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Let $(M,g)$ be a semi-Riemannian manifold with Levi-Civita connection $\nabla$. Fix $x \in M$ and a convex neighborhood $U$ of $x$. Then for each $y \in U$ there exists a unique geodesic $\gamma_{xy}$ connecting $x$ and $y$. Let $(s_1, \dots, s_n)$ be a pseudo-orthogonal basis of $T_xM$. Define the local pseudo-orthogonal frame $(S_1, \dots, S_n)$ in $U$ by parallel transport of the $s_i$ along these unique geodesics.

The $S_i$ will satisfy $\nabla _X S_i = 0$ for all $X \in T_xM$ and $1 \leq i \leq n$.

We then have for the $(3,1)$-curvature tensor $R$ in the point $x$: $R(s_i,s_j,s_k)= \nabla _{s_i} \nabla _{s_j } s_k - \nabla _{s_j} \nabla _{s_i } s_k - \nabla _{[s_i,s_j]} s_k =0$.

And because $R$ is tensorial, we also have $R(X,Y,Z)=0$ for all $X,Y,Z \in T_xM$. In other words, $M$ is flat.

Obviously, this cannot be the case. Where have I gone wrong? Thank you for the help. (This is kind of a follow up to the post Existence of a local geodesic frame)

tortemath
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1 Answers1

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The equation $\nabla_X S_i = 0$ is only satisfied at $x$, so you cannot differentiate this equation to conclude that $R$ is identically zero.