A Characterization of Moments of Inertia of
a Solid Body

Introduction

This blog post is a rewriting of this paper. A short version has been also published as a note to the Comptes-Rendus de l'Académie des Sciences: P. Barbaroux, Characterization of moments of inertia of a solid body, C. R. Acad. Sci. Paris, Ser. I 334 (2002) 1067-1070. (Note presented par Alain Connes). Let \(\E\) be is a euclidian affine space of dimension \(n≥3,\) with associated euclidian vector space \(E.\) Let us consider a solid body \(\S,\) given by the data consisting of a real measure \(\rho\) on the Lebesgue \(\sigma\)-algebra of \(\E.\) We take \(\rho\) with compact support to avoid convergence problems.They do not have much interest here: we are mainly concerned in the algebraic aspects. We can consider, for each affine subspace \(\F\) of \(\E,\) the moment of inertia \[I_{\cal S}({\cal F})=\int_{\cal E}d(m,{\cal F})^2\d\rho(m)\] of \(\S\) with respect to the affine subspace \(\F.\) The map \(f:\F\mapsto I_{\S}(\F)\) satisfies the following three conditions:

1. Perpendicular additivity For all perpendicular subspaces \({\F},\) \(\G\) (i.e. \(\F^\perp\perp\G^\perp),\) \[f(\F\cap \G)=f(\F)+f(\G)\]
2. Huygens' theorem For \(x\perp\F,\) we have, when noting \(M\) for the total mass of \(\S,\)
    a) If \(M≠0\) and \(\F\) contains the center of inertia of \(\S,\) then \[f(\F+x)=f({\F})+M||x||^2\,;\]  b) If \(M=0,\) the vector \(a=\int_{\cal E}\vecteur{Am}\d\rho(m)\) does not depend on the point \(A,\) and \[f({\cal F}+x)=f(\F)-2(a\ps x).\]
3. Continuity \(f\) is a continuous map for the Grassmannian topology on the set of affine subspaces of \(\E.\)

Conversely, let's now forget the solid body and keep only its inertial trace on the affine subspaces: consider a map \(f\) defined on the set of affine subspaces of \(\E\) satisfying the three conditions above. We show that there then exists a solid body \(\S\) whose moments of inertia are represented by \(f.\) This result constitutes in some way the affine version of Gleason's theorem.Andrew M. Gleason, Measures on the closed subspaces of a Hilbert space. J. math. Mech. 6. (1957), 885-893. We also show that condition (ii) is useless, in other words: perpendicular additivity and continuity imply Huygens' theorem. Moreover we can always take for \(\S\) a finite set of point masses, of which we are able to express the minimum number in geometric terms.

Preliminary: the linear case

Lemma 1 allows to define the dual quadratic form of inertia of a solid body at some point:

We have \(\displaystyle q(x)=||x||^2I_{\cal S}(O+x^\perp)\) \(\displaystyle=||x||^2\int_{\cal E}d(m,O+x^\perp)^2\d\rho(m) \) \(\displaystyle=\int_{\cal E}(\vecteur{Om}\ps x)^2\d\rho(m)\,,\) therefore \(q\) is the quadratic form coming from the bilinear form \(B\) defined by \[B(x,y)=\int_{\cal E}(\vecteur{Om}\ps x)(\vecteur{Om}\ps y)\d\rho(m)\,.\] Moreover, for every linear subspace \(F\) of \(E,\) if we note \((e_i)\) an orthonormal basis of \(F^\perp,\) perpendicular additivity gives \(\displaystyle I_{\cal S}(O+F)=\sum_i I_{\cal S}(O+e_i^\perp)\) \(\displaystyle=\sum_i q(e_i)=\tr(q|_{F^\perp}).\)

Note that any quadratic form \(q\) on \(E\) comes in this way from a solid body. Indeed, by considering a basis of \((e_i)\) of \(E\) that is both orthonormal and \(q\)-orthogonal, one has \(q(x)={\displaystyle\sum_i} q(e_i)(e_i\ps x)^2,\) so that taking for \(\S\) the solid body obtained by placing at each point \(O+e_i\) a point mass \(m_i\) equal to \(q(e_i),\) one has \(q(x)=||x||^2I_{\S}(O+x^\perp).\)

Gleason's theorem, in the special case of a Euclidean space \(E,\) asserts that for any bounded measure \(\mu\) (orthogonally additive real function on the set of linear subspaces of \(E),\) there exists a quadratic form \(q\) such that \(\mu(F)=\tr(q|_F).\) Switching to the orthogonal subspace we get:

