Given a Riemannian manifold $(M, g)$, the scalar curvature of the metric $g$ is $g^{i j} Ric_{i j}$ in local coordinates. Now I want to calculate the laplacian of the scalar curvature in local coordinates and did something like $g^{i j} \nabla_{i} \nabla_{j} g^{p q}Ric_{pq} = g^{i j} \nabla_{i} (g^{pq} \nabla_{j} Ric_{pq}) = g^{i j} g^{p q}\nabla_{i} \nabla_{j} Ric_{pq}$. Is this correct? I have two concerns
- $\nabla_{i} \nabla_{j}$ should not be interpreted as $\nabla_{i}(\nabla_{j} g^{p q}Ric_{pq})$ because it is actually the $(i, j)$-th component of $\nabla \nabla$. However my impression is that when we are doing this kind of calculation, you do calculate first $\nabla_{j}g^{pq}Ric_{pq}$, why?
- I am using the fact that $\nabla_{j} g^{pq} = 0$, which presumably is derived from the fact that $\nabla g = 0$. $\nabla g = 0$ means that when you take the covariant derivative of a (0, 2) two tensor you get 0, but in the expression of the scalar curvature I do not think that $g^{pq}$ can be interpreted as the component of a (0, 2) tensor. To put it another way, the product rule works for covariant derivative on tensor products by definition but how do we interpret $g^{pq}Ric_{pq}$ as a tensor product?
Best Answer
The Laplacian of a scalar function $u$ on a Riemannian manifold $(M, g)$ is defined to be $$ \Delta u = \text{tr}(\text{Hess}(u)). $$ I am using the notation Hess$(u)$ to denote the Hessian of $u$, which is a 2-covariant symmetric tensor field on $M$ defined by $$ \text{Hess}(u)(X, Y) = (\nabla_X (du))(Y) \hspace{5mm} \forall X, Y \in T_p M. $$ This is sometimes written as $\nabla^2 u$. A headache can arise from this notation in general; as alluded to in the previous answer/comments, $\nabla_X \nabla_Y u \neq (\nabla^2 u)(X, Y)$, as long as the left hand side is interpreted as $\nabla_X (\nabla_Y u)$. (A quick way to see that this equality does not hold in general is to note that the left hand side is not tensorial in $Y$, but the right hand side is.)
One can compute in arbitrary local coordinates $(x^\alpha)$ to see that $$ \text{Hess}(u)_{\alpha\beta} = \partial_\alpha \partial_\beta u - \Gamma_{\alpha\beta}^\gamma \partial_\gamma u. $$ Thus in general the Hessian on a Riemannian manifold has a first-order term which does not appear in flat space (!)
The trace of a 2-tensor $T$ is the contraction of $T$ with the metric, i.e. $\text{tr}(T) = g^{\alpha\beta}T_{\alpha\beta}$. Thus, $$ \Delta u = g^{\alpha\beta}\text{Hess}(u)_{\alpha\beta} = g^{\alpha\beta}\partial_\alpha \partial_\beta u - g^{\alpha\beta}\Gamma_{\alpha\beta}^\gamma \partial_\gamma u. $$ One then proceeds with $u = R$, the scalar curvature.
More to your questions:
I have seen the expression $\nabla_i \nabla_j T$ used tensorially, i.e. to mean the $i, j$ component of $\nabla^2 T$. To my mind, this is not typical; see e.g. the Wikipedia article https://en.wikipedia.org/wiki/Second_covariant_derivative. However, one can in general stop worrying about all of this by choosing at the point of computation a frame $e_i$ such that $\nabla_i e_j|_p = 0$. Then $\nabla_i \nabla_j T = \nabla^2_{i,j} T$.
The expression $g^{ij}\text{Ric}_{ij}$ is the contraction of the tensor product $g^{-1} \otimes \text{Ric}$. Then one can use the tensor product Leibniz rule (as well as the fact that the connection commutes with contractions, and that $\nabla g = 0$) to obtain $\nabla_X (g^{ij}\text{Ric}_{ij}) = g^{ij}\nabla_X \text{Ric}_{ij}$. In practice to verify this one can introduce (cumbersome) notation like $C_{i,j}$ for contraction in indices $i, j$ and write $g^{ij}\text{Ric}_{ij}$ as $C_{1, 3} C_{2, 4}g^{-1} \otimes \text{Ric}$... This can be a pain, but is one way of verifying the manipulations common to this subject.
Also, brief remark that was already addressed in the previous answer: $\nabla g = 0$ is saying only that the covariant derivative of the metric tensor $g$ is zero, not of any 2-tensor; also, $g^{ij}$ is the component of a 2 (contravariant) tensor, namely $g^{-1}$.