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I came across the following statement in a paper:

On $S^3$, the eigenvalues of the vector Laplacian on divergenceless vector fields is $(\ell + 1)^2$ with degeneracy $2\ell(\ell+2)$ with $\ell \in \mathbb{ Z}$.

Is it possible to prove the spectrum and degeneracy using the representation theory of $SO(4)$? Perhaps there is a general result for the n-sphere.

The paper then proceeds to make the non-sense statement (RHS is divergent):

$$ \det \big(-\Delta + a\big) = \prod_{\ell=1}^\infty \big((\ell + 1)^2 + a \big)^{2\ell(\ell+2)} $$

How do we make sense of the determinant of the Laplacian on the space of divergenceless vector fields?

cactus314
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  • My question is how the spectra are calculated in the first place - using harmonic analysis - it was the 1st of several spectra in the paper. Then there is a separate question about regularizing the infinite product. Hardy published a book on divergent series. – cactus314 Aug 13 '13 at 15:30
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    Yes, the repn theory of $SO(4)$ is useful to determine the spectrum. For example, the functions on the 3-sphere are functions that descend to $SO(4)/SO(3)$. From the regular repn of $L^2(SO(4))$, functions on $S^3$ decompose as the sum of $\pi^{SO(3)}\otimes \check{\pi}$ where $\pi$ runs over irreducibles. This gives multiplicities in terms of those dimensions. The eigenvalues are eigenvalues of Casimir. – paul garrett Aug 13 '13 at 16:13
  • @paulgarrett OK. Then I have decompose (divergenceless) vector fields on $S^3$ - which I guess is not $L^2(S^3)\oplus L^2(S^3)$ - into eigenspaces. – cactus314 Aug 13 '13 at 16:25

1 Answers1

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The spectral determinant of $-\Delta+a$ is defined as $e^{-\zeta_{-\Delta+a}'(0)}$ where

$$\zeta_{-\Delta+a}(s) = \sum_{n = 0}^{\infty} (\lambda_n+a)^{-s} $$ is the zeta function defined in a region of the complex plane, it admits analytic continuation to negative values of s and this is the way to write the "nonsense" infinite product. To compute the RHS you need to find the way to define a function which gives the analytic continuation of the series (usually is similar as in the Riemann zeta function, otherwise is too difficult), and then find the derivative at $s = 0$.

For rigorous detais on convergence I recommend Collected Papers volume V - J. Jorgenson, S. Lang.

For the physics behind these definitions (most of them related with quantum field theory and string theory) see Quantum Fields and Strings: A Course for Mathematicians: Volume 2: (American Mathematics Society non-series title) for example.

Guille Aparicio
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    So here $\zeta_{-\Delta+ a}(s) = \sum_{l=0}^\infty 2l(l+2) ((l+1)^2+a)^{-s}$ whose analytic continuation is obtained from $((l+1)^2+a)^{-s} = (l+1)^{-2s} \sum_{k=0}^\infty {-s \choose k} a^k (l+1)^{-2k}$ ie. $\zeta_{-\Delta+ a}(s)= \sum_{k=0}^\infty {-s \choose k} a^k (\zeta(s+2k-2)+2\zeta(s+2k-1)-2)$. The question is then what $e^{\zeta_{-\Delta+ a}(s)'(0)}$ the zeta regularization of the above product is supposed to help for ? – reuns Oct 29 '17 at 11:58
  • Usually, in analytic continuation of zeta functions, Theta functions (defined as the trace of the Heat semigroup) are used. Furthermore, the Mellin transform of the theta function gives the analytic continuation of the zeta function. So to obtain the values of the zeta function it is more convenient to expand the theta series in a power series and then with the Mellin transform you get a useful expression of the zeta function which you can also take derivatives at s = 0. I hope this helps. In your case is complicated because your analytic continuation involves a series on zeta functions. – Guille Aparicio Oct 31 '17 at 07:39
  • Right $\sum_n e^{-\lambda_n t}\phi_n(x)$ is the heat kernel. The method I used gives the meromorphic continuation and works for any polynomials. – reuns Oct 31 '17 at 07:58