TODO make this 'examples' in the differential geometry worksheets.

Gauge covariant derivative:
$D_u = \partial_u - i q A_u$

Such that $D_u \phi = \partial_u \phi - i q A_u \phi$.

Covariant derivative definition:
$\nabla_a e_b = {\Gamma^c}_{ab} e_c$
$\nabla_a (v^u e_u) = e_a (v^u) e_u + v^u {\Gamma^b}_{au} e_b$
$= (e_a (v^u) + v^v {\Gamma^u}_{av} ) e_u$

First let's assume $e_a \approx \partial_a$, so $D_u (T) = e_u (T) - i q A_u \cdot T$.

Next, for the gauge covariant derivative, we find:
${\Gamma^a}_{bc} = -i q \delta^a_c A_b$
Such that $D_u (v^v e_v) = (e_u (v^v) - i q A_u v^v) e_v = (e_u (v^v) + v^a {\Gamma^v}_{ua} ) e_v = \nabla_u (v^v e_v)$

Commutation is not specified yet: ${c_{ab}}^c e_c = [e_a, e_b]$

But what is its torsion?
$T(x,y) = \nabla_x y - \nabla_y x - [x,y]$

Torsion components evaluate to be:
${T^a}_{bc} = 2 {\Gamma^a}_{[bc]} + {c_{cb}}^a$
${T^a}_{bc} = -2 i q \delta^a_{[c} A_{b]} + {c_{cb}}^a$.

Next, Riemann curvature:
$R(x,y) z = ([\nabla_x, \nabla_y] - \nabla_{[x,y]} ) z$

Has components:
${R^c}_{dab} = 2 e_{[a} ( {\Gamma^c}_{b]d} ) + 2 {\Gamma^c}_{[a| u} {\Gamma^u}_{b] d} - {\Gamma^c}_{ud} {c_{ab}}^u$
${R^c}_{dab} = 2 e_{[a} ( -i q A_{b]} \delta^c_d ) - 2 q^2 \delta^c_u A_{[a} A_{b]} \delta^u_d - i q A_u \delta^c_d {c_{ab}}^u$
${R^c}_{dab} = -2 i q e_{[a} ( A_{b]} ) \delta^c_d - i q \delta^c_d {c_{ab}}^u A_u$
${R^c}_{dab} = -i q \delta^c_d (2 e_{[a} ( A_{b]} ) + {c_{ab}}^u A_u )$

Let $F_{ab} = e_a (A_b) - e_b (A_a) + {c_{ab}}^u A_u$.
(or is the ${c_{ab}}^u$ included?)

${R^c}_{dab} = -i q \delta^c_d F_{ab}$

But then what about the antisymmetry on the first two indexes? For this connection, the Riemann curvature tensor is just zero.
And what of the symmetry between 12 and 34? That ... nope.

Ricci curvature: $R_{ab} = {R^c}_{acb} = -i q \delta^c_a F_{cb} = -i q F_{ab}$

And of course this means $R = 0$