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Going by 2012 Frankel's chapter on Lie derivatives, and taking tips from the Lie Derivative wikipedia page:

Let $\mathcal{M}$ be our manifold.
Let $p \in \mathcal{M}$ be a point on the manifold,
Let $T_p(\mathcal{M})$ be the tangent space on $\mathcal{M}$ at point $p$.
Let $TM = \underset{p \in \mathcal{M}}{\cup} T_p$ be the tangent bundle, i.e. the union of all tangent spaces on the manifold.

TODO Definition of Vector Field ... vs Vector Space... vector-space has $\cdot$ scalar operator with reals, while vector field has $\times$ operator with vectors?
Wikipedia on vector fields (source plz?) "a vector field on $\mathcal{M}$ is an assigment of a tangent vector to each point on $\mathcal{M}$".
"More precisely, a vector field F is a mapping from M onto the tangent bundle TM so that $p \circ F$ is the identity mapping where p denotes the projection TM to M."
"In other words, a vector field is a section of the tangent bundle."

Flow:

(Wiki article on mathematics Flow cites D.V. Anosov (2001) [1994], "Continuous flow", Encyclopedia of Mathematics, EMS Press)
TODO now I have to talk about tangent-space (tangent-vector-space at a single point 'p' in manifold $\mathcal{M}$) versus tangent-bundle (collection of all tangent-spaces at all points on $\mathcal{M}$) ... probably all the way back at section 0
Let $v : \mathbb{R} \times \mathcal{M} \rightarrow TM$ is a time-dependent vector-field on $\mathcal{M}$, i.e. for $t \in \mathbb{R}$ and $p \in \mathcal{M}$, we have $v(t, p) \in T_p(\mathcal{M})$. So $v(t, p) \in T_p(\mathcal{M})$ produces the vector (within the tangent-space at point p) that the field is pointing at time t at point p.
For interval $I \subseteq \mathbb{R}$, $0 \in I$, we define the flow of $v$ is the function $f : I \times \mathcal{M} \rightarrow \mathcal{M}$ such that, $\forall p \in \mathcal{M}$, $\forall s, t \in \mathbb{R}$ the following are true:
I want to write this in interval form, but the input of the derivative would be the previous increment of the interval...
$\int_{t=t_0}^{t=t_1} v(t, ... f(t, p)) dt = f(t_1, p)$ ... idk ...

So $f(t, p)$ gives back the point that starts at 'p' after time 't' of movement.
This is the function notation , meanwhile there is also a popular group action notation: $f^t(p)$
Then the two rules become:

Lie derivative:

Let $\phi : \mathcal{M} \rightarrow \mathbb{R}$ be a scalar function of the manifold,
Let $p \in \mathcal{M}$ be a point on the manifold.
Let $v : \mathcal{M} \rightarrow TM$ be a vector field on the manifold. So $v(p) \in T_p(\mathcal{M})$. Is v time-varying like in the vector-field definition? I suspect no need to make it time-varying.
Let $f(t, p)$ be a flow on the manifold based on vector field $v(t, p)$.
$\mathcal{L}_v(\phi)|_p = \underset{t \rightarrow 0}{lim} \frac{\phi(f(t, p)) - \phi(p)}{t}$
$\mathcal{L}_v(\phi)|_p = \frac{d}{dt} \left( \phi(f(t, p)) \right) |_{t=0}$
$\mathcal{L}_v(\phi)|_p = \frac{\partial \phi}{\partial p} ( f(t, p) ) \cdot \frac{d}{dt} \left( f(p) \right) |_{t=0} $
... using $f(0, p) = f(p)$ and using $\frac{d}{dt} f = v$...
$\mathcal{L}_v(\phi)|_p = \frac{\partial \phi}{\partial p} ( p ) \cdot v(t, p) |_{t=0} $
... let $p = u(x^i)$ for some chart $u(x), u : U \subseteq \mathbb{R}^n \rightarrow \mathcal{M}$.
$\mathcal{L}_v(\phi) = \frac{\partial \phi}{\partial x^i} \cdot v^i$ ... hmm ... where did those indexes come from ...

