I am working in the riemannian setting with Levi-Civita connection and
I would like to prove using local coordinates that the laplacian is the trace of the Hessian i.e.
$$ \text{tr}_g(\nabla^2 f)=\sum_{i,j} g^{ij} \bigl ( f_{ij} – \sum_k\Gamma_{ij}^k f_k \bigr ) = \sum_{i,k} \bigl ( \partial_i (g^{ki} f_k) + \sum_{j}g^{kj} \Gamma_{ji}^i f_k \bigr )= \Delta f$$
Expanding summations and using Liebniz I obtain
$$ \sum_{i,j} g^{ij} f_{ij} – \sum_{i,j,k}g^{ij}\Gamma_{ij}^k f_k = \sum_{i,k} g^{ki}f_{ik} + \sum_{i,k} \partial_i(g^{ki})f_k+ \sum_{i,j,k} g^{kj}\Gamma_{ji}^if_k$$
and then to prove the equality it is enough to show
$$ -\sum_{i,j,k}g^{ij}\Gamma_{ij}^kf_k = \sum_{i,k} \partial_i(g^{ki})f_k+ \sum_{i,j,k}g^{kj}\Gamma_{ji}^if_k$$
i.e. that for every $k$ it holds
$$ -\sum_{i,j}g^{ij} \Gamma_{ij}^k = \sum_{i}\partial_i (g^{ki}) + \sum_{i,j}\Gamma_{ji}^ig^{kj}$$
I do not know how to proceed from this point.
Best Answer
While I don't know what definition of Laplacian you are using (normally it is defined just as the trace of Hessian), for the last equation there is the general formula $\partial_i g^{ks}=-g^{kp}(\partial_i g_{pq})g^{qs}$, as you can check using $\partial_i(g_{pq}g^{qs})=\partial_i\delta_p^s=0$. Let $s=i$ and then put in $\partial_i g_{pq}=\Gamma_{ip}^rg_{rq}+\Gamma_{iq}^rg_{pr}$ you will get that last equation.