2.1 Existence of free bases

Let A be a countably infinite set and let α = (A _i) be an increasing sequence of finite subsets of A which cover A (i.e. A _i ⊂ A _i+1 for all $i\in {{\mathbb {N}}}$ and $A=\cup _{i\in {{\mathbb {N}}}} A_i)$. For each $i\in {{\mathbb {N}}}$, let $\pi _i:X^{A}\to X^{A_i}$ denote the natural restriction map. Let W be an infinite dimensional vector subspace of X ^A and let W _i = π _i(W) for each $i\in {{\mathbb {N}}}$. For all $i, j\in {{\mathbb {N}}}$ with j ≥ i, let π _j,i:W _j → W _i be the natural restriction maps. Note that the pair ((W _i), (π _j,i)) is an inverse sequence of finite dimensional vector spaces.

Given W and α as above, denote by $\varprojlim W_i$ the inverse limit of the inverse sequence ((W _i), (π _j,i)). Thus $\varprojlim W_i$ is the vector space of all (f _i) in the Cartesian product $\Pi _{i\in {{\mathbb {N}}}} W_i$ with the property that π _j,i(f _j) = f _i whenever j ≥ i. In the following, we consider the linear map $\Phi _\alpha : X^{A}\to \Pi _{i\in {{\mathbb {N}}}} W_i$, $f\mapsto (\pi _i(f))$.

Given f ∈ X ^A we write $\|f\|_\infty = \sup _{a\in A} \|f(a)\|$. A mapping f ∈ X ^A is bounded if $\|f\|_\infty <\infty$. Otherwise, f is said to be unbounded.

Proof. By Theorem 2.9, there exists a free basis S for W with the property that $W={{\mathcal {M}}}(S)$. Since S is countably infinite, we may choose a sequence (h _k) in S and a sequence (a _k) in A so that

\[ h_k(a_k) \neq 0\text{ and } h_k(a_j)=0, \quad \text{for } 1 \leq j \leq k-1. \]

Replacing h _k by a scalar multiple of h _k, we may assume for convenience that $\|h_k(a_k)\|=1$ for all k. Choose non-zero scalars (α _1,n) successively such that α _1,1 = 1 and $|\alpha _{1,n}|\geq n+\|\sum \nolimits _{k=1}^{n-1} \alpha _{1,k} h_k(a_n)\|$ for each n ≥ 2. Let $g_1= \sum \nolimits _{k=1}^{\infty } \alpha _{1,k}h_{k}\in W$ and note that for each $n\in {{\mathbb {N}}}$,

\[ \|g_1(a_n)\| = \|\sum_{k=1}^{n} \alpha_{1,k} h_{k}(a_n)\| \geq \|\alpha_{1,n} h_n(a_n)\| - \|\sum_{k=1}^{n-1} \alpha_{1,k} h_k(a_n)\| \geq n. \]

Thus, g ₁ is unbounded. For m ≥ 2, similarly define $g_m = \sum \nolimits _{k=m}^{\infty } \alpha _{m,k} h_{k}$ with non-zero scalars (α _m,k) chosen so that g _m is unbounded. To see that the set $\{g_m:m\in {{\mathbb {N}}}\}$ is linearly independent note that if $\sum \nolimits _{m=1}^{d}\lambda _mg_{m}=0$ for some $\lambda _1, \ldots , \lambda _d\in {{\mathbb {K}}}$ then $\lambda _1\alpha _{1,1}h_{1}(a_{1})=\sum \nolimits _{m=1}^{d}\lambda _mg_{m}(a_{n1}) = 0$ and so λ ₁ = 0. By similar arguments it follows that λ _j = 0 for each j = 2, …, d.

2.2 Application to bar-joint frameworks

The results of the previous section apply to the infinitesimal flex spaces of countable bar-joint frameworks (and in particular, to crystallographic bar-joint frameworks). Let G = (V, E) be a simple graph with vertex set V and edge set E. A bar-joint framework for G in ${{\mathbb {R}}}^{d}$ is a pair ${{\mathcal {G}}}=(G, p)$ where $p: V\to {{\mathbb {R}}}^{d}$, $v \mapsto p_v$, is an injective map. The points p _v are referred to as joints of ${{\mathcal {G}}}$ and the line segments (p _v, p _w), where vw ∈ E, are referred to as bars. Let ${{\mathcal {V}}}({{\mathcal {G}}};{{\mathbb {K}}})=X^{A}$ where A = p(V) and $X={{\mathbb {K}}}^{d}$ (for ${{\mathbb {K}}}={{\mathbb {R}}}$ or ${{\mathbb {C}}}$). The elements of ${{\mathcal {V}}}({{\mathcal {G}}};{{\mathbb {K}}})$ are referred to as velocity fields for ${{\mathcal {G}}}$ over ${{\mathbb {K}}}$. If $u\in {{\mathcal {V}}}({{\mathcal {G}}};{{\mathbb {K}}})$ then u(p _v) is referred to as the velocity vector for u at p _v. An infinitesimal flex (or first order flex) for ${{\mathcal {G}}}$ over ${{\mathbb {K}}}$ is a velocity field $u\in {{\mathcal {V}}}({{\mathcal {G}}};{{\mathbb {K}}})$ with the property that 〈p _v − p _w, u(p _v)〉 = 〈p _v − p _w, u(p _w)〉 for each edge vw ∈ E. The set of infinitesimal flexes for ${{\mathcal {G}}}$ is a closed vector subspace of ${{\mathcal {V}}}({{\mathcal {G}}};{{\mathbb {K}}})$, denoted ${{\mathcal {F}}}({{\mathcal {G}}};{{\mathbb {K}}})$. The vector subspace of bounded infinitesimal flexes for ${{\mathcal {G}}}$ is denoted ${{\mathcal {F}}}_\infty ({{\mathcal {G}}};{{\mathbb {K}}})$.

