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arXiv · 1506.02447

Inaudibility of sixth order curvature invariants

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 $2k$.

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Teresa Arias-Marco, Dorothee Schueth. 2015-06-08. Inaudibility of sixth order curvature invariants. https://arxiv.org/abs/1506.02447

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