# Two questions regarding Spivak's Configuration Space

The following is from the fifth Chapter Rigid Bodies of Spivak's Physics for Mathematicians. The post consists of a statement Spivak makes -with no proof- that I do not understand. For clarity, I've tried including the definitions and lemma that I believe may be relevant in regards to the result Spivak states.

Definition: let $$b_1, \ldots ,b_K$$ be a collection of points which we regard as a single object $$b=(b_1,\ldots ,b_K)$$ and let $$F=(F_1,\ldots ,F_K)$$ be a collection of forces where we regard $$F_i$$ as acting on $$b_i$$, then $$b$$ is in rigid equilibrium under the forces $$F$$ if there exist internal forces $$F_{ij}=-F_{ji}$$ which are multiples of $$c_i-c_j$$ so that $$F_i=-\sum_jF_{ij}.$$

Definition: a rigid motion $$f$$ of $$b$$ is a curve $$t\to A(t)$$ where $$A(t)$$ is an isometry of $$\mathbb{R}^3$$ such that

1. The position function $$c_i(t) = A(t)(b_i)$$ of $$b_i$$ is smooth with $$c_i(0)=b_i$$.
2. We have $$\langle c_i(t) - c_j(t), c_i(t) - c_j(t)\rangle \ \text{ stays constant} \tag{1}$$ so that the distance between the points $$b_i$$ and $$b_j$$ stays constant.

Definition: the configuration space $$\mathcal{M}_b$$ of $$b$$ is given by $$\big\{ \big(A(b_1),\ldots ,A(b_K)\big) : A \text{ is an orientation preserving isometry of }\mathbb{R}^3\big\}.$$

Lemma: we have that $$\langle v_i(0) - v_j(0), b_i - b_j\rangle = 0$$

Proof: Follows by differentiating $$(1)$$ and evaluating at zero. $$\square$$

Theorem: if we define the linear functions $$\phi_{ij}$$ on $$(\mathbb{R}^3)^K$$ by $$\phi_{ij}(v_1,\ldots ,v_K) = \langle v_i - v_j, b_i - b_j\rangle$$ then $$\mathcal{M}_b\subseteq \bigcap_{i,j}\ker \phi_{ij}.\tag{2}$$

It is this last Theorem that Spivak states without proof, and I'm confused on two grounds:

1. In what sense are the functions $$\phi_{ij}$$ linear? If we substitute any of the arguments $$v_i$$ by $$\mu v_i$$ for some scalar $$\mu$$, the resulting output will not get scaled by $$\mu$$, so I do not understand how the $$\phi_{ij}$$ can be linear nor multilinear.

2. How is $$(2)$$ derived? The elements of $$\mathcal{M}_b$$ are supposed to represent positions, not velocities. Thus, for an element $$m=\big( A(b_1), \ldots ,A(b_K)\big)\in \mathcal{M}_b$$ we get $$\phi_{ij}(m) = \langle A(b_i) - A(b_j), b_i - b_j\rangle$$ which need not be zero (compare with the Lemma).

• Commented Jul 16, 2022 at 13:42

1. The $$\phi_{ij}$$ are linear in the usual sense. I don't see any issues with it. For any $$v=(v_1,\dots, v_K), w=(w_1,\dots, w_K)\in (\Bbb{R}^n)^K$$, and any scalar $$\mu\in\Bbb{R}$$, \begin{align} \phi_{ij}\left(\mu v+w\right)&:=\phi_{ij}\left((\mu v_1+w_1,\dots, \mu v_K+w_K)\right)\\ &:=\langle (\mu v_i+w_i)-(\mu v_j+w_j), b_i-b_j\rangle\\ &=\mu\langle v_i-v_j,b_i-b_j\rangle + \langle w_i-w_j,b_i-b_j\rangle \\ &=\mu\phi_{ij}(v) + \phi_{ij}(w), \end{align} where we have used bilinearity of the inner product $$\langle\cdot,\cdot\rangle$$.
2. I just read the corresponding text in Spivak, and you must have mistyped what he wrote. He says that the configuration space of the point $$b=(b_1,\dots, b_K)\in(\Bbb{R}^n)^K$$ is denoted as $$\mathcal{M}$$. In the theorem that $$\mathcal{M}_b\subset\bigcap_{ij}\ker\phi_{ij}$$, the symbol $$\mathcal{M}_b$$ refers to the tangent space of $$\mathcal{M}$$ at the point $$b$$ (this is his notation for the tangent space to a manifold at a given point, and he also uses it in his other texts such as Calculus on Manifolds and his 5-Volume tome A Comprehensive Introduction to Differential Geometry). Just to be clear, $$\mathcal{M}$$ is an embedded submanifold of $$(\Bbb{R}^n)^K$$, so the tangent space at any given point, a-priori has an abstract definition, but because this is embedded in $$(\Bbb{R}^n)^K$$, we can use the identity chart to view this tangent space as an actual subspace of $$(\Bbb{R}^n)^K$$ (see this MSE answer of mine for more details) and hence it makes sense to say it is contained in the intersection of the kernels of the linear maps $$\phi_{ij}$$.
• I have to admit, when I first read your post, I was confused as well, because it seemed to be mixing up positions and velocities, something completely uncharacteristic for Spivak, and only then did I read the corresponding portion of the text and realized it's his notation which is tripping you up; $\mathcal{M}_b\equiv T_b\mathcal{M}$. Commented Jul 16, 2022 at 15:53