Projections of a Regular Polyhedron

Paul Barbaroux August 01, 2025

Introduction

Il this post we focus on the following exercise Readers are invited to look for a solution, based on concepts rather than computations.:

— In the usual \(3\)-dimensional space, consider a regular polyhedron, the sum \(S\) of squared lengths of the edges, and the sum \(S_\P\) of squared lengths of the orthogonal projections of the edges on a plane \(\P.\) Prove that \(S_\P/S=2/3.\) Then generalize.

Before we give a solution, let's begin with a simpler statement which will allow to understand the main idea: sphericity of an ellipsoïd of inertia can be viewed as the geometric form of Schur's Lemma.

— In the usual \(3\)-dimensional euclidian space, let \(\S\) be the set of vertices of a regular polyhedron, and \(\D\) a line passing through the center. Show that \(\dsum_{s\in S}d(s,\D)^2\) does not depend on \(\D\). The proof must use abstract tools rather than analytic computations.

A physicist would give a one-line solution of ex. 2 by noticing that \(\sum_{s\in \S}d(s,\D)^{2}\)is the moment of inertia of \(\S\) with respect to \(\D\) (after providing each vertex with a point mass of value 1), then arguing that the ellipsoïd of inertia is a sphere from reasons of symmetry. We will first formalize correctly the symmetry argument, that is in an intrinsic way (without use of a coordinates system).

We will replace points by vectors by setting an origin at the center of the polyhedron. In this note the ambiant space will be a euclidian vector space Real finite-dimensional vector space endowed with an inner product. \(E\) of dimension \(n≠0.\) The inner product will be denoted by \((\cdot\ps\cdot).\) Given a vector subspace \(F,\) \(p_F\) denotes the orthogonal projection on \(F\). Given a set \(A\subset E,\) we say that an isometry \(g\) is an isometry of \(A\) if the set \(A\) is globally invariant under \(g.\) The set of isometries of \(A\) is a subgroup of \(O(E).\)

First, the symmetry argument relies on the fact that no direction is particular, Mathematically, this leads to the following definition: we say that a set \(a\) of vectors of a euclidian space has enough symmetries if the group of isometries of \(A\) is irreducible We recall that a subgroup \(G\) of \(\GL(E)\) is irréducible if the only subspaces of \(E\) stable under the action of \(G\) are trivial..

It turns out that this is actually the case for a regular polyhedron:

— In a \(3\)-dimensional euclidian space, the set \(\S\) of vertices of a regular polyhedron has enough symmetries.

The group \(G\) of isometries of \(\S\) contains at least two rotations of distinct axis and angles not multiple of \(\pi\) (consider rotations stabilizing two adjacent faces). Let \(F\) be a \(G\)-stable subspace of \(E\), \(F≠\{0\}.\) There exists a vector \(a\in F\setminus\{0\}.\) The axis of at least one of the rotations, call it \(r\), does not contain \(a.\) Hence \(a\) is not invariant by \(r\), neither anti-invariant (since \(\dim(E)=3\) the only rotations with an anti-invariant vector \(a≠0\) have angle \(\pi.\)) Hence \(a\) and \(r(a)\) are linearly independant and \(\dim F≥2.\) But \(F^\perp\) is also \(G\)-stable, hence \(\dim F^\perp≥2\) or \(F^\perp=\{0\}.\) Since \(\dim E=3\) the only possibility is \(F^\perp=\{0\}\) that is \(F=E,\) hence \(G\) is irreducible.

We can observe that if \(F\) is a subspace of \(E\) and \(x\in E\) then \(d(x,F)=||p_{F^\ortho(x)}||.\) When \(\dim E=3\) the orthogonal of a line is a plane, thus statement of ex. 2 is a special case of the following theorem, which additionally gives the value \(\textstyle\sum_{s\in \S}d(s,\D)^2=(2/3)\times \sum_{s\in \S}||s||^2.\)

— Let \(E\) be a euclidian vector space and \(A\subset E\) a finite set. If \(A\) has enough symmetries, then for any subspace \(F\subset E\) one has \[\dsum_{a\in A}||p_F(a)||^2={\dim F\over \dim E}\>\dsum_{a\in A}||a||^2\,.\]