1. a As for Gleason's theorem, the result fails if \(E\) is only 2-dimensional: being given an orthonormal basis \((i,j)\) of \(E,\) put \(u_\theta=(\cos\theta)\, i+(\sin\theta)\, j.\) Let \(f\) be any \(\pi\)-périodic function bounded on \(\rmat\) such that \(f(x+\pi/2)+f(x)\) is a fixed value \(\lambda.\) The function \(h\) defined by \(h(\rmat u_\theta)=f(\theta),\) \(h(\{0\})=\lambda,\) and \(h(E)=0,\) satisfies the conditions of theorem 1 but in general does not come from a solid body.
2. b Boundedness is also an essential condition, as shown by considering \(h:F\mapsto\varphi(I_{\cal S}(O+F)),\) where \(\cal S\) consists of a single non-zero point mass placed elsewhere than at \(O\) and \(\varphi:\rmat\longrightarrow\rmat\) any arbitrary non-continuous \(\qmat\)-linear map.

The affine case: statement of main result

Let \(h\) be a real function defined on the set of vector (resp. affine) subspaces of \(E\) (resp. \(\cal E\hbox{).}\) A subspace \(\cal F\) is said to be total for \(h\) if it satisfies the following two conditions:

1. For every subspace \(\cal G\) containing \(\cal F,\) \(h({\cal G})=0\,;\)
2. For every subspace \(\cal G\) and every linear (resp. affine) isometry \(\Phi\) that fixes each point of \(\cal F,\) one has \(h(\Phi({\cal G}))= h({\cal G}).\)

The smallest dimension of a total subspace for \(h\) will be called the rank of \(h\) and denoted by \(\rg(h).\) Note that if \(h\) has the property of perpendicular additivity, then the entire space is total for \(h,\) hence \(\rg(h)\) is well defined.

In the case of a map \(g\) defined on the set of affine subspaces of \(\E,\) we will denote for simplicity \(g(M)\) instead of \(g(\{M\}),\) and the map on \(\cal E\) induced by \(g\) will be denoted by \(g_0.\) For \(A\in\cal E,\) we denote \(g^A\) the restriction of \(g\) to the set of affine subspaces containing \(A.\)

The remainder is dedicated to the proof of the following result.

As a consequence of the proof and of Huygens' theorem, all possible situations will be described by the following four cases, that correspond to four types of solid bodies:

1. a \(g_0(O+x)=g_0(O)+M||x||^2\) (\(M\neq0):\) body with total mass \(M\neq0\,;\)
2. b \(g_0(O+x)=(a\ps x)\) (\(a\neq 0):\) body such that \(M=0\) and \(a=-2\int_{\cal E}\vecteur{Am}\d\rho(m)\neq0.\) This is the case, for example, of a system of two opposite non-zero masses placed at two distinct points, or of two masses of value \(1\) and a mass of value \(-2,\) placed elsewhere than the middle;
3. c Trivial case \(g=0.\) A sufficient condition is (see lemma 3) \(g_0\) being constant and that there exists \(A\in{\cal E}\) such that \(g^A=0.\) This is the case of a system satisfying \(M=0,\) \(a=0,\) and having null moments of inertia with respect to the subspaces passing through a given point. In this case the system behaves like an empty body. This is for example the case of a system made of six masses equaling 1, -2, 1, 1, -2, 1, placed on a straight line at abscissas -7, -5, -1, 1, 5, 7, which therefore does not have any inertia.
4. d \(g_0\) is constant but \(g\neq0:\) system satisfying \(M=0,\) \(a=0,\) and at least one nonzero moment of inertia. For example, the system \(\cal S\) made up of \(2n+1\) point masses obtained by placing, at each point \(O\pm e_i\) (\(O\in\cal E,\) \((e_i)\) orthonormal basis of \(\cal E),\) a mass of value \(1/2,\) and at \(O\) a mass of value \(-n,\) satisfies \(I_{\cal S}({\cal F})=\codim\ {\cal F}\) and is reduced, by virtue of the theorem, to \(n+2\) masses (of which \(n+1\) are placed, for symmetry reasons, at the vertices of a regular simplex).

It will also follow from the proof that in case where \(g\) is positive (which can only happen in the cases a, c, d), all the masses, except one in the case d), can be chosen to be positive.

Some useful lemmas

By perpendicular additivity one has \[g({\cal F}+\rmat x)+g({\cal H})=g({\cal F})\] since \({\cal F}+\rmat x\) and \(\cal H\) are perpendicular and their intersection is \(\cal F.\) Hence \(\displaystyle g({\cal F})-g({\cal H})=g({\cal F}+\rmat x)=g({\cal F}+x+\rmat x)\) \(\displaystyle =g({\cal F}+x)-g({\cal H}+x),\) and the result is obtained by replacing \(\cal F\) by \(\cal G\) in the previous relation and subtracting.

