## 3 Projective Rigidity
### Infinitesimal and Static Rigidity in Projective Space
The basic observation is that for any quadrilateral of bars, \(a_{1}a_{2}b_{2}b_{1}\), the relative center of \((a_{2},b_{2})\) relative to \((a_{1},b_{1})\) is a multiple of the point of intersection \((a_{1}a_{2})\wedge(b_{1}b_{2})=c_{12}\), and the center of \((a_{1},b_{1})\) relative to \((a_{2},b_{2})\) is the same projective point with a negative weight (Figure 6a). For a cycle of 4 quadrilaterals, the count is \(|E|=12=16-4<2|V|-3\) revealing a non-trivial infinitesimal motion. This infinitesimal motion can be described by an affine combination of the four centers around the cycle. If these four centers are collinear (Figure 6b), then there will be two independent affine (or projective) combinations, and therefore an additional non-trivial infinitesimal motion. This extra infinitesimal motion corresponds to a drop in rank of the rigidity matrix which implies an equilibrium stress. With a general cycle of quadrilaterals of length \(n\) (Figure 6c), the analysis gives \(|E|=2|V|-n\) and \(n-3\) degrees of freedom. However if the \(n\) centers are collinear, then there will be \(n-2\) independent projective combinations of the relative centers. This implies that the collinearity is sufficient (and necessary) for an equilibrium stress in these under-braed frameworks. The collinearity of the centers, along a line of perspective of the inside and outside polygon [61], creates an image that (correctly) suggests we can hold one polygon flat in the plane and tilt the other one up into 3-space, lifting vertices vertically. With this image we can'see' a spatial polyhedron, as workers in rigidity since at least the time of J.C. Maxwell did [3, 62, 63]. We will encounter these connections in detail in an exploration of'reciprocal diagrams' in our companion paper [27]. Note that the entire analysis applies if some or all of the relative centers happen to lie on the projective line at infinity (as relative centers for relative translations). If the two edges of a quadrilateral are parallel, then their lines will intersect 'at infinity' and we need to include centers at infinity. This analysis is a fundamentally projective tool, based on projective constructions. We will examine the full inclusion of vertices at infinity (or'slide joints') in Section 6. Note that, in the projective rigidity matrix \(\tilde{\mathbf{R}}(G,\hat{p})\), the weight of each projective point \(\tilde{p}_{i}\) is 1. We can, of course, change this weight to an arbitrary non-zero number \(\lambda_{i}\) for each \(\tilde{p}_{i}\) by simply multiplying the column for \(i\) by \(\frac{1}{\lambda_{i}}\), the rows \(ij\) by \(\lambda_{i}\lambda_{j}\), and the row for \(i\) by \(\lambda_{i}^{2}\). These row and column multiplications do not change the rank of the matrix, or the dimension of the kernel or cokernel. Since the solutions \(M_{i}\) depend on the weight \(\lambda_{i}\) assigned to each vertex, the name we often use for the solution set \(M\) is (projective) _momenta_ (as in velocity times mass). Note that we have focused our discussion on finite  
Figure 6: Each plane quadrilateral has a relative center of motion of opposite edges (how one moves when the other is held still): a multiple of the intersection point of the other two sides (**a**). For a larger ring of quadrilaterals, the collinearity of these relative centers will guarantee extra infinitesimal motions (**b**,**c**). projective points (i.e., points with nonzero weight) so far. We will discuss how to deal with infinite projective points in Section 6.2. Since row rank equals column rank, for an infinitesimally rigid framework, the row rank of \(\hat{\mathbf{R}}(G,\bar{p})\) is \((d+1)|V|-\binom{d+1}{2}\). We need to confirm this is equivalent to static rigidity for the framework by connecting linear combinations of the rows with resolutions of equilibrium loads. Let us now consider static rigidity in the projective setting.