Difference between $\Delta_g \phi$ and $\Delta \phi ~$

Question

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:

1. What is the difference between $\Delta_g \phi$ and $\Delta \phi ~$ ?

2. 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.

Answer 1

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}