7
$\begingroup$

We know that the eigenvalues of the Laplacian contains a lot of information of a Riemannian manifold, but they do not determine the full information ( Hearing the shape of a drum). And the eigenfunctions of the Laplacian seem to have much more information (see the reference). Now my question is that whether the eigenfunctions of the Dirac operator would contain more information than that of Laplacian, since it seems to me that the Dirac operator is a more refined version of the Laplacian ( a Riemmanian manifold could have several spin structure). Here the Laplacian means the Laplace-Beltrami operator and the Dirac operator means the Dirac operator on the spinor bundle. Thank you.

$\endgroup$
6
$\begingroup$

No, we cannot (completely) hear the shape of a drum, even if it is spinorial. Two metric fields with the same collection of eigenvalues are called isospectral. There exist Dirac isospectral deformations; continuous 1-parameter families of mutually non- isometric metrics with the same Dirac spectrum have been constructed. They are of the form $M_s = G/F_s$, $s \in \mathbb{C}$, with $G$ a nilpotent group (e.g. the Heisenberg group) and $F_s$ a nilpotent subgroup. Also, there exist known examples of Laplace-isospectral 4-dimensional flat tori which are also Dirac-isospectral (at least for the trivial spin structure).

Noncommutative Geometry and the Standard Model of Elementary Particle Physics (page 304).

$\endgroup$
  • 9
    $\begingroup$ (+1) That said, by a result of Connes, if one knows not only the spectrum of the Dirac operator $D$ as an operator on the separable Hilbert space $H$ of $L^2$ spinor fields, but also the relative position within the von Neumann algebra $B(H)$ of bounded operators on $H$ of the von Neumann subalgebra of $B(H)$ generated by the functional calculus of $D$ (i.e., the datum of the actual diagonalisation of $D$) and $L^\infty(M)$, viewed as multiplication operators on $B(H)$, then you can indeed pin down your Riemannian geometry. $\endgroup$ – Branimir Ćaćić May 13 '17 at 17:38
  • $\begingroup$ @Carlo Beenakker. The examples of 4-dimensional tori follow from [8] Schiemann A. Ein Beispiel positiv definiter quadratischer Formen der Dimension 4 mit gleichen Darstellungszahlen // Arch. Math. 1990. V. 54. P. 372–375 and Conway J. H., Sloane N. J. A. Four-dimensional lattices with the same theta series // International Mathematics Research Notices. 1992. V. 4. P. 93–96. Every pair of Laplace-isospectral tori qith the trivial spin structure is also $\endgroup$ – Bernd Ammann May 21 '17 at 5:34
  • $\begingroup$ @Carlo Beenakker: However, I do not know any reference where Dirac-isospectrals pairs buidling on Heisenberg manifolds are worked out. $\endgroup$ – Bernd Ammann May 21 '17 at 5:44
  • 1
    $\begingroup$ @BranimirĆaćić: Sorry, I don't know much about the noncommutative geometry. Just wonder, is there any similar results formulate in the classical language or is it really necessary to go deep into the noncommutative geometry? Thank you. $\endgroup$ – Z. Ye Jun 3 '17 at 17:37
2
$\begingroup$

Most of the questions raised above are answered in the article "The Dirac Operator on Nilmanifolds and Collapsing Circle Bundles" by Christian Bär and myself published in Annals of Global Analysis and Geometry June 1998, Volume 16, Issue 3, pp 221–253. http://link.springer.com/article/10.1023/A:1006553302362. A preprint version is available on the arxive as https://arxiv.org/abs/math/9801091.

All examples of Laplace-isospectral tori are also Dirac-isospectral for the trivial spin structure. In the reference above we constructed Dirac-isospectral families of 3-step nilmanfolds, inspired by a construction by Ruth Gornet. There are also families which are Dirac-isospectral for some spin structures and not Dirac-isospectral for other spin structures.

$\endgroup$
  • $\begingroup$ Nice answer! I still want to know if Dirac operator gives a bit more information. From your answer it's pretty clear that the Dirac operator can distinguish different spin structures. But for a trivial spin structure, it seems that the spectrum of Dirac operator will not give more information than Laplaician? Thank you very much. $\endgroup$ – Z. Ye Jun 2 '17 at 9:40
  • 1
    $\begingroup$ For the Laplace-Beltrami operator the answer is given in Theorem 5.6 in link. There are manifolds isospectral for the Laplace-Beltrami, but not for the Dirac operator. One can also construct manifolds which are isospectral for the Hodge-Laplacian, but not for the Dirac operator by changing the spin structure, this works even on tori. I expect that examples isospectral for the Hodge-Laplacian, but non-Dirac-isospectral for any choice of spin structure should exist, but I do not think that anyone has constructed such pairs so far. $\endgroup$ – Bernd Ammann Jun 4 '17 at 10:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.