## 3 Projective Rigidity
### Infinitesimal and Static Rigidity in Projective Space
An Euclidean force \(f=(f_{1},\ldots,f_{d})^{T}\) that is applied to an Euclidean point \(p=(p_{1},\ldots,p_{d})^{T}\) in \(\mathbb{R}^{d}\) can be written in the projective space \(\mathbb{P}^{d}\) as the 2-extensor given by the join of the projective points \(\hat{f}=(f_{1},\ldots,f_{d},0)^{T}\) and \(\bar{p}=(p_{1},\ldots,p_{d},1)^{T}\). For an appropriate choice of basis, the first \(d\) coordinates of the \(\binom{d+1}{2}\)-dimensional vector \(F_{i}=\hat{f}\vee\bar{p}\) is the free vector \((f_{1},\ldots,f_{d})\), and the remaining \(\binom{d}{2}\) coordinates may be interpreted as the moment of the force about the various coordinate axes. If we have a set of forces (2-extensors) \(F_{i}\), then the composition \(F\) of the \(F_{i}\) is defined as \(F=\sum_{i}F_{i}\) (where the sum is obtained by adding the corresponding minors). This composition is in general not a new single force (or 2-extensor) but a _wrench_[53]. However, if a set of forces \(F_{i}=\hat{f}_{i}\vee\bar{p}\) is applied to the same point \(\bar{p}\) (i.e., all forces \(F_{i}\) are on lines through \(\bar{p}\)), then we obtain the resultant force \(G=\sum F_{i}=\sum_{i}(\hat{f}_{i}\vee\bar{p})=(\sum_{i}\hat{f}_{i})\vee\bar{p}\). **Example 3**. : _Two opposite forces on parallel lines form a static couple (see Figure 7a). In the projective plane, the forces add up to an extensor on the line at infinity. After a projective transformation brings this line into the finite plane, the sum looks like (b) or (c). These are equivalent as the same force \(F_{1}\) can be drawn anywhere along its line._  
If \(f\) is an equilibrium load on a framework \((G,p)\) in Euclidean \(d\)-space which assigns the force \(f_{i}\) to the point \(p_{i}\), then in the projective space \(\mathbb{P}^{d}\), this _equilibrium load_ is given by the assignment of the force \(\hat{f}_{i}\vee\bar{p}_{i}\) to each point \(\bar{p}_{i}\) so that \(\sum_{i\in V}\hat{f}_{i}\vee\bar{p}_{i}=0\). A stress \(\rho\) resolves this equilibrium load if  
\[\sum_{j:j\in E}\rho(ij)\,\bar{p}_{j}\vee\bar{p}_{i}=-\hat{f}_{i}\vee\bar{p}_{ i}\qquad\text{for all }i\in V. \tag{5}\]  
As mentioned in Section 2, a framework is _statically rigid_ if it can resolve every equilibrium load. Moreover, the resolution of the zero force is an _equilibrium stress_ (or _self-stress_). The set of Equations (5) can be written in matrix form as  
Figure 7: Two opposite forces on parallel lines (**a**) form a couple. They will add up to a force along the line at infinity. After a projective transformation, they appear as (**b**) or equivalently (**c**). Thus, the row space of \(\tilde{\mathbf{R}}(G,\tilde{p})\) is the space of equilibrium loads. We conclude that a framework is statically rigid if and only if \(\tilde{\mathbf{R}}(G,\tilde{p})\) has rank \((d+1)|V|-\binom{d+1}{2}\). This completes the equivalence of static and infinitesimal rigidity. (see also ([29], Section 5.2))  
**Remark 1**. : _If we allow infinite graphs where every vertex has finite degree, then it turns out that infinitesimal rigidity is no longer equivalent to static rigidity, since for infinite-dimensional matrices the row rank is no longer equal to the column rank. Figure 8b shows an example of an infinite framework on the line which is statically but not infinitesimally rigid._  
_The line framework in Figure 8a is connected and therefore infinitesimally rigid. The framework in Figure 8b is disconnected and hence infinitesimally (and finitely) flexible, with the velocities of a non-trivial infinitesimal motion shown. Figure 8c shows a resolution of a force applied to one part of the framework, and Figure 8d shows the resolution of another force applied to the framework. Note that these are not equilibrium loads; this framework resolves all loads that can be applied (with no conditions for equilibrium) and hence it is statically rigid. The framework in (a) is also statically rigid, but with an equilibrium stress (which has the same stress coefficient on each edge).