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Covariant derivative: $\nabla$

Covariant derivative basii decomposition: $\nabla = e^a \nabla_{e_a} = e^a \nabla_a$

Covariant derivative of scalar functions:
$\nabla \phi = e_a (\phi)$

Covariant of scalar multiplication:
$\nabla (\phi w) = (\nabla \phi) w + \phi \nabla w = e(\phi) w + \phi \nabla w$

Covariant of dot, inner, outer products:
$\nabla_{sx} = s \nabla x$ for function $s$ and vector $x$
$\nabla_x (a \cdot b) = (\nabla_x a) \cdot b + a \cdot (\nabla_x b)$ for vectors $a,b$.
$\nabla_x a(b) = \nabla_x a (b) + a(\nabla_x b)$ for one-form $a$ and vector $b$
$\nabla_x a \otimes b = (\nabla_x a) \otimes b + a \otimes (\nabla_x b)$

Covariant derivative of vector and one-form basii:
$0 = \nabla_b \delta^a_c = \nabla_b e^a (e_c) = (\nabla_b e^a) (e_c) + e^a (\nabla_b e_c)$
$(\nabla_b e^a) (e_c) = -e^a (\nabla_b e_c)$

Let ${\Gamma^a}_{bc} e_a = {\nabla_b} e_c$ be the linear relation of how a covariant derivative of one basis changes as it is moved along another basis.
$(\nabla_b e^a) (e_c) = -e^a ({\Gamma^d}_{bc} e_d)$ by substitution
$(\nabla_b e^a) (e_c) = -{\Gamma^d}_{bc} e^a (e_d)$ by linearity.
$(\nabla_b e^a) (e_c) = -{\Gamma^d}_{bc} \delta^a_d$ by orthogonality.
$(\nabla_b e^a) (e_c) = -{\Gamma^a}_{bc}$ by orthogonality.

Let ${\omega^a}_b = {\Gamma^a}_{cb} e^c$
${\omega^a}_b = -(\nabla_c e^a) (e_b) e^c$

So $e^a(\nabla_b e_c) = {\Gamma^a}_{bc}$ is the connection coefficients describing $\nabla_b e_c$ in terms of $e_a$.
And ${\omega^a}_b = -(\nabla_c e^a) (e_b) e^c = {\Gamma^a}_{cb} e^c$ are the one-forms of the connections.

Covariant derivative of a (p, q) tensor:
$\nabla_x y = \nabla_{x^u e_u} ({y^{a_1 ... a_p}}_{b_1 ... b_q} e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q})$
$= x^u \nabla_u ({y^{a_1 ... a_p}}_{b_1 ... b_q} e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q})$
$= x^u ( \nabla_u ({y^{a_1 ... a_p}}_{b_1 ... b_q}) e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q} + {y^{a_1 ... a_p}}_{b_1 ... b_q} e_{a_1} \otimes ... \otimes \nabla_u e_{a_i} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q} + {y^{a_1 ... a_p}}_{b_1 ... b_q} e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes \nabla_u e^{b_j} \otimes ... \otimes e^{b_q} )$
$= x^u ( e_u ({y^{a_1 ... a_p}}_{b_1 ... b_q}) e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q} + {y^{a_1 ... a_p}}_{b_1 ... b_q} e_{a_1} \otimes ... \otimes {\Gamma^c}_{u a_i} e_c \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q} + {y^{a_1 ... a_p}}_{b_1 ... b_q} e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes (- {\Gamma^{b_j}}_{uc} e^c) \otimes ... \otimes e^{b_q} )$
$= x^u ( e_u ({y^{a_1 ... a_p}}_{b_1 ... b_q}) e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q} + {\Gamma^{a_i}}_{uc} {y^{a_1 ... c ... a_p}}_{b_1 ... b_q} e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q} - {\Gamma^c}_{u b_j} {y^{a_1 ... a_p}}_{b_1 ... c ... b_q} e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q} )$
$= x^u ( e_u ({y^{a_1 ... a_p}}_{b_1 ... b_q}) + {\Gamma^{a_i}}_{uc} {y^{a_1 ... c ... a_p}}_{b_1 ... b_q} - {\Gamma^c}_{u b_j} {y^{a_1 ... a_p}}_{b_1 ... c ... b_q} ) e_{a_1} \otimes ... \otimes e_{a_p} \otimes e^{b_1} \otimes ... \otimes e^{b_q}$
This demonstrates the "add contravariant, subtract covariant" rule for the covariant derivative in index notation.
In multi-index notation: $\nabla_x y = \nabla_x ({y^A}_B {e_A}^B) = x^u ( e_u ({y^A}_B) + \underset{a_i \leftrightarrow c}{{y^A}_B} {\Gamma^{a_i}}_{uc} - \underset{b_j \leftrightarrow c}{{y^A}_B} {\Gamma^c}_{u b_j}) {e_A}^B$

