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I am studying gauge theory, and we derive the Lagrangian for electrodynamics by wanting the Lagrangian to be gauge invariant under U(1) symmetry group. That is, invariant under the phase rotation, $$\psi(x) \rightarrow e^{i\alpha(x)} \psi(x).$$ We derived the covariant derivative, $$D_{\mu} = \partial_{\mu} + A_{\mu},$$ where $A_\mu$ is the gauge field, and transforms as $$A_\mu \rightarrow A_\mu -\partial_\mu \alpha(x).$$ We then found the electromagnetic field tensor by saying that $[D_\mu, D_\nu]$ must be gauge invariant also. When we substitute $D_\mu = \partial_\mu + A_\mu$ into $[D_\mu, D_\nu] \psi(x)$, we can derive, $$[D_\mu, D_\nu] = \partial_\mu A_\nu - \partial_\nu A_\mu.$$ This is in fact the electromagnetic field tensor, $$F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu.$$ How does this derivation show that the electromagnetic field tensor therefore has geometrical origins, and is fundamentally geometric in nature? Is this because the commutator of the covariant derivative is the comparison of parallel transporting an object in one direction, and then the opposite direction? Could there be a pictorial visualization of what is going on here somehow? Would the $F_{\mu \nu}$ being geometric in this way account for why it is called a curvature 2-form, since it is related to the curvature of the manifold it is defined over? I want to gain an intuitive understanding of what is happening here, so that the link to the geometry is clear.

I also wanted to ask if I am missing some factors of i here? I am unsure if a few of my derivations are wrong or not.

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    You might want to think carefully about what you mean when you say "geometric" here. There are two obvious notions of "geometry" at play: The pseudo-Riemannian geometry of the spacetime manifold $M$, and the geometry of the $U(1)$ gauge theory defined on $M$ (generally formulated in terms of a principal $U(1)$-bundle $P\to M$ iirc). The field strength has much more to do with the latter than the former. – Kajelad Feb 15 '22 at 03:25
  • Okay, but I thought that $[D_\mu, D_\nu]$ had more to do with the geometry of the spacetime manifold than the $U(1)$ principal bundle. How is the electromagnetic field tensor related to the principal U(1) bundle? –  Feb 15 '22 at 20:22
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    See Electrodynamics in general spacetime. We first learn about electric and magnetic fields, then we learn about the "vector potential" $A$, and its exterior derivative $F=dA$, and finally as curvature of a connection. Note that you seem to think curvature is a property of a manifold, when strictly speaking, it is a property of the connection (hence more to do with the bundle than the base $M$, as mentioned above). Finally, regarding a picture, perhaps the one at the bottom here might help. – peek-a-boo Feb 15 '22 at 20:56
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    In the mathematician's formulation, the potential $A$ is a $\mathfrak{u}(1)$-valued connection 1-form on $P$, and the the field strength $F$ is the corresponding $u(1)$-valued curvature 2-form on $P$. These are defined without reference to any Riemannian structure on $M$. This maps onto the physicist's formulation in a somewhat complicated way, but the basic idea is the same. If I'm interpreting your notation correctly, then it should be meaningful that neither the metric $g_{\mu\nu}$ not the Christoffel symbols $\Gamma^\mu{}_{\nu\lambda}$ appear. – Kajelad Feb 15 '22 at 20:57
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    (in the second link, make the analogy $A_i$ with $\Gamma_i$ and $R_{ij}$ with $F_{ij}$). – peek-a-boo Feb 15 '22 at 20:57
  • One potential stumbling block is that the pseudo-Riemannian and $U(1)$ gauge geometries both have their own notions of connection and curvature. These are different objects, and they don't interact with each other until you start dealing with Lagrangians/field equations. – Kajelad Feb 15 '22 at 21:05

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