Exercise 4.12

Let $M$ be a riemannian manifold, ∇ a linear connection and $γ : I \to M$ a curve. Let $D_{t}$ denote the corresponding covariant derivative operator. Define for all $t_{0}, t_{1} \in I$ $P_{t_{0} t_{1}} (V_{0}) = V (t_{1})$ , where $V$ is the parallel translate of $V_{0} \in T_{γ (t_{0})} M$ along γ. Then the following formula holds:

$D_{t} V (t_{0}) = lim_{t \to t_{0}} \frac{({P_{t_{0} t}}^{-1} (V (t))) - V (t_{0})}{t - t_{0}}$

Proof:

\partial_{k}

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P_{t_{0} t} \partial_{k}

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Let $\partial_{k} (t)$ be the standard basis frame of $T_{γ (t)} M$ . Then $P_{t_{0} t} \partial_{k} (t_{0})$ is also a basis of $T_{γ (t)} M$ . Hence for each $t \in I$ , we can write $\partial_{k} (t) = \sum_{l} a_{k, l} (t) (P_{t_{0} t} \partial_{l} (t_{0}))$ for some smooth functions $a_{k, l}$ .

$V (t) = \sum_{k} V^{k} (t) \partial_{k} (t) = \sum_{k, l} a_{k, l} (t) V^{k} (t) (P_{t_{0} t} \partial_{l} (t_{0}))$

${P_{t_{0} t}}^{-1} V (t) = \sum_{k} V^{k} (t) \partial_{k} (t) = \sum_{k, l} a_{k, l} (t) V^{k} (t) (\partial_{l} (t_{0}))$

$\frac{{P_{t_{0} t}}^{-1} V (t) - V (t_{0})}{t - t_{0}} = \frac{1}{t - t_{0}} \sum_{l} (\sum_{k} a_{k, l} (t) V^{k} (t) - V^{l} (t_{0})) \partial_{l} (t)$

$\frac{{P_{t_{0} t}}^{-1} V (t) - V (t_{0})}{t - t_{0}} = \sum_{l} (\frac{V^{l} (t) - V^{l} (t_{0})}{t - t_{0}} + \sum_{k} (V^{k} (t) \frac{a_{k, l} (t) - δ_{k, l}}{t - t_{0}})) \partial_{l} (t)$

Note that $a_{k, l} (t_{0}) = δ_{k, l}$ and the functions $a_{k, l}$ and $V^{l}$ are smooth. Hence after renaming the indices we get almost the formula expected:

$lim_{t \to t_{0}} \frac{({P_{t_{0} t}}^{-1} (V (t))) - V (t_{0})}{t - t_{0}} = \sum_{k} ({\dot{V}}^{k} (t_{0}) + \sum_{j} V^{j} (t_{0}) ({\dot{a}}_{j, k} (t_{0}))) \partial_{k} (t_{0})$

If $a^{i, j}$ are the coefficient of the inverse matrix of $a_{i, j}$ then $\sum_{k} a_{i, k} a^{k, j} = δ_{i, j}$ so $\sum_{k} {\dot{a}}_{i, k} (t) a^{k, j} (t) + a_{i, k} (t) {\dot{a}}^{k, j} (t) = 0$ . At $t = t_{0}$ , $a_{k, l} (t_{0}) = a^{k, l} (t_{0}) = δ_{k, l}$ so this formula becomes: ${\dot{a}}_{i, j} (t_{0}) + {\dot{a}}^{i, j} (t_{0}) = 0$ .

Recall that $\partial_{k} (t) = \sum_{l} a_{k, l} (t) (P_{t_{0} t} \partial_{l} (t_{0}))$ i.e. $P_{t_{0} t} \partial_{k} (t) = \sum_{l} a^{k, l} (t) (\partial_{l} (t_{0}))$ . But for all $k$ , $P_{t_{0} t} \partial_{k} (t)$ is by definition a parallel vector field along γ, so its coordinates satify the equation:

${\dot{a}}^{k, l} (t) + \sum_{i, j} (a^{k, j} (t) {\dot{γ}}^{i} (t) Γ_{i, j}^{l} γ (t))) = 0$

Again, when evaluating this expression at $t = t_{0}$ and we get:

${\dot{a}}_{k, l} (t_{0}) = - {\dot{a}}^{k, l} (t_{0}) = \sum_{i} ({\dot{γ}}^{i} (t_{0}) Γ_{i, k}^{l} (γ (t_{0})))$

And finally:

$lim_{t \to t_{0}} \frac{({P_{t_{0} t}}^{-1} (V (t))) - V (t_{0})}{t - t_{0}} = \sum_{k} ({\dot{V}}^{k} (t_{0}) + \sum_{i, j} V^{j} (t_{0}) {\dot{γ}}^{i} (t_{0}) Γ_{i, j}^{k} (γ (t_{0}))) \partial_{k} (t_{0}) = D_{t} V (t_{0})$

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