Back
Reminder: different representations of a p-form:
$w$
$= w_{|a_1 ... a_p|} dx^{|a_1} \wedge ... \wedge dx^{a_p|}$
$= \frac{1}{p!} w_{a_1 ... a_p} dx^{a_1} \wedge ... \wedge dx^{a_p}$
$= \frac{1}{p!} w_{[a_1 ... a_p]} dx^{[a_1} \wedge ... \wedge dx^{a_p]}$
$= w_{[a_1 ... a_p]} dx^{[a_1} \otimes ... \otimes dx^{a_p]}$
Exterior derivative in a coordinate basis
Rule of exterior derivative: $d(dx^a) = d^2 x^a = 0$
Exterior derivative in a coordinate basis:
$d = dx^a \wedge \partial_a(...)$
Exterior derivative of a scalar:
$d (\phi) = \partial_a (\phi) dx^a$
Exterior derivative of one-form in a coordinate basis:
$dw = d(w_b {dx}^b)$
$= dw_b \wedge {dx}^b + w_b d^2{x}^b$
using $d^2 x^a = 0$
$= dw_b \wedge {dx}^b$
$= \partial_a w_b {dx}^a \wedge {dx}^b$
$= (\partial_a w_b - \partial_b w_a) {dx}^{|a} \wedge {dx}^{b|}$
Exterior derivative of a p-form:
$d w$
$= d (w_{|a_1 ... a_p|} dx^{|a_1} \wedge ... \wedge dx^{a_p|})$
$= \frac{1}{p!} d (w_{a_1 ... a_p} dx^{a_1} \wedge ... \wedge dx^{a_p})$
$= \frac{1}{p!} (
(d w_{a_1 ... a_p}) dx^{a_1} \wedge ... \wedge dx^{a_p}
+ (-1)^{k-1} w_{a_1 ... a_p} dx^{a_1} \wedge ... \wedge d^2x^{a_k} \wedge ... \wedge dx^{a_p}
)$
using $d^2 x^a = 0$
$= \frac{1}{p!} (d w_{a_1 ... a_p}) dx^{a_1} \wedge ... \wedge dx^{a_p}$
using exterior derivative of a scalar:
$= \frac{1}{p!} \partial_b w_{a_1 ... a_p} dx^b \wedge dx^{a_1} \wedge ... \wedge dx^{a_p}$
$= \frac{(p+1)!}{p!} \partial_{|b} w_{a_1 ... a_p|} dx^{|b} \wedge dx^{a_1} \wedge ... \wedge dx^{a_p|}$
$= p \partial_{|b} w_{a_1 ... a_p|} dx^{|b} \wedge dx^{a_1} \wedge ... \wedge dx^{a_p|}$
... a better looking result would be:
$= \partial_b w_{|a_1 ... a_p|} dx^b \wedge dx^{|a_1} \wedge ... \wedge dx^{a_p|}$
NOTICE this definition is a bit circular, it uses the definition of $d (a \wedge b)$ for p-forms a and b to define $da$.
Equivalently the definition of $da$ is used to prove $d (a \wedge b)$.
This circular reference can be ignored because we are assuming there is some linearity involved in applying the exterior derivative to wedges of the 0-form coefficients and the 1-form basis that is wedged together to form the p-form basis.
From there, we are applying the exterior derivative to each of the one-form basis and using $d^2 x^a = 0$ to eliminate all these terms, so any coefficient in front of the (which happens to be $(-1)^{p-1}$ is irrelevant.
