Six-connected -covered graphs
are rigid in three dimensions

Proof from Theorem 1.1. The completed spanning graph is -covered and has rigidity rank at most that of . Theorem 1.1 makes it rigid, and hence makes rigid. □

Proof. For there is equality, so assume . The coordinate functionals on form a rank- representable matroid of cogirth at least . The supported subspace is the kernel of projection onto the coordinates outside , so its dimension is . Conditional on the marker, is an independent random subset with a uniform inclusion probability . Give the elements of independent rate-one exponential clocks, and let be the rank still missing at time . The clock subset has inclusion probability ; changing variables in the integral over therefore gives

If is the number of nonloops, then ; extending independent elements to a hyperplane gives a cocircuit of size at most . The first nonloop clock has rate , and its index is uniform. Before that clock rings, the missing rank is ; afterwards the remaining process is the one for the contraction by its element. Memorylessness therefore gives

Contraction preserves the lower bound on cogirth. Induction on now applies, and the last expression decreases with , since . Consequently

All inequalities are equalities for the uniform matroid , represented over an infinite field by an MDS code. □

Proof. The new supported stresses modulo the old ones inject into the global star image, with support among the preceding neighbors. A nonzero equilibrium stress at a generic vertex has at least four nonzero star coordinates, so Lemma 2.1 applies. For the full stress space, the star image has dimension , while the expected number of newly inserted edges is . □

Proof. Completing a rigid cluster preserves rank, six-connectivity and coverage. Edge maximality gives the first assertion. Every vertex has degree at least six. If were simplicial, then would remain six-connected; after deleting at most five other vertices, the surviving neighbors of form a nonempty clique, so adding cannot join different components. The graph is also -covered, since the neighbor clique replaces any covering tetrahedron that used . Vertex minimality would make rigid, and three generic bars from would make rigid.

The same separator argument shows that is six-connected. Completing the neighbor clique preserves coverage, including for new edges. Thus is rigid by minimality. The nonsimplicial case of [1, Lemma 8.23] gives . Every neighbor of lies with in a body. A vertex in just one body would therefore be simplicial. Finally, two rigid bodies sharing three generic vertices have rigid union, so maximality bounds their intersection by two. □

Proof. Each is a maximal planar rigid component of the neighborhood; a larger one would cone with to a larger rigid cluster. A circuit in the union of the chosen bases is planar rigid. Whenever it meets a basis in an edge, it glues rigidly to that component along the edge’s endpoints. It cannot lie in one basis, and meeting two would merge distinct maximal components. The union is therefore independent, whence . □

Proof. The neighborhood is not planar rigid, since its cone would make simplicial. Its planar rank is thus at most eight. By Lemma 3.2, each incident body contributes at least three independent neighborhood edges, so there are at most two bodies. There are at least two by Lemma 3.1. Their neighbor cliques cover six vertices and meet in at most one. Their sizes are therefore with empty intersection, or with one common vertex. □

Proof. Write . If the Pfaffian is not identically zero on , its zero locus in has dimension at most . Since each singular form has a kernel line, the incidence of pairs with has dimension at most . The rank hypothesis instead gives fibers of dimension at least over generic , hence dimension at least .

Thus is totally isotropic, so . Such a space of decomposable two-forms either consists of forms containing a fixed covector, or is for a three-dimensional covector space . Indeed, start with and . A form wedging to zero with both either contains or belongs to ; if one does not contain , all belong to this exterior square. The first alternative has all kernel points in the hyperplane and cannot account for generic . The second is , with the annihilator of .

For , the projective singular forms have dimension at most one. Their kernel lines fill a set of dimension at most two. □

Proof. At a type-A vertex , work over the coordinate field with omitted. A motion of restricts to rigid motions on its two incident body triangles. Subtract the first and retain the relative motion of the second. This map, modulo common rigid motions, is injective; if both triangles move together, all neighbors move rigidly, and the rigidity of makes the entire motion common. Call its image .