Example 2.11 Consider the crystallographic bar-joint framework ${{\mathcal {C}}}_\textrm {kite}$ in the Euclidean plane which is defined by the kite-shaped motif indicated in Figure 1 consisting of two joints and five bars. We assume that the vertical crossbar is closer to the apex joint, positioned to the left, than it is to the tail joint on the right. The figure also indicates a pair of periodicity vectors {a ₁, a ₂} and four velocity vectors at the four joints of a kite subframework of ${{\mathcal {C}}}_\textrm {nkite}$. These velocity vectors are indicative of an infinitesimal rotational flex of this subframework about the centre of the crossbar.

Figure 1. A motif for the bar-joint framework ${{\mathcal {C}}}_\textrm {kite}$ together with four velocity vectors from the unbounded infinitesimal flex a _kite.

Let $\vec {x}, \vec {y}$ be a choice of non-zero infinitesimal translation flexes of ${{\mathcal {C}}}_\textrm {kite}$, for the x and y directions, respectively, and let $\vec {r}$ be a non-zero infinitesimal rotation flex of ${{\mathcal {C}}}_\textrm {kite}$. Additionally, note that there is an unbounded infinitesimal flex a _kite in ${{\mathcal {F}}}({nn{\mathcal {C}}}_\textrm {kite};{{\mathbb {R}}})$ which restricts to alternating rotations of the 5-bar kite framework and its translates, each of these restriction flexes being rotational about the centre of the crossbar. The magnitude of these rotations tends to infinity exponentially in the positive x-direction and is constant in magnitude in the y-direction.

It is straightforward to show that the finite set $\{\vec {x}, \vec {y}, \vec {r}, a_\textrm {kite}\}$ is a vector space basis for ${{\mathcal {F}}}({{\mathcal {C}}}_\textrm {kite};{{\mathbb {R}}})$. Indeed, let ${{\mathcal {G}}}$ be a finite subframework of ${{\mathcal {C}}}_\textrm {kite}$ formed by 2 kite subframeworks which meet at a common joint. The infinitesimal flex space of ${{\mathcal {G}}}$ is four-dimensional and is spanned by the restrictions of the quartet $\vec {x}, \vec {y}, \vec {r}, a_\textrm {kite}$. Thus, if u is an infinitesimal flex of ${{\mathcal {C}}}_\textrm {kite}$ then we may subtract from u a linear combination of this quartet to obtain a flex u ₁ whose restriction to ${{\mathcal {G}}}$ is 0. One can see readily from the geometry of ${{\mathcal {C}}}_\textrm {kite}$ and the rigidity of the kite subframeworks that this implies that u ₁ = 0.

Let ${{\mathcal {G}}}=(G, p)$ be a finite or countable bar-joint framework whose joints do not lie in a hyperplane. The space ${{\mathcal {F}}}_\textrm {rig}({{\mathcal {G}}};{{\mathbb {R}}})$ of trivial infinitesimal flexes, or rigid motion infinitesimal flexes, is the vector subspace of ${{\mathcal {F}}}({{\mathcal {G}}};{{\mathbb {R}}})$ consisting of real infinitesimal flexes of the complete graph bar-joint framework (K _V, p). Here K _V is the complete graph on the vertex set V. The bar-joint framework ${{\mathcal {G}}}$ is infinitesimally rigid if ${{\mathcal {F}}}({{\mathcal {G}}};{{\mathbb {R}}}) = {{\mathcal {F}}}_\textrm {rig}({{\mathcal {G}}};{{\mathbb {R}}})$ and boundedly infinitesimally rigid if ${{\mathcal {F}}}_\infty ({{\mathcal {G}}};{{\mathbb {R}}}) \subset {{\mathcal {F}}}_\textrm {rig}({{\mathcal {G}}};{{\mathbb {R}}})$.

Several examples of free bases for the infinitesimal flex spaces of crystallographic bar-joint frameworks are presented in § 5.

