# Laplace-Beltrami vs d'Alembert operators in flat vs curved space-time

I am confused with the difference between Laplace-Beltrami (LB) and d'Alembert operators in flat/curved space-time. d'Alembert operator in flat space-time (Minkowski) is defined as $$\Box= \partial^\mu \partial_\mu = g^{\mu\nu} \partial_\nu \partial_\mu = \frac{1}{c^{2}} \frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} - \frac{\partial^2}{\partial z^2}$$ i.e. it is the standard Euclidean Laplace operator ($\Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$) + time derivative (with the opposite sign). However, since scalar wave equation in curved space-time can be written using d'Alembert operator $$\Box\phi \equiv \frac{1}{\sqrt{-g}} \partial_{\mu} \left(\sqrt{-g}g^{\mu\nu}\partial_{\nu}\phi \right),$$ and Laplace-Beltrami operator is defined as $$\nabla^2 f = \frac{1}{\sqrt {|g|}} \partial_i \left(\sqrt{|g|} g^{ij} \partial_j f \right)$$ is then LB operator just d'Alembert operator in "3D"?

Mathematically speaking they are the same operator. Usually we reserve the d'Alembertian for 3+1 dimensional spacetime (so in absence of curvature it takes the form $\partial_0^2 - \nabla^2$), while the Laplace-Beltrami operator is defined for an aribtrary dimensional manifold with arbitrary signature. The only possible difference is that sometimes (not always, though), $\Box$ is defined as $\partial_0^2 - \partial_1^2 - \partial_2^2 - \partial_3^2$ independently of signature, so if your metric is $(-+++)$ then you will have $\Box = -\nabla^2$, where $\nabla^2$ here means the LB operator.