Suppose that $M$ is a Riemannian manifold with Levi-Cevita connection, $\nabla$ and a parallel global orthonormal frame $\{X_1,\ldots,X_n\}$. This seems to imply that the Riemannian curvature endomorphism, $R(X_i,X_j)X_k$ vanishes by simple reasoning that $\nabla_{X_i}\nabla_{X_j}X_k = 0$ as the frame is parallel and similarly $\nabla_{[X_i,X_j]}X_k = 0$. By linearity of the curvature endomorphism and the fact that the $X_i$'s form a frame this implies the curvature endomorphism vanishes on all of $M$.
On the other hand, a Lie group with a bi-invariant metric exhibits such an orthonormal frame by pushing forward an orthonormal basis by left-multiplication. This resulting orthonormal frame, $\{X_1,\ldots,X_n\}$, appears to be parallel since the connection defined by $\nabla_{Y}(a^iX_i) = Y(a^i)X_i$ ($a^i$ are smooth component functions) by a quick calculation looks to be g-compatible and torsion free so that $\nabla X_i = 0$? As Lie groups under bi-invariant metrics can have positive sectional curvature this contradicts the reasoning in the previous paragraph.
A second reformulation of the question is that symmetry of the connection and parallelism implies that
$0 = \nabla_{X_i}X_j – \nabla_{X_j}X_i = [X_i,X_j]$. The vanishing of these Lie brackets then implies that there exist global coordinates of $M$, $x^i$, whose coordinate vector fields are the orthonormal $X_i$'s which again further implies the metric is flat.
My guess is that having a parallel frame doesn't imply a manifold is flat but a parallel ON frame does. One cannot use Gram-Schmidt on a parallel non-ON frame to get an orthonormal one as this ruins the parallelism. The question remains as to why the Lie group example is not flat in general; are the left invariant vector fields provided not actually parallel? Thanks for your help.
Best Answer
I am not compeltely sure how to interpret your question, but there is an error in your question, which probably causes the problem: In the setting of an invariant metric on a connected Lie group, the connection that you describe always has vanishing curvature, but the torsion vanishes if and only if the group is commutatitve. In your notation $[X_i,X_j]=\sum_kc^k_{ij}X_k$, where the $c^k_{ij}$ are the structure constants of your Lie algebra, and the torsion vanishes if and only if this coincides with $\nabla_{X_i}X_j-\nabla_{X_j}X_i=0$. So you get an orthonormal frame which is parallel for a connection that preserves the metric but has torsion and thus is different from the Levi-Civita connection. But flatness of the metric is defined via the Levi-Civita connection. To get the Levi-Civita connection, you have to change your connection by a left invariant tensor field (or purely algebraic origin), which then gives an algebraic expression for the Riemann curvature tensors, which is again left invariant.