TODO here, show where the Lie derivative rules come from ...
1) $\mathcal{L}_v (\phi) = v(\phi) = v_i e_i(\phi)$
2) $\mathcal{L}_v (S \otimes T) = (\mathcal{L}_v S) \otimes T + S \otimes (\mathcal{L}_T)$ for tensor fields S and T.
3) $\mathcal{L}_v (T(y_1, ..., y_n)) = (\mathcal{L}_v T)(y_1, ..., y_n) + T(\mathcal{L}_v (y_1), ..., y_n) + T(y_1, ..., \mathcal{L}_v (y_n))$, for tensor T and input vectors $y_1 ... y_n$.
4) $[\mathcal{L}_v, d] = 0$

Lie derivative of scalar / (0 0) tensor in a coordinate basis:
$\mathcal{L}_v \phi = v \lrcorner d\phi$
$= v \lrcorner \partial_a (\phi) dx^a$
$= \partial_a (\phi) dx^a (v)$
$= \partial_a (\phi) dx^a (v^b \partial_b)$
$= v^b \partial_a(\phi) dx^a(\partial_b)$
$= v^b \partial_a(\phi) \delta^a_b$
$= v^a \partial_a(\phi)$

Lie derivative of a scalar / (0 0) tensor in a non-coordinate basis?
$\mathcal{L}_x \phi = x \lrcorner d\phi$
$= x \lrcorner e_a(\phi) e^a$
$= e_a(\phi) e^a(x)$
$= e_a(\phi) e^a(x^b e_b)$
$= x^b e_a(\phi) e^a(e_b)$
$= x^b e_a(\phi) \delta^a_b$
$= x^a e_a(\phi)$

Lie derivative of p-form (Cartan homotopy formula):
$\mathcal{L}_x w = x \lrcorner dw + d(x \lrcorner w)$

Lie derivative of a 1-form coordinate dual-basis element:
$\mathcal{L}_x dx^i = x \lrcorner d^2 x^i + d(x \lrcorner dx^i)$
using $d^2x^i = 0$
$= d(dx^i (x^a \partial_a))$
$= d(x^a dx^i(\partial_a))$
$= d(x^a \delta^i_a)$
$= d(x^i)$
$= \partial_a(x^i) dx^a$

Linearity of Lie derivative and a 1-form coordinate dual-basis element:
$\mathcal{L}_x (e^i)$
$= \mathcal{L}_x ({e^i}_a dx^a)$
$= \mathcal{L}_x ({e^i}_a) dx^a + {e^i}_a \mathcal{L}_x (dx^a)$
$= x^b \partial_b( {e^i}_a ) dx^a + {e^i}_a \partial_b(x^a) dx^b$
$= (x^a \partial_a( {e^i}_b ) + {e^i}_a \partial_b(x^a) ) dx^b$
$= (x^a \partial_a( {e^i}_b ) + {e^i}_a \partial_b(x^a) ) \delta^b_c dx^c$
$= (x^a \partial_a( {e^i}_b ) + {e^i}_a \partial_b(x^a) ) {e_j}^b {e^j}_d dx^d$
$= (x^a \partial_a( {e^i}_b ) {e_j}^b + {e^i}_a {e_j}^b \partial_b(x^a) ) e^j$
$= (x^k e_k( {e^i}_b ) {e_j}^b + {e^i}_a e_j(x^a) ) e^j$
...
would be nice if this was equal to:
$= e_j(x^i) e^j$

