Local coordinate expression for the equations of motion in gauge theory

Let's assume $$P$$ is a principal bundle, $$F^A \in \Omega^2(M,Ad(P))$$ the curvature 2-form, $$Ad(P)$$ the adjoint bundle. $$d_A$$ the covariant differential. For sections in the associated bundle $$E=P \times_{(G, \rho)} V$$, $$d_A$$ is just the covariant differential. In local coordinates it is of the form $$d_A \rightarrow \partial_{\mu}+\rho_*(A_{\mu})$$. $$\phi$$ is a section in the associated bundle and in local coordinates takes the form $$[s(x),\varphi(x)]$$ where $$s:U \rightarrow P$$ is a section in the principal bundle and $$\varphi:U \rightarrow V$$. The Yang-Mills-Higgs action is $$$$\mathcal{S}_{Y K}: \mathcal{C}(P) \times \Gamma(E) \rightarrow \mathbb{R}, \quad \mathcal{S}_{Y K}[A, \phi]=\int_{M}\left(-\frac{1}{2}\left\langle F^{A}, F^{A}\right\rangle_{\mathrm{Ad}(P)}+\left\langle d_{A} \phi, d_{A} \phi\right\rangle_{E}-m^{2}\langle\phi, \phi\rangle_{E}\right) d \nu_{g}$$$$ The variation $$A\mapsto A+\omega$$ gives the equations of motion $$$$\delta_{A} F^{A}=j$$$$ $$$$\delta_{A} d_{A} \phi + m^{2} \phi=0$$$$ with the codifferential $$\delta_A$$ and $$j \in \Omega^{1}(M , \operatorname{Ad}(P))$$ implicitly defined by $$$$\langle j, \omega\rangle_{\mathrm{Ad}(P)}=-2 \operatorname{Re}\left(\left\langle d_{A} \phi, \rho_{*}(\omega) \phi\right\rangle_{E}\right)\quad\text{for all }\omega.$$$$ In physics, the current is defined by $$$$j_{\nu}^{a}=-i\left(\left(D_{\nu} \varphi_{i}\right)^{\dagger}\left(T_{a}^{r} \varphi\right)_{i}-\left(T_{a}^{r} \varphi\right)_{i}^{\dagger} D_{\nu} \varphi^{j}\right)$$$$ where $$T_a$$ is a basis of the Lie algebra and $$T_a^r=\rho_*(T_a)$$, $$D_{\nu}=\partial_{\nu}+A_{\nu}^aT_a^r$$. $$\varphi_i$$ is just the $$i$$-th component of $$\varphi$$. The $$i$$'s come into play due to the definition of physicists that every Lie algebra element is multiplied with $$I$$.

$$\mathbf{Question}$$: How exactly can one derive the physical local coordinate expression from the mathematical definition?

• You need to be a bit more precise in your question. What exactly prevents you from expanding everything in local coordinates, a basis of $\mathfrak{g}$, a local frame of $E$ and the local gauge potentials $A_\nu^a$. Moreover, what is $\omega$ which you pair the current $j$ with in its definition? Jul 5, 2021 at 12:40
• @nicrot000 I posted an answer to ma own question to show my progress. I would really like to hear your thoughts. Jul 5, 2021 at 15:00
• By the way, I think this gauge theory/Yang-Mills-Higgs theory stuff should be started from the point of view that $E$ is some given vector bundle and then you go through the theory defining the frame bundle, the gauge freedom etc. From this perspective it has indeed applications in what physicists call the gauge principle (from where they start their perspective on gauge theory, motivated by relativity principles). Starting with the principal bundle makes it somewhat unnecessarily abstract, e.g. your local representation of sections $\phi$ and so on. Jul 5, 2021 at 15:38

Okay, I think most of the question was answered in the comment sections.

However, I have the impression that this question (and also part of your other questions I came across before) rises and falls by keeping very precisely track of the spaces where your objects live.

