Problem: Let $h:J\to I$ reparaemtrizing a curve $\alpha:I\to M$.
If $Z\in\mathscr{X}(\alpha)$ then $Z\circ h\in \mathscr{X}(\alpha\circ h)$
and a) $(Z\circ h)'=\frac{dh}{dt}Z'\circ h$
b)$(Z\circ h)''=\frac{d^2 h}{dt^2}Z'\circ h+\left(\frac{dh}{dt}\right)^2 Z''\circ h $
I think the following results are useful to state:
Proposition 3.8. Let $\alpha: I \rightarrow M$ be a curve in a semi-Riemannian manifold. There is a unique function
$$
\mathscr{X}(\alpha) \rightarrow \mathscr{X}(\alpha) \quad \text { denoted by } Z \mapsto Z^{\prime}
$$
which satisfies
(1) $\mathbb{R}$-linearity: $\left(a Z_{1}+b Z_{2}\right)^{\prime}=a Z_{1}^{\prime}+b Z_{2}^{\prime}$, where $a, b \in \mathbb{R}$,
(2) a product rule: $(h Z)^{\prime}=h^{\prime} Z+h Z^{\prime}$, where $h \in C^{\infty}(I)$,
(3) agreement with $\nabla:\left(V_{\alpha}\right)^{\prime}(t)=\nabla_{\alpha^{\prime}(t)} V$, where $t \in I$ and $V \in \mathscr{X}(M)$,
(4) compatibility with the metric: $\left\langle Z_{1}, Z_{2}\right\rangle^{\prime}=\left\langle Z_{1}^{\prime}, Z_{2}\right\rangle+\left\langle Z_{1}, Z_{2}^{\prime}\right\rangle$.
For the proof of the above proposition of the induced covariant derivative can be expressed as:
$$
Z^{\prime}(t)=\left.\sum\left(Z^{i}\right)^{\prime}(t) \partial_{i}\right|_{\alpha(t)}+\sum Z^{i}(t) \nabla_{\alpha^{\prime}(t)} \partial_{i} \quad \text { for } t \text { near } t_{0}
$$
I wanted to start with the following:
$(Z\circ h)'=\left.\sum\left((Z\circ h)^{i}\right)^{\prime}(t) \partial_{i}\right|_{\alpha(t)}+\sum (Z\circ h)^{i} (t) \nabla_{(\alpha\circ h)^{\prime}(t)} \partial_{i} \quad \text { for } t \text { near } t_{0}$
Howeve the Levi-Civita connection has no property on how to handle compositions.
Question:
How should I prove this?
Thanks in advance.
Best Answer
$h$ is a reparametrization hence a diffeomorphism. Let $X = Z \circ h$. You can get back $Z$ from $X$ with $Z = X \circ h^{-1}$. Proposition $a)$ is equivalent to, $$ X' = h'(X \circ h^{-1})' \circ h $$ Let $D$ be the operator of $\mathcal{X}(\alpha \circ h)$ defined by $D(X) = h'(X \circ h^{-1})' \circ h$. If we prove that $D$ verfies the properties $(1),(2),(3),(4)$, then, by uniqueness, $D = \cdot'$, which proves $a)$.
When something is defined as "the only thing that verifies such properties", it is rarely a good idea to look for closed expression and you would better use the properties defining implicitely this object.
$(1)$ Trivial by $\mathbb{R}$-linearity of $\cdot'$, of multiplication by $h'$ and of the composition on the right.
$(2)$ Let $f : J \rightarrow \mathbb{R}$ smooth. \begin{align*} D(fX) & = h'((fX) \circ h^{-1})' \circ h\\ & = h'((f \circ h^{-1})(X \circ h^{-1}))' \circ h\\ & = h'((f \circ h^{-1})'(X \circ h^{-1}) + (f \circ h^{-1})(X \circ h^{-1})') \circ h\\ & = h'((h^{-1})'(f' \circ h^{-1})(X \circ h^{-1})) \circ h + h'((f \circ h^{-1})(X \circ h^{-1})') \circ h\\ & = h'((h^{-1})' \circ h)f'X + fh'(X \circ h^{-1}) \circ h\\ & = f'X + fD(X). \end{align*}
I let you prove $(3)$ and $(4)$ but it is the same idea, you need to use the fact that $(3)$ (resp. $(4)$) is true for $\cdot' : \mathcal{X}(\alpha) \rightarrow \mathcal{X}(\alpha)$ to prove that it is also true for $D : \mathcal{X}(\alpha \circ h) \rightarrow \mathcal{X}(\alpha \circ h)$.
Once $a)$ is proved, you easily get $b)$ with, \begin{align*} (Z \circ h)'' & = (h'Z' \circ h)'\\ & = h''Z' \circ h + h'(Z' \circ h)'\\ & = h''Z' \circ h + (h')^2Z'' \circ h. \end{align*}