Geodesic trace formula for the Laplace-Beltrami operator ($\Delta f(p)={\sum }_i(f\circ {\gamma }_i{)}^{\prime \prime }(0)$)

Choose an orthonormal basis $e_i$ of the tangent space at $p$ and the unit-speed geodesics with initial velocities $e_i$. Their covariant accelerations vanish, so $(f\circ {\gamma }_i{)}^{\prime \prime }(0)={\mathrm{Hess}}_{g}f(e_i,e_i)$. Taking the metric trace of the Riemannian Hessian proves the formula for $\Delta =\mathrm{div}\nabla$. The positive Laplace-Beltrami operator has the opposite sign.