22
$\begingroup$

Assume that $(M,g)$ is a Riemannian manifold.

Is there any relation between the sequence of eigenvalues of Laplace operator acting on the space of smooth functions and the sequence of eigenvalues of Laplace operator acting on the space of smooth $k$-forms for $k\neq 0$?

$\endgroup$
  • 5
    $\begingroup$ It is not directly related to your question, BUT read $\text{Analytic torsion}$, $\text{Ray–Singer torsion}$, or if you want to have more fantasy ;) BCOV Invariants read this interesting paper arxiv.org/abs/math/0601411 I read it several months ago . Note that BCOV invariant play an important role in degeneration of KE metrics $\endgroup$ – user21574 Dec 11 '17 at 14:13
  • 4
    $\begingroup$ See uni-regensburg.de/Fakultaeten/nat_Fak_I/Bunke/sixtorsion.pdf and p.16 of my master thesis in Aix-Marseille in south of France arxiv.org/ftp/arxiv/papers/1211/1211.4171.pdf $\endgroup$ – user21574 Dec 11 '17 at 14:43
  • 1
    $\begingroup$ See Minakshisundaram formula in p.12 math.mcgill.ca/toth/spectral%20geometry.pdf we have same type formula when you work on $p$-forms $\endgroup$ – user21574 Dec 11 '17 at 16:11
  • 2
    $\begingroup$ Let me give more philosophical picture about my first comment: Bershadsky–Cecotti-Ooguri-Vafa invariant is a Ray–Singer torsion(Analytic torsion) used in mirror symmetry and play an important role in the degeneration of Calabi-Yau varieties. The good thing is that we can extend BCOV invariant for varieties of general type which give naïve connection to Ginzburg–Landau theory and study of Analytic torsion. So the same nature of arxiv.org/abs/math/0601411 can be extended when fibers admit Kahler-Einstein metrics of negative Ricci curvature and geometry of Quillen metrics $\endgroup$ – user21574 Dec 11 '17 at 16:38
  • 2
    $\begingroup$ @HassanJolany Many thanks for your very interesting information and links. $\endgroup$ – Ali Taghavi Dec 12 '17 at 12:53
6
$\begingroup$

It sounds to me like you have heard, and are trying to remember the statement of, the following fact about the Hodge Laplacian $d^*d+dd^*$ on forms.

Slogan: eigenforms come in pairs.

Sketch proof: If $(d^*d+dd^*)\eta=-\lambda\eta$, then $(d^*d+dd^*)(d\eta)=dd^*d\eta=d(d^*d+dd^*)\eta=-\lambda d\eta$.

Statement: Recall the Hodge decomposition $\Gamma(M,\Lambda^kM)=A_k\oplus B_k\oplus H_k=\operatorname{im}(d^*_{k+1})\oplus\operatorname{im}(d_{k-1})\oplus (\operatorname{ker}(d_k^*)\cap\operatorname{ker}(d_k))$, such that for all $k$, $d_k:A_k\to B_{k+1}$ is an isomorphism. Then for each $\lambda$, $d_k$ sends the $\lambda$-eigenspace of $A_k$ to the $\lambda$-eigenspace of $B_{k+1}$.

Special case: Each nonzero eigenvalue of the Laplace operator on functions is also an eigenvalue of the Laplace operator on 1-forms.

Also, if you take any sensible choice whatsoever of Laplacian on $\Gamma(M,\Lambda^kM)$, then the asymptotic distribution (à la Weyl) of the eigenvalues will be the same as the asymptotic distribution of the eigenvalues of the Laplacian on scalars (up to multiplication by the rank of $\Lambda^kM$). See, eg, Theorem 2.41 of Berline-Getzler-Vergne.

$\endgroup$
  • 7
    $\begingroup$ Weyl is German, so why "à la Weyl"?!!! $\endgroup$ – user21574 Dec 11 '17 at 14:34
  • 1
    $\begingroup$ @HassanJolany English borrowed many foreign expressions. See en.wiktionary.org/wiki/%C3%A0_la#English $\endgroup$ – ACL Dec 12 '17 at 18:05
  • 1
    $\begingroup$ @ACL , à la, borrowed also In my grand mothers language means "handsome " and my fathers language means "this side" . I know that in Turkish language means God , so I believe you! $\endgroup$ – user21574 Dec 12 '17 at 18:43
  • 2
    $\begingroup$ @HassanJolany Good point! The expression exists in French, meaning “in the style of” and the similar one — alla — exists in Italian. One more letter closer to God. $\endgroup$ – ACL Dec 12 '17 at 19:45
  • 2
    $\begingroup$ @ACL Many thanks for your very interesting information and explanation $\endgroup$ – user21574 Dec 12 '17 at 20:51

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.