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There are many similarities between the gauging of a $U(1)$ symmetry to obtain the physics of electromagnetism and the gauging of diffeomorphisms to obtain the physics of general relativity. In particular, in both cases one can obtain a covariant derivative: for electromagnetism,

$$ \nabla_\mu = \partial_\mu + i q A_\mu $$

where $A_\mu$ is the gauge field and for general relativity

$$ D_\mu = \partial_\mu + \Gamma_\mu^{\;ij}f_{ji} $$ where $\Gamma_\mu^{\;ij}$ is the connection and $f_{ji}$ is the generator of rotations.

However, these covariant derivatives behave quite differently: whilst in general relativity we have a nice 'distributive' property,

\begin{align} D_\mu a_\alpha b_\beta &= \partial_\mu a_\alpha b_\beta + \Gamma_{\mu\alpha}^{\;\;\gamma}a_\gamma b_\beta + \Gamma_{\mu\beta}^{\;\;\gamma}a_\alpha b_\gamma \\ &= (D_\mu a_\alpha) b_\beta + a_\alpha (D_\mu b_\beta) \end{align} which echoes the behaviour of normal derivatives, we don't have a similar relationship in electromagnetism:

\begin{align} \nabla_\mu a_\alpha b_\beta &= \partial_\mu a_\alpha b_\beta + i q A_\mu a_\alpha b_\beta \\ &= (\nabla_\mu a_\alpha) b_\beta + a_\alpha \partial_\mu b_\beta \\ &= a_\alpha (\nabla_\mu b_\beta ) + (\partial_\mu a_\alpha) b_\beta \\ &\neq (\nabla_\mu a_\alpha) b_\beta + a_\alpha (\nabla_\mu b_\beta) \end{align}

This makes me curious about the definition of the electromagnetism covariant derivative as a derivative at all. Does anyone have any insight they could share into the differences between these covariant derivatives, and in particular this lack of distributive property in electromagnetism?

I could imagine defining an operator $\mathcal{I}$ that has the same distributive property as $\partial_\mu$ and $f_{ji}$, and so $\nabla_\mu = \partial_\mu + i q A_\mu \mathcal{I}$ is distributive - is there a reason the covariant derivative isn't defined that way (or is it implicitly defined like that, and no one bothers to write it out fully)?

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    $\begingroup$ You may have some misunderstanding about how the covariant derivatives work. Your EM example is incorrect, see physics.stackexchange.com/questions/543482/… $\endgroup$ – Eletie Mar 3 at 20:05
  • $\begingroup$ All derivatives obey the product rule. They are all derivations. $\endgroup$ – G. Smith Mar 3 at 20:07
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    $\begingroup$ You cannot write $\partial_\mu a_\alpha b_\beta$ and expect readers to assume that it means $\partial_\mu (a_\alpha b_\beta)$. It is ambiguous. $\endgroup$ – G. Smith Mar 3 at 20:15
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Your calculation of the gauge covariant derivative is incorrect. The $\nabla$ operator changes depending on what it acts upon, just like the covariant derivative in GR. In general a gauge covariant derivative changes based on the representation under which the object it acts upon transforms. This is actually the same as for the covariant derivative in GR as well...there the group is basically $GL(4,\mathbb{R})$ and the analog of the vector potential is the Christoffel symbols (to make the two formalisms really precisely identical though one should introduce vierbeins and a spin connection).

If you go and learn about non-abelian gauge theories, this point becomes much clearer because there we speak of fields directly as transforming under some representation or another while in Maxwell theory (which has gauge group $U(1)$), it's the charge that labels the representation so some of this is easily muddled. If you have some familiarity with differential geometry (differentials and so on) and Lie algebras, I can recommend section 2 of this paper (there may be better sources, but this is the first that comes to mind for this particular issue). The goal in that paper is anomalies in quantum field theory, but section 2 is a lightning review of precisely the issues I'm trying to point out here. My recommendation is to try and prove to yourself, using the machinery in the linked paper, how the product rule works out using the fact that a product of fields transforms under the tensor product representation of the representations of the factors in the product (for the GR analog, this is the statement that $a_\mu b_\nu$ transforms as a tensor). For more introductory sources on the matter, essentially any introductory text on QFT should point these things out in varying levels of detail.

For the sake of this answer, however, let me point out the following: suppose that $a_\alpha$ has charge $q_a$, meaning it transforms as $a_\alpha\rightarrow e^{iq_a\Lambda}a_\alpha$ where $\Lambda=\Lambda(x)$ is the gauge parameter (in the sense $A_\mu\rightarrow A_\mu+\partial_\mu\Lambda$ up to conventions about where the gauge coupling appears). Similarly suppose $b_\beta$ has charge $q_b$.

It then follows that $a_\alpha b_\beta$ has charge $q_a+q_b$ since if we were to apply a gauge transformation we would find $$ a_\alpha b_\beta\rightarrow e^{iq_a\Lambda}a_\alpha e^{iq_b\Lambda}b_\beta=e^{i(q_a+q_b)\Lambda}a_\alpha b_\beta. $$ It therefore follows that the appropriate covariant derivative acting on the product is (as pointed out in some of the comments under OP, it's a good idea to make explicit where your derivatives act, by the way) $$ D_\mu(a_\alpha b_\beta)=\partial_\mu(a_\alpha b_\beta)+i(q_a+q_b)A_\mu. $$ The product rule now follows trivially from the product rule of the coordinate derivative and regrouping terms.

Again, I want to stress that this is really exactly the same as the statement that the expression for $\nabla_\mu V_\nu$ and $\nabla_\mu T_{\nu\lambda}$ are not the same...they have different additions of Christoffel symbols because the index structures are different, and the index structures are identical here to the group representations we are working with.

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  • $\begingroup$ Thanks @RichardMyers, it was the clarity that the charge is labelling the representation (and so taking the place of the generator of the symmetry) that I was looking for $\endgroup$ – AnotherShruggingPhysicist Mar 4 at 9:25

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