Exterior derivative of a (p q) tensor:
$d ( {T^{a_1 ... a_p}}_{b_1 ... b_q} \partial_{a_1} \otimes ... \otimes \partial_{a_p} \otimes dx^{b_1} \otimes ... \otimes dx^{b_q} )$
$= \partial_a ( {T^{a_1 ... a_p}}_{b_1 ... b_q} ) dx^a \wedge (\partial_{a_1} \otimes ... \otimes \partial_{a_p} \otimes dx^{b_1} \otimes ... \otimes dx^{b_q})$
TODO what is the wedge of a vector basis and a one-form basis element? Leaning towards it being an outer product. Especially since the wedge of a vector and a vector is symmetric. The wedge of a form and a form is symmetric. And the exterior derivative of the vector basis 0-form is defined as the connection, so as to not introduce unnecessary symmetries (a 0-form wedge is symmetric) or antisymmetries (if we define the wedge of a vector and a form to be symmetric, as the vector-and-vector and form-and-form are), all that's left is defining it to be the outer product.
Exterior derivative of wedge product of p-form $a$ and q-form $b$:
$d( a \wedge b ) $
expand both:
$ = d ( a_{i_1... i_p} dx^{i_1} \wedge ... \wedge dx^{i_p} \wedge b_{j_1 ... j_q} dx^{j_1} \wedge ... \wedge dx^{j_q})$
move coefficients to the left
$ = d ( a_{i_1... i_p} b_{j_1 ... j_q} dx^{i_1} \wedge ... \wedge dx^{i_p} \wedge dx^{j_1} \wedge ... \wedge dx^{j_q})$
distribute exterior derivative through coefficients and through the form basis (is this using this definition to prove this definition?):
$ = d ( a_{i_1... i_p} b_{j_1 ... j_q}) \wedge dx^{i_1} \wedge ... \wedge dx^{i_p} \wedge dx^{j_1} \wedge ... \wedge dx^{j_q}
+ (-1)^{k-1} a_{i_1... i_p} b_{j_1 ... j_q} dx^{i_1} \wedge ... \wedge ddx^{i_k} \wedge ... \wedge dx^{i_p} \wedge dx^{j_1} \wedge ... \wedge dx^{j_q}
+ (-1)^{p+l-1} a_{i_1... i_p} b_{j_1 ... j_q} dx^{i_1} \wedge ... \wedge dx^{i_p} \wedge dx^{j_1} \wedge ... \wedge ddx^{j_l} \wedge ... \wedge dx^{j_q}$
substitute $d^2 x^a = 0$, substitute definition of exterior derivative for scalars
$ = \partial_k ( a_{i_1... i_p} b_{j_1 ... j_q}) dx^k \wedge dx^{i_1} \wedge ... \wedge dx^{i_p} \wedge dx^{j_1} \wedge ... \wedge dx^{j_q}$
distribute partial derivative between the scalar product
$ = \left(
\partial_k(a_{i_1... i_p}) b_{j_1 ... j_q}
+ a_{i_1... i_p} \partial_k(b_{j_1 ... j_q})
\right)
dx^k \wedge dx^{i_1} \wedge ... \wedge dx^{i_p} \wedge dx^{j_1} \wedge ... \wedge dx^{j_q}$
distribute wedges through the parenthesis
$ = \partial_k(a_{i_1... i_p}) b_{j_1 ... j_q}
dx^k \wedge dx^{i_1} \wedge ... \wedge dx^{i_p} \wedge dx^{j_1} \wedge ... \wedge dx^{j_q}
+ a_{i_1... i_p}
\partial_k(b_{j_1 ... j_q})
dx^k \wedge dx^{i_1} \wedge ... \wedge dx^{i_p} \wedge dx^{j_1} \wedge ... \wedge dx^{j_q}$
for the right sum, exchange the partial's associated one-form basis p times to the right using $dx^a \wedge dx^b = -dx^b \wedge dx^a$, to separate the 'b' term.
