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I asked this question https://physics.stackexchange.com/questions/820924/is-tetrad-postualte-independent-of-gauge-field

Here is what I know, $g_{\mu \nu} = e^{a}_{\mu} e^{b}_{\nu} \eta_{ab}$ and the tetrad postulate is \begin{equation} D_{\rho} e^{a}_{\nu} = \partial_{\rho} e^{a}_{\mu} - \Gamma^{\lambda}_{\rho \nu}e^{a}_{\lambda}+ \omega^{a}_{b \mu} e^{b}_{\nu}=0 \end{equation} But in case of abelian or non abelian Yang mills, $D_{\mu}$ is different and that should affect the tetrad postulate. In the same spirit what becomes of the statement that gamma matrices are covariantly constant. The covariant derivative in flat space-time for abelian gauge field is $D = \gamma^{\mu}(\partial_{\mu} + iA_{\mu})$ and $D \gamma^{\nu} =0$ here using $\partial_{\mu}\gamma^{\nu}=0$ we get $\gamma^{\mu} A_{\mu} \gamma^{\nu}=0$ what does it mean? I don't understand the underlying mathematical structure.

The answer I received is that for objects inavriant under gauge transformation, their covariant derivative does not contain gauge field so $D_{\mu} \gamma^{\nu}= \partial^{\mu} \gamma^{\nu} =0$. It makes sense but now I am looking for a more precise explanation about the gauge bundle we are working with and how connections act on the relevant objects.

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    $\begingroup$ You posted on physics stackexchange two hours ago. Give people a chance to reply for a couple of days before asking on another forum. Otherwise, you risk duplication of effort (we're not going to click back and forth all the time to check what's happening in the other forum), which is not fair to people investing time to think about your question. $\endgroup$ Commented Jul 9 at 14:02
  • $\begingroup$ hello, I understand but I thought requesting a more mathematical answer is more appropriate here. @MichaelEngelhardt $\endgroup$
    – trying
    Commented Jul 9 at 14:04

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For the first part of your post, related to tetrads, a tetrad is basically a smooth locally defined orthonormal coframe with respect to $g_{\mu \nu}$. More precisely, it consists of $4$ (assuming the dimension of spacetime is $4$) locally defined $1$-forms, $e^1_\mu, \dots, e^4_\mu$ which are orthonormal with respect to $g_{\mu \nu}$. Taking into account that the metric has Lorentzian signature, this precisely amounts to the first equation you wrote down, defining a tetrad.

You can also think of a tetrad $e^a_\mu$ at a point $p$ as defining a linear isometry map from $T_p$, with the metric $g_{\mu \nu}$, to $T_p$, with the Minkowski metric $\eta_{ab}$.

When you differentiate a tetrad, you need to use the Levi-Civita connection of $g_{\mu \nu}$ for the $g$ index of $e$ (namely the lower index of the tetrad) and you have a minus sign because you are differentiating a $1$-form, and you need to use the Levi-Civita connection of the Minkowski $\eta_{ab}$ metric to differentiate the $\eta$ index (the upper index of the tetrad). Note that if we were using local coordinates, then $\omega$ would be $0$ but, because we are using a tetrad instead, $\omega$ is in general non-zero.

For the second part of your post, to differentiate the $\gamma$s, note that each $\gamma$ is a linear map from $T_p$ to $\operatorname{End}(\mathscr{S}_p)$, the space of linear endomorphisms of the spinor bundle $\mathscr{S}_p$ at $p$. Thus each $\gamma$ has a "vector" type of index and two "spinor" type of indices. To differentiate the "vector" type index you use the Levi-Civita connection of $g_{\mu \nu}$, and to differentiate the two "spinor" type indices you use the spin connection.

As you can see, in both these cases we did not use the gauge connection. So, when do we use the gauge connection? A gauge connection is a connection on principal fiber bundle $E$ on spacetime. You use it to differentiate sections of "associated" bundles of $E$.

Edit: the OP seems to be interested in differentiating a composite object, i.e. a field, say $\phi$, which has different types of indices. For example, let us say that $\phi^{\nu j}$, where $\nu$ is a spacetime index and $j = 1, 2$ is an index corresponding to the trivial bundle $M \times \mathbb{C}^2$, corresponding to the gauge group $G = SU(2)$, where $M$ denotes our spacetime manifold, with metric $g_{\mu \nu}$. Assume that we have a gauge connection $A_\mu$ on the principal $SU(2)$ bundle $M \times SU(2)$. Note that, for each $\mu$, $A_\mu$ is an element in the Lie algebra of $SU(2)$ (more precisely in a bundle of Lie algebras of $SU(2)$, called the adjoint bundle, but this is a technicality, which we can omit for the time being). So, for each $\mu$, $A_{\mu}$ is a complex $2$ by $2$ skew-hermitian matrix for mathematicians, though for physicists $A_\mu$ would be a complex $2$ by $2$ hermitian matrix, but physicists would then write $i A_{\mu}$, so the two notations are equivalent, because $i$ times a hermitian matrix is a skew-hermitian matrix. But my point is, for each $\mu$, $A_\mu$ is a $2$ by $2$ matrix, and so it has two extra indices, say $j$ and $k$, with each of them having only two possible values, say $1$ and $2$.

With these remarks out of the way, differentiating the field $\phi$ using the covariant derivative $D$ gives the following.

$$D_\mu \phi^{\nu j} = \partial_\mu \phi^{\nu j} + \omega_{\mu ab} (\sigma^{ab})^\nu_{\phantom{\nu} \rho} \phi^{\rho j} + i (A_\mu)^j_{\phantom{j}k} \phi^{\nu k}.$$

I tried to use a notation close to the OP's notation, but then again, I don't know the author's conventions, so there may be silly factors missing from what I wrote. I only wrote it to hopefully help with the OP's understanding.

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  • $\begingroup$ thanks, I understand. So $D A_{\nu}$ for gauge field will have $D =\gamma^{\mu}( \partial_{\mu} + \omega_{\mu a b} \sigma^{ab} + i A_{\mu})$. $\endgroup$
    – trying
    Commented Jul 15 at 15:15
  • $\begingroup$ Hmm, there are really many connections appearing in your setting. We often use the same letter, such as $D$, to denote various different connections. Please distinguish between them. To differentiate a spinor, you use the spin connection, to differentiate a tensor, you use the Levi-Civita connection, to differentiate a section of a vector bundle, associated to a gauge group $G$, you use a gauge connection. We may use the same letter $D$ in all 3 cases, but they are really different connections/objects. $\endgroup$
    – Malkoun
    Commented Jul 15 at 17:54
  • $\begingroup$ The $\gamma^\mu \partial_\mu$ is (up to a factor of $i$) the Dirac operator. Please distinguish between all of these. $\endgroup$
    – Malkoun
    Commented Jul 15 at 17:57
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    $\begingroup$ Regarding the Dirac operator vs covariant derivative, well, the Dirac operator is $i$ times the Gamma matrices contracted with the covariant derivative. It does not have any remaining free index, unlike the covariant derivative, and is usually denoted by $D$ with a slash through it. $\endgroup$
    – Malkoun
    Commented Jul 17 at 18:24
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    $\begingroup$ Regarding the $\sigma^{ab} = [\gamma^a, \gamma^b]$, for $1 \leq a < b \leq 4$, those form a basis for the Lie algebra of $O(1, 3)$, but note for each $(a, b)$ with $a < b$, $\sigma^{ab}$ is a $4$ by $4$ matrix, so it has two additional spacetime indices, which are hidden from the notation. $\endgroup$
    – Malkoun
    Commented Jul 17 at 18:48

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