# (Induced) Covariant Derivative

Let $\alpha : I \longrightarrow M$ be a curve and $Z \in \mathcal{X} (\alpha)$ be an element of the set of all smooth vector fields on $\alpha$.

Let $h : J \longrightarrow I$ re-parameterize the curve.

Then $Z \circ h : J \longrightarrow TM$ and

$(Z \circ h)f(p) = Z(h(p))f = Zf(h(p)) = (Zf \circ h)(p)$.

Thus $(Z \circ h)f = Zf \circ h$ which is smooth since the composition of smooth functions is a smooth function and therefore $Z \circ h \in \mathcal{X}(\alpha \circ h)$.

By definition

$(\partial_i\mid_\alpha)\prime (s) = D_{\alpha\prime (t)}(\partial_i)$

So

$(\partial_i\mid_{\alpha \circ h})\prime (s) = D_{{(\alpha \circ h)}\prime (t)}(\partial_i)$

Also by definition

$\alpha\prime (t) = d\alpha (d/du\mid_t)$

So, using the chain rule

$(\alpha \circ h)\prime (s) = d(\alpha \circ h) (d/du\mid_s) = d\alpha (dh (d/du\mid_s))$

$= d\alpha (\frac{dh}{du} \frac{d}{dt}|_{h(s)}) = \frac{dh}{du} \alpha\prime(h(s))$

Which gives

$(\partial_i|_{\alpha \circ h})\prime (s) = D_{\frac{dh}{du}\alpha\prime(h(s))}(\partial_i) = \frac{dh}{du}D_{\alpha\prime(h(s))}(\partial_i) = \frac{dh}{du}(\partial_i|_\alpha)\prime (h(s))$

Thus

$(\partial_i|_{\alpha \circ h})\prime = \frac{dh}{du}(\partial_i|_\alpha)\prime \circ h$