# Is $∂_\mu + i e A_\mu$ a “covariant derivative” in the differential geometry sense?

I have heard the expression "$$∂_\mu + i e A_\mu$$" referred to as a "covariant derivative" in the context of quantum field theory. But in differential geometry, covariant derivatives have an ostensibly different meaning. While the $$i$$-th component of the ordinary derivative in the $$j$$-th direction of a vector $$v$$ is $$∂_j v^{\,i}$$, the covariant derivative is $$∂_j v^{\,i} + v^{\,k}\Gamma^i_{\,k\,j}$$, where the $$\Gamma^i_{\,k\,j}$$ are the Christoffel symbols that encode the connection of a manifold.

How closely related are these two meanings of "covariant derivative"? Is it fairly superficial, in that both contexts have a type of derivative that is covariant under some form of coordinate transformation (arbitrary for differential geometry, Lorentz for QFT)? Or is it more deep, in that the "$$+ieA_\mu$$" term genuinely represents the connection coefficients/Christoffel symbols of differential geometry in some direct way?

• It is the same (except for physicists conventions to put the coupling $e$ in front of $A_\mu$, whereas mathematicians would put it in front of the field strength, $\frac{1}{g^2}F\wedge *F$). You may want to look at Nakahara's book, for instance. – user178876 Oct 5 '18 at 4:30

It is the same. Mathematicians would refer to a gauge theory setting as a fiber bundle. $$A_\mu$$ takes the role of a connection on the fibre bundle. Of course, out of the three indices the Cristoffel symbols $$\Gamma^i{}_{jk}$$ have, two indices "live" in the fiber. You can appreciate this more easily if you look at non-Abelian gauge theories, in which $$(A_\mu)_{ij}=A_\mu^a T_{ij}^a$$.

1. Why would one care? If you are only interested in computing the NNNLO correction in some QFT process, and you are in a rush, you may postpone looking into these things. Otherwise IMHO you greatly benefit from looking at the geometrical description, which yields a universal description of gravity, electromagnetism, and the weak and strong forces.
2. Is this interpretation new? Not at all. Recall that the old Kaluza-Klein idea already made this connection. The U(1) of electromagnetism was "derived" from the symmetries of a compact $$\mathbf{S}^1$$. Of course, in this case the compact space is 1-dimensional, so it is harder to appreciate that $$A$$ has two indices in that space.
3. What is this good for? At the very moment you want to understand why, say, an instanton number is an integer, this is the IMHO easiest way to go. All boils down to some rather easy topology statements.
4. Where can one read more? There are many sources, one of them being Nakahara's textbook on "Geometry, topology and physics", and if you do not want to buy a book, you may look e.g. at these notes.
• In your opinion, based on your comment, is there much to be gained by studying fibre bundles if I've already covered the same material in a basic QFT course? Just curious, and a bit short on time. Thanks – user207480 Oct 5 '18 at 13:03
• @StudyStudyStudy Depends on what you want to do. If you just want to do some QFT computations very quickly, you may want to postpone this. On the long run, I do recommend to have a look at differential geometry. – user178876 Oct 5 '18 at 15:50
• @marmot (1) Does this mean that two indices of the Christoffel symbols represented by $A_\mu$ are contracted, and (2) Does this mean that $A_\mu$, and hence electromagnetism, can be regarded as a geometric phenomenon, similar to gravity? – WillG Oct 5 '18 at 16:04
• @WillG (1) No. That's why I gave the non-Abelian example, in which you see that $(A_\mu)_{ij}$ does have three indices, the only difference to the $\Gamma$s is that two are in the fiber. (2) Absolutely. All known forces of Nature can. (Some will jump at me and say that the Higgs also mediates a force. That is true, but in many string-derived models the Higgs comes from extra components of the gauge bosons, and hence the statement is true in these. Whether or not you can cook up a UV complete model in which the fermion-Higgs interactions have no geometric interpretation, I don't know.) – user178876 Oct 5 '18 at 16:15

I'll expand on the existing answer with a few points explained in more detail here; I'll refer to section numbers therein as appropriate, and you'll forgive my switching to its notation.

Yang-Mills theory's gauge covariant derivative $$D_\mu$$ is constructed so that $$\psi\to U\psi$$ wth $$U$$ in the Lie group imposes $$D_\mu\psi\to UD_\mu\psi$$, and we can prove $$[D_\mu,\,D_\nu]=igF_{\mu\nu}$$ (1.2.2). The analogous treatment of gravity takes general coordinate transformations (GCT) as the gauge transformations (1.3.1), so the GCT $$T_{\mu\nu\cdots}^{\rho\sigma\cdots}\to \tilde{T}_{\mu\nu\cdots}^{\rho\sigma\cdots}$$ imposes $$\nabla_\lambda T_{\mu\nu\cdots}^{\rho\sigma\cdots}\to \nabla_\lambda \tilde{T}_{\mu\nu\cdots}^{\rho\sigma\cdots}$$. While the Christoffel symbols play the role of $$A_\mu$$, the Riemann tensor is analogous to $$F_{\mu\nu}$$ since $$[\nabla_\mu,\,\nabla_\nu]A_\rho=R_{\mu\nu\rho\sigma}A^\sigma$$.

We can “derive” general relativity from gauge-theoretic principles, just as we can electromagnetism (1.1); that gravity is attractive precludes a spin-1 graviton, and that it lenses photons precludes a spin-0 graviton, so we must have a spin-2 graviton instead. This dissimilarity to the Standard Model's gauge bosons suggests gravity is a “squared” gauge theory, an idea formalised in the KLT-relations (9.1), which historically originated in string theory but can be derived just from the usual treatment of Feynman-diagram amplitudes.

Apart from that, however, an effective field theory can be obtained (8) despite 4-dimensional quantum gravity's nonrenormalisability. Treating the graviton's trace as a scalar field fixes the EFT's parameters (8.4.1), resulting in quantum corrections to the Newtonian potential (8.5) and Reissner–Nordström metric (8.6).