For \(x\in E\) and \(A\subset E\) a finite set, let \[u_A(x)=\dsum_{a\in A}(x\ps a)\,a\,.\] \(u_A\) is a self-adjoint Actually \(u_A\) is the dual operator of inertia of \(A:\) if \(||x||=1\) then \((u_A(x)\ps x)\) is the moment of inertia of \(A\) w.r. to the hyperplane \(x^\perp\). operator, hence has an eigenvalue \(\lambda\). But \(u_A\) commutes commutes with all isometries of \(A:\) if \(g\) is such an isometry, then \(\displaystyle g(u_A(x))=\dsum_{a\in A}(x\ps a)g(a)\) \(\displaystyle =\dsum_{a\in A}(g(x)\ps g(a))g(a)\) \(\displaystyle =\dsum_{b\in A}(g(x)\ps b)b=u_A(g(x))\,.\) Therefore, \(G\) stabilizes the eigenspace \(\ker(u_A-\lambda\id)\). But \(G\) is irreducible, hence \(\ker(u_A-\lambda\id)=E\) that is \(u_A=\lambda\id\)Which can be viewed as Schur's Lemma applied to \(G\) and \(u_A.\).
Now for every \(e\in E\) such that \(||e||=1,\) one has \(\displaystyle\dsum_{a\in A}(e\ps a)^2=\dsum_{a\in A}(e\ps a)(a\ps e)\) \(\displaystyle=(u_A(e)\ps e)=(\lambda e\ps e)=\lambda||e||^2=\lambda\,.\) By denoting \((e_i)_{1≤i≤p}\) an orthonormal basis of the subspace \(F\), we obtain \(\displaystyle\dsum_{a\in A}||p_F(a)||^2=\dsum_{a\in A}\dsum_{i=1}^p(e_i\ps a)^2\) \(\displaystyle=\dsum_{i=1}^p\dsum_{a\in A}(e_i\ps a)^2=\dsum_{i=1}^p\lambda=\lambda\dim F\,.\) The special case \(F=E\) then gives \[\dsum_{a\in A}||a||^2=\lambda \dim E\,,\] which gives the value of \(\lambda\) and completes the proof.

Th. 1 can now be easily extended to a more general result that allows to treat not only ex. 2 but also ex. 1 as well as several other geometric situations. We just have to replace the use of a finite set of vectors by the one of a finite family. Let \(I\) be a set (of indices). We suppose given an action \((g,i)\mapsto g\cdot i\) of the orthogonal group \(O(E)\) on \(I\). We say that a family of vector \((x_i)_{i\in I}\) is admissible if \(g(x_i)=x_{g\cdot i}\) for every \(i\in I\) and \(g\in O(E).\) As before an isometry of a set \(J\subset I\) is an isometry that leaves \(J\) globalement unchanged, and we say that \(J\) has enough symmetries if its group of isometries is irreducible. Then we have

— Let \(f=(x_i)_{i\in I}\) be an admissible family of vectors in a euclidian vector space \(E\) and \(J\subset I\) a finite set having enough symmetries. Then for every subspace \(F\subset E\) one has: \[\dsum_{i\in J}||p_F(x_i)||^2={\dim F\over \dim E}\>\dsum_{i\in J}||x_i||^2\,.\]

The proof works in the same way as for th. 1 by now considering \(u_f(x)=\sum_{i\in I}(x\ps x_i)\,x_i\) instead of \(u_A(x)=\dsum_{a\in A}(x\ps a)\,a.\)

a Th. 2 gives back th. 1, when applied to \(I=E\) and the identity-indexed family.
b Ex. 2 is solved by taking
- \(I=E\times E\) and, for \((a,b)\in I,\) and \(g\in O(E),\) \(g\cdot (a,b)=(g(a),g(b)).\)
- \(x_{(a,b)}=b-a.\) The admissibility of \((x_{(a,b)})\) comes from \(g\)'s linearity.
- \(J=\{(a,b)\in E\times E\,,\>\{a,b\}\) is an edge of the polyhedron\(\}\) (hence two ordered pairs \((a,b)\) and \((b,a)\) are given for each edge \(\{a,b\}\) such that \(a≠b\)). Obviously \(J\) has enough symmetries, since any isometry of the polyhedron sends every edge to an edge, hence is an isometry of \(J.\)
cA polyhedron does not need to be regular in order to ex. 1 to hold: it just has to have the same isometry group as the set vertices of a regular polyhedron It is even enough for its isometry group to contain the isometry group of some regular polyhedron.. One can obtain such polyhedra by truncating regular polyhedra (for instance, conclusion of ex. 1 remains true for archimedeanSee e.g. the page about archimedean solids on Wikipedia or Mathcurve. solids) or, on the opposite, by having other polyhedra grow on faces. For instance, one can glue on each face of a regular polyhedron a pyramid or a prism of given height.
dLet's call a squeleton any finite set \(J\subset E\times E.\) Starting with skeletons of edges of regular polyhedra, or given by remark c), one can easily construct many other ones satisfying the statement of ex. 1 by noticing that the set of such skeletons is stable under translations and disjoint unions.