*6.1. Affine Covariant Derivative*

We have built our setup by taking *ρ* to be the fundamental representation of O(3, 1), *P* = *FO*(*M*), and V = *FO*(*M*) ×*<sup>ρ</sup> V* to be the Minkowski bundle, as in (*ii*) of Observations 4. Therefore, as mentioned above, the isomorphism between Ω*<sup>k</sup> G* (*P*, *V*) and Ω*<sup>k</sup>* (*M*, *P* ×*<sup>ρ</sup> V*) allows to define an exterior covariant derivative of forms in Ω<sup>1</sup> (*M*, V):

$$d\_A: \Omega^k(M, \mathcal{V}) \to \Omega^{k+1}(M, \mathcal{V}).\tag{41}$$

In this way, we have a covariant differentiation for V-valued differential forms on *M* and, thus, also for tetrads, since *<sup>e</sup>* <sup>∈</sup> <sup>Ω</sup><sup>1</sup> (*M*, V).

Since we note that *<sup>d</sup><sup>ρ</sup>* induces a one form *<sup>d</sup>ρ*(*A*) <sup>∈</sup> <sup>Ω</sup><sup>1</sup> *M*, End(*V*) , we can further define another kind of derivative that "takes care" of internal indices only; in particular, this will not be necessarily a map between differential forms.

This derivative in components reads, for *<sup>φ</sup>* ∈ <sup>Γ</sup>(V),

$$(D\_A \phi)^a\_\mu = (\partial\_\mu \phi^a + A^{ac}\_\mu \eta\_{cb} \phi^b) \tag{42}$$

and, for *<sup>α</sup>* <sup>∈</sup> <sup>Ω</sup>*<sup>k</sup>* (*M*, V),

$$(D\_A \mathfrak{a})^{a}\_{\mu \nu\_1 \dots \nu\_k} = (\partial\_{\mu} \mathfrak{a}^{a}\_{\nu\_1 \dots \nu\_k} + A^{ac}\_{\mu} \eta\_{cb} \mathfrak{a}^{b}\_{\nu\_1 \dots \nu\_k})\_{\prime} \tag{43}$$

which shows that it does not map *α* to a differential form.

Now, we immediately apply the inverse of a tetrad to *DAφ* and identify it with ∇.

In fact, we take a vector field *<sup>X</sup>* <sup>∈</sup> <sup>Γ</sup>(*TM*), feed the tetrad *<sup>e</sup>* with it, then apply<sup>17</sup> *<sup>D</sup><sup>A</sup>* to get *<sup>D</sup>A*(*ιXe*), and finally pull it back with the inverse of the tetrad *e*¯.

In components, this reads as follows:

$$\left(D\_A(\iota\_X e)\right)^a\_\mu = D\_\mu(e^a\_\nu X^\nu) = \partial\_\mu(e^a\_\nu X^\nu) + \omega^{ab}\_\mu \eta\_{bc} e^c\_\nu X^\nu,\tag{44}$$

where, for reasons of metric compatibility<sup>18</sup> with *η*, we have the only antisymmetric part of the gauge field, for which we used the notation *A a <sup>µ</sup><sup>c</sup>* = *ωab <sup>µ</sup> ηbc*.

**Be aware:** Do not get confuse. We shall refer to *<sup>ω</sup>* <sup>∈</sup> <sup>Ω</sup><sup>1</sup> (*M*, <sup>Λ</sup>2V) as the *spin connection*. To stress that we want *D<sup>A</sup>* to depend on the spin connection only, we shall denote it with *Dω*.

Pulling back via *e*¯, we obtain

$$
\overline{e}\_a^{\sigma} \left( D\_\mu (e\_\nu^a X^\nu) \right) = \overline{e}\_a^{\sigma} \left( \partial\_\mu (e\_\nu^a X^\nu) + \omega\_\mu^{ab} \eta\_{bc} e\_\nu^c X^\nu \right). \tag{45}
$$

We define the Christoffel symbols Γ *σ µν* as

$$\begin{split} \Gamma^{\mathcal{T}}\_{\mu\nu} &= \overline{e}^{\mathcal{T}}\_{a} (D\_{\mu} e^{a}\_{\nu}) \\ &= \overline{e}^{\mathcal{T}}\_{a} (\partial\_{\mu} e^{a}\_{\nu} + \omega^{ab}\_{\mu} \eta\_{bc} e^{c}\_{\nu}) \end{split} \tag{46}$$

<sup>17</sup> Here, we use the so-called *interior product*, i.e., a map *ι<sup>ξ</sup>* : Ω*<sup>k</sup>* (*M*) <sup>→</sup> <sup>Ω</sup>*k*−<sup>1</sup> (*M*), such that (*ιξα*)(*X*1, ..., *Xk*−1) = *α*(*ξ*, *X*1, ..., *Xk*−1), for vector fields *ξ*, *X*1, ...*Xk*−<sup>1</sup> . Furthermore it respects *ι<sup>ξ</sup>* (*α* ∧ *β*) = (*ιξα*) ∧ *β* + (−1) *<sup>k</sup><sup>α</sup>* <sup>∧</sup> (*ι<sup>ξ</sup> <sup>β</sup>*), where *<sup>α</sup>* <sup>∈</sup> <sup>Ω</sup>*<sup>k</sup>* (*M*). Therefore, it forms an antiderivation. The relation with the Lie derivative is given by the formula L*ξα* = *d*(*ιξα*) + *ιξdα*, called the Cartan identity. The interior product of a commutator satisfies *ι* [*X*,*Y*] = [L*X*, *ιY*], with *X* and *Y* vector fields.

