(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

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