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Lie bracket
Lie bracket of two vectors:
$\mathcal{L}_x y = [x, y] = x y - y x = x (y^c e_c) - y (x^c e_c)$
Note that vectors are not linear wrt scalar(function) multiplication like one-forms are (are one-forms?)
$= x^a e_a (y^c e_c) - y^a e_a (x^c e_c)$
$= x^a e_a (y^c) e_c + x^a y^c e_a e_c - y^a e_a (x^c) e_c - y^a x^c e_a e_c$
$= (x^a e_a (y^c) - y^a e_a (x^c)) e_c + x^a y^b (e_a e_b - e_b e_a)$
$= (x^a e_a (y^c) - y^a e_a (x^c)) e_c + x^a y^b {c_{ab}}^c e_c$
$= (x^a e_a(y^c) - y^a e_a(x^c) + x^a y^b {c_{ab}}^c) e_c$
Is the Lie bracket linear?
$[s x, y] f = s x( y (f)) - x (s y (f))$
$ = s x^a e_a ( y^b e_b (f)) - y^a e_a (s x^b e_b (f))$
$ = s x^a (e_a y^b) e_b (f) + s x^a y^b e_a e_b (f) - y^a (e_a s) x^b e_b (f) - y^a s (e_a x^b) e_b (f) - y^a s x^b e_a e_b (f)$
$ = s x^a {y^b}_{,a} e_b (f) - y^a s_{,a} x^b e_b (f) - s y^a {x^b}_{,a} e_b (f) + s x^a y^c {c_{ac}}^b e_b (f)$
$ = (s x^a {y^b}_{,a} - s y^a {x^b}_{,a} + s x^a y^c {c_{ac}}^b - y^a s_{,a} x^b) e_b (f)$
$ = s [x, y] - y(s) x$
...I think I got my Lie derivative / bracket definition backwards, maybe $\mathcal{L}_x y = [y, x]$?
TODO name this one "structure coefficients" instead, since "structure constants" usually aren't constant.
Structure constants / commutation coefficients
Structure constants / commutation coefficients:
$[e_a, e_b] = e_a (e_b) - e_b (e_a) = {c_{ab}}^c e_c$
Antisymmetry of Lie bracket of vector basis and therefore of structure constants.
$[e_a, e_b] = e_a e_b - e_b e_a = -(e_b e_a - e_a e_b) = -[e_b, e_a]$
Therefore ${c_{ab}}^c = -{c_{ba}}^c$
Structure constants in a coordinate basis:
$[e_a, e_b] = [\partial_a, \partial_b] = \partial_a \partial_b - \partial_b \partial_a = 0$, therefore ${c_{ab}}^c = 0$
Structure constants in a non-holonomic basis:
If $e_a = {e_a}^{\bar{a}} \partial_\bar{a}$ and $e^a = {e^a}_{\bar{a}} dx^\bar{a}$ are linear combinations of a coordinate basis $\partial_\bar{a}$ and dual $dx^\bar{a}$ then we can calculate the following:
Using the identities that $\delta^a_b = {e^a}_\bar{u} {e_b}^\bar{u}$ and $\delta^\bar{a}_\bar{b} = {e_u}^\bar{a} {e^u}_\bar{b}$
$[e_a, e_b] = e_a(e_b) - e_b(e_a)$
$ = {e_a}^\bar{a} \partial_\bar{a} ( {e_b}^\bar{b} \partial_\bar{b} )
- {e_b}^\bar{b} \partial_\bar{b} ( {e_a}^\bar{a} \partial_\bar{a} )
$
$ = {e_a}^\bar{a} \partial_\bar{a} ({e_b}^\bar{b}) \partial_\bar{b}
+ {e_a}^\bar{a} {e_b}^\bar{b} \partial_\bar{a} \partial_\bar{b}
- {e_b}^\bar{b} \partial_\bar{b} ({e_a}^\bar{a}) \partial_\bar{a}
- {e_b}^\bar{b} {e_a}^\bar{a} \partial_\bar{b} \partial_\bar{a}
$
$ = {e_a}^\bar{a} \partial_\bar{a} ({e_b}^\bar{b}) \partial_\bar{b}
- {e_b}^\bar{b} \partial_\bar{b} ({e_a}^\bar{a}) \partial_\bar{a}
$
$ = (
{e_a}^\bar{u} \partial_\bar{u} ({e_b}^\bar{c})
- {e_b}^\bar{u} \partial_\bar{u} ({e_a}^\bar{c})
) {e^c}_\bar{c} e_c
$
Therefore ${c_{ab}}^c = (
{e_a}^\bar{u} \partial_\bar{u} ({e_b}^\bar{c})
- {e_b}^\bar{u} \partial_\bar{u} ({e_a}^\bar{c})
) {e^c}_\bar{c}
= 2 {e_{[a}}^\bar{u} \partial_\bar{u} ({e_{b]}}^\bar{c}) {e^c}_\bar{c}
= 2 e_{[a} ({e_{b]}}^\bar{c}) {e^c}_\bar{c}
$
Transforming components into the coordinate basis:
${c_{\bar{a}\bar{b}}}^\bar{c}$
$ = {e^a}_\bar{a} {e^b}_\bar{b} {e_c}^\bar{c} (
{e_a}^\bar{u} \partial_\bar{u} ({e_b}^\bar{v})
- {e_b}^\bar{u} \partial_\bar{u} ({e_a}^\bar{v})
) {e^c}_\bar{v}
$
$ = {e^b}_\bar{b} \partial_\bar{a} ({e_b}^\bar{c})
- {e^a}_\bar{a} \partial_\bar{b} ({e_a}^\bar{c})
$
Trace of 1st and 3rd in a linear non-coordinate basis:
${c_{ab}}^a$
$= 2 e_{[a} ({e_{b]}}^\bar{a}) {e^a}_\bar{a}$
$= (
e_a ({e_b}^\bar{a})
- e_b ({e_a}^\bar{a})
) {e^a}_\bar{a}$
$=
e_a ({e_b}^\bar{a}) {e^a}_\bar{a}
- e_b ({e_a}^\bar{a}) {e^a}_\bar{a}
$
$=
- {e_b}^\bar{a} e_a ({e^a}_\bar{a})
+ {e_a}^\bar{a} e_b ({e^a}_\bar{a})
$
Note that $\delta^a_b = {e_b}^\bar{a} {e^a}_\bar{a}$ and $n = \delta^a_a = {e_a}^\bar{a} {e^a}_\bar{a}$.
Using Jacobi identity (TODO define that one much earlier than worksheet #13 ... would fit well after worksheet #5 on Levi-Civita permutation tensor):
$=
- {e_b}^\bar{a} e_a ({e^a}_\bar{a})
+ \frac{1}{e} e_b (e)
$
TODO maybe not just use $g = det(g_{ab})$ and $e = det({e^a}_\bar{a})$?
structure constants and wedge product
Structure constant two-form:
Let $c^u = {c_{ab}}^u e^a \otimes e^b = \frac{1}{2} {c_{ab}}^u e^a \wedge e^b$
structure constant definition using wedge product:
$c^c e_c (\phi)$
$= {c_{ab}}^c e_c (\phi) e^a \otimes e^b$
$= ( e_a (e_b (\phi)) - e_b (e_a(\phi)) ) e^a \otimes e^b$
$= e_a (e_b (\phi)) (e^a \otimes e^b - e^b \otimes e^a)$
$= e_a (e_b (\phi)) e^a \wedge e^b$
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