$ = \partial_k(a_{i_1... i_p})
dx^k \wedge dx^{i_1} \wedge ... \wedge dx^{i_p}
\wedge b_{j_1 ... j_q} dx^{j_1} \wedge ... \wedge dx^{j_q}
+ (-1)^p a_{i_1... i_p} dx^{i_1} \wedge ... \wedge dx^{i_p} \wedge
\partial_k(b_{j_1 ... j_q})
dx^k \wedge dx^{j_1} \wedge ... \wedge dx^{j_q}$
substitute definitions of $a, b, da, db$
$= da \wedge b + (-1)^p a \wedge db$
Second exterior derivative in a coordinate basis
TODO should I put this up higher so I can make use of it in other places?
second exterior derivative of a p-form in a coordinate basis:
$ {d}({dw}) = {dd} ({w}_{a_1 ... a_p} {dx}^{a_1} \wedge ... \wedge {dx}^{a_p})$
$= {d} \partial_b ({w}_{a_1 ... a_p}) {dx}^b \wedge {dx}^{a_1} \wedge ... \wedge {dx}^{a_p}$
$= \partial_c \partial_b ({w}_{a_1 ... a_p}) {dx}^c \wedge {dx}^b \wedge {dx}^{a_1} \wedge ... \wedge {dx}^{a_p}$
Now we have the sum over the symmetric $\partial_c \partial_b$ and the antisymmetric ${dx}^b \wedge {dx}^c$, so our sum is zero.
${ddw} = 0$
Exterior derivative in a non-coordinate basis
Exterior derivative of a scalar:
$d (\phi) = \partial_\tilde{a} (\phi) dx^\tilde{a}$
$= {e_\tilde{a}}^a e_a (\phi) {e^\tilde{a}}_b e^b$
$= \delta^a_b e_a (\phi) e^b$
$= e_a (\phi) e^a$
Exterior derivative of a non-coordinate basis one-form:
(Treating the one-form basis as a linear combination of a coordinate basis)
$d e^u = d ({e^u}_\tilde{b} dx^\tilde{b})$
$= d ({e^u}_\tilde{b}) \wedge dx^\tilde{b} + {e^u}_\tilde{b} d(dx^\tilde{b})$
$= \partial_\tilde{a} ({e^u}_\tilde{b}) dx^\tilde{a} \wedge dx^\tilde{b}$
$= e_a ({e^u}_\tilde{b}) {e_b}^\tilde{b} e^a \wedge e^b$
$= e_a ({e^u}_\tilde{b}) {e_b}^\tilde{b} (e^a \otimes e^b - e^b \otimes e^a)$
$= (e_a ({e^u}_\tilde{b}) {e_b}^\tilde{b}
- e_b ({e^u}_\tilde{b}) {e_a}^\tilde{b}
) e^a \otimes e^b$
$= -(e_a ({e_b}^\tilde{b}) - e_b ({e_a}^\tilde{b})) {e^v}_\tilde{b} e^a \otimes e^b$
$= -{c_{ab}}^u e^a \otimes e^b$
$= -\frac{1}{2}{c_{ab}}^u e^a \wedge e^b$
$= -c^u$
$d e^u + c^u = 0$
Exterior derivative of a p-form, starting with the coordinate basis definition, expressed in a non-coordinate basis:
$d w$
$= d (w_{|a_1 ... a_p|} dx^{|a_1} \wedge ... \wedge dx^{a_p|})$
$= \frac{1}{p!} d (w_{a_1 ... a_p} dx^{a_1} \wedge ... \wedge dx^{a_p})$
$= \frac{1}{p!} (\partial_b w_{a_1 ... a_p} dx^b \wedge dx^{a_1} \wedge ... \wedge dx^{a_p})$
TODO - continue on to rewrite the basis in terms of $e^a$
$d T = (d T)_{i_1 ... i_{p+1}} e^{i_1} \otimes ... \otimes e^{i_{p+1}}$
$= e_{j_1} (T_{j_2 ... j_{p+1}}) \delta^{j_1 ... j_{p+1}}_{i_1 ... i_{p+1}} e^{i_1} \otimes ... \otimes e^{i_{p+1}}$
$= (p+1)! \cdot e_{[i_1} ( T_{i_2 ... i_{p+1}]} ) e^{i_1} \otimes ... \otimes e^{i_{p+1}}$
$= e_{i_1} ( T_{i_2 ... i_{p+1}} ) e^{i_1} \wedge ... \wedge e^{i_{p+1}}$
$= e^{i_1} \wedge e_{i_1} ( T_{i_2 ... i_{p+1}} ) e^{i_2} \wedge ... \wedge e^{i_{p+1}})$
$= e^a \wedge e_a (T)$
TODO again ... $e_a (T) = e_a (T_{i_1 ... i_p} e^1 \otimes ... \otimes e^p)$ ... $ = e_a (T_{i_1 ... i_p}) e^1 \otimes ... \otimes e^p$.
