Back

second exterior derivative of a vector basis:
$d e^a = \Theta^a - {\omega^a}_b \wedge e^b$
so $d^2 e^a = d \Theta^a - d( {\omega^a}_b \wedge e^b )$
$= d \Theta^a - d {\omega^a}_b \wedge e^b + {\omega^a}_b \wedge d e^b$
$= d \Theta^a - d {\omega^a}_c \wedge e^c + {\omega^a}_b \wedge (\Theta^b - {\omega^b}_c \wedge e^c)$
$= d \Theta^a + {\omega^a}_b \wedge \Theta^b - {\Omega^a}_c \wedge e^c$
if $d^2 e^a = 0$ then we get
$d \Theta^a + {\omega^a}_b \wedge \Theta^b = {\Omega^a}_b \wedge e^b$
This is the first Bianchi identity.

Exterior covariant derivative of torsion two-form (Bianchi's first identity using an exterior covariant derivative):
$D \Theta^a = {\Omega^a}_b \wedge e^b$

First Bianchi identity, expanding terms:
${\Omega^a}_b \wedge e^b = d \Theta^a + {\omega^a}_b \wedge \Theta^b$
$\frac{1}{2} {R^a}_{bcd} e^c \wedge e^d \wedge e^b = \frac{1}{2} d ({T^a}_{uv} e^u \wedge e^v) + \frac{1}{2} {\Gamma^a}_{cb} {T^b}_{uv} e^c \wedge e^u \wedge e^v $
$\frac{1}{2} {R^a}_{bcd} e^b \wedge e^c \wedge e^d = ( \frac{1}{2} e_b ({T^a}_{cd}) - \frac{1}{4} {c_{bc}}^u {T^a}_{ud} - \frac{1}{4} {c_{bd}}^u {T^a}_{cu} + \frac{1}{2} {\Gamma^a}_{bu} {T^u}_{cd} )e^b \wedge e^c \wedge e^d $
${R^a}_{bcd} e^b \wedge e^c \wedge e^d = ( e_b ({T^a}_{cd}) - {c_{bc}}^u {T^a}_{ud} + {\Gamma^a}_{bu} {T^u}_{cd} ) e^b \wedge e^c \wedge e^d $
${R^a}_{bcd} e^b \wedge e^c \wedge e^d = ( e_b ({T^a}_{cd}) - {\Gamma^u}_{bc} {T^a}_{ud} + {\Gamma^u}_{cb} {T^a}_{ud} + {\Gamma^a}_{bu} {T^u}_{cd} + {T^u}_{bc} {T^a}_{ud} ) e^b \wedge e^c \wedge e^d $
${R^a}_{bcd} e^b \wedge e^c \wedge e^d = ( e_b ({T^a}_{cd}) + {\Gamma^a}_{bu} {T^u}_{cd} - {\Gamma^u}_{bc} {T^a}_{ud} - {\Gamma^u}_{bd} {T^a}_{cu} + {T^u}_{bc} {T^a}_{ud} ) e^b \wedge e^c \wedge e^d $
Now I'm using $\nabla_b$ to represent the covariant derivative applied to the index notation representation (and not to the geometric objects)
${R^a}_{bcd} e^b \wedge e^c \wedge e^d = ( \nabla_{[b} ({T^a}_{cd]}) + {T^u}_{bc} {T^a}_{ud} ) e^b \wedge e^c \wedge e^d $
The definition here gets around that by defining a 'cyclic permutation tensor' function, which must be the geometric object equivalent of the brackets in index notation.
But if I was going to use $\nabla$ as a covariant derivative of geometric objects, then where exactly would they go?
Especially the $e_a$ which must be present for the correct definition of the covariant derivative?
${R^a}_{bcd} e_a \otimes (e^b \wedge e^c \wedge e^d) = e_a \otimes (( \nabla_{[b} ({T^a}_{cd]}) + {T^u}_{bc} {T^a}_{ud} ) e^b \wedge e^c \wedge e^d) $
It looks as if the $e_a$ goes on the lhs but it must be to the right of the $e^b$ in order for the covariant derivative $\nabla_b$ to apply to it (right?)
${R^a}_{bcd} e_a \otimes (e^b \wedge e^c \wedge e^d) = e^b \otimes ( 6 \nabla_{[b} ({T^a}_{cd]} e_a \otimes e^c \otimes e^d)) + {T^u}_{bc} {T^a}_{ud} e_a \otimes (e^b \wedge e^c \wedge e^d) $
... unless $e^a \otimes e_b = e_b \otimes e^a$, only for mixed co/contra-variance ... but I can't imagine that's true ...
but this would be true if we were using $\wedge$ instead, and $e^a \wedge e_b = e_b \wedge e^a$, as I proposed a few pages back...

