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IV. Spin Connection
The coderivative of a tensor in the coordinate basis is given by its
partial derivative plus correction terms. The same procedure applies
to the tetrad basis $\left\{ {{{\mathbf{e}}_a}} \right\} = \left\{
{{{\mathbf{e}}_0},{{\mathbf{e}}_1},
{{\mathbf{e}}_2},{{\mathbf{e}}_3}} \right\}$, but the $\Gamma
$-connection coefficients (1.15) are replaced by the spin
connection coefficients denoted $\Sigma {{_\mu}{^b}}{_a}$:
\[ {D_\mu
}{{\mathbf{e}}_a} := \Sigma {{_\mu}{^b}}{_a}
{{\mathbf{e}}_b}{\quad}{D_\mu }{{\mathbf{e}}^a} := - \Sigma
{{_\mu}{^a}}{_b} {{\mathbf{e}}^b} \]
\[ \Sigma {{_\mu}{^b}}{_a}
= {{\mathbf{e}}^b} \cdot ({D_\mu }{{\mathbf{e}}_a}): =
{{\mathbf{e}}^b} \cdot {\Sigma _\mu }({{\mathbf{e}}_a}) \]
The name ‘spin connection’ comes from the fact that this formalism
can be used to take coderivatives of spinors. The generic name is ‘Lorentz
connection’ in recognition of the underlying Lorentz symmetry
of the tetrad basis. The index convention is taken from [Wikipedia:
Spin Connection].
By definition the spin connection coefficients are anti-symmetric,
${\Sigma _\mu }^{ab} = - {\Sigma _\mu }^{ba}$, in their internal
indices. This can be understood as following from the requirement
that the internal Minkowski metric be compatible with the
coderivative:
\[0 = {D_\mu }{\eta ^{ab}}: =
{D_\mu }({{\mathbf{e}}^a} \cdot {{\mathbf{e}}^b}) = - {\Sigma _\mu
}^{ab} - {\Sigma _\mu }^{ba}\]
The anti-symmetry reflects the fact that the spin connection is the
generator of a Lorentz transformation for each $\mu $.
Given that tetrads, at any given point, are orthonormal by
definition, the relationship between two tetrad frames is
necessarily a transformation of the group $\text{SO}(1,3)$.
Therefore, the basis vectors at two nearby events must be related to
each other by an infinitesimal Lorentz transformation:
\[ {{\mathbf{e'}}_a}(x + dx) =
{{\mathbf{e}}_a}(x) + d{x^\nu } {\Sigma
{{_\nu}{^b}}{_a}}(x){{\mathbf{e}}_b}(x)\]
which can be thought of as a rotation in spacetime keeping lenths
and angles the same: ${{\mathbf{e'}}_a}(x + dx) \cdot
{{\mathbf{e'}}_b}(x + dx) = {\eta _{ab}}$.