Inaudibility of sixth order curvature invariants

TL;DR

The paper demonstrates that sixth order curvature invariants are not spectrally determined on Riemannian manifolds, showing examples where isospectral manifolds differ in higher order curvature invariants, thus extending known inaudibility results.

Contribution

It proves that sixth order curvature invariants are not determined by the spectrum and provides explicit examples of isospectral manifolds with differing curvature invariants.

Findings

Spectral data does not determine sixth order curvature invariants.

Two isospectral Heisenberg-type nilmanifolds are isometric iff they share $| abla R|^2$.

Existence of non-curvature equivalent manifolds indistinguishable by invariants up to order $2k$.

Abstract

It is known that the spectrum of the Laplace operator on functions of a closed Riemannian manifold does not determine the integrals of the individual fourth order curvature invariants $\operatorname{scal}^2$, $|\operatorname{ric}|^2$, $|R|^2$, which appear as summands in the second heat invariant $a_2$. We study the analogous question for the integrals of the sixth order curvature invariants appearing as summands in $a_3$. Our result is that none of them is determined individually by the spectrum, which can be shown using various examples. In particular, we prove that two isospectral nilmanifolds of Heisenberg type with three-dimensional center are locally isometric if and only if they have the same value of $|\nabla R|^2$. In contrast, any pair of isospectral nilmanifolds of Heisenberg type with centers of dimension $r>3$ does not differ in any curvature invariant of order six, actually not in any curvature invariant of order smaller than $2r$. We also prove that this implies that for any $k\in\Bbb N$, there exist locally homogeneous manifolds which are not curvature equivalent but do not differ in any curvature invariant of order up to $2k$.