In a torsion-free coordinate system, the first Bianchi identity looks like:
${R^a}_{bcd} e^b \wedge e^c \wedge e^d = 0$
${R^a}_{[bcd]} = 0$

Notice: the exterior covariant derivative of the connection is not the curvature:
$D {\omega^a}_b = d {\omega^a}_b + {\omega^a}_c \wedge {\omega^c}_b - {\omega^c}_b \wedge {\omega^a}_c$
$= d {\omega^a}_b + {\omega^a}_c \wedge {\omega^c}_b + {\omega^a}_c \wedge {\omega^c}_b$
$= d {\omega^a}_b + 2 {\omega^a}_c \wedge {\omega^c}_b$
$= {\Omega^a}_b + {\omega^a}_c \wedge {\omega^c}_b$

${\Omega^a}_c \wedge {\omega^c}_b = (d {\omega^a}_c + {\omega^a}_d \wedge {\omega^d}_c) \wedge {\omega^c}_b$
$= d {\omega^a}_c \wedge {\omega^c}_b + {\omega^a}_d \wedge {\omega^d}_c \wedge {\omega^c}_b$

${\omega^a}_c \wedge {\Omega^c}_b = {\omega^a}_c \wedge (d {\omega^c}_b + {\omega^c}_d \wedge {\omega^d}_b)$
$= {\omega^a}_c \wedge d {\omega^c}_b + {\omega^a}_c \wedge {\omega^c}_d \wedge {\omega^d}_b$
$= {\omega^a}_c \wedge d {\omega^c}_b + {\omega^a}_d \wedge {\omega^d}_c \wedge {\omega^c}_b$

${\Omega^a}_c \wedge {\omega^c}_b - {\omega^a}_c \wedge {\Omega^c}_b$
$ = d {\omega^a}_c \wedge {\omega^c}_b + {\omega^a}_d \wedge {\omega^d}_c \wedge {\omega^c}_b - {\omega^a}_c \wedge d {\omega^c}_b - {\omega^a}_d \wedge {\omega^d}_c \wedge {\omega^c}_b$
$ = d {\omega^a}_c \wedge {\omega^c}_b - {\omega^a}_c \wedge d {\omega^c}_b$

Exterior derivative of curvature two-form:
(Bianchi's second identity, using exterior derivatives)
$d {\Omega^a}_b$
$= d( d {\omega^a}_b + {\omega^a}_c \wedge {\omega^c}_b)$
$= d^2 {\omega^a}_b + d ({\omega^a}_c \wedge {\omega^c}_b)$
$= d^2 {\omega^a}_b + d {\omega^a}_c \wedge {\omega^c}_b - {\omega^a}_c \wedge d {\omega^c}_b$
$ = d^2 {\omega^a}_b + {\Omega^a}_c \wedge {\omega^c}_b - {\omega^a}_c \wedge {\Omega^c}_b$
Assume $d^2 {\omega^a}_b = 0$
$d {\Omega^a}_b = {\Omega^a}_c \wedge {\omega^c}_b - {\omega^a}_c \wedge {\Omega^c}_b$
And that's the second Bianchi identity

Exterior derivative of curvature two-form, expanded in index notation:
(Bianchi's second identity using index notation)
$e_a \otimes e^b \otimes d {\Omega^a}_b = e_a \otimes e^b \otimes ({\Omega^a}_c \wedge {\omega^c}_b - {\omega^a}_c \wedge {\Omega^c}_b)$
$e_a \otimes e^b \otimes (e^e \wedge e^c \wedge e^d) e_e ( {R^a}_{bcd} ) = e_a \otimes e^b \otimes (e^u \wedge e^d \wedge e^e {R^a}_{cud} {\Gamma^c}_{eb} - {\Gamma^a}_{ec} {R^c}_{bud} e^e \wedge e^u \wedge e^d)$
$e_a \otimes e^b \otimes (e^e \wedge e^c \wedge e^d) e_e ( {R^a}_{bcd} ) = e_a \otimes e^b \otimes (e^e \wedge e^c \wedge e^d) ( {R^a}_{uec} {\Gamma^u}_{db} - {\Gamma^a}_{eu} {R^u}_{bcd})$
$e_{[e} ( {R^a}_{b|cd]} ) = {R^a}_{u[ec} {\Gamma^u}_{d]b} - {\Gamma^a}_{[e|u} {R^u}_{b|cd]}$
$e_{[e} ( {R^a}_{b|cd]} ) = {R^a}_{u[cd} {\Gamma^u}_{e]b} - {R^u}_{b[cd} {\Gamma^a}_{e]u} $
TODO ... expand, and rewrite in terms of covariant derivative, and maybe you'll get $\nabla_{[e} ( {R^a}_{b|cd]}) = 0$, ideally with no torsion or commutation.
TODO then ... trace and get a better definition of $R_{[ab]}$
(Between this and the exterior derivative definition, I'm starting to notice a basis-elements-go-to-the-left pattern emerging...)

Exterior covariant derivative of curvature two-form:
(Bianchi's second identity, using exterior covariant derivative)
$D {\Omega^a}_b$
$= d {\Omega^a}_b + {\omega^a}_c \wedge {\Omega^c}_b - {\omega^c}_b \wedge {\Omega^a}_c$
$= d {\Omega^a}_b + {\omega^a}_c \wedge {\Omega^c}_b - {\Omega^a}_c \wedge {\omega^c}_b$
$= d (d {\omega^a}_b + {\omega^a}_c \wedge {\omega^c}_b) + {\omega^a}_c \wedge (d {\omega^c}_b + {\omega^c}_d \wedge {\omega^d}_b) - (d {\omega^a}_c + {\omega^a}_d \wedge {\omega^d}_c) \wedge {\omega^c}_b$
$= d^2 {\omega^a}_b + d {\omega^a}_c \wedge {\omega^c}_b - {\omega^a}_c \wedge d {\omega^c}_b + {\omega^a}_c \wedge d {\omega^c}_b + {\omega^a}_c \wedge {\omega^c}_d \wedge {\omega^d}_b - d {\omega^a}_c \wedge {\omega^c}_b - {\omega^a}_d \wedge {\omega^d}_c \wedge {\omega^c}_b $
$= d^2 {\omega^a}_b$
...assuming $d^2 = 0$
$D {\Omega^a}_b = 0$

Back