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For a connected, finite graph $G$, let $\lambda_1, \ldots, \lambda_n$ denote the nonzero eigenvalues of the graph Laplacian. We define $\zeta_G = \Sigma_{i = 1}^n \lambda_i^s$.

Then Kirkoffs Matrix-Tree theorem can be reformulated as saying that $e^{ - \zeta_G'(0)} / |G| = \tau(G)$, where $\tau$ is the number of trees of the graph $G$. (See here for more: http://www.math.ucsd.edu/~fan/lattice.pdf )

If $M$ is a Riemannian manifold, and $\Delta$ is it's Laplacian, we can define a similar zeta function: $\zeta_{\Delta}$. ( https://en.wikipedia.org/wiki/Minakshisundaram%E2%80%93Pleijel_zeta_function )

Question: Is there a reasonable meaning that can be assigned to $e^{- \zeta_{\Delta}'(0)} / Vol(M)$?

I think (but am not sure) that this quantity doesn't make much sense as stated. The reason is because:

$\zeta_{\Delta}(s) \sim (4 \pi s)^{-n/2} \Sigma_{m = 0}^{\infty} a_m s^m$ (see application one on the wikipedia page) , which has no derivative at zero.

Thank you!

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    $\begingroup$ The analogy is good, and this is known as the analytic torsion of the Riemannian manifold. It is, of course, a regularized value of the determinant of the Laplace operator. What you look for may be the Ray-Singer conjecture (Adv. Math., 1971) and its proofs by Cheeger (Analytic torsion and Reidemeister torsion, 1979) and Muller (Analytic torsion and R-torsion of Riemannian manifolds, Adv. Math., 1978). $\endgroup$
    – Vesselin Dimitrov
    Commented Jul 4, 2018 at 22:06
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    $\begingroup$ I think there is something wrong with the wikipedia page. Two times, the notation Z(s) of the zeta function is used in place of the heat kernel K(t). In particular, the asymptotic expansion is for K(t) and not Z(s) ! In the correct first paragraph, it is explained that Z(s) as an analytic continuation, without pole at s=0 and so the derivative at 0 makes sense. $\endgroup$
    – user25309
    Commented Jul 4, 2018 at 22:06
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    $\begingroup$ I have corrected the wikipedia page. $\endgroup$
    – user25309
    Commented Jul 4, 2018 at 22:23
  • $\begingroup$ (I have to correct myself in that the analytic torsion is an alternating combination of these quantities by considering the action of $\Delta$ on the $i$-forms of $M$ for $i = 0,1,\ldots$. The wikipedia page is en.wikipedia.org/wiki/Analytic_torsion . Your quantity is the $i = 0$ case and known simply as the determinant of the Laplacian. It is the alternating combination that has a geometric meaning, in Ray and Singer's conjecture.) $\endgroup$
    – Vesselin Dimitrov
    Commented Jul 4, 2018 at 22:32

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If I am not mistaken, this cannot be true in general. For example, one may scale the metric for $\mathbb{S}^{1}$ proportionally and the determinant of the Laplacian remains the same independent of the length of the circle. For dimension 2, the determinant of the Laplacian obtains its maximum when $M$ has constant sectional curvature. If we rescale the metric proportionally at all places, similar phenomenon happens. One may work with a sphere or a torus as $\mathbb{R}^{2}/L$ with $L$ being a lattice to check explicitly how the $\zeta$-function behaves under dilation.

There is (allegedly) a related conjecture by Nicolas Bergeron, Mehmet Haluk Sengun, Akshay Venkatesh, which claims analytic torsion grows exponentially with respect to the covering of 3-manifolds. I am not entirely sure whether the statement is true (I heard people found counter-examples and the conjecture has to be modified). I think the motivation of $\tau(S)$ may be from elsewhere (like Weyl's paper).

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  • $\begingroup$ Thanks! This paper on constant sectional curvature is very helpful. I'm wondering if something like the classical planar isoperimetric inequality can be read from their theorem 1c. In particular, if $M$ is a closed disc, then 1c says that among metrics on M in a given conformal class, with fixed perimeter and non-positive average Gaussian curvature, the flat metric with constant geodesic curvature boundary maximizes the determinant - and (I think) this is the disc with induced metric. Are all topological closed discs in $\mathbb{R}^2$ (with the induced metric) in the same conformal class? $\endgroup$
    – Elle Najt
    Commented Jul 5, 2018 at 15:11
  • $\begingroup$ I am not entirely sure what you are talking about. What you asked follows from uniformization theorem and no torsion is involved. If I am not mistaken, your interest is on the discrete vs continuous analogy for analytic torsion. $\endgroup$
    – Bombyx mori
    Commented Jul 5, 2018 at 20:38
  • $\begingroup$ I'm asking if their theorem implies the following: among all topological discs in the plane with fixed perimeter, the one that maximizes the determinant is the usual disc. $\endgroup$
    – Elle Najt
    Commented Jul 5, 2018 at 22:13
  • $\begingroup$ I think the point I'm stuck on is that I've only ever seen uniformization for closed Riemann surfaces (or surfaces with Riemannian metric). I guess there is a statement also about any metric on the closed disc being conformally equivalent to the usual flat disc with circular boundary? $\endgroup$
    – Elle Najt
    Commented Jul 6, 2018 at 1:07
  • $\begingroup$ Okay, I think I understand how this works (modulo details about extending Riemann mapping theorem to the boundary). I have a follow up question on the discrete vs. continuous analogy: In the graph case the Laplacian determinant measures the size of a parameter space of geometric objects. Is there a similar statement for manifolds? I.e. the Laplacian determinant measures the volume of the parameter / moduli space of something? $\endgroup$
    – Elle Najt
    Commented Jul 6, 2018 at 4:53

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