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I am reading the John Lee's Riemannian Manifold, Theorem 11.8. ( p.201 ) and some question arises.

Let $M$ be a Riemannian manifold with the Riemannian connection. Let $\gamma : I \to M$ be a a minimizing unit speed geodesic segment.

Then does there exists a parallel orthonormal frame $(E_1, \dots , E_n)$ along $\gamma$ such that $E_n = \dot{\gamma}$ , and each $E_1 ,\dots E_{n-1}$ are normal? Here, a vector field $V$ along a curve $\gamma$ is said to be parallel along $\gamma$ with respect to a linear connection if $D_t V =0$ ( the covariant derivative of $V$ along $\gamma$ ).

I am trying to use next proposition 5.5 of the John Lee's Introduction to Riemannian manifolds, p.118 :

Proposition 5.5 ( Characterizations of Metric Connections). Let $(M,g)$ be a Riemannian or pseudo-Riemannian manifold (with or without boundary), and let $\nabla$ be a connection on $TM$. The following conditions are equivalent: (a) $\nabla$ is compatiable with $g$. (b), (c) , (d), (e), (f) : Omitted. (g) : Given any smooth curve $\gamma$ in $M$, every orthonormal basis at a point of $\gamma$ can be extended to a parallel orthonormal frame along $\gamma$.

Choose $t_0 \in I$ and orthonormal basis $V_1, V_2 , \dots , V_{n-1}, V_n := \dot{\gamma}(t_0)$ of $T_{\gamma(t_0)}M$. And use the Proposition 5.5 and obtain extended parallel orthonormal frame $(E_1, \dots , E_n)$ along $\gamma$. An issue that makes me difficult is possibility of $E_n \neq \dot{\gamma}$. Can we choose a parallel orthonormal frame $(E_1, \dots E_n)$ such that $E_n = \dot{\gamma}$ ? Can anyone helps?

Plantation
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    You can only have $\dot\gamma$ parallel along $\gamma$ if $\gamma$ is a geodesic. This is the typical definition of a geodesic, in fact. So your “smooth curve or a … geodesic” doesn’t work for general curves. No need for it to be distance-minimizing, however. – Ted Shifrin Jun 25 '23 at 05:33
  • And How about exsistence of such parallel orthonormal frame satisfying $E_n=\dot{\gamma}$? :) – Plantation Jun 25 '23 at 06:00
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    Sure. You can parallel translate any orthonormal frame along any smooth curve. Remember that parallel translation preserves lengths and angles. As long as the curve is a geodesic, you can specify that $E_n = \dot\gamma$. – Ted Shifrin Jun 25 '23 at 06:01
  • Thanks. And..how can we use the parallel translate to obtain such orthonormal frame? Your comment sounds plausible, but I don't know how to make a rigorous proof. – Plantation Jun 25 '23 at 06:22
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    Just parallel translate each $E_i$ one at a time. Then use my original comment. – Ted Shifrin Jun 25 '23 at 06:27

1 Answers1

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O.K. I leave answer as a record.

Choose $t_0 \in I$ and orthonormal basis $V_1, V_2, \dots , V_{n-1}, V_n := \dot{\gamma}(t_0)$ of $T_{\gamma(t_0)}M$. We can extend each $V_i$ by parallel transport to obtain a smooth parallel vector field $E_i$ along $\gamma$ ( C.f. his book p.60, Theorem 4.11). Furhermore, by the uniqueness of parallel translation, $E_n = \dot{\gamma}$. ( $\because$ $\dot{\gamma}$ is parallel along $\gamma$ since $\gamma$ is geodesic). Since parallel transport is a linear isometry, the resulting $n$-tuple $(E_i)$ is an orthonormal basis as all points of $\gamma$ ( C.f. John Lee's Introduction to smooth manifolds, p.118, proof of Proposition 5.5, (f)$\Rightarrow$(g) ).

And, furthurmore, note that $V_1, V_2, \dots, V_{n-1}$ are orthogonal to $\dot{\gamma(t_0)}$. So we can prove that each $E_{1}, E_2 ,\dots , E_{n-1}$ are normal. (C.f. For a geodesic $γ:I→M$ with $a∈I$, a jacobi field $J$ along $γ$ is normal if and only if $J(a)⊥ \dot{γ}(a)$ and $D_tJ(a)⊥\dot{γ}(a)$ (his book, Lemma 10.6), and use the theorem 10.2 (The Jacobi Equation) in his book etc.. ).

Plantation
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