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Consider on the circle $S^1$ the operator $$L := - \frac{\partial^2}{\partial \theta^2} + c$$ for some constant $c \in \mathbb{R}$.

What is its $\zeta$-regularized determinant?

This should be well-known, I suppose, but I didn't find a reference.

Some background: The eigenvalues of $L$ are $n^2+c$ for $n \in \mathbb{Z}$ (with multiplicity two each), and therefore the zeta-function for positive $c$ is given by $$\zeta_c(s) = 2\sum_{n=0}^\infty \frac{1}{(n^2+c)^s}.$$ Now the $\zeta$-regularized determinant is defined by $$\det(L) = e^{-\zeta^\prime(0)}.$$ How does one compute such a thing?

\Edit: The hint to look for "thermal zeta functions" was good: Here is a paper that computes the determinant in question: http://arxiv.org/pdf/hep-th/9505154v1.pdf

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  • $\begingroup$ It seems to me that a variant of Riemann's integral representation of $\zeta$ using $\theta$, the latter's functional equation from Poisson summation, gives the meromorphic continuation... but, unlike zeta itself, superficially it appears difficult to get information about the derivative at $0$, but is that what's really desired, or is it more the residue at the pole at $s=1$? Can you clarify "in terms of what" you'd really like the outcome? $\endgroup$
    – paul garrett
    Commented Jul 27, 2015 at 18:18
  • $\begingroup$ Oh, it would be terrific if the outcome was an explicit expression, depending on $c$, in terms of elementary functions or so. At least something which you could numerically evaluate. But you seem to think this is not realistic? $\endgroup$
    – Matthias Ludewig
    Commented Jul 27, 2015 at 18:33
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    $\begingroup$ I don't have the paper in front of me because it's behind a pay wall, but I believe that this calculation has been done in the 1977 paper by Gary Gibbons called "Thermal zeta functions". At least for $c>0$, the operator in the question is associated to the harmonic oscillator and being on the circle corresponds to the harmonic oscillator at finite temperature, hence the "thermal" in the title: sciencedirect.com/science/article/pii/0375960177900263 $\endgroup$
    – José Figueroa-O'Farrill
    Commented Jul 27, 2015 at 20:34

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Let me write a bit of what seems to me that natural/naive approach, and perhaps the questioner can comment on the direction...

From $\int_0^\infty y^s\,e^{-ty}\;{dy\over y}=t^{-s}\,\Gamma(s)$, writing $Z(s)$ for your zeta, $$ \pi^{-s/2}\,\Gamma(s/2)\,(Z(s)-{1\over c^s}) \;=\; \int_0^\infty y^{s/2}\,\sum_{n\in \mathbb Z} e^{-\pi (n^2+c)y}\;{dy\over y} $$ Unlike Riemann's argument, the presence of $c>0$ will cause Poisson summation to produce an expression that behaves well, without breaking the integral into two pieces. Namely, the Fourier transform of $x\to e^{-\pi (x^2+c)y}$ is ${1\over \sqrt{y}} e^{-\pi (x^2/y +cy)}$, so this becomes $$ \sum_{n\in \mathbb Z} \int_0^\infty y^{{s-1\over 2}} e^{-\pi (n^2/y+cy)}\;{dy\over y} $$ The $n=0$ term should be taken out, and is elementary. For the rest, replace $y$ by $\sqrt{c} y/n$, to get $$ \pi^{-s/2}\,\Gamma(s/2)\,(Z(s)-{1\over c^s}) \;=\; (n=0) + \sum_{n\not=0}(\sqrt{c}/n)^{(s-1)/2} \int_0^\infty y^{{s-1}\over 2} e^{-\pi \sqrt{c}\,n\,(y+1/y)}\;{dy\over y} $$ The right-hand side seems to be entire in $s$, so the factor of $\Gamma(s/2)$ on the left-hand side means that the factor $Z(s)-c^{-s}$ vanishes at $s=0$. Thus, the right-hand side is close to computing the derivative of $Z(s)-c^{-s}$ at $s=0$.

But/and this makes me wonder whether about that extra term $c^{-s}$. It is needed to make Poisson summation work, but then it doesn't seem to me that $Z(s)=0$, so $Z'(0)$ would not be the leading term, which I would have thought would have been the object of interest.

In any case, the right-hand side is a sum of values of Bessel functions, but not obviously (to me) further simplifiable at $s=0$. At $s=1$, the sum over $n$ admits summing as geometric series, so at least the outcome is just an integral in $y$... though it seems not elementary, still.

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  • $\begingroup$ Also, an Euler-MacLaurin summation approach might work as well as anything, depending, ... $\endgroup$
    – paul garrett
    Commented Jul 27, 2015 at 20:06
  • $\begingroup$ Some minor computational boo-boos in the above: should be $e^{-\pi(n^2+c)y}$, of course, and pursuant... doesn't change it qualitatively. $\endgroup$
    – paul garrett
    Commented Jul 28, 2015 at 12:04
  • $\begingroup$ Hey, thank you! Indeed, you were on the right track. I found a paper that finishes the calculation with some crazy identities of Bessel functions: arxiv.org/pdf/hep-th/9505154v1.pdf In the end, the outcome is very simle: $-log(2sinh(..))$ $\endgroup$
    – Matthias Ludewig
    Commented Jul 29, 2015 at 18:02
  • $\begingroup$ Repaired the computational errors since the question got bumped upward anyway... $\endgroup$
    – paul garrett
    Commented Jul 29, 2015 at 21:02

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