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. 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 $\zeta(0)=1/6$.

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 $\zeta(0)=1/6$.

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.

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. 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 $\zeta(0)=1/6$.