try again, using the k-form definition (is that definition only for forms that are wedges of basis elements?):
$\mathcal{L}_x e^i = x \lrcorner d e^i + d(x \lrcorner e^i)$
$= x \lrcorner (-{\omega^i}_j \wedge e^j + \Theta^i ) + d(x^i)$
$= x \lrcorner (-2 {\Gamma^i}_{[kj]} e^k \otimes e^j + T^i_{[kj]} e^k \otimes e^j ) + e_j(x^i) e^j$
$= x \lrcorner (-{\omega^i}_j e^j + \Theta^i ) + d(x^i)$
$= -{\Gamma^i}_{[kj]} x^k e^j + T^i_{[kj]} x^k e^j + e_j(x^i) e^j$
...
$= x \lrcorner (-{c_{[kj]}}^i e^k \otimes e^j ) + e_j(x^i) e^j$
$= {c_{jk}}^i e^j x^k + e_j(x^i) e^j$
$= e^i(e_j (x) - x(e_j)) + e_j(x^i) e^j$ ... ?
...hmm, another extra term...

Would also be nice if $\mathcal{L}_x (e_i) = -e_i(x^a) e_a$

But what if they're not? Especially if you consider $e_i = {e_i}^a \partial_a$ as a half-weight tensor density, in which case the Lie derivative (p q) tensor density definition has an added term which I haven't included yet.

Lie derivative of outer products:
$\mathcal{L}_x (a \otimes b) = (\mathcal{L}_x a) \otimes b + a \otimes (\mathcal{L}_x b)$

Lie derivative of a (p q) tensor:
$\mathcal{L}_x T$
$= \mathcal{L}_x ({T^{i_1 ... i_p}}_{j_1 ... j_q} e_{i_1} \otimes ... \otimes e_{i_p} \otimes e^{j_1} \otimes ... \otimes e^{j_q})$
$= \mathcal{L}_x ({T^I}_J) {e^I}_J + {T^I}_J \mathcal{L}_x (e^I) \otimes e_J + {T^I}_J e^I \otimes \mathcal{L}_x (e_J) $
$= \mathcal{L}_x ({T^I}_J) {e^I}_J + {T^I}_J (e^{i_1} \otimes ... \otimes \mathcal{L}_x (e^{i_k}) \otimes ... \otimes e^{i_p}) \otimes e_J + {T^I}_J e^I \otimes (e_{j_1} \otimes ... \otimes \mathcal{L}_x (e_{j_k}) \otimes ... \otimes e_{j_q}) $
...
$= (x^a e_a ({T^I}_J) - e_a(x^{i_k}) \underset{i_k \leftrightarrow a}{{T^I}_J} + e_{j_k}(x^a) \underset{j_k \leftrightarrow a}{{T^I}_J}) {e^I}_J$

Lie derivative along a coordinate-aligned vector:
Let vector $t^u = \delta^u_t$ point along the distinct coordinate 't',
such that $t^u e_u(t) = \delta^u_t \delta^t_u = 1$
The Lie derivative of a (p q) tensor T along t is:
$\mathcal{L}_\vec{t} {T^{a_1 ... a_p}}_{b_1 ... b_q} = \mathcal{L}_\vec{t} {T^A}_B = t^u e_u( {T^A}_B ) - \underset{a_k \leftrightarrow c}{{T^A}_B} e_c ( t^{a_k} ) + \underset{b_k \leftrightarrow c}{{T^A}_B} e_{b_k} ( t^c ) $
... using $t^u = \delta^u_t$ and $e_v ( t^u ) = e_v ( \delta^u_t ) = 0$
$\mathcal{L}_\vec{t} {T^{a_1 ... a_p}}_{b_1 ... b_q} = \delta^u_t e_u ( {T^A}_B ) - \underset{a_k \leftrightarrow c}{{T^A}_B} \cdot 0 + \underset{b_k \leftrightarrow c}{{T^A}_B} \cdot 0 $
$\mathcal{L}_\vec{t} {T^{a_1 ... a_p}}_{b_1 ... b_q} = e_t ( {T^A}_B ) $

