# Existence of a Parallel Orthonormal Frame Implies Manifold is Flat?

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.

• You have parallel in the title but omit to repeat it in the statement of the question in the first paragraph. Commented May 3, 2020 at 18:48
• You're right, thanks
– amc
Commented May 3, 2020 at 18:52
• Thanks. My standard way to do this would be to observe that the connection $1$-forms $\omega^j_i$ defined by $\nabla X_i = \sum \omega^j_i \otimes X_j$ are all $0$ because the frame is parallel. So of course the curvature matrix vanishes. Commented May 3, 2020 at 18:53

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.

• Thanks, I think I now see why the torsion of the connection provided doesn't vanish. My calculations were incorrectly presupposing the conclusion that $[X_i,X_j] = 0$. I'm still not exactly sure what it meant by changing the connection by a left invariant tensor field however.
– amc
Commented May 3, 2020 at 19:11
• It is a general result that the difference between two linear connection on $TM$ is a $\binom12$-tensor field on $M$. Conversely, if you add such a tensor field to a linear connection, then you again get a linear connection. Moreover, it is easy to see what happens to torsion in this process. For the special case of connections preserving a Reimannian metric, the map between the change of connection and change of torsion is bijective. Thus you can compute the Levi-Civita connection given any metric connection and its torsion, and this leads to a left-invariant $\binom12$-tensor field. Commented May 4, 2020 at 12:50
• @AndreasCap Does this mean that I can't define an orthonormal frame on say $S^{3}=SU(2)$ having nonzero curvature, but rather that I need to define it instead as a nonzero torsion? Commented Apr 13, 2021 at 21:25
• @R.Rankin: I am not sure what you mean (a frame doesn't "have" curvature or torsion). If you put a left invariant metric on $SU(2)$ then you get golobal orthonormal frames using left invariant vector fields. However, these frames are not parallel for the Levi-Civita connection. There is a unique linear connection on $SU(2)$, which is metric and for which these frames are parallel, which implies that this connection is flat. However, it has non-zero torsion given by the Lie bracket on $\mathfrak{su}(2)$ and thus is different from the Levi-Civita connction. Commented Apr 14, 2021 at 6:17
• @AndreasCap Thank you, I was confused as from the three-sphere perspective I'd expect the Levi-Cevita connection to have curvature. I do understand however that curvature and torsion can be interchanged in a consistent manner with mathematical equivalence (In physics for example this might be called the teleparallel equivalent of General relativity). I think this is what you mean? Commented Apr 14, 2021 at 8:16