If the star of has a nonzero stress image, then . Consequently , and reintroducing the generic pin imposes conditions on . Lemma 3.4 gives and . At a type-B vertex, the relative motion in is a rotation about the common hinge. The same injectivity gives .

Suppose . There are no type-B vertices, and each degree-six vertex has . Its incident edges are rigidity-matroid coloops, so . If , Corollary 2.2 and give

Summing gives , a contradiction. □

Proof. As in Proposition 3.5, the relative-motion map is injective and has dimension . Generic evaluation has rank two, so Lemma 3.4 applies before is chosen. If were another actual vertex of either body, that vertex would be fixed relative to the other body in every motion. It would enlarge a maximal rigid cluster, contrary to closure. Finally, for every precisely when is parallel to . □

Proof. Its dimension is . Generic equilibrium excludes support at most three. If a stress uses exactly four edges at , retain those edges and all of . This extension has rank increment three. By [1, Lemma 8.3], its four neighbors form a rigid cluster in . Adding gives a rigid after closure, contrary to type A. □

Proof. Put . The angular projection is an isomorphism, so gives local coordinates. Infinitesimal fixation of says

Differentiating gives . Write in angular Maurer–Cartan directions. Then

The three pairs of coordinate vectors force the off-diagonal entries of to vanish and its diagonal entries to be pairwise negatives. Thus , and . □

Proof. Locally write and . The angular Maurer–Cartan equations, restricted to with angular components , give

up to a common sign convention. Write . As an operator on , the left side is , of rank at most one. Its symmetric part forces , since a rank-one matrix and its transpose have sum of rank at most two. A nonzero skew matrix of order three has rank two, so also . The three-dimensional case has zero right-hand side. □

Proof. If , the Maurer–Cartan equations imply, on the four-dimensional twist image , an identity

for a linear functional . An abelian subspace of has angular projection of dimension at most one. If that projection is nonzero, its translational kernel lies on the rotation axis, so the subspace has dimension at most two. Hence the only three-dimensional abelian subspace is the translation space. This excludes . If , its three-dimensional kernel must be the translation space. Choosing with would make act on all translations as the nonzero scalar , which is impossible. □

Proof. Let the centers be . Both angular projections of the tangent image are surjective, since they contain nonzero vectors parallel to and for generic . If the combined image had dimension three, its projections would be isomorphisms; a linear map would send to a vector parallel to identically in . Comparing quadratic and linear terms makes that map scalar and gives . Thus the image has dimension four.

Write , where are matrices of rank two. The new-pin condition gives

The entries of are relatively prime affine linear polynomials. It follows that for a nonzero scalar . Hence and . Writing , every tangent space has the form

The fixed ambient rotation fields and are tangent everywhere, so their flows preserve the manifold locally. Let be a small nontrivial rotation about , and flow in the first field. In spatial angular coordinates, the differential sends to . The relation on is consequently transported to . At the new point it must have the form . A small nontrivial planar rotation is not a scalar operator, giving the contradiction. □

Proof of Proposition 4.1. Ground one incident body after deleting . No other incident body has zero relative motion in the surviving flex of , since it would merge with the base body. Each relative-body projection therefore has dimension at least three; by Lemma 3.4, a smaller space cannot have a nonzero member annihilating generic .

If , the grounded fiber has dimension three. Every nonbase projection is an isomorphism onto some . Make the centers finite. For two such projections, the common-pin kernel gives an invertible angular map with

identically in . Its quadratic terms make scalar and its linear terms give ; after writing , the identity is , whose homogeneous quadratic part forces every vector to be an eigenvector of . The full twists are therefore proportional. Lemma 4.3 makes them equal, contradicting maximality of the bodies.

Suppose now that . The grounded fiber has dimension four, and each nonbase projection has rank three or four.