Lie derivative of any (0 2) tensor $T_{ab}$ for arbitrary commutation and torsion:
TODO this should go after defining nabla, T, etc
$\mathcal{L}_\vec{v} T_{ab} + {T^d}_{ac} v^c T_{db} + {T^d}_{bc} v^c T_{ad} $
using ${T^a}_{bc} + {c_{bc}}^a = {\Gamma^a}_{bc} - {\Gamma^a}_{cb}$
$ = v^c e_c(T_{ab}) + T_{cb} e_a(v^c) + T_{ac} e_b(v^c) + ({\Gamma^d}_{ac} - {\Gamma^d}_{ca}) v^c T_{db} + ({\Gamma^d}_{bc} - {\Gamma^d}_{cb}) v^c T_{ad} $
$ = v^c (e_c (T_{ab}) - {\Gamma^d}_{ca} T_{cb} - {\Gamma^d}_{cb} T_{ad}) + T_{cb} (e_a(v^c) + {\Gamma^c}_{ad} v^d) + T_{ac} (e_b(v^c) + {\Gamma^c}_{bd} v^d) $
$ = v^c \nabla_c T_{ab} + T_{cb} \nabla_a v^c + T_{ac} \nabla_b v^c $

Therefore, for holonomic basis and torsion-free connections:
$\mathcal{L}_\vec{v} T_{ab} = v^c \nabla_c T_{ab} + T_{cb} \nabla_a v^c + T_{ac} \nabla_b v^c $
Therefore, for the Lie derivative acting on a (0 2) tensor, on a holonomic basis with torsion-free connection, you can swap the partial derivatives with covariant derivatives.

Lie derivative of a vector:
$\mathcal{L}_\vec{n} v^a = n^u e_u ( v^a ) - v^u e_u ( n^a ) = [n, v]^u$
$= n^u e_u ( v^a ) + n^u {\Gamma^a}_{uv} v^v - n^u {\Gamma^a}_{uv} v^v - v^u e_u ( n^a ) - v^u {\Gamma^a}_{uv} v^v + v^u {\Gamma^a}_{uv} v^v $
$= n^u \nabla_u v^a - v^u \nabla_u n^a - {\Gamma^a}_{uv} n^u v^v + {\Gamma^a}_{uv} v^u n^v $
$= n^u \nabla_u v^a - v^u \nabla_u n^a + {\Gamma^a}_{uv} v^u n^v - {\Gamma^a}_{vu} v^u n^v $
$= n^u \nabla_u v^a - v^u \nabla_u n^a + ({T^a}_{uv} + {c_{uv}}^a) v^u n^v $
... assuming we are using a torsion-free holonomic basis:
$= n^u \nabla_u v^a - v^u \nabla_u n^a $

Lie derivative of a form:
$\mathcal{L}_\vec{n} w_a = n^u e_u (w_a) + w_u e_a(n^u)$
$ = n^u e_u (w_a) - n^u {\Gamma^v}_{ua} w_v + n^u {\Gamma^v}_{ua} w_v + w_u e_a(n^u) + w_u {\Gamma^u}_{av} n^v - w_u {\Gamma^u}_{av} n^v $
$ = n^u \nabla_u w_a + w_u \nabla_a n^u + ({\Gamma^u}_{va} - {\Gamma^u}_{av}) w_u n^v $
$ = n^u \nabla_u w_a + w_u \nabla_a n^u + ({T^u}_{va} + {c_{va}}^u) w_u n^v $
... assuming we are using a torsion-free holonomic basis:
$ = n^u \nabla_u w_a + w_u \nabla_a n^u$

TODO verify:
Either only true for p > 0, or if you want it to be a generalization of the scalar case then you have to define $d(x \lrcorner \phi) = 0$
Same as $[\mathcal{L}_x, d] = 0$?

TODO show for vectors $[x,y] \lrcorner w = [\mathcal{L}_x, y \lrcorner] (w)$

TODO show $d \mathcal{L}_x (w) = \mathcal{L}_x (dw)$

TODO show $\mathcal{L}_{\phi x} w = \phi \mathcal{L}_x w + d\phi \wedge (x \lrcorner w)$

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