Curvature tensor and Levi-Civita connection

Question

Assuming $M$ is a Riemannian manifold and $p$ is a point on $M$, consider a local orthonormal frame around $p$ denoted by $\{e_1,\dots,e_n\}$. Suppose that we are given the components $R_{ijk} = R(e_i,e_j)e_k$ for every $1 \leq i,j,k\leq n$. The question is whether there exists a formula for $\nabla_{e_i}e_j$ in terms of $R_{ijk}$ alone?

Answer 1

In general this isn't possible.

For $n = 2$, consider the metric $$g = e^{2 \lambda(x, y)} (dx^2 + dy^2)$$ for some function $\lambda$. (In fact, every surface metric can be locally written this way, and we call $(x, y)$ isothermal coordinates for $g$.) Then, $$(E_1, E_2) := (e^{-\lambda} \partial_x, e^{-\lambda} \partial_y)$$ is an orthonormal frame. Computing gives that the Levi-Civita connection $\nabla$ of $g$ is characterized by $$\nabla_{E_1} E_1 = -\lambda_y e^{-\lambda} E_2, \qquad \nabla_{E_1} E_2 = \lambda_y e^{-\lambda} E_1, \qquad \nabla_{E_2} E_1 = \lambda_x e^{-\lambda} E_2, \qquad \nabla_{E_2} E_2 = -\lambda_x e^{-\lambda} E_1 ,$$ and the curvature components $R_{ijk}$ with respect to $(E_1, E_2)$ are determined by $$R_{1212} = -e^{-2 \lambda} \Delta \lambda ,$$ where $\Delta$ is the Laplacian operator, $\Delta := \nabla^i \nabla_i$.

So if we define a metric $g$ by picking a nonconstant harmonic function $\lambda$ (i.e., one satisfying $\Delta \lambda = 0$), then $g$ and the flat metric $\bar g = dx^2 + dy^2$ have the same curvature, $R = 0$, but different connections $\nabla$.