gauge covariant derivative

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

Gauge covariant derivative:

D_{u} = \partial_{u} - i q A_{u}

Such that

D_{u} ϕ = \partial_{u} ϕ - i q A_{u} ϕ

.

Covariant derivative definition:

\nabla_{a} e_{b} = {Γ^{c}}_{a b} e_{c}

\nabla_{a} (v^{u} e_{u}) = e_{a} (v^{u}) e_{u} + v^{u} {Γ^{b}}_{a u} e_{b}

= (e_{a} (v^{u}) + v^{v} {Γ^{u}}_{a v}) 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:

{Γ^{a}}_{b c} = - i q δ_{c}^{a} 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} {Γ^{v}}_{u a}) e_{v} = \nabla_{u} (v^{v} e_{v})

Commutation is not specified yet:

{c_{a b}}^{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}}_{b c} = 2 {Γ^{a}}_{[b c]} + {c_{c b}}^{a}

{T^{a}}_{b c} = - 2 i q δ_{[c}^{a} A_{b]} + {c_{c b}}^{a}

.

Next, Riemann curvature:

R (x, y) z = ([\nabla_{x}, \nabla_{y}] - \nabla_{[x, y]}) z

Has components:

{R^{c}}_{d a b} = 2 e_{[a} ({Γ^{c}}_{b] d}) + 2 {Γ^{c}}_{[a | u} {Γ^{u}}_{b] d} - {Γ^{c}}_{u d} {c_{a b}}^{u}

{R^{c}}_{d a b} = 2 e_{[a} (- i q A_{b]} δ_{d}^{c}) - 2 q^{2} δ_{u}^{c} A_{[a} A_{b]} δ_{d}^{u} - i q A_{u} δ_{d}^{c} {c_{a b}}^{u}

{R^{c}}_{d a b} = - 2 i q e_{[a} (A_{b]}) δ_{d}^{c} - i q δ_{d}^{c} {c_{a b}}^{u} A_{u}

{R^{c}}_{d a b} = - i q δ_{d}^{c} (2 e_{[a} (A_{b]}) + {c_{a b}}^{u} A_{u})

Let

F_{a b} = e_{a} (A_{b}) - e_{b} (A_{a}) + {c_{a b}}^{u} A_{u}

.
(or is the

{c_{a b}}^{u}

included?)

{R^{c}}_{d a b} = - i q δ_{d}^{c} F_{a b}

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_{a b} = {R^{c}}_{a c b} = - i q δ_{a}^{c} F_{c b} = - i q F_{a b}

And of course this means

R = 0