Exterior derivative of a p-form:
$d w = d (\frac{1}{p!} w_{i_1 ... i_p} e^{i_1} \wedge ... \wedge e^{i_p})$
$= \frac{1}{p!} d (w_{i_1 ... i_p}) \wedge e^{i_1} \wedge ... \wedge e^{i_p}
+ (-1)^{j-1} \frac{1}{p!} w_{i_1 ... i_p} e^{i_1} \wedge ... \wedge d(e^{i_j}) \wedge ... \wedge e^{i_p}
$
...in a coordinate basis, where $e^\tilde{a} = dx^\tilde{a}$ and $d(dx^\tilde{a}) = 0$:
$= \frac{1}{p!} \partial_j (w_{i_1 ... i_p}) dx^j \wedge dx^{i_1} \wedge ... \wedge dx^{i_p}$
$= \frac{(p+1)!}{p!} \partial_j (w_{i_1 ... i_p}) dx^{[j} \otimes dx^{i_1} \otimes ... \otimes dx^{i_p]}$
...in a non-coordinate basis:
$dw = e_k (\frac{1}{p!} w_{i_1 ... i_p}) e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p} + (-1)^{j-1} w_{i_1 ... i_p} e^{i_1} \wedge ... \wedge d(e^{i_j}) \wedge ... \wedge e^{i_p}$
$ = \frac{1}{p!} e_k (w_{i_1 ... i_p}) e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p}
+ (-1)^{j-1} \frac{1}{p!} w_{i_1 ... i_p} e^{i_1} \wedge ... \wedge (-\frac{1}{2} {c_{ab}}^{i_j} e^a \wedge e^b) \wedge ... \wedge e^{i_p}
$
$ = \frac{1}{p!} e_k (w_{i_1 ... i_p}) e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p}
- \frac{1}{2} \frac{1}{p!} {c_{ab}}^{i_j} w_{i_1 ... i_p} e^a \wedge e^{i_1} \wedge ... \wedge e^b \wedge ... \wedge e^{i_p}
$
$ = \frac{1}{p!} (
e_k (w_{i_1 ... i_p})
- \frac{1}{2} {c_{k i_j}}^b w_{i_1 ... i_{j-1} b i_{j+1} ... i_p}
) e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p}
$
Exterior derivative of a wedge of basis forms:
$d (e^{i_1} \wedge ... \wedge e^{i_p})$
$= (-1)^{j-1} e^{i_1} \wedge ... \wedge d(e^{i_j}) \wedge ... \wedge e^{i_p} $
$= (-1)^{j-1} e^{i_1} \wedge ... \wedge e^{i_{j-1}} \wedge c^{i_j} \wedge e^{i_{j+1}} \wedge ... \wedge e^{i_p}$
$= (-1)^{j-1} e^{i_1} \wedge ... \wedge e^{i_{j-1}} \wedge (-\frac{1}{2} {c_{ab}}^{i_j} e^a \wedge e^b) \wedge e^{i_{j+1}} \wedge ... \wedge e^{i_p}$
$= -\frac{1}{2} {c_{ab}}^{i_j} (-1)^{j-1} e^{i_1} \wedge ... \wedge e^{i_{j-1}} \wedge e^a \wedge e^b \wedge e^{i_{j+1}} \wedge ... \wedge e^{i_p}$
$= \frac{1}{2} {c_{ab}}^{i_j} (-1)^j e^{i_1} \wedge ... \wedge e^{i_{j-1}} \wedge e^a \wedge e^b \wedge e^{i_{j+1}} \wedge ... \wedge e^{i_p}$
$= \frac{1}{2} {c_{ab}}^{i_j} (-1)^{2j-1} e^a \wedge e^{i_1} \wedge ... \wedge e^{i_{j-1}} \wedge e^b \wedge e^{i_{j+1}} \wedge ... \wedge e^{i_p}$
$= \frac{1}{2} {c_{ab}}^{i_j} (-1)^{3j-2} e^a \wedge e^b \wedge e^{i_1} \wedge ... \wedge e^{i_{j-1}} \wedge e^{i_{j+1}} \wedge ... \wedge e^{i_p}$
