Exercise 9.5 (Translation lemma)

Question

\paragraph{Lemma 9.3 (Rescaling Lemma).} Let $V$ be a smooth vector field on a smooth manifold $M$, let $J \subseteq \mathbb{R}$ be an interval, and let $\gamma: J ightarrow M$ be an integral curve of $V$. For any $a \in \mathbb{R}$, the curve $	ilde{\gamma}: 	ilde{J} ightarrow M$ defined by $	ilde{\gamma}(t)=\gamma(a t)$ is an integral curve of the vector field a $V$, where $	ilde{J}=\{t:$ at $\in J\}$.\

Proof: See page 208

\paragraph{Lemma 9.4 (Translation Lemma).} Let $V, M, J$, and $\gamma$ be as in the preceding lemma. For any $b \in \mathbb{R}$, the curve $\hat{\gamma}: \widehat{J} ightarrow M$ defined by $\hat{\gamma}(t)=\gamma(t+b)$ is also an integral curve of $V$, where $\widehat{J}=\{t: t+b \in J\}$.

\paragraph{Exercise 9.5.} Prove the translation lemma.

Elu Thingol · Accepted Answer

\begin{proof} We wish to demonstrate that for all $t_0 \in \widehat{J}$:
\begin{equation*}
    \widehat{\gamma}^{\prime}(t_0) = V_{\widehat{\gamma}(t_0)}.
\end{equation*}

For any $f \in C^{\infty}(M)$, we have
\begin{align*}
    \widehat{\gamma}^{\prime}(t_0) 
    &= \left.\dfrac{d}{dt}ight|_{t_0} (f \circ \widehat{\gamma})(t)\
    &= \left.\dfrac{d}{dt}ight|_{t_0} (f \circ \gamma)(t+b)
\end{align*}
Now apply the (Euclidean) chain rule of differentiation $[(f \circ \gamma) \circ \operatorname{tr}]^{\prime} = (f \circ \gamma)^{\prime} (\operatorname{tr}) \cdot \operatorname{tr}^{\prime}$ to the composition $(f \circ \gamma) \circ \operatorname{tr}$ where $\operatorname{tr}(t) = t+b$:
\begin{align*}
    \widehat{\gamma}^{\prime}(t_0)  &= (f \circ \gamma)^{\prime}(t_0
+b) \cdot 1\
    &= \gamma^{\prime}(t_0 + b) f\
    &= V_{\widehat{\gamma}(t_0)}.
\end{align*}
\end{proof}

Exercise 9.5 (Translation lemma)

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Exercise 9.5 (Translation lemma)

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