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In class, we have seen that the covariant derivative of some form $R$ can be written as:

$$DR = dR + [A, R] = dR + A\wedge R - R\wedge A \tag1$$

Here, $d$ represents the external derivative over forms and $A$ is the local connection defined via the pull-back of a section $S: U_i \in M \rightarrow P(M, G)$ where $P(M, G)$ is the principal bundle with $M$ the base space and $G$ the Lie group that plays the fiber role. Therefore, $A = S^*\omega$, with $\omega \in \Omega^1(P)\otimes T_eG$ and $\Omega^1(P)$ the set of 1-forms in $P(M, G)$. So while $\omega$ is a connection for all $P$, $A$ is just over $U_i$

So by Eq. (1) we can write:

$$D = d + [A,\ ·\ ] \tag2$$

Eq. (2) is pretty similar to the one used in QFT:

$$D_\mu = \partial_\mu + igA_\mu \tag3$$

$g$ is just the coupling constant of the interaction, so $igA_\mu$ is somehow equivalent to the connection $A$ of the Eq. (1). I understand that the index $\mu$ comes out from the fact that in Eq. (1) we work with forms, so

$$A\sim A_\mu dx^\mu \tag4$$

But, what I don't see is how to make the relation between the commutator in Eq. (2) and the simple form $igA_\mu$.

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  • $\begingroup$ Are there any further assumptions concering $R$? Is it just any form, is it a differential form, does it take values in a specific vector space? $\endgroup$
    – Creo
    Nov 11, 2018 at 16:05
  • $\begingroup$ Nothing else, just $R \in \Omega^p\otimes T_eG$, so $R = R^a\otimes T_a$ (sum over $a$ implicit) where $R_a$ is a $p$-form and $T_a$ a generator of the Lie algebra of the Lie group $G$ $\endgroup$
    – Vicky
    Nov 11, 2018 at 16:14
  • $\begingroup$ Then there is no immediate relation between the two versions, because allowing $R$ to only take values in $T_eG$ is to specific. You'd have to let it take values in an arbitrary vectorspace $V$ with a given representation $\rho:G \to GL(V)$, but then equation (1) no longer holds. And I suppose in eq. (3) you are working in QED? $\endgroup$
    – Creo
    Nov 11, 2018 at 17:20
  • $\begingroup$ Not necessarily because I can make $A_\mu \rightarrow A_\mu^iT_i$ for any symmetry group with generators $T_i$. About your comment: there must be a way to relate Eq. (2) and Eq. (3), because it is the same concept. Or are there many ways to do the covariant derivative? $\endgroup$
    – Vicky
    Nov 11, 2018 at 17:55
  • $\begingroup$ Ah, okay. (The QED case doesn't really matter though). Yes, there is a way to relate them, but that's because they are two special cases of the general notion of a covariant derivative on a Principal Bundle with given connection. I can try to illustrate this, but for that I'd have to know if you are familiar with: 1.) Horizontal Vector spaces; 2.) Vertical Vector spaces; 3.) horizontal differential forms $\endgroup$
    – Creo
    Nov 11, 2018 at 18:06

1 Answer 1

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The formal definition of horizontal and vertical spaces are not to important right now, for us they will just serve as a tool. The Idea is that by means of horizontal spaces we can talk about how a connection (i.e, a background field) changes the differentiation rule (you can compare that with general relativity, where curvature of spacetime does, indeed, force you to use the covariant differentiatial operator $\nabla_{\mu}$ (with respect to some coordinate system). Let me try to make this claim precise!

Formal prelimarys

Throughout this section, we will fix a representation $\rho:G \to GL(V)$ of the structure group on some finite dimensional vector space $V$ and a connection form $A$ on $P$.