3.1 Lattice group actions

Let $L=\{\sum \nolimits _{i=1}^{m} n_i b_i:n_i\in {{\mathbb {Z}}}\}$ be a rank m lattice in ${{\mathbb {R}}}^{d}$ determined by linearly independent vectors $b_1, \ldots , b_m\in {{\mathbb {R}}}^{d}$. For each b ∈ L, let $T_b:{{\mathbb {R}}}^{d}\to {{\mathbb {R}}}^{d}$ denote the translation T _b(x) = x + b. The translation group associated with L is the multiplicative abelian group ${{\mathcal {T}}}=\{T_b:b\in L\}$. A multilattice on L is a set $A = A_1 \cup \ldots \cup A_s$ where A ₁, …, A _s are pairwise disjoint translates of L. Define a free group action $\theta _L:{{\mathcal {T}}}\times A\to A$ by setting θ _L(T, a _i) = T(a _i) for a _i ∈ A _i, 1 ≤ i ≤ s. The lattice group action on X ^A induced by θ _L is the group action $\pi _L:{{\mathcal {T}}}\times X^{A}\to X^{A}$ with π _L(T, f) = f° T ⁻¹. Note that the quotient set $A/{{\mathcal {T}}}$ is finite and a translation $T_b\in {{\mathcal {T}}}$ is a geometric direction for f ∈ X ^A if and only if there exists a non-zero scalar λ such that f(a − b) = λf(a) for all a ∈ A. If $\Gamma _f\not =\{1\}$ then the sublattice L _f = {b ∈ L:T _b ∈ Γ_f} is referred to as the lattice of geometric directions for f.

If S ⊂ X ^A then, for each a ∈ A, let $S_a=\{f\in S:f(a)\not =0\}$.

In the following, ${{\mathbb {R}}}^{d}$ is endowed with a norm and $\operatorname {dist}(a, E)= \inf _{x\in E} \|a-x\|$ denotes the distance between a point $a\in {{\mathbb {R}}}^{d}$ and a subset $E\subset {{\mathbb {R}}}^{d}$.

A subset S of X ^A is bounded if each element of S is bounded and $\sup _{f\in S}\|f\|_\infty <\infty$.

In the following, a set $S\subset X^{A}\backslash \{0\}$ has the local basis property if, for each a ∈ A, the set {f(a):f ∈ S _a} is a basis for X. Also, W _∞ denotes the set of bounded elements in a subspace W of X ^A.

Theorem 3.13 Let A be a multilattice on L and let $S=\{f_n:n\in {{\mathbb {N}}}\}$ be a free spanning set for an infinite-dimensional subspace W of X ^A with the following properties.

1. (i) The lattice group action π _L acts on S.

2. (ii) S is bounded and has the local basis property.

3. (iii) No element of S is finitely supported.

4. (iv) $\inf \{\|g(a)\|:a\in A, g\in S_a\}>0$.

Then,

\[ W_\infty =\{f\in W: f = \sum_{n=1}^{\infty} \alpha_nf_n \mbox{ and } (\alpha_n)\in \ell^{\infty}({{\mathbb{N}}})\}. \]

5.1 The frameworks ${{\mathcal {C}}}_{{{\mathbb {Z}}}^{2}}$ and ${{\mathcal {C}}}_\textrm {hex}$

Let ${{\mathcal {C}}}_{{{\mathbb {Z}}}^{2}}$ be the grid framework in two dimensions, that is, the framework with joints located on the ${{\mathbb {Z}}}^{2}$ lattice and bars between nearest neighbours. Let ${{\mathcal {S}}}_\textrm {grid} = \{u_n, v_n: n\in {{\mathbb {Z}}}\}$ be a set of velocity fields with u _n (respectively v _m) supported by the joints on the line y = n (respectively x = m) and with unit velocities in the direction of the support line. Then it is elementary to check, by an exhaustion argument in the style of the proofs below, that ${{\mathcal {S}}}_\textrm {grid}$ is a crystal basis for the space ${{\mathcal {F}}}({{\mathcal {C}}}_{{{\mathbb {Z}}}^{2}};{{\mathbb {R}}})$.

Let ${{\mathcal {C}}}_\textrm {hex}$ be the honeycomb framework associated with the regular hexagonal tiling of the plane. Note that any hexagon ring subframework, ${{\mathcal {H}}}$ say, is the support of a (normalised) local infinitesimal flex, $u_{{\mathcal {H}}}$ say, which acts as infinitesimal rotation of ${{\mathcal {H}}}$. Let ${{\mathcal {S}}}_\textrm {hex}$ be a set of identical non-zero flexes of this type for all the honeycomb cells. We claim this is a crystal flex spanning set. We give the argument for this since it is typical of the simple exhaustion argument needed to show that a given set is a free spanning set or a free basis. Note also that $\sum \nolimits _{{\mathcal {H}}}u_{{\mathcal {H}}}$ is the zero infinitesimal flex of ${{\mathcal {C}}}_\textrm {hex}$ so ${{\mathcal {S}}}_\textrm {hex}$ is not a free basis. On the other hand, as the proof below shows, it has a curious minimal redundancy property, in the sense that the removal of any flex from ${{\mathcal {S}}}_\textrm {hex}$ gives a free basis, although not a crystal basis.

From these observations and Proposition 5.2, it follows that $cpx({{\mathcal {C}}}_{{{\mathbb {Z}}}^{2}})=cpx({{\mathcal {C}}}_\textrm {hex}) =1$.

Proof. Let z be an infinitesimal flex of ${{\mathcal {C}}}_\textrm {hex}$. Subtracting a linear combination of the local flexes for the cells A and B indicated in Figure 2, we may assume that the velocity field for z at the origin is 0. Consider the infinite path of framework points to the right of the origin, lying on cells 1, 2, 3, …, as indicated in Figure 2. We may subtract a multiple of the local flex for cell 1 to ‘fix’ the first joint, that is, to create a flex with zero velocities at O and at the first joint of the path. We may continue similarly to see that there is an infinite linear combination w of the local flexes for cells 1, 2, 3, … such that z − w has zero velocity at each joint of the infinite path.

