The Ricci decomposition of the inertia tensor for a rigid body in arbitrary spatial dimensions

TL;DR

This paper extends the Ricci decomposition to the inertia tensor of rigid bodies in arbitrary dimensions, revealing that the inertia tensor's Weyl component is always zero, simplifying its geometric characterization.

Contribution

It provides the first calculation of the Ricci decomposition of the inertia tensor in any dimension, showing the absence of a Weyl component unlike in the Riemann tensor.

Findings

Inertia tensor's Weyl tensor is always zero in any dimension.

The inertia tensor is fully characterized by its Ricci contraction.

No new phenomenology arises from the Weyl tensor in higher-dimensional rigid-body dynamics.

Abstract

The rotations of rigid bodies in Euclidean space are characterized by their instantaneous angular velocity and angular momentum. In an arbitrary number of spatial dimensions, these quantities are represented by bivectors (antisymmetric rank-2 tensors), and they are related by a rank-4 inertia tensor. Remarkably, this inertia tensor belongs to a well-studied class of algebraic curvature tensors that have the same index symmetries as the Riemann curvature tensor used in general relativity. Any algebraic curvature tensor can be decomposed into irreducible representations of the orthogonal group via the Ricci decomposition. We calculate the Ricci decomposition of the inertia tensor for a rigid body in any number of dimensions, and we find that (unlike for the Riemann curvature tensor) its Weyl tensor is always zero, so the inertia tensor is completely characterized by its (rank-2) Ricci contraction. So unlike in general relativity, the Weyl tensor does not cause any qualitatively new phenomenology for rigid-body dynamics in $n \geq 4$ dimensions.