Covariant derivative of vectors, specifically:
$\nabla_u v = \nabla_{u^a e_a} (v^b e_b) = u^a \nabla_a (v^b e_b) = u^a (e_a(v^b) + {\Gamma^b}_{ac} v^c) e_b$

Covariant derivative of one-forms:
$\nabla_u v = \nabla_{u^a e_a} (v_b e^b) = u^a \nabla_a (v_b e^b) = u^a (e_a(v_b) - {\Gamma^c}_{ab} v_c) e_b$

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Exterior derivative of a vector basis:
Consider vectors to be vector-valued 0-forms, i.e. 0-form scalars multiplied by a vector basis.
Then an exterior derivative of a vector is a vector-valued 1-form.
The coefficients of the relation of the exterior derivative of a vector basis to the basis is arbitrary, but we will define it to be the same as the covariant vector derivative coefficients:
$d e_b = e_a \wedge {\omega^a}_b$
Now we have previously defined the wedge product of a vector and a vector to be antisymmetric, and of a 1-form and a 1-form to be antisymmetric. But in the case of the wedge of a vector and a wedge product, we'll define it to be an outer product so that it is not symmetric (as is the case of the wedge of a 0-form (scalar) and a 1-form) nor antisymmetric (as is the cases we just described).
$d e_b = e_a \otimes {\omega^a}_b$
...and last, we equate the coefficients of our exterior derivative of a vector basis, our vector-valued 1-form, with our connection coefficients:
$= e_a \otimes {\Gamma^a}_{cb} e^c $
I put the vector basis on the left side, it'll come up later in my exterior derivative of a 1-1 basis element in the worksheet on torsion.

Also notice that now our exterior derivative needs to be equipped with an associated connection. There has been no specification as to what specific connection is needed. You don't necessarily have to use the Levi-Civita torsion-free connection with your exterior derivative. Should I start denoting which connection my exterior derivative is using? This means that just as you can have separate covariant derivatives with associated connection coefficients, you can also have separate exterior derivatives with their respective connection coefficients.

Exterior derivative of a vector:
$dv = d(v^a e_a) = dv^a \wedge e_a + v^a de_a$
using the rule that the wedge of a 1-form and a vector-valued 0-form is the outer product:
$= dv^a \otimes e_a + v^a de_a$
$= e_b(v^a) e^b \otimes e_a + v^a {\Gamma^b}_{ca} e_b \otimes e^c$
$= (e_b(v^a) + v^u {\Gamma^a}_{bu}) e_a \otimes e^b$
$= \nabla_b (v^a e_a) \otimes e^b$
$= \nabla v$

TODO nah put the basis on the right, and then you can only define $d (e^u \wedge e_u)$ instead of vice versa to be the torsion.
Or maybe the wedge of a vector valued 0 form is symmetric, and the inter-ordering of vectors and forms doesn't matter, only the ordering of vectors with respect to eachother and ordering of forms with respect to eachother?

Exterior derivative of a vector and p-form:
For vector $v$ and p-form $\alpha$,
let $d(v\alpha) = (dv) \wedge \alpha + v \otimes d\alpha$

Exterior and covariant derivative for 2-forms:
TODO explain this more.
$dF = \nabla \cdot\star F$, $d$ $\star{F} = \nabla \cdot F$

exterior derivative of metric tensor:
$ dg_{ij} = d (e_i \cdot e_j)$
$ = {de}_i \cdot e_j + e_i \cdot {de}_j$
$ = {\omega^k}_i \otimes e_k \cdot e_j + e_i \cdot {\omega^k}_j \otimes e_k$
$ = {\Gamma^k}_{li} e^l \otimes e_k \cdot e_j + e_i \cdot {\Gamma^k}_{lj} e^l \otimes e_k$
$ = {\Gamma^k}_{li} e^l g_{kj} + g_{ik} {\Gamma^k}_{lj} e^l$
$ = (\Gamma_{jki} + \Gamma_{ikj}) e^k$
$ = e_k (g_{ij}) e^k$