Figure 2. Part of ${{\mathcal {C}}}_\textrm {hex}$.

In the same manner, we can subtract an infinite linear combination of the local flexes to the left of cells A and B to obtain a new flex which fixes all the framework points on the two-way infinite horizontal path, π say, through O. This may be achieved without making use of the local flex u _C for the other hexagon incident to the origin. We may assume then that z has zero velocities on the joints on π. The line of joints on y = 1 may be now be ‘fixed’ by subtraction of a unique infinite linear combination of the local flexes for the next horizontal line of cells, a, b, c, … etc. At this point, the next horizontal line of joints is necessarily fixed by z, in view of the flex condition. Continuing this process with the horizontal hexagonal strips above and below π we see that there is an infinite linear combination of the local flexes which is equal to the original flex. Also, the coefficients of this infinite linear combination are determined uniquely, with the proviso that the local flex u _C is not used in the representation. It follows that ${{\mathcal {S}}}_\textrm {hex}$ is a crystal flex spanning set.

5.2 The 2D kagome framework

We next consider the kagome framework in two dimensions, part of which is indicated in Figure 3. Let a, b, c be the vertices of a triangular subframework of ${{\mathcal {C}}}_\textrm {kag}$, with horizontal base edge [p _a, p _b], and let ${{\mathcal {L}}}_u^{0}$ be the (infinite) linear subframework which contains this edge. Note that there is an evident one-dimensional subspace of infinitesimal flexes of ${{\mathcal {C}}}_\textrm {kag}$ that are supported on this linear subframework. Consider the non-zero element u ₀ in this space which acts on alternate joints of ${{\mathcal {L}}}_u^{0}$ with unit norm velocity fields, with

\[ u_0(p_a)=(\cos \pi/6, -\sin\pi/6),\quad u_0(p_b)= (\cos \pi/6, \sin\pi/6). \]

Let u _n, where $n\in {{\mathbb {Z}}}$, be the parallel translates of u, naturally labelled, with u ₁ supported by the first linear subframework ${{\mathcal {L}}}_u^{1}$ above ${{\mathcal {L}}}_u^{0}$. Also, let $\{v_n:n\in {{\mathbb {Z}}}\}$ (respectively $\{w_n:n\in {{\mathbb {Z}}}\}$) be obtained from $\{u_n:n\in {{\mathbb {Z}}}\}$ by rotation about the centroid of the triangle abc by 2π/3 (respectively by 4π/3).

Figure 3. The horizontal band-limited flexes u _n for $n \in {{\mathbb {Z}}}$.

The next theorem is due to A. Sait [Reference Kasim Sait9] from which it follows that $cpx({{\mathcal {C}}}_\textrm {kag})=1$.

Proof. Since the space group acts on the set it will be sufficient to show that ${{\mathcal {B}}}_\textrm {kag}$ is a free basis. Write ${{\mathcal {L}}}_u^{n}$ (respectively ${{\mathcal {L}}}_v^{n}, {{\mathcal {L}}}_w^{n}$), for $n \in {{\mathbb {Z}}}$, for the supporting linear subframeworks for u _n (respectively v _n, w _n). Let z be an infinitesimal flex of ${{\mathcal {C}_{\textrm {kag}}}}$. By subtracting an appropriate linear combination of the three infinitesimal flexes u ₀, v ₀, w ₀, we may assume that the three velocity vectors z _a, z _b, z _c for the joints p _a, p _b, p _c of a central triangle subframework are zero. Subtracting an appropriate multiple of w ₁ we may then arrange z _d = 0, where d is the next vertex in the direction from a to b. Following this, subtracting an appropriate multiple of v ₋₁, we may arrange z _e = 0 for the next vertex. Continuing in this way, we obtain an infinite linear combination

\[ z' = \sum_{n\in {{\mathbb{Z}}}}\beta_nv_n + \gamma_nw_n \]

such that the infinitesimal flex z ^″ = z − z′ imparts only zero velocities to the joints of ${{\mathcal {L}}}_u^{0}$. From the flex condition and the rigidity of triangles, we deduce that the flex velocities are also zero on the apex vertices for the triangle subframeworks, such as def, which are horizontal translates of abc. Now subtract an appropriate multiple of u ₁ so that the resulting flex is zero on ${{\mathcal {L}}}_u^{1}$. Continuing upwards in this manner, and similarly downwards, we obtain an infinite sum representation for z ^″ in terms of the infinitesimal flexes u _n, $n\in {{\mathbb {Z}}}$. Thus, the original flex z is an infinite sum of the basis vectors. Also, the representation is unique and so the proof is complete.

Combining this result with Theorem 4.4(iii), we obtain the following description of the space of bounded infinitesimal flexes.

