Trace of the Hessian is the Laplacian(divergence of a gradient field) in Riemannian geometry Here are two equivalent definitions of Hessian. The trace of a (0,2)-tensor $h$ can be defined in coordinates by
$$\text{tr}_g(h)=\sum_{i,j}g^{ij}h_{ij},$$
where $g^{ij}$ is the component of the inverse matrix $(g_{ij}).$
We know that in calculus, the trace of Hessian of a function $u$ is clearly the Laplacian $\Delta u$(In Riemannian geometry, it is defined to be the divergence of a gradient field). But I don't know how to prove this for general Riemannian manifolds(coordinate free or not). 
 A: $\def\tr{\mathrm{tr}}\def\dv{\mathrm{div}}\def\gr{\mathrm{grad}}\def\o{\circ}\def\n{\nabla}$You have a small error in your question: the trace you are describing is the trace of a $(0,2)$-tensor $h$ with respect to the metric $g$, which I write $\tr_gh$, reserving $\tr$ for the invariant trace $\tr\, T = T^i_i$ of a $(1,1)$-tensor $T$.
The equality of these two definitions of the Laplacian essentially just falls out of the definitions. The divergence operator is defined on vector fields as the contraction of the covariant derivative, so $\dv = \tr \o \n$. Thus $$\Delta := \dv \o \gr = \tr \o \n \o \gr.$$ Since the gradient is just what we get by raising an index of the covariant derivative using the metric, this becomes $$\Delta = \tr_g \o (\n \o \n) = \tr_g \o \mathrm{Hess},$$ where we used the metric-compatibility of $\nabla$ to commute the metric past the derivative. In abstract index notation this whole calculation is just $$\Delta f = \nabla_i (g^{ij} \nabla_j f)=g^{ij}\nabla_i\nabla_j f.$$