A rank-three projection exists. Its image is by Lemma 3.4, and its local configuration image has fixed finite center by Lemma 4.2. Some second projection is nonzero on its one-dimensional kernel; otherwise all neighbor bodies would be fixed there, contrary to rigidity of . The combined projection to these two bodies is consequently injective.

If the second projection also has rank three, its center is fixed. Lemma 4.5 excludes . For , the same two-by-two determinant calculation gives

The rank-three distribution has proportional spherical twists. The generic-new-pin kernel lies in it, since both angular vectors are parallel to whereas is fixed as varies. Lemma 4.3 makes the two twists equal in the flex of , again merging the bodies.

If the second projection has rank four, identify with its alternating-form image , and let be the first angular projection. Generic contraction on has rank three, so there is a unique projective kernel form . The common-pin condition says that is nonzero and parallel to . The image closure of has dimension two, with fibers the kernel lines of its forms. For generic in that image, all generic points of its kernel line yield the same . Since is fixed, that line passes through . The image is therefore the whole plane , and . On , the map is a nonzero scalar multiple of the identity on angular vectors. Lemma 4.3 applies to its rank-three preimage in , with the same contradiction.

Every nonbase projection has rank four. Choose two, with images and matching isomorphism . Generic contraction on again has a unique projective kernel form . Its image closure is an irreducible surface, since a fiber is an open part of a kernel line. The Pfaffian does not vanish identically on the four-dimensional space , so is a component of its quadric section in . For generic , every generic point of also belongs to . Rank-two alternating forms with the same kernel are proportional; hence on .

If spans , then . Irreducibility places in one of the finitely many eigenspaces of , so on . Lemma 4.4 gives . Otherwise is a plane of decomposable forms. The classification in Lemma 3.4 excludes the plane whose kernel points lie in a fixed hyperplane; therefore and . Make finite. Its rank-three preimage in satisfies Lemma 4.3, even if the center varies and the distribution is not integrable. Again . In both cases the surviving flex makes the two bodies identical, contrary to maximality. This excludes . □

Proof. One, two and three generic pins impose respectively three, five and six conditions on a relative alternating form. For any partition , motions constant on its blocks therefore have dimension at least the expression in (1). This gives the lower bound.

For the upper bound, stratify twists by their equality partition, with distinct blocks, and ground one block. A stratum dominating generic pin space has no pair of blocks sharing three pins; a nonzero singular alternating difference has only a projective kernel line. Hence crossing multiplicities are one or two. Let be the simple support graph on the blocks, let be the number of crossing pins, and put

Each crossing pin will cost two dimensions in the incidence count, because it must lie on a kernel line in three-dimensional pin space. The number is the additional codimension needed in the twist variables. We obtain it from independent Pfaffian equations. Since the crossing multiplicities are one or two, their total constraint count is ; this is the relation between the defect formula and the following sparse-graph calculation. We claim that contains a -sparse set of edges. The two-forest matroid-union formula is

For a partition into parts, write and for the numbers of internal and crossing support edges. If , the defect bound at the original equality partition gives

If , then , while subadditivity of at the coarsened partition gives

Adding yields the same inequality. Every term of the matroid-union formula is thus at least , proving the claim.

Apply the codimension theorem to this -edge set. The grounded distinct-twist space has dimension at most . Each crossing pin has affine fiber dimension at most one, and each noncrossing pin has three free coordinates. If there are pins in total, the incidence stratum has dimension at most

There are finitely many equality partitions; the all-equal stratum has grounded dimension zero. The generic compatible-motion space therefore has dimension at most .

For an optimal partition, its block-constant motion subspace already has the full dimension , so it is the entire motion space. Bodies in one block then have identical twists in every motion and form a rigid cluster. If no bodies merge, the partition is singleton. For simple , its value is , proving the last assertion. □

Proof. Select the bodies with a nonzero coefficient. For an active neighbor , let count its nonzero copies and its total multiplicity among these bodies. The number of active incidences satisfies

by Lemma 3.2. But each active body needs at least four generic directions for equilibrium, so . □