Corollary 5.4 With ${{\mathcal {B}}}_\textrm {kag} = \{u_n, v_n, w_n: n\in {{\mathbb {Z}}}\}$ as above,

\begin{align*} & {{\mathcal{F}}}_\infty({{\mathcal{C}}}_\textrm{kag};{{\mathbb{R}}}) \\ & \quad = \left\{u\in {{\mathcal{F}}}({{\mathcal{C}}}_\textrm{kag};{{\mathbb{R}}}): u = \sum_{n\in {{\mathbb{Z}}}}(\alpha_nu_n+\beta_nv_n + \gamma_nw_n) : (\alpha_n), (\beta_n), (\gamma_n)\in \ell^{\infty}({{\mathbb{Z}}})\right\}. \end{align*}

5.3 The octahedron crystal framework ${{\mathcal {C}}}_\textrm {Oct}$

We now consider some crystal frameworks in three dimensions.

Write ${{\mathcal {C}}}_\textrm {Oct}$ for the crystal framework of corner-connected regular octahedra, in the symmetric placement for which there is one translation class for them. The construction of a crystal flex basis for ${{\mathcal {C}}}_\textrm {Oct}$ may be determined by viewing ${{\mathcal {C}}}_\textrm {Oct}$ as the union of countably many copies of the 2D grid framework whose orientations and joint connections are consistent with Figure 4.

Figure 4. A motif of 3 joints and 12 bars for the framework ${{\mathcal {C}}}_\textrm {Oct}$.

To be precise, assume that the period vectors for ${{\mathcal {C}}}_\textrm {Oct}$ are (2, 0, 0), (0, 2, 0), (0, 0, 2) and let ${{\mathcal {C}}}_z$ denote the grid framework in the xy-plane which contains the adjacent joints p ₁ = (0, − 1, 0), p ₂ = (1, 0, 0), p ₃ = (0, 1, 0) and p ₄ = ( − 1, 0, 0). These joints lie on a 4-cycle of bars in the xy-plane. Similarly, let ${{\mathcal {C}}}_x$ denote the 2D grid framework in the yz-plane which contains the adjacent joints p ₁, p ₅ = (0, 0, − 1), p ₃ and p ₆ = (0, 0, 1), and let ${{\mathcal {C}}}_y$ denote the 2D grid framework parallel to the zx-plane which contains the adjacent joints p ₂, p ₅, p ₄ and p ₆.

Let ${{\mathcal {C}}}^{n}_x$ be the translated bar-joint frameworks ${{\mathcal {C}}}_x + (2n, 0, 0),$ for $n\in {{\mathbb {Z}}}$, and similarly define ${{\mathcal {C}}}^{n}_y$ and ${{\mathcal {C}}}^{n}_z$. Then ${{\mathcal {C}}}_\textrm {Oct}$ is the union of all of these frameworks, that is, it is the bar-joint framework whose joint set is the union of the joints (without multiplicity) and whose set of bars is the union of all the bars.

Let us also define ${{\mathcal {C}}}_\textrm {Oct}^{+}$ as the augmented framework in which each regular octahedron is augmented by three bars parallel to the coordinate axes. In view of the infinitesimal rigidity of a convex octahedron it follows that the vector spaces ${{\mathcal {F}}}({{\mathcal {C}}}_\textrm {Oct};{{\mathbb {R}}})$ and ${{\mathcal {F}}}({{\mathcal {C}}}_\textrm {Oct}^{+};{{\mathbb {R}}})$ are equal.

Let us write ${{\mathcal {C}}}_\textrm {sq}$ for a 2D bar-joint framework composed of corner-connected rigid squares. This is obtained from the 2D grid framework by adding an edge to each alternate square. Let ${{\mathcal {C}}}_\textrm {sq}^{+}$ be the related framework which has both cross diagonals added to the rigid squares. We may thus view ${{\mathcal {C}}}_\textrm {Oct}^{+}$ as the union of copies of ${{\mathcal {C}}}_\textrm {sq}^{+}$, where these copies are augmentation frameworks, $\tilde {{{\mathcal {C}}}}^{n}_x$, $\tilde {{{\mathcal {C}}}}^{n}_y$ and $\tilde {{{\mathcal {C}}}}^{n}_z$ say, of the frameworks ${{{\mathcal {C}}}}^{n}_x$, ${{{\mathcal {C}}}}^{n}_y$ and ${{{\mathcal {C}}}}^{n}_z$. It follows immediately that the alternation flex a of $\tilde {{{\mathcal {C}}}}^{n}_x$ extends to a flex $a^{x}_n$ of ${{\mathcal {C}}}_\textrm {Oct}^{+}$ which has zero velocities on all the other vertices. Let us similarly define the ‘local alternation flexes’ $a^{y}_n$ and $a^{z}_n$ for $n\in {{\mathbb {Z}}}$.

Let r ^x, r ^y, r ^z be infinitesimal flexes for axial rotations of ${{\mathcal {C}}}_\textrm {Oct}$ about the principal axes of the central octahedron. These infinitesimal flexes act on the entire framework and are unbounded flexes. Also, we assume the normalization such that for σ = x, y or z the restrictions of $a^{\sigma }_n$ to the octahedron meeting the σ-axis agrees with the restriction of r ^σ.

Finally, let $\vec {x}$ be the velocity field in ${{\mathcal {F}}}({{\mathcal {C}}}_\textrm {Oct};{{\mathbb {R}}})$ with joint velocities (1, 0, 0) and let $\vec {y}$ and $\vec {z}$ be analogous velocity fields for the y and z directions.

