟨G⟩(p)]∩U′ are C1-submanifolds ofXV(G)), and both have
As
f⟨−G1⟩[f⟨G⟩(p)]∩U′ ⊆fG−1[fG(p)]∩U′ ⊆XV(G)
and both are C1-submanifolds of XV(G), the inclusion map
f⟨−G1⟩[f⟨G⟩(p)]∩U′ ,→fG−1[fG(p)]∩U′
is a C1-embedding, thus f−1
⟨G⟩[f⟨G⟩(p)]∩U′ is a C1-submanifold of fG−1[fG(p)]∩U′. As
both have the same tangent space at p, there exists an open neighbourhood U of p
such that fG−1[fG(p)]∩U =f⟨−G1⟩[f⟨G⟩(p)]∩U as required.
Theorem 4.4.6. Suppose (G, p) is infinitesimally rigid and generic in a normed space X with an open set of smooth points, then (G, p) is weakly continuously rigid.
Proof. Suppose there exists a proper flex α : (−δ, δ) → XV(G) of (G, p), then there
ϵ >0 andv, w∈V(G) such that
∥αv(t)−αw(t)∥ ̸=∥pv−pw∥
for all t ∈ (−ϵ,0)∪(0, ϵ). Let ((Gn, pn))n∈N be a complete relatively rigid tower of
(G, p), then for each n ∈N,
F(Gn+1, pn+1) = F(Gn+1∪KV(Gn), pn+1).
As (G, p) is generic it is also completely well-positioned, thus by Lemma 4.1.19
D Gn+1E p = D Gn+1∪KV(Gn) E p
4.4 Continuous rigidity for countable frameworks 177
for all n ∈N. As ((Gn, pn))
n∈N is a complete tower of (G, p), there exists m∈N such
that v, w∈V(Gm).
By Lemma 4.4.5, there exists an open neighbourhood U of pm+1 such that
fG−m1+1[fGm+1(pm+1)]∩U =f−1 ⟨Gm+1∪K V(Gm)⟩ f⟨Gm+1∪K V(Gm)⟩(p m+1)∩U.
As α|V(Gm+1) is a finite flex of (Gm+1, pm+1) it now follows that for some ϵ′ >0,
∥αv(t)−αw(t)∥=∥pv−pw∥
for all t ∈(−ϵ′, ϵ′), a contradiction.
Remark 4.4.7. Theorem 4.4.6 requires that our framework is generic; for instance,
figure 6 is shown in [38, Example 6.4] to be infinitesimally rigid but weakly continuously flexible.