I read recently in a textbook that Laplacian of an invariant $\phi$ is given by $$\nabla^2\phi=\frac{1}{\sqrt{g}}\frac{\partial}{\partial x^i}\bigg(\sqrt{g}g^{ij}\frac{\partial\phi}{\partial x^j}\bigg)$$ where $g_{ij}$ is the metric tensor and $g=|g_{ij}|$. I went through the proof and what it stated is $$\nabla^2\phi=\text{div}(\text{grad} \ \phi)$$ Now they stated without any reasoning that $\boxed{\text{grad} \ \phi=A^i}$ for some arbitrary contravariant vector $A^i$ and defined it as $\displaystyle A^i=g^{ij}\frac{\partial\phi}{\partial x^j}$. Then applied the formula for divergence of contravariant vector as $$\text{div}(\text{grad} \ \phi)=\text{div}(A^i)=A^i_{,i}=\frac{1}{\sqrt{g}}\frac{\partial}{\partial x^i}\bigg(\sqrt{g}A^i\bigg)$$ Finally using $A^i$ as above we get $$\nabla^2\phi=\frac{1}{\sqrt{g}}\frac{\partial}{\partial x^i}\bigg(\sqrt{g}g^{ij}\frac{\partial\phi}{\partial x^j}\bigg)$$
What I can't get in this proof is that why they have taken $\text{grad} \ \phi$ as a contravariant vector only and not a covariant vector. What would be the problem if a covariant vector was taken? Any help is appreciated.
Best Answer
It all comes down to the coordinates being $x^i$. The outermost operator uses $\frac{\partial}{\partial x^i}=\partial_i$, so this needs to contracted with an inner $\nabla^i$. (I'm putting aside the $|g|^{\pm\frac12}$ factors for the moment, but these result from the identity $\nabla_iV^i=|g|^{-\frac12}\partial_i(|g|^{\frac12}V_i)$.) Since $\phi$ has no spacetime indices, $\nabla^i\phi=g^{ij}\nabla_j\phi=g^{ij}\partial_j\phi=g^{ij}\frac{\partial\phi}{\partial x^j}$.