$= \frac{1}{2} {c_{ab}}^{i_j} (-1)^j e^a \wedge e^b \wedge e^{i_1} \wedge ... \wedge e^{i_{j-1}} \wedge e^{i_{j+1}} \wedge ... \wedge e^{i_p}$
$= -\frac{1}{2} {c_{ab}}^{i_j} (-1)^{j-1} e^a \wedge e^b \wedge e^{i_1} \wedge ... \wedge e^{i_{j-1}} \wedge e^{i_{j+1}} \wedge ... \wedge e^{i_p}$
$= (-1)^{j-1} c^{i_j} \wedge e^{i_1} \wedge ... \wedge e^{i_{j-1}} \wedge e^{i_{j+1}} \wedge ... \wedge e^{i_p}$
Exterior derivative of wedge product of p-form $a$ and q-form $b$:
$d( a \wedge b ) $
$ = d ( a_{i_1... i_p} e^{i_1} \wedge ... \wedge e^{i_p} \wedge b_{j_1 ... j_q} e^{j_1} \wedge ... \wedge e^{j_q})$
$ = d ( a_{i_1... i_p} b_{j_1 ... j_q} e^{i_1} \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge e^{j_q})$
$ = d ( a_{i_1... i_p} b_{j_1 ... j_q}) \wedge e^{i_1} \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge e^{j_q}
+ (-1)^{k-1} a_{i_1... i_p} b_{j_1 ... j_q} e^{i_1} \wedge ... \wedge de^{i_k} \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge e^{j_q}
+ (-1)^{p+l-1} a_{i_1... i_p} b_{j_1 ... j_q} e^{i_1} \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge de^{j_l} \wedge ... \wedge e^{j_q}$
$ = e_k ( a_{i_1... i_p} b_{j_1 ... j_q}) e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge e^{j_q}
+ (-1)^{k-1} a_{i_1... i_p} b_{j_1 ... j_q} e^{i_1} \wedge ... \wedge (-\frac{1}{2} {c_{ab}}^{i_k} e^a \wedge e^b) \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge e^{j_q}
+ (-1)^{p+l-1} a_{i_1... i_p} b_{j_1 ... j_q} e^{i_1} \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge (-\frac{1}{2} {c_{ab}}^{j_l} e^a \wedge e^b) \wedge ... \wedge e^{j_q}$
$ =
\left( e_k(a_{i_1... i_p}) b_{j_1 ... j_q}
+ a_{i_1... i_p} e_k(b_{j_1 ... j_q}) \right)
e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge e^{j_q}
- \frac{1}{2} {c_{ab}}^{i_k} a_{i_1... i_p} b_{j_1 ... j_q} e^a \wedge e^{i_1} \wedge ... \wedge e^b \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge e^{j_q}
- \frac{1}{2} {c_{ab}}^{j_l} a_{i_1... i_p} b_{j_1 ... j_q} e^a \wedge e^{i_1} \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge e^b \wedge ... \wedge e^{j_q}$
$ = (
e_k(a_{i_1... i_p}) b_{j_1 ... j_q}
+ a_{i_1... i_p} e_k(b_{j_1 ... j_q})
- \frac{1}{2} {c_{k i_k}}^b a_{i_1 ... i_{k-1} b i_{k+1} ... i_p} b_{j_1 ... j_q}
- \frac{1}{2} {c_{k j_l}}^b a_{i_1... i_p} b_{j_1 ... j_{l-1} b j_{l+1} ... j_q}
) e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge e^{j_q}$
