From: greg@math.ucdavis.edu (Greg Kuperberg) Subject: This week in the mathematics arXiv (3 Jul - 7 Jul) Date: 13 Jul 2000 10:53:29 -0700 Newsgroups: sci.math.research Summary: [missing] [deletia --djr] In May Carolyn Gordon contributed "Isospectral deformations of metrics on spheres" [math.DG/0005156]. The article is on the differential geometry question, "Can one hear the shape of a drum?", posed by Mark Kac (in exactly these words) in 1966. In other words, can two disks in the plane have the same Laplace spectrum? The original question was eventually solved by Gordon, Webb, and Wolpert [Invent. Math. 110 (1992), no. 1, 1--22]. Along the way it became clear that the right question is "When can you hear the shape of a Riemannian manifold?". In particular Sunada [Ann. of Math. (2) 121 (1985), no. 1, 169--18] constructed many manifolds which are isospectral because they are (subtly different) covering spaces of the same base manifold. According to the Math Review, Sunada borrowed a method from number theory! In his analogy, manifolds correspond to number fields, covering spaces correspond to field extensions, and the Laplace spectrum corresponds to the Dedekind zeta function. Gordon, Webb, and Wolpert extended Sunada's idea to orbifolds and with other tricks whittled the counterexamples down to make them planar disks. The message of these first constructions was that manifolds can be isospectral for global reasons if they are locally the same. E.g. two Sunada-type manifolds are necessarily "scissors congruent". This is a little like saying that two orchestras may sound the same if they have the same musicians seated differently, so it may not seem counterintuitive once you accept the basic idea. (It is an imperfect analogy since musicians make sound separately, while adjacent regions in a manifold are coupled under the Laplace operator.) math.DG/0005156 explains that you can do even better: Many manifolds admit isospectral deformations. The article's new result is that in high dimensions (n>=8), even a round sphere S^n has this property. By constract, negatively curved closed manifolds are isospectrally rigid. I wonder if the round 2-sphere S^2 is known to be isospectrally rigid. If so, in what dimension does rigidity disappear? "This week in the mathematics archive" may be freely redistributed with attribution and without modification. [deletia --djr] /\ Greg Kuperberg (UC Davis) / \ \ / Visit the Math ArXiv Front at http://front.math.ucdavis.edu/ \/ * All the math that's fit to e-print * ============================================================================== From: sprouse@cims.nyu.edu () Subject: Re: This week in the mathematics arXiv (3 Jul - 7 Jul) Date: Fri, 14 Jul 2000 07:51:12 GMT Newsgroups: sci.math.research In article <8kkvmp$360$1@manifold.math.ucdavis.edu>, Greg Kuperberg wrote: [snipped] >you can do even better: Many manifolds admit isospectral deformations. >The article's new result is that in high dimensions (n>=8), even a round >sphere S^n has this property. By constract, negatively curved closed >manifolds are isospectrally rigid. I wonder if the round 2-sphere >S^2 is known to be isospectrally rigid. If so, in what dimension does >rigidity disappear? There are no isospectral deformations of the round metric on the sphere, in *all* dimensions (S. Tanno, Math. Z. 175, '80). I think what this article does is construct new metrics, that can be taken arbitrarily close to the metric on the sphere, so that these metrics admit nontrivial isospectral deformations. Any manifold that has the spectrum of S^2 is isometric to S^2. This was proved by Berger, who deserves some credit for really getting the study of the Laplace spectra of Riemannian manifolds started. To do this, he just computed the second heat invariant, which in 2 dimensions is some constant multiple of the integral of the square of the Gaussian curvature. The zeroth and first heat invariants are the volume, and an integral of the Gaussian curvature (not squared). So the spectrum determines all of these things. You then just have to use the Cauchy-Schwarz inequality to show that the curvature has to be constant. Tanno computed the third heat invariant, and showed that anything with the spectrum of a sphere must be a sphere in dimensions < 7 (Tohoku Math J. 25 '73). This seems like it should be true in higher dimensions, but the technique of just using the heat invariants gets way too hard. I think that the only manifolds known at this point to be completely determined by their spectra are these low dimensional spheres, and two-dimensional flat tori. (Unless you make some other conditions like Kaehler, etc.) Chad