Definition 1. $\quad$ Given a PFB $P \to M$ over a (spacetime) $M$ and a vector space $V$ we define the space $\Omega^k_{\mathrm{hor}}(P,V)^{(G, \rho)}$ as the space of all $V$ valued $k$ forms $\omega \in \Omega^k(P,V)$ which satisfy $$1. \quad \omega_p(X_1,...,X_n) = 0, \text{if any of the} \ X_i \ \text{is vertical};$$ $$2. \quad R^*_g\omega_p = \rho(g^{-1}) \circ \omega_{p}, \ \text{for all} \ g \in G, \ \text{where $R_g$ denotes the right translation by $g$}.$$ We will call differential forms which satisfy 1. horizontal and forms which satisfy 2. of type $\rho.$

Definition 2. $\quad$ The covariant derivative of $A$ is given by $$D_{A} \omega_p(X_1,...,X_k) := d \omega_p (pr_hX_1,...,pr_hX_k)$$ for any $k$ form $\omega \in \Omega^k(P,V)$, where $p \in P$ and $pr_h$ denotes the projection onto the horizontal subspace of $T_pP.$

Then you get:

Theorem. $\quad$ On $\Omega^{k}_{\mathrm{hor}}(P,V)^{(G. \rho)}$ we have: $$D_{A}\omega = d \omega + \rho_{*}A \wedge \omega,$$ where $\rho_*$ denotes the differential of $\rho$ at $e \in G$ and the expression $\rho_* A \wedge \omega$ is defined via $$\rho_*A \wedge \omega (X_0,...,X_k) := (-1)^i \rho_*A(X_i) \omega(X_0,...,\hat{X_i},...,X_k),$$ the hat denotes that this vector is omitted and we employed the summation convention.

Relating the two derivatives:

(1.) $\quad$ In your first equation, you are looking at horizontal 1 forms of type $Ad,$ where $Ad:G \to GL(T_eG)$ is the adjoint representation of the Lie Group $G$ on its Lie Algebra. Since $Ad_*(X) = [X, \ . \ ],$ you get your equation as a special case of the preeceding theorem (after everything is pulled back to a suitable set $U \subset M$ via a section $s: U \to P$).

(2.) $\quad$ Now, lets take any smooth function $\psi: P \to V$, which satisfies $\psi(pg)= \rho(g^{-1})\psi$. If you pull it back via the section $s:U \to P$ you get a smooth function $\psi': U \to V$ wich you can think of as a gauged (fermion) field. These are the functions on which the derivative induced by the connections which correspond to gauge Bosons are meant to act. Observe that if you choose a different gauge $s': U \to V$ then you can find a function $\mu: U \to G$ such that $\psi'' := \psi \circ s' = \rho(\mu( \ \ )^{-1}) \psi'$ which is why you want to look at those kind of functions. They precisely capture the transformation property of your fields under a given gauge transformation. Now, on $\psi$ we have, since $\psi \in \Omega^0_{\mathrm{hor}}(P,V)^{(G, \rho)}$: $$D_A \psi = d\psi + \rho_*A \psi$$ which, after pulling it back via $s$ and then writing it in coordinates, becomes: $$D_{\mu} \psi' = \partial_{\mu}\psi' + \rho_*A^{s}_{\mu} \psi' $$ where $A^s$ is the pulled back connection, i.e your ''gauge potential'' and the index $\mu$ denotes it's components in your coordinate system of choice. The factor $ig$ is, as far as I know, convention to emphazise that the representation $\rho$ is not trivial, i.e $\rho_* \neq 0$ and the mentioning of the explizit representation is most of the time omitted in physical literature.

Concluding Remarks:

For further reading, I'd recommend ''Gauge Theory and Variational Principles'' by d. Bleecker. This is not the easiest read, but he has many physical examples and I think it'll help you. Furthermore, the preeceding is technical but that's really it! I didn't write everything out in detail, since I think it'll be a good exercise to verify it yourself (mainly because it doesn't seem to ''brutal'' anymore, after you've done a few calculations with it).

Edit: Maybe Bleeckers book is overkill, I think any book that's about connections on principal fibre bundles will do the job. Be also aware that what I've written is meant pre second quantisation.

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