$ = (e_k(a_{i_1... i_p})
- \frac{1}{2} {c_{k i_k}}^b a_{i_1 ... i_{k-1} b i_{k+1} ... i_p}
) b_{j_1 ... j_q}
e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge e^{j_q}
+ a_{i_1... i_p} (
e_k(b_{j_1 ... j_q})
- \frac{1}{2} {c_{k j_l}}^b b_{j_1 ... j_{l-1} b j_{l+1} ... j_q}
) e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p} \wedge e^{j_1} \wedge ... \wedge e^{j_q}$
$ = (e_k(a_{i_1... i_p})
- \frac{1}{2} {c_{k i_k}}^b a_{i_1 ... i_{k-1} b i_{k+1} ... i_p}
)
e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p}
\wedge b_{j_1 ... j_q} e^{j_1} \wedge ... \wedge e^{j_q}
+ (-1)^p a_{i_1... i_p} e^{i_1} \wedge ... \wedge e^{i_p} \wedge (
e_k(b_{j_1 ... j_q})
- \frac{1}{2} {c_{k j_l}}^b b_{j_1 ... j_{l-1} b j_{l+1} ... j_q}
) e^k \wedge e^{j_1} \wedge ... \wedge e^{j_q}$
$= da \wedge b + (-1)^p a \wedge db$
Exterior derivative of a (p q) tensor:
$d ( {T^{a_1 ... a_p}}_{b_1 ... b_q} e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q} )$
$= e_a ( {T^{a_1 ... a_p}}_{b_1 ... b_q} ) e^a \wedge (e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q})$
TODO what about subsequent applications to $d e^{a_i}$ and $d e_{b_j}$?
Exterior derivative of inner product:
TODO under what context is this defined? Two vectors, i.e. vector-valued 0-forms? Or as two one-forms, therefore using the hodge dual definition of inner product?
What exactly is an inner product, and should I define inner- as between vectors ... vs interior- as between forms and vectors, ... vs what (inner again?) between two forms?
$ d(u \cdot v) = du \cdot v + u \cdot dv$
Second exterior derivative in a non-coordinate basis:
Second exterior derivative of a 1-form in a non-coordinate basis:
$d^2 e^a$
$= d (-\frac{1}{2} {c_{bc}}^a e^b \wedge e^c)$
$= -\frac{1}{2} \left(
d ({c_{bc}}^a) e^b \wedge e^c
+ {c_{bc}}^a d (e^b \wedge e^c)
\right)$
$= -\frac{1}{2} \left(
d ({c_{bc}}^a) e^b \wedge e^c
+ {c_{bc}}^a (d e^b \wedge e^c - e^b \wedge de^c)
\right)$
$= -\frac{1}{2} \left(
\partial_d ({c_{bc}}^a) dx^d \wedge e^b \wedge e^c
+ {c_{bc}}^a (
(-\frac{1}{2} {c_{de}}^b e^d \wedge e^e) \wedge e^c
- e^b \wedge (-\frac{1}{2} {c_{de}}^c e^d \wedge e^e)
)
\right)$
$= -\frac{1}{2} (
e_d ({c_{ec}}^a)
- {c_{bc}}^a {c_{de}}^b
) e^d \wedge e^e \wedge e^c$
$= -3 (
e_{[d} ({c_{ec]}}^a)
+ {c_{[de}}^b {c_{c]b}}^a
) e^d \otimes e^e \otimes e^c$
...but in our coordinate basis, $d^2 w = 0$ for p-form w...
