Is the structure constant additive on connected components?

Question

This is the reanimation of a question which already got an answer, that I did not fully understand. Coming back to it, after let it sit in a corner for some time, I keep not getting the point. I would be most grateful for clarifications!

Definition of the Structure Constant Let $M$ be a possibly non-compact Riemann surface and $\mu$ a smooth metric on it; let $\Delta_{\mu,\,M}$ be the Laplacian on $M$ induced by $\mu$ and $\mathrm{det}^*(\Delta_{\mu,\,M})$ its $\zeta$-regularized determinant. The structure constant is defined by $$c_{\Delta,\,\mu}(M) := \log \left(\frac{\mathrm{det}^*(\Delta_{\mu,\,M})}{\mathrm{vol}_\mu(M)} \right).$$

A precise statement for the Additivity on Connected Components is the following. Let $(C,\mu_C) = (A\sqcup B,\mu_A + \mu_B)$ be the disjoint union of the metrized Riemann surfaces $(A,\mu_A)$ and $(B,\mu_B)$, then $$c_{\Delta,\,\mu_C}(C) = c_{\Delta,\,\mu_A}(A) + c_{\Delta,\,\mu_B}(B).$$

Does the additivity on connected components hold?

Here follow some observations:

1. By direct examination of its construction it seems to me that the logarithm of the regularized determinant is already additive on connected components.

2. As pointed out by Carlo Beenakker in his answer, if we scale the metric by a positive factor $\gamma^2$ then the regularized determinant scales as $$\mathrm{det}^*(\Delta_{\gamma^2\cdot\mu,\,M}) = \gamma^{-\mathcal{X}(M)/3} \mathrm{det}^*(\Delta_{\mu,\,M}),$$ where $\mathcal{X}$ is the Euler characteristic. Given the additivity on connected components of the Euler characteristic, this is not in contradiction with the observation above.

3. Also in his answer, Carlo Beenakker mentions that the role of the volume in the definition of the structure constant is to "normalize to unit area". My lack of understanding of this point is twofold: on one hand, since $$c_{\Delta,\,(\mathrm{vol}_\mu(M))^{-1}\cdot\mu}(M) = \log \left(\frac{\mathrm{det}^*(\Delta_{(\mathrm{vol}_\mu(M))^{-1}\cdot\mu,\,M})}{\mathrm{vol}_{(\mathrm{vol}_\mu(M))^{-1}\cdot\mu}(M)} \right) = \log \left(\frac{\mathrm{det}^*(\Delta_{\mu,\,M})}{\mathrm{vol}_{\mu}(M)^{-\mathcal{X}(M)/6}} \right),$$ I do not get the idea why this is a meaningful renormalization for the regularized determinant. On the other hand I do not see why this renormalization should ensure the additivity on connected components rather than spoiling it.

All in all it appears I'm pretty confused about the whole business. Thanks for your support!

Answer 1

The need for a volume (or surface area) normalization appears because the eigenvalues of the Laplacian scale inversely proportional to the area. More precisely, if you scale the metric as $\mu'=\gamma^2\mu$, then the Laplacian determinant scales with a factor $\gamma^{-\chi/3}$, with $\chi$ the Euler characteristic of the surface. So unless $\chi=0$, you need to normalize to unit area.**

If I am not mistaken, the normalization actually ensures the additivity you are seeking, rather than spoiling it.

** To elaborate, take a look at section III of Conformal invariants for determinants of Laplacians on Riemann surfaces. There the area normalization is worked out for several surfaces. For example, in the case of a flat disk of radius $r$, the eigenvalues of the Laplacian are $l(l+1)/r^2$ with multiplicity $l=1,2,\ldots$. The zeta-regularized Laplacian determinant has a factor $\exp[-2\zeta(0)\ln r]$, with $\zeta(s)=\sum_{l=1}^{\infty}l[l(l+1)]^{-s}$ and $\zeta(0)=1/6$.