I am trying to understand classical gauge theory from a differential geometric point of view, but there is something I don't particularily understand.

I think the most appropriate answer as to why we usually view the $\text{U}(1)$ connection of QED (and any other gauge connection as well) as a connection on a principal fiber bundle as opposed to a vector bundle is because any field that is invariant under an $\text{U}(1)$ transformation will interact with the same electromagnetic field $A_\mu$. Complex scalar fields, Dirac fields, quark fields etc, they interact with the same photon field $A_\mu$. If we considered $A_\mu$ to be a connection form of a connection on the vector bundle whose sections are complex scalar fields, then we'd have no guarantee that the $A_\mu$ of a Dirac field would represent the same field.

If we view the connection to reside on a principal bundle, then it will induce connections on all associated vector bundles, so if we view the vector bundles of charged scalar fields, Dirac fields, quark fields etc. as all being associated vector bundles of a single $\text{U}(1)$ principal bundle, then this problem is gone.

The problem:

Consider the electron field $\psi_e$, and the connection acting on these fields as $$ D_\mu\psi_e=\partial_\mu\psi_e+\mathcal{A}_\mu\psi_e. $$

The field $\mathcal{A}_\mu$ is an $\mathfrak{u}(1)$-valued field, and we can identify $\mathfrak{u}(1)$ with $i\mathbb{R}$, so we can split off the imaginary unit, and also a coupling constant since why the hell not and we get $\mathcal{A}_\mu=iqA_\mu$, where $A_\mu$ is now a real valued field and we identify $q=-e$ as the charge of the electron.

But if we consider $\psi_u$ the field for the up quark, we have $$ D_\mu\psi_u=\partial_\mu\psi_u+\tilde{\mathcal{A}}_\mu\psi_u, $$ and we make a split $\tilde{\mathcal{A}}_\mu=i\tilde{q}A_\mu$ where now $\tilde{q}=\frac{2}{3}e$ is a different charge.

Usually we identify the electromagnetic field as the "realified" field $A_\mu$ and we can take these to be the same for both $\psi_e$ and $\psi_u$, but the actual connection forms $\mathcal{A}$ and $\tilde{\mathcal{A}}$ are different.

The question:

What makes a difference between these two fields? I mean, if we consider general relativity/Riemannian geometry, the same $\text{O}(1,3)$ connection on the orthonormal frame bundle induces the connection forms $\omega_{\mu\ \ b}^{\ a}$ on the tangent bundle as $\nabla_\mu X^a=\partial_\mu X^a+\omega_{\mu\ \ b}^{\ a}X^b$ and the connection forms $-\omega_{\mu\ \ b}^{\ a}$ on the cotangent bundle as $\nabla_\mu X_b=\partial_\mu X_b-\omega_{\mu\ \ b}^{\ a}X_a$, but this difference arises naturally from the requirement that the covariant derivative commutes with contractions.

No such consideration exists for the case of the $\psi_e$ and $\psi_u$ fields.

Of course, I guess, the almighty answer is that the electron has a charge of $-1$ and the up quark has a charge of $+2/3$ and so that's how it needs to be done, but I am curious if there is a more in-depth mathematical answer.

I do know that the local connection forms depend on how the structure group is represented on the model fiber of the associated vector bundles, however we can view the "Étale space" of up quark fields as the threefold direct sum of a "Dirac bundle" with itself, and as such if a representation is given on the model fiber for the Dirac bundle then there is a very natural direct sum representation on the "up quark bundle".

Does this mean that the group $\text{U}(1)$ is represented differently on the "up quark bundle" than the "Dirac bundle"? If so, how are the two representations different?


1 Answer 1


Notice that the one dimensional representations $U(1)\times\mathbb{C}\to\mathbb{C}$ of $U(1)$ are of the form $(g,z) \mapsto g^q z$, where $q$ is a real number. The representation of $U(1)$ for the fibers of the bundle of the electron is the one with $q=-1$, whereas for the up quark it is $q=2/3$.

The connection differs from one to the other because they are different bundles, but they are both induced by the same connection $A$ in the principal bundle.


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