<sup>18</sup> In fact, imposing the condition of *DAη* = 0 implies the antisymmetry of the gauge field.

and, thus, we get

$$\nabla\_{\mu}X^{\sigma} := \overline{e}\_{a}^{\sigma} \left( D\_{\mu} (e\_{\nu}^{a}X^{\nu}) \right) = \partial\_{\mu}X^{\sigma} + \Gamma\_{\mu\nu}^{\sigma}X^{\nu} \,, \tag{47}$$

which is the covariant derivative well known in General Relativity.

We can also see what the curvature form is in terms of the commutator of two derivatives, given by the only antisymmetric part of the connection.

Then,

$$F\_{A^{anti}} := F\_{\omega} \tag{48}$$

and it is given by

$$\left(D\_{\left[\mu\right.}D\_{\left.\nu\right]}\Phi\right)^{a} = \left(\partial\_{\left[\mu\right.}\omega\_{\left.\nu\right]}\;^{ab} + \omega\_{\left[\mu\right.}^{ad}\eta\_{dc}\omega\_{\left.\nu\right]}\;^{cb}\right)\eta\_{bc}\Phi^{c} = F\_{\mu\nu}^{ab}\eta\_{bc}\Phi^{c},\tag{49}$$

where *A*[*µBν*] = *AµB<sup>ν</sup>* − *AνB<sup>µ</sup>* is our convention for the antisymmetrization. The fact that *F<sup>ω</sup>* is a 2-form shows that *F ab µν* = −*F ab νµ*; furthermore, metric compatibility ensures also *F ab µν* = −*F ba µν*, therefore *F<sup>ω</sup>* ∈ Ω2 *G* (*M*, <sup>Λ</sup>2V).

**Observation 10:** We see<sup>19</sup> that, here, the bundle metric *η* acts as a map *η* : Ω<sup>2</sup> (*M*, <sup>Λ</sup>2V) <sup>→</sup> <sup>Ω</sup><sup>2</sup> (*M*, End(*V*)) isomorphically; thus, it permits to identify elements of the second exterior power <sup>Λ</sup>2<sup>V</sup> with linear maps given by the fundamental representation of the algebra g = so(3, 1). We can introduce the notation for the wedge product in the fundamental representation as ∧*<sup>f</sup>* ; namely for, say, an *<sup>α</sup>* <sup>∈</sup> <sup>Ω</sup><sup>1</sup> (*M*, V), we have (*ω* ∧*<sup>f</sup> α*) *<sup>a</sup>* <sup>=</sup> *<sup>ω</sup>abηbc* <sup>∧</sup> *<sup>α</sup> c* .

### *6.2. Riemann Curvature Tensor*

We can now consider the commutator of two affine covariant derivatives and use Equation (47), getting

$$(\nabla\_{\left[\mu\right.} \nabla\_{\left.\nu\right]} X)^{\sigma} = \overline{e}\_{a}^{\sigma} \left( D\_{\left[\mu\right.} D\_{\left.\nu\right]} (\iota\_{X} e) \right)^{a} = \overline{e}\_{a}^{\sigma} F\_{\mu\nu}^{ab} \eta\_{bc} e\_{\omega}^{c} X^{\omega}. \tag{50}$$

We identify the Riemann tensor

$$R\_{\mu\nu\omega}{}^{\sigma} = \overline{e}\_a^{\sigma} F\_{\mu\nu}^{ab} \eta\_{b\varepsilon} \overline{e}\_{\omega\nu} \tag{51}$$

the Ricci curvature tensor

$$R\_{\mu\omega} = R\_{\mu\sigma\omega}{}^{\sigma} = \overline{e}\_{a}^{\sigma} F\_{\mu\sigma}^{ab} \eta\_{bc} e\_{\omega}^{c} \tag{52}$$

and thus the Ricci scalar

$$R = \mathcal{g}^{\mu\omega} R\_{\mu\omega} = \overline{e}\_d^{\mu} \overline{e}\_e^{\omega} \eta^{dc} \overline{e}\_a^{\sigma} F\_{\mu\sigma}^{ab} \eta\_{bc} e\_\omega^c = -\overline{e}\_a^{\mu} \overline{e}\_b^{\omega} F\_{\mu\omega}^{ab}. \tag{53}$$

It follows the antisymmetry of the Riemann tensor in the indices *µν* and *ωσ*, but it is important to note that we cannot ensure any symmetry in the Ricci curvature instead due to the presence of torsion.

### **7. Torsion**

Here, we start focusing on the importance of torsion, which arises quite naturally as curvature does.

<sup>19</sup> We denote the metric acting on the bundle and the metric acting on the fibers in the same way; however, we will usually deal with elements of the fibers.