That is, for example, a covariant derivative is a map $$\nabla:\mathfrak{X}(M)\times\Gamma(E)\to\Gamma(E),~(X,s)\mapsto \nabla_Xs,$$ so it gets a vector field, making one covariant space-time index, and a section, making one covariant "E-frame index" and it returns a section, making a contravariant "E-frame index". For a local frame the respective trivial covariant derivative is just $$d$$ on the expansion coefficients, making it, in coordinates, $$\partial_\mu$$ or more precisely $$\delta_a^b\partial_\mu$$. Every other covariant derivative (particularly the one induced by your $$A$$) is then representable as this trivial one, plus a connection form, so in coordinates we have $$d_A\to\delta_a^b\partial_\mu+{A_\mu}_a^b$$ for some $$\text{End}(E)$$-valued differential form $${A_\mu}_a^b$$, usually already represented in a local frame of $$\text{End}(E)$$, $${A_\mu}_a^b=A_\mu^j(T_j)_a^b$$, or rearranged in a matrix $$A_\mu^jT_j$$ again, as you wrote it.

However, if you keep track of these indices really carefully, almost every physicist's local formula is just this expansion and nothing more, and I can barely come up with an example where there is a trick or something. So I have the impression that many of your questions could be solved by carefully keeping track of the indices, or in which spaces equations hold and so on.

And moreover, it might be helpful to follow my previous comment on that sometimes it is better to start with a vector bundle $$E$$ and from this point start the theory. Then some difficulties might vanish, e.g. then $$\text{Ad}(P)$$ is just (canonically isomorphic) to $$\text{End}(E)$$ (or, depending on your gauge group, some section space of skew-adjoint/skew-symmetric matrices and so on), and $$\rho_*(A_\mu)$$ is just what I wrote above. Afterwards you can still make it more complicated and start from the principal bundle side again.

• Thanks for the answer. I will accept it as soon as I think the problem is solved. I edited my answer. And I think I got the current now. However, the expression does not look like what I thought I would get. Now I am confused if it is wrong. Jul 6, 2021 at 15:06

$$\omega$$ id an Ad$$(P)$$-valued form I use to take the variation. I.e. I plug in $$A+t \omega$$ to vary. It holds $$\begin{equation*} \langle j, \omega \rangle_{\operatorname{Ad}(P)}=-2 \operatorname{Re}( \langle d_A \phi, \rho_*(\omega) \phi \rangle_{E}) \end{equation*}$$

I use the following inner product: $$\langle \cdot , \cdot \rangle_E \rightarrow (\cdot)^{\mu \dagger}(\cdot)_{\mu}$$ and write $$\rho_*(\omega_{\mu})=\omega_{\mu}$$. With that I get \begin{align*} j^{\mu}\omega_{\mu} &= -2 \operatorname{Re}( (D^{\mu} \varphi)^{\dagger} \omega_{\mu} \varphi )\\ &=- \left( (D^{\mu} \varphi)^{t} (\omega_{\mu} \varphi)^*+ (D^{\mu} \varphi)^{\dagger} \omega_{\mu} \varphi \right)\\ &=-\left( (D^{\mu} \varphi)^{t} \omega_{\mu} \varphi^*+ (D^{\mu} \varphi)^{\dagger} \omega_{\mu} \varphi \right) \end{align*}

This is my process. But how is this equivalent to $$\begin{equation*} j^{\mu}\omega_{\mu}=-\left( \varphi^{\dagger} (D^{\mu} \varphi) \omega_{\mu}- (D^{\mu} \varphi)^{\dagger} \varphi \omega_{\mu} \right) \end{equation*}$$ My first problem is, how do I get the $$\varphi$$ to the left side of $$(D_{\mu} \varphi)$$. The second problem is the minus sign in the middle. But maybe above one needs the imaginary part. But in the book Mathematical gauge theory by Hamilton it says real part.

$$\mathbf{Edit:}$$ I think we have to take the adjoