Difference between $\Delta_g \phi$ and $\Delta \phi ~$?
Let $\phi$ be a smooth function on a Riemannian manifold $(M^n,g)$.
Now, we could write $\nabla_g \phi$ as
$$\nabla_g \phi = \sum^n_{i,j=1} g^{ij} \frac{\partial \phi}{\partial x^i} \frac{\partial}{\partial x^j} = g^{ij} \partial_i \phi ~ \partial_j \tag{1}$$
and
$$\text{div}X = \frac{1}{\sqrt{|g|}} \partial_i \left({\sqrt{|g|} }X^i \right) \tag{2}$$
where $X^i \in T_p{M}$ in local coordinate.
Thus
$$\Delta_g \phi = \text{div} \nabla_g \phi = \frac{1}{\sqrt{|g|}} \partial_i \left({\sqrt{|g|} }g^{ij}\partial_j \phi \right) \tag{3}$$
We also have
$$\Delta \phi = \nabla^i \nabla_i \phi = g^{ij} \nabla_i \nabla_j \phi \tag{4}$$
Questions:
What is the difference between $\Delta_g \phi$ and $\Delta \phi ~$ ?
Could I rewrite (4) as
$$\Delta \phi = \nabla^i \nabla_i \phi = \frac{1}{\sqrt{|g|}} \partial_i \left({\sqrt{|g|} }g^{ij}\partial_j \phi \right)? \tag{5}$$
Thank you.
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$\begingroup$If $\nabla$ is the Levi-Civita connection, then they are the same (on scalar functions). This follows from the identity $$\Gamma^i_{ij}=|g|^{-1/2}\partial_j|g|^{1/2}$$ for the contracted Christoffel symbols. We therefore obtain \begin{equation} \begin{split} g^{ij}\nabla_i\nabla_j\phi&=g^{ij}\nabla_i\partial^j\phi \\ &=\nabla_i\partial^i\phi=\partial_i\partial^i\phi+\Gamma^i_{ij}\partial^j\phi \\ &=\partial_i\partial^i\phi+\big(|g|^{-1/2}\partial_i|g|^{1/2}\big)\partial^i\phi\\ &=|g|^{-1/2}\big(\partial_i\partial^i\phi\big) |g|^{1/2} +\big(|g|^{-1/2}\partial_i|g|^{1/2}\big)\partial^i\phi\\ &=|g|^{-1/2}\partial_i\big(|g|^{1/2} \partial^i\phi\big). \end{split} \end{equation}
$\endgroup$ 0 $\begingroup$They are the same as shown in S.Surace's answer.
One can calculate at the center of a normal coordinates, at this center we have
$$ \Delta \phi = \Delta_g \phi = \sum_{i=1}^n \frac{\partial^2 \phi}{\partial x_i^2}$$
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