...therefore:
$
e_{[d} ({c_{ec]}}^a)
+ {c_{[de}}^b {c_{c]b}}^a
= 0$
$e_{[d} ({c_{ec]}}^a)
= - {c_{[de}}^b {c_{c]b}}^a
= {c_{b[c}}^a {c_{de]}}^b
$
TODO isn't there another property like Jacobi's identity that proves this as well?
If it's not ... then does this mean there is a separate definition for exterior derivatives in coordinate and in non-coordinate basis?
Just like the exterior derivative on vectors must be defined to be associated with a specific connection, does the exterior derivative also need to be associated with a specific (non-coordinate) basis?
Second exterior derivative of a p-form in a non-coordinate basis:
$d^2(w) = d(
(
e_k (w_{i_1 ... i_p})
- \frac{1}{2} {c_{k i_j}}^b w_{i_1 ... i_{j-1} \ b \ i_{j+1} ... i_p}
) e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p}
)$
$=
d (
e_k (w_{i_1 ... i_p})
- \frac{1}{2} {c_{k i_j}}^b w_{i_1 ... i_{j-1} \ b \ i_{j+1} ... i_p}
) \wedge e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p}
+ (
e_k (w_{i_1 ... i_p})
- \frac{1}{2} {c_{k i_j}}^b w_{i_1 ... i_{j-1} \ b \ i_{j+1} ... i_p}
) d(e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p})
$
$=
e_l (
e_k (w_{i_1 ... i_p})
- \frac{1}{2} {c_{k i_j}}^b w_{i_1 ... i_{j-1} \ b \ i_{j+1} ... i_p}
) e^l \wedge e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p}
+ (
e_k (w_{i_1 ... i_p})
- \frac{1}{2} {c_{k i_j}}^b w_{i_1 ... i_{j-1} \ b \ i_{j+1} ... i_p}
) d(e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p})
$
$=
(
e_l (e_k (w_{i_1 ... i_p}))
- \frac{1}{2}
e_l ({c_{k i_j}}^b)
w_{i_1 ... i_{j-1} \ b \ i_{j+1} ... i_p}
- \frac{1}{2}
{c_{k i_j}}^b
e_l (w_{i_1 ... i_{j-1} \ b \ i_{j+1} ... i_p})
) e^l \wedge e^k \wedge e^{i_1} \wedge ... \wedge e^{i_p}
+ (
e_k (w_{i_1 ... i_p})
- \frac{1}{2} {c_{k i_j}}^b w_{i_1 ... i_{j-1} \ b \ i_{j+1} ... i_p}
)
(
-\frac{1}{2} {c_{mn}}^k e^m \wedge e^n \wedge e^{i_1} \wedge ... \wedge e^{i_p}
- e^k \wedge d (e^{i_1} \wedge ... \wedge e^{i_p})
)
$
TODO finishme.
I bet you end up with $e_i (e_j(...)) - {c_{ij}}^k ...$, the first turns to c's, and somehow they all cancel out.
Also maybe $e(c)$ goes away because it is really just $e_a([e_b,e_c]) e^a \wedge e^b \wedge e^c = (e_a(e_b(e_c)) - e_a(e_c(e_b))) e^a \wedge e^b \wedge e^c$ so we have symmetric and antisymmetric summed which should cancel.
TODO TODO what about exterior derivatives of vectors?
Are they antisymmetric, as wedge product of two vectors (vector-valued 0-forms) is antisymmetric?
Or are they symmetric, as the connection is symmetric?
Does the definition of $d e_a = {\omega^b}_a \wedge e_b = {\Gamma^b}_{ca} e^c \wedge e_b$ allow for symmetry?
What's the (anti)symmetry of a vector and a one-form? $e^a \wedge e_b = (-1)^? e_b \wedge e^a$? What does the generalized Kronecker delta definition of wedge product say about this?
The definition of (p q) tensor a wedge (r s) tensor b is $a \wedge b = (-1)^{pr + qs} b \wedge a$ would imply that a one-form wedge a vector is symmetric... and therefore $de$ is symmetric as well...
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