# (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

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

# Product Rule for Tensors

$\mathcal{D}(A(X_1,\ldots,X_s)) = \mathcal{D}(C^2 (A \otimes X_1 \otimes X_2)) =$

$C^2(\mathcal{D} (A \otimes X_1 \otimes X_2)) =$

$C^2(((\mathcal{D}A) \otimes X_1 \otimes X_2) + (A \otimes \mathcal{D} X_1 \otimes X_2) + (A \otimes X_1 \otimes \mathcal{D} X_2)) =$

$(\mathcal{D}A)(X_1,X_2) + A(\mathcal{D} X_1,X_2) + A(X_1,\mathcal{D} X_2)$

# Tensor Derivations

Define $(\mathcal{D}_\mathcal{U}B)_p = \mathcal{D}(fB)_p$. This is well defined since if $g$ is another bump function then

$\mathcal{D}(fgB)_p = \mathcal{D}(f)_p(gB)(p) + f(p)\mathcal{D}(gB)_p = \mathcal{D}(gB)_p$

and reversing $f$ and $g$ gives

$\mathcal{D}(gfB)_p = \mathcal{D}(g)_p(fB)(p) + g(p)\mathcal{D}(fB)_p = \mathcal{D}(fB)_p$