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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).

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  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 $\phi_{ij}$.

Of course, the theorem follows immediately from the preceding discussion in the text (particularly the lemma you quote).

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  • $\begingroup$ 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; $\mathcal{M}_b\equiv T_b\mathcal{M}$. $\endgroup$
    – peek-a-boo
    Commented Jul 16, 2022 at 15:53

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