By considering \(X/\ker Q,\) we can assume that \(Q\) is non-degenerate and \(u\neq 0.\) We start by finding a \(Q\)-orthogonal basis \((\eta_i)\) such that \[\sum_i Q(\eta_i)=Q(u).\] For this, let \((\xi_i)\) be an arbitrary \(Q\)-orthogonal basis. If \(u\) is isotropic, since \(Q\) is non-degenerate we have \(\exists i,j,\> Q(\xi_i)Q(\xi_j)<0,\) and we can always multiply \(\xi_i\) ou \(\xi_j\) by a real number \(\lambda>0\) so as to obtain a basis \((\eta_i)\) satisfying \(\sum_i Q(\eta_i)=0=Q(u).\) Otherwise, we can assume \(Q(u)>0,\) and then \(I=\{i\ps Q(\xi_i)>0\}\neq\vide.\) Define \((\eta_i)\) by \(\eta_i=\lambda\xi_i\) if \(i\in I,\) and \(\eta_i=\xi_i\) if not. Then relation \(\sum_i Q(\eta_i)=Q(u)\) can be written \[\lambda^2 \sum_{i\in I} Q(\xi_i)+\sum_{i\notin I} Q(\xi_i)=Q(u),\] equation which admits a solution \(\lambda\neq0.\)

We conclude using Witt's theorem: if \(Q(u)=Q(v)\) there exists a \(Q\)-isometry \(\phi\) of \(X\) such that \(u=\phi(v).\) Just apply it to \(v=\sum\eta_i\) and take \(\varepsilon_i=\phi(\eta_i).\)

This is a very classic undergraduate exercise when we take a continuity assumption.See for example the following two references in French:
• Revue des mathématiques de l'enseignement supérieur (RMS), 3-4 (2000). Réponse R364 de H. Pépin p. 530-533.
• S. Francinou, H. Gianella, S. Nicolas, Exercices de mathématiques Oraux X-ENS : Algèbre 3, ex. 16. Cassini, Paris (2008). The solution can be easily extended to the present case of a function that is bounded in a neighbourhood of \(0.\)

Proof of main result

Now let's proceed to the proof of Theorem 2. First of all, the given conditions are clearly necessary. Moreover, the given number of masses cannot be improved: a system \(\cal S\) of \(k\) masses \((k\geq 1)\) generates an affine subspace of dimension at most \(k-1,\) total for \(g=I_{\cal S},\) which gives \(k\geq \rg(g)+1.\) If additionally \(g_0\) is constant and \(g\) not identically zero, then the total mass is zero and one of the masses is non-zero, and case b) of Huygens' theorem then shows that the places of the point masses are affinely dependent, hence \(k\geq \rg(g)+2.\)

Conversely, let \(g\) satisfy (i), (ii) et (iii). Let \(A\in\cal E,\) in a neighbourhood of which \(g_0\) is bounded. We set, for all \(x\in E,\) \(\phi(x)=g(A+x)-g(A).\) For \(x,y\in E,\) by lemma 3 we then have, as soon as \(x\perp y,\) \(\phi(x)+\phi(y)=\phi(x+y).\) But \(\phi\) is bounded in a neighbourhood of \(0\) since \(g_0\) is bounded in a neighbourhood of \(A.\) By lemma 5 there exists \((M,a)\in\rmat\times E\) such that \(\displaystyle \forall x\in E,\ g(A+x)-g(A)=\phi(x)\) \(\displaystyle=M||x||^2+(a\ps x).\) Il follows that the map \(g_0\) is an affine quadratic form, and \(\displaystyle \forall P\in{\cal E}\,,\quad\exists b\in E\,,\quad \forall x\in E,\quad\) \[\kern-1.5em g(P+x)-g(P)=M||x||^2+(b\ps x)\,.\tag{1}\] Let \({\cal F}_0\) be a total subspace for \(g\) of least dimension \(\rg(g),\) and \(F_0\) its direction.