Exterior derivative of connection one-form:
$d {\omega^c}_d = d ({\Gamma^c}_{bd} e^b)$
$= (d {\Gamma^c}_{bd}) \wedge e^b + {\Gamma^c}_{bd} d e^b$
$= e_a ({\Gamma^c}_{bd}) e^a \wedge e^b - {\Gamma^c}_{bd} \cdot \frac{1}{2} {c_{uv}}^b e^u \wedge e^v$
$= (e_a ({\Gamma^c}_{bd}) - \frac{1}{2} {\Gamma^c}_{ud} {c_{ab}}^u) e^a \wedge e^b$
$= (2 e_{[a} ({\Gamma^c}_{b]d}) - {\Gamma^c}_{ud} {c_{ab}}^u) e^a \otimes e^b$

Second exterior derivative of connection one-forms:
$d^2 {\omega^c}_d = d (d {\omega^c}_d )$
$= d( (e_a ({\Gamma^c}_{bd}) - \frac{1}{2} {\Gamma^c}_{ud} {c_{ab}}^u) e^a \wedge e^b )$
$= e_e (e_a ({\Gamma^c}_{bd}) - \frac{1}{2} {\Gamma^c}_{ud} {c_{ab}}^u) e^e \wedge e^a \wedge e^b + (e_a ({\Gamma^c}_{bd}) - \frac{1}{2} {\Gamma^c}_{ud} {c_{ab}}^u) (de^a \wedge e^b - e^a \wedge de^b)$
$= e_e (e_a ({\Gamma^c}_{bd})) e^e \wedge e^a \wedge e^b - \frac{1}{2} e_e( {\Gamma^c}_{ud} {c_{ab}}^u) e^e \wedge e^a \wedge e^b + ( e_a ({\Gamma^c}_{bd}) - \frac{1}{2} {\Gamma^c}_{ud} {c_{ab}}^u ) \cdot (-\frac{1}{2} {c_{pq}}^a e^p \wedge e^q) \wedge e^b - ( e_a ({\Gamma^c}_{bd}) - \frac{1}{2} {\Gamma^c}_{ud} {c_{ab}}^u ) e^a \wedge (-\frac{1}{2} {c_{pq}}^b e^p \wedge e^q) $
$= ( \frac{1}{2} e_f ({\Gamma^c}_{bd}) {c_{ea}}^f - \frac{1}{2} e_e( {\Gamma^c}_{ud}) {c_{ab}}^u - \frac{1}{2} e_e( {c_{ab}}^u) {\Gamma^c}_{ud} ) e^e \wedge e^a \wedge e^b - \frac{1}{2} {c_{pq}}^a ( e_a ({\Gamma^c}_{bd}) - \frac{1}{2} {\Gamma^c}_{ud} {c_{ab}}^u ) e^p \wedge e^q \wedge e^b + \frac{1}{2} {c_{pq}}^b ( e_a ({\Gamma^c}_{bd}) - \frac{1}{2} {\Gamma^c}_{ud} {c_{ab}}^u ) e^a \wedge e^p \wedge e^q $
$= ( \frac{1}{2} e_u ({\Gamma^c}_{rd}) {c_{pq}}^u - \frac{1}{2} e_p( {\Gamma^c}_{ud}) {c_{qr}}^u - \frac{1}{2} e_p( {c_{qr}}^u) {\Gamma^c}_{ud} - \frac{1}{2} e_a ({\Gamma^c}_{rd}){c_{pq}}^a + \frac{1}{2} e_p ({\Gamma^c}_{bd}){c_{qr}}^b - \frac{1}{4} {c_{qr}}^b {\Gamma^c}_{ud} {c_{pb}}^u + \frac{1}{4} {c_{pq}}^a {\Gamma^c}_{ud} {c_{ar}}^u ) e^p \wedge e^q \wedge e^r $
$= ( - \frac{1}{2} e_p( {c_{qr}}^u) + \frac{1}{2} {c_{pq}}^v {c_{vr}}^u ) {\Gamma^c}_{ud} e^p \wedge e^q \wedge e^r $
Using $ e_{[p} ({c_{qr]}}^u) + {c_{[qr}}^v {c_{p]v}}^u = 0$
$= 0$

Exterior covariant derivative:
Let $D = d + \omega$
i.e. $D T = d T + \omega \wedge T$
...implicit wedge with p-forms...
...implicit composition with vectors...
...implicit sum when $\omega$ spans a basis...

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