I'm trying to understand in more detail some of the justifications for a proof of the second derivative test for Riemannian manifolds, given below:
I've never seen the Laplacian interpreted as an average like the one the author describes. What does this mean exactly? I guess what I'm looking for here is a formula that showcases this precisely.
The result is quite easy to prove if $M = \mathbb{R}^n$, but I haven't been able to flesh out the details for the argument the author proposes for general Riemannian manifolds (differentiate $t \mapsto f(\text{exp}_p(tv))$ at $t = 0$... I'm having some difficulty applying the chain rule here, what expression will we actually get computing this, and how can we relate it to the gradient and laplacian?). I'd appreciate some help here.
I'll provide an alternate argument for my second question:
Since the problem is of a local nature, we can work in local coordinates $(x^{i})$ around $p$ to do the computations. The vanishing of the gradient is then immediate, as a consequence of $\nabla_{i} u = \frac{\partial u}{\partial x^{i}} = 0$ (by the second derivate test in $\mathbb{R}^n$). The entries of the Hessian are all positive and given by: $$\nabla_{i} \nabla_{j} u=\frac{\partial^{2} u}{\partial x^{i} \partial x^{j}}- \Gamma_{i j}^{k} \frac{\partial u}{\partial x^{k}} \geq 0$$ by the second derivate test in $\mathbb{R}^n$ (and because the second term above is $0$ at a local minimum). Since the Laplacian is the trace of the Hessian, the desired conclusion follows.
I'll take a look at the links in the comments, I think they'll be enough to solve my first question as well.