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Let $(M,g)$ be a (say closed) Riemannian manifold. One can try to understand the geometry/topology of $(M,g)$ by studying the eigenvalues of the Laplacian (this I guess has two versions: when considering the Laplacian on functions only, or on differential forms. feel free to answer about whichever).

Sometimes (for example for nice bounded planar domains) the topology of the manifold is completely determined by the spectrum of the Laplacian (on functions).

Also, by the Hodge theorem, for an arbitrary closed Riemannian manifold all of the Betti numbers can be read off the spectrum of the Laplacian (on forms). Sadly, it is well known that there are examples of isospectral but not isometric Riemannian manifolds.

Q1: Are there examples of pairs of Riemannian manifolds which are isospectral but not diffeomorphic/homeomorphic/homotopy equivalent? If so, what is the simplest one?

assuming the answer to Q1 is "Yes" or "It is not known":

Q2: What are the best positive results in this direction? How much can I know about the topology of $M$ only from my knowledge of the spectrum of the Laplacian?

I'm sorry if the question is too vague, but still any information will be nice.

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    $\begingroup$ Q1 is answered here, I think: mathoverflow.net/q/97083/11260 $\endgroup$ – Carlo Beenakker May 17 '18 at 13:25
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    $\begingroup$ Already Vigneras's 1980 paper in Annals included an example of an isospectral pair with non-isomorphic fundamental groups. Regarding Q2 a good start would be to have a look into the 1997 conference volume Progress in Inverse Spectral Geometry (ed. S. Anderson and M. Lapidus), published by Springer in its "Trends In Mathematics" series. $\endgroup$ – Vesselin Dimitrov May 17 '18 at 13:39
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There are examples due to Ikeda of isospectral Lens spaces which are not homotopy equivalent.

Likeliest the simplest examples are the compact connected 3-dimensional flat manifolds which are a tetracosm and didicosm. These are isospectral but not homotopy equivalent. I'm not sure if they are known to have the same spectrum for the Laplace-de Rham operator (i.e. on forms). Presumably this could be computed explicitly.

There are many other examples of non-isometric manifolds which are strongly isospectral (having the same spectrum on forms) using Sunada's method, and arithmetic techniques (originally due to Vignéras) for locally symmetric spaces. Coupled with the Mostow rigidity theorem for locally symmetric spaces, this implies that these manifolds are not homotopy equivalent.

In 1-dimension, the spectrum determines the manifold.

For connected two-dimensional surfaces, the Laplace-de Rham spectrum determines the betti numbers, and hence the topological type (the surface $S$ is orientable iff $b_2(S)=1$, and then $b_1(S)$ determines the topological type if it is orientable or non-orientable). (For surfaces with boundary, an example of Bérard-Webb shows that the Neumann spectrum does not determine orientability.) Thus, the 3-dimensional examples coming from Sunada's construction or Vignéras' are minimal dimensional examples which are strongly isospectral. I don't know if the Laplace spectrum determines the topological type of closed surfaces.

See section 5 of the survey paper by Carolyn Gordon for more information on how topology cannot be detected by the spectrum and references. Note also that Lubotzky-Samuels-Vishne showed that there are isospectral arithmetic lattices which are not commensurable (so not obtained by Sunada's method or Vigneras' criterion).

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    $\begingroup$ All the pairs here have the same universal covering space. The article "On the dimension datum of a subgroup and its application to isospectral manifolds" by Jinpeng An, Jiu-Kang Yu, and Jun Yu (<zbmath.org/?q=ai:yu.jun+ai:an.jinpeng>) shows the first pair of simply connected isospectral closed Riemannian manifolds that are not homeomorphic. The smallest dimension is 26. $\endgroup$ – emiliocba May 18 '18 at 3:09
  • $\begingroup$ See also this paper for finiteness results regarding isospectral manifolds: mathscinet.ams.org/mathscinet-getitem?mr=1155747 $\endgroup$ – Ian Agol May 22 '18 at 14:25
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I'm not entirely certain if this is applicable to what you are asking, but you mention Laplacian and eigenvalues... PBS Infinite Series on YouTube did a video related to this. Its titled 'Can We Hear Shapes' and in it they address the question 'Can we hear the shape of a drum?' as posed by Mark Kac. In short, she concludes that for a 2 dimensional surface, no, you cannot, as the frequency it would produce does not uniquely define it. What you can know, is the area and perimeter length, and if the topology has holes, what genus it is. But because two distinct surfaces can have the same area and parameter, we cannot determine its shape. Now, does this apply to higher dimensions, or what you are talking about? I am uncertain...

Regardless, here is the link to the PBS's Infinite Series video on YouTube: https://www.youtube.com/watch?v=bkJZbyrfx-Q

I hope I may have been some help.

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    $\begingroup$ The present question is about hearing the drum's topology, not its geometry. (This is the difference between Riemannian manifolds being isometric versus merely being homeomorphic, or even only homotopy equivalent.) The reason OP asks the question at all is because it is already well-known we cannot hear a drum's geometry, which by itself still leaves open whether or not we can hear its topology. $\endgroup$ – anon May 17 '18 at 20:55
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    $\begingroup$ @anon That depends on the meaning on well-known, though :). That one cannot hear the fundamental group (so, a fortiori, the topology) might surely be regarded as classical, too. Many of the constructions, such as the one of Vigneras from 1980, include this feature. $\endgroup$ – Vesselin Dimitrov May 17 '18 at 21:42
  • $\begingroup$ The real question here, of course, is what is the topology of one hand, and can you hear it clapping? $\endgroup$ – Adam White May 18 '18 at 10:05

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