In the next proof, we use the following elementary flex projection principle. If the bar [p _a, p _b] lies in a plane ${{\mathcal {P}}}$ of ${{\mathbb {R}}}^{3}$ and if the joint velocities v _a, v _b in ${{\mathbb {R}}}^{3}$ give an infinitesimal flex of the bar [p _a, p _b] then the ${{\mathcal {P}}}$ components v _a′, v _b′ of v _a, v _b also give an infinitesimal flex of the bar. We say that such a flex is an in-plane flex when the plane in question is understood.

Theorem 5.5 The set ${{\mathcal {S}}}$ of velocity fields

\[ \{r^{x}, r^{y}, r^{z}\} \cup \{\vec{x}, \vec{y}, \vec{z}\}\cup \{a^{x}_n, a^{y}_n, a^{z}_n: n\in {{\mathbb{Z}}}\} \]

is an essential crystal flex basis for ${{\mathcal {F}}}({{\mathcal {C}}}_\textrm {Oct};{{\mathbb {R}}})$.

Proof. The subset ${{\mathcal {S}}}_0\subseteq {{\mathcal {S}}}$ of non-rotational flexes has the crystallographic group action property and so it will suffice to show that ${{\mathcal {S}}}$ is a free basis for ${{\mathcal {C}}}_\textrm {Oct}$. Equivalently, we show that ${{\mathcal {S}}}$ is a free basis for ${{\mathcal {C}}}_\textrm {Oct}^{+}$.

Let z be an infinitesimal flex in ${{\mathcal {F}}}({{\mathcal {C}}}_\textrm {Oct}^{+};{{\mathbb {R}}})$. There is a linear combination z _rig of $\vec {x}, \vec {y}, \vec {z}, r^{x}, r^{y}, r^{z}$ which agrees with z on the joints p ₁, …, p ₆. Replacing z by z − z _rig, for some rigid motion infinitesimal flex z _rig, we may assume that these velocities for z are zero.

We now make use of the flex projection principle. Note that the velocity field z _xy given by the xy-plane projection of the joint velocities z(p), for joints in $\tilde {{{\mathcal {C}}}}_{z}^{0}$, is an infinitesimal flex of $\tilde {{{\mathcal {C}}}}_{z}^{0}$. Since z has zero velocity vectors on the central octahedron it follows that the xy-plane projection of z also has zero velocities on the central square of $\tilde {{{\mathcal {C}}}}_{z}^{0}$. It follows that this in-plane flex is equal to the restriction of a scalar multiple of $a_0^{z}-r^{z}$. In this way, we obtain scalar multiples $\alpha _0 (a_0^{z}-r^{z}), \beta _0 (a_0^{x}-r^{x})$ and $\gamma _0 (a_0^{y}-r^{y})$ which provide the in-plane flexes of z for the planes z = 0, x = 0 and y = 0.

Consider now the tower subframework given by the tower of octahedra whose connecting joints lie on the z-axis. Since z is zero on the central octahedron supported by p ₁, …, p ₆, denoted O _(0,0,0), it follows that the z component of the joint velocity for a joint on this line is zero. It also follows that there is an infinitesimal flex $\beta _0 (a_0^{x}-r^{x}) + \gamma _0 (a_0^{y}-r^{y})$ with joint velocities agreeing with those of z for the joints on the axial line. It follows similarly that there is a flex of the form

\[ w= \alpha (a_0^{z}-r^{z}) + \beta_0 (a_0^{x}-r^{x}) +\gamma_0 (a_0^{y}-r^{y}) \]

with this agreement property for the three axial lines through O _(0,0,0).

Replacing z by z − w, we may assume that z is zero on O _(0,0,0) and on the joints of the three axial lines of O _(0,0,0). Note that the restriction of such a flex z to any other octahedron O with an axis on the coordinate axes must be an infinitesimal rotation flex of the octahedron about this axis. Also each such flex of an individual octahedron O, on the σ-axis say, agrees with the restriction of a scalar multiple of the local alternation flex $a^{\sigma }_n$, for some n ≠ 0. Evidently, these infinitesimal flexes act on distinct octahedra on the axial lines.

It follows that there is an infinite linear combination of these flexes, w ₂ say, whose restriction to any octahedron on a coordinate axis is equal to the restriction of z. Replacing z by z − w ₂, we may assume that z is zero on this triple tower, ${{\mathcal {T}}}_{Oct}$ say. We now observe that any infinitesimal flex which is zero on ${{\mathcal {T}}}_{Oct}$ is the zero flex. This follows from the fact that the entire framework may be built up from ${{\mathcal {T}}}_{Oct}$ by attaching octahedra in groups of four such that at each stage every flex which is zero on ${{\mathcal {T}}}_{Oct}$ is the zero flex. It follows that z must be identically zero. Thus, every velocity field z in ${{\mathcal {F}}}({{\mathcal {C}}}_\textrm {Oct};{{\mathbb {R}}})$ is an infinite linear combination of the vectors in the set ${{\mathcal {S}}}$.

