How does the spin covariant derivative $\nabla^S_{\mu}$ act on gamma matrices satisfying: $\{\gamma^{\mu},\gamma^{\nu}\} = g^{\mu\nu}$, i.e. $$\nabla^S_{\mu}\gamma^{\nu} = ?$$ where $\nabla^S := \partial + \omega^S$ is the spin covariant derivative associated to the spin connection $\omega^S$ on the spinor bundle $S \rightarrow M$.

3$\begingroup$ the answer you received on physics.stackexchange is not satisfactory? $\endgroup$– Carlo BeenakkerCommented Feb 11, 2018 at 23:42

$\begingroup$ @CarloBeenakker didn't get any, that's why posted here $\endgroup$– phydevCommented Feb 12, 2018 at 11:31
2 Answers
The question is not clearly stated, but if appropriately interpreted it does make sense.
Let $S$ be a spinor bundle over a pseudoRiemannian manifold $(M,g)$, whose bundle of Clifford algebras is denoted by $Cl(M,g)$. Since $S$ is a spinor bundle, there exists a morphism of bundles of associative, unital algebras:
$\gamma \colon Cl(M,g)\to End(S)$
to the endomorphisms of $S$. Let $\left\{ e_{\mu}\right\}$ be a local orthonormal frame of $TM$ over $U\subset M$. Using the injection $TM\subset Cl(M,g)$ we can apply $\gamma$ to $\left\{ e_{\mu}\right\}$:
$\gamma_{\mu} := \gamma(e_{\mu})\in End_U(S)\, ,$
obtaining what the physicists usually call the gamma matrices $\gamma_{\mu}$. Therefore, the gamma matrices are nothing but local sections of the endomorphism bundle of the spinor bundle obtained by Clifford multiplication with respect to an orthonormal frame.
Now, let $\nabla^S$ be a connection on $S$. To the question "how does $\nabla^S$ acts on the gamma matrices?" the answer is simple: it acts through the connection induced by $\nabla^S$ on the endomorphism bundle of $S$ by the usual formula, since we just saw that the gamma matrices are local sections of the endomorphism bundle of $S$. A more interesting question would be: Is $\nabla^S$ a module derivation with respect to Clifford multiplication?, that is, is the following formula satisfied:
$\nabla^S (\gamma(x) (\xi)) = (\nabla^S \gamma(x)) (\xi) + \gamma(x) (\nabla^S \xi) $
for all $x\in \Gamma(Cl(M,g))$ and $\xi\in \Gamma(S)$?. Here I am denoting by the same symbol the connection in $S$ and the induced connection on the endomorphism bundle. A more interesting situation happens when $\nabla^S$ is the lift of a metric connection $\nabla$ on $(M,g)$, which is the usual case in many applications. In this case, the natural question would be: is the following formula
$\nabla^S (\gamma(v) (\xi)) = ( \gamma(\nabla v)) (\xi) + \gamma(v) (\nabla^S \xi) $
satisfied for all $v\in \mathfrak{X}(M)\subset \Gamma(Cl(M,g))$ and $\xi\in \Gamma(S)$? That is, we wonder if $\nabla^S$ is a module derivation with respect to Clifford multiplication and in addition:
$\nabla^S \gamma(v) = \gamma(\nabla v)$
for all $v\in \mathfrak{X}(M)$. If both these conditions are satisfied, $\nabla^S$ is usually called a Clifford connection. For the case of $S$ being a spinor bundle associated to a spin structure and $\nabla^S$ being the lift of the LeviCivita connection $\nabla$, one can see that $\nabla^S$ is indeed a Clifford connection and thus:
$\nabla^S\gamma_{\mu} = \nabla^S\gamma(e_{\mu}) = \gamma(\nabla e_{\mu}) = \mathcal{A}_{\mu\nu} \otimes \gamma( e_{\nu}) = \mathcal{A}_{\mu\nu} \otimes \gamma_{\nu}$
where $\nabla e_{\mu} = \mathcal{A}_{\mu\nu} \otimes e_{\nu}$. Hence, the gamma matrices behave as vectors (or oneforms) with respect the LeviCivita connection when applying $\nabla^S$ and this tells you how the "spin covariant derivative" $\nabla^S$ acts on gamma matrices in the case of a Clifford connection lifting the LeviCivita connection, which is probably the situation of interest for the OP.
You may take a look at this paper (https://inspirehep.net/literature/144424), especially relations (6.7) and (6.10) where discuss covariant derivative of gamma matrices.

$\begingroup$ Hi Mojtaba, and welcome to math.MO Please avoid linkonly answers, since the link may disappear at some point. $\endgroup$ Commented Jul 27, 2020 at 14:43