Case 1: \(g_0\) is not an affine form. By changing \(g\) to \(-g\) if needed, we can assume \(M>0.\) The the map \(g\) reaches a global minimum, at a point, which we will denote \(O,\) that satisfies \[\kern-1.5em \forall x\in E,\> g(O+x)-g(O)=M||x||^2\,.\tag{2}\] Let \(B\in{\cal E}\) such that \(g^B\) is bounded. By applying relation \((1)\) to the point \(B\) and using lemma 3, \(g^P\) is bounded for every point \(P.\) As a special case \(g^O\) is bounded. By theorem 1, there exists a quadratic form \(q\) and a solid body \(\cal S\) such that for every linear subspace \(F,\) \(g({O+F})=\tr(q|_{F^\perp})=I_{\cal S}(O+F).\) By relation \((2)\), lemma 3, and Huygens' theorem, it remains to be seen that one can impose the mass of the system to be \(M\) and that its center of inertia to be \(O,\) while preserving the moments of inertia with respect to all subspaces containing \(O,\) by using only \(\rg (g)+1\) point masses. For that, consider the quadratic form \(Q\) defined on \(E\times \rmat\) by \(Q(x,t)=q(x)+M\,t^2.\) The vector \(u=(0,1)\) is not in \(\ker Q,\) since \(Q(u)=M>0.\) By lemma 4, there exists a \(Q\)-orthogonal basis \((\varepsilon_i)\) such that \(u=\sum \varepsilon_i,\) that is \(\varepsilon_i^*(u)=1.\) But every linear form \(\varepsilon_i^*\) is of the form \(\varepsilon_i^*(x,t)=(a_i\ps x)+\lambda_it,\) where \((a_i,\lambda_i)\in E\times \rmat.\) Then \(\varepsilon_i^*(u)=1\) gives \(\lambda_i=1,\) hence for all \((x,t)\in E\times \rmat,\) \(\displaystyle q(x)+M\,t^2=Q(x,t)\) \(\displaystyle=\sum_{i=1}^{n+1}Q(\varepsilon_i) (\varepsilon_i^*)^2(x,t)\) \(\displaystyle=\sum_{i=1}^{n+1}Q(\varepsilon_i) ((a_i\ps x)+t)^2\,.\) By setting \(m_i=Q(\varepsilon_i)\) and developing the square, an immediate argument makes it possible to identify and we get: \(\displaystyle a)\>q(x)=\sum m_i(a_i\ps x)^2\ ; \) \(\qquad\displaystyle b)\> \sum m_ia_i=0\ ;\) \(\displaystyle\qquad c)\> \sum m_i=M.\) It is then enough to consider the system \(\cal S',\) obtained by placing, for each \(i\) such that \(m_i\neq0,\) the mass \(m_i\) at \(O+a_i.\) By using relation \((2)\) and the fact that \({\cal F}_0\) is total, the latter contains \(O\) (consider the mirror image of \(O\) with respect to \({\cal F}_0),\) and its direction \(F_0\) is then total for \(F\mapsto g(O+F).\) By lemma 2 \((\ker q)^\perp\subset F_0,\) and the number of point masses is then \(\displaystyle r=\rg (Q)=\rg (q)+1\) \(\displaystyle\leq(\dim{\cal F}_0)+1=\rg (g)+1.\)
Case 2: \(g_0\) is an affine form. This time there exists \(a\in E\) such that for every \(P\in{\cal E}\) and \(x\in{E}\) we have \[g(P+x)-g(P)=(a\ps x)\,.\tag{3}\]

• If \(a≠0,\) we place ourselves at a point \(O\in{\cal F}_0\) that is arbitrary for the moment, and we set \(Q(x,t)=q(x)-2(a\ps x)\,t.\) Then \((0,1)\) is still not in \(\ker Q\) car \(a\neq 0.\) We get: \(\displaystyle a)\>q(x)=\sum m_i(a_i\ps x)^2\ ; \) \(\qquad\displaystyle b)\> \sum m_ia_i=-a\ ;\) \(\displaystyle\qquad c)\> \sum m_i=0,\) and this time it is the case b) of Huygens' theorem which then allows us to conclude. Concerning the number of masses we used, a priori, \(\rg (Q)\leq\rg (q)+2.\) But by noting \(q_O\) the quadratic form associated with \(g\) at point \(O,\) relation \((3)\), equality \(q_O(x)=||x||^2g(O+x^\perp)\) and lemma 3 give, for \(b\in E,\) \[\kern-1.5em q_{O+b}(x)=q_O(x)+(a\ps x)(b\ps x)\,.\tag{4}\] It is the enough to choose \(O\in{\cal F}_0\) in which \(q=q_O\) is of maximal rank. Then \(a\in(\ker q)^\perp,\) since otherwise by replacing \(O\) with \(O+a\) (\({\cal F}_0\) is total and hence one can easily see, using relation \((3)\), that \(a\in F_0),\) the rank of \(q\) would increase, since according to \((4)\) one has \(q_{O+a}(x)=q_O(x)+(a\ps x)^2\,.\) But \(a\in(\ker q)^\perp\) gives \(\rg (Q)\leq\rg (q)+1,\) and we conclude as before.
• If \(a=0,\) that is \(g_0\) is constant, we come back to case 1 by adding the moment of inertia of a non-zero point mass placed anywhere in \({\cal F}_0,\) which then remains total, and we recover the initial map \(g\) by adding to the new system the opposite mass, which gives at most \(\rg (g)+2\) point masses.