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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.

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  • $\begingroup$ You have parallel in the title but omit to repeat it in the statement of the question in the first paragraph. $\endgroup$ Commented May 3, 2020 at 18:48
  • $\begingroup$ You're right, thanks $\endgroup$
    – amc
    Commented May 3, 2020 at 18:52
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    $\begingroup$ 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. $\endgroup$ Commented May 3, 2020 at 18:53

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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.

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  • $\begingroup$ 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. $\endgroup$
    – amc
    Commented May 3, 2020 at 19:11
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    $\begingroup$ 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. $\endgroup$ Commented May 4, 2020 at 12:50
  • $\begingroup$ @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? $\endgroup$
    – R. Rankin
    Commented Apr 13, 2021 at 21:25
  • $\begingroup$ @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. $\endgroup$ Commented Apr 14, 2021 at 6:17
  • $\begingroup$ @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? $\endgroup$
    – R. Rankin
    Commented Apr 14, 2021 at 8:16

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