Note that ${{\mathcal {S}}}$ is a countable set of velocity fields which tend to zero strictly and is a free spanning set for a vector space of infinitesimal flexes. Also, the scalar coefficients in the identifications above are determined uniquely by the joint velocities of the flex z. Thus, ${{\mathcal {S}}}$ is a free infinitesimal flex basis, as required.

Remark 5.6 From a simple adaptation of the above proof, it follows that the space ${{\mathcal {F}}}_\infty ({{\mathcal {C}}}_\textrm {Oct};{{\mathbb {R}}})$ of bounded infinitesimal flexes is the space of infinitesimal flexes of the form

\[ u = \alpha\vec{x}+ \beta\vec{y}+ \gamma\vec{z} + \sum_{n\in {{\mathbb{Z}}}} \alpha_na^{x}_n +\beta_na^{y}_n+\gamma_n a^{z}_n, \quad (\alpha_n), (\beta_n), (\gamma_n) \in \ell^{\infty}({{\mathbb{Z}}}) \]

The octahedral framework serves to model the rigid unit atomic structure of the crystal perovskite. Its zero mode (or RUM mode) phonon spectrum was among early examples to be computed by experiment and simulation. Other examples were quartz and cristobalite. See Giddy et al. [Reference Giddy, Dove, Pawley and Heine5]. The RUM spectrum for crystobalite has complete linear structure in [0, 2π)³, being the union of the three line segments (t, 0, 0), (0, t, 0), (0, 0, t), for 0 ≤ t < 2π. The existence of this linear structure in the RUM spectrum also follows from the band-limited flexes in ${{\mathcal {S}}}$.

5.4 The 3D kagome framework

The kagome framework in three dimensions, ${{\mathcal {C}}}_\textrm {Kag}$, has the structure of the kagome net, a network of regular tetrahedral units connected pairwise at their vertices. It may be constructed from the 2D kagome net by first completing its triangles to tetrahedra with alternating up and down orientations to create a horizontal layer framework, and then joining translational copies of such layers. The period vectors determine a parallelepiped unit which is occupied by a single tetrahedron and one can determine a motif for ${{\mathcal {C}}}_\textrm {Kag}$ with a corresponding 12-by-12 matrix-valued symbol function on the 3-torus [Reference Owen and Power14]. However, the simple layer structure description, together with the symmetry of ${{\mathcal {C}}}_\textrm {Kag}$, are sufficient to determine a crystal basis by an exhaustion argument, as before, as we now sketch.

For $m\in {{\mathbb {Z}}}$, let $\{u_n^{m}, v_n^{m}, w_n^{m}:n\in {{\mathbb {Z}}}\}$ be the crystal flex bases for the in-plane infinitesimal flex space of the m ^th layer and note that the infinitesimal flexes in these sets extend (by zero velocity specifications) to infinitesimal flexes of ${{\mathcal {C}}}_\textrm {Kag}$. If w is an arbitrary infinitesimal flex of ${{\mathcal {C}}}_\textrm {Kag}$, we may subtract an appropriate infinite linear combination of these flexes to replace w by a flex w ₁ with the property that the velocity vectors at each joint in the horizontal layers are in the vertical direction. Let us make a distinction between the ‘thick’ horizontal layer frameworks and their ‘thin’ kagome subframeworks (which support the added tetrahedra). For layer m = 0, fix a triangle subframework in the thin layer to be a base triangle. Its tetrahedron has three additional bars. Each of these bars indicates the direction of a linearly localized flex, each of which is a space group element image of $u_0^{0}$. We may choose a linear combination of these flexes and subtract them from w ₁ to obtain a flex w ₂ so that the tetrahedron has zero velocities (under w ₂) at all four of its joints.

Continuing in this way we see that, as in the 2D kagome, there is a crystal flex basis with a single generator and it consists of the space group element images of $u_0^{0}$. In particular, $cpx({{\mathcal {C}}}_\textrm {Kag}) = 1$.

5.5 The bipyramid framework ${{\mathcal {C}}}_\textrm {Bipyr}$

We next define the bipyramid crystal framework ${{\mathcal {C}}}_\textrm {Bipyr}$ and compute the transfer function associated with the natural primitive periodic structure. An appropriate basis of period vectors takes the form

\[ (1,0,0),\quad (1/2,\sqrt{3}/2,0), \ (0,0,2 h) \]

and we consider a motif (F _v, F _e) where F _v consists of the two joints p ₁ = (0, 0, 0), p ₂ = (1/2, α, − h) and F _e consists of the nine bars p _ip _j associated with the nine directed bars indicated in Figure 5. Here $h=\sqrt {2/3}$ is the height of a regular tetrahedron of unit sidelength and $\alpha = \sqrt {3}/6$ is the distance from B to C, that is, the distance of the centroid of a unit sidelength equilateral triangle to the triangle boundary.

Figure 5. A motif of 2 joints and  9 bars for the bipyramid framework.

The following data for the motif edges will suffice for the computation of the associated transfer function Ψ_Bipyr(z).

A factor-periodic infinitesimal flex, with factor $\omega =(\omega _1, \omega _2, \omega _3)\in {{\mathbb {C}}}_*^{3}$, has the form

\[ u = (u_{\kappa,k}) = (\omega^{k}{a}) = (\omega_1^{k_1}\omega_2^{k_2}\omega_3^{k_3}a), \quad k\in {{\mathbb{Z}}}^{3}, \]

where $a=(u_{1,x}, u_{1,y}, u_{1,z}, u_{2,x}, u_{2,y}, u_{2,z}) \in {{\mathbb {R}}}^{6}$ is the velocity field for the motif joints, being the list of the coordinate velocities of u at the joints p ₁ and p ₂. We have,

The submatrix of Ψ_Bipyr(z ⁻¹) for columns 3, 4, 5 is a 9 × 3 function matrix. We note that the first three rows are zero. Indeed, the three edges e ₁ = (v ₁, (0, 0, 0))(v ₁, (1, 0, 0)), e ₂ = (v ₁, (1, 0, 0))(v ₁, (0, 1, 0)) and e ₃ = (v ₁, (0, 0, 0))(v ₁, (0, 0, 1)) are of v = w type and so the entries for columns 4, 5, 6 are zero. Also, the entries for column 3 are zero since the z-component is zero for the vectors p(e _i), for i = 1, 2, 3.

We next determine the set of factors $\omega \in {{\mathbb {C}}}_*^{3}$ for which Ψ_Bipyr(ω ⁻¹)a = 0 for some non-zero vector of the form a = (0, 0, u _1,z, u _2,x, u _2,y, 0). In doing so, we shall determine the factor periodic infinitesimal flexes u with the property that their velocity fields impart only vertical velocities to the joints that lie in the horizontal copies of the fully triangulated framework ${{\mathcal {C}}}_\textrm {tri}$ and impart only horizontal velocities to the other joints, the polar joints of the constituent bipyramids. In view of the latter condition, we refer to these infinitesimal flexes simply as sheering flexes. The required solutions for ω are the values of z = (z ₁, z ₂, z ₃) for which the submatrix Ψ_sub(z ⁻¹) has non-zero kernel, where

Noting the similarity between the rows for e ₆ and e ₉, it follows that if z ₃ is not equal to −1 then the kernel is trivial. On the other hand if z ₃ = −1 then the rows for e ₇, e ₈, e ₉ are the negatives of the rows for e ₄, e ₅, e ₆. We conclude that (ω ₁, ω ₂, ω ₃) is a periodicity multifactor for a sheering flex u if and only if ω ₃ = −1 and the determinant of the 3 × 3 submatrix for the first 3 rows is zero at ω ₁, ω ₂. This determinant is αh(1 + z ₁ + z ₂). We conclude that when z ₁ and z ₂ are unimodular then there are exactly two solutions and so the RUM spectrum $\Omega ({{\mathcal {C}}}_\textrm {Bipyr})$ contains the set

\[ \{(1,1,1), (e^{2\pi i/3},e^{4\pi i/3},e^{\pi i}),\quad (e^{4\pi i/3},e^{2\pi i/3},e^{\pi i})\}. \]

We also note that there are unbounded geometric flexes associated with the solutions (ω ₁, ω ₂) of the equation of 1 + z ₁ + z ₂ = 0, where ω ₁ = r, ω ₂ = −(1 + r), and 0 < r < 1. These infinitesimal flexes bear some analogy with the unbounded flex of the kite framework considered in Example 2.11. Since there are uncountably many linearly independent unbounded flexes of this type the infinitesimal flex space ${{\mathcal {F}}}({{\mathcal {C}}}_\textrm {Bipyr}; {{\mathbb {R}}})$ is infinite dimensional.

5.6 Two crystal extensions of ${{\mathcal {C}}}_\textrm {Bipyr}$

Consider the crystal framework ${{\mathcal {C}}}_\textrm {Bipyr}^{e}$ which is obtained from ${{\mathcal {C}}}_\textrm {Bipyr}$ by adding bonds of length 1 between polar joints whenever this is possible. It is straightforward to see that ${{\mathcal {C}}}_\textrm {Bipyr}^{e}$ is sequentially infinitesimally rigid [Reference Kitson and Power12, Reference Owen and Power14] and so is infinitesimally rigid in the strongest possible sense. Note that every 2D subframework parallel to the xy-plane is a copy of the fully triangulated framework ${{\mathcal {C}}}_\textrm {tri}$. Let ${{\mathcal {C}}}_\textrm {Bipyr}^{+}$ be obtained from ${{\mathcal {C}}}_\textrm {Bipyr}$ by adding edges and vertices so that every triangle in the horizontal copies of ${{\mathcal {C}}}_\textrm {tri}$ in ${{\mathcal {C}}}_\textrm {Bipyr}$ is the equator of a bipyramid. This framework may be described as having horizontal layers of maximally packed bipyramids. The joints are once again of two types, with polar joints having degree 6, as before, and equatorial joints having degree 18. One can observe that in fact the two sheering RUMs of ${{\mathcal {C}}}_\textrm {Bipyr}$ extend to this framework and so, despite the edge rich structure, ${{\mathcal {C}}}_\textrm {Bipyr}^{+}$ is not infinitesimally rigid.

Added in proof. The geometric spectrum has been used by the third author and E. Kastis [Reference Kastis and Power10, Reference Kastis and Power11] to obtain necessary and sufficient conditions for the infinitesimal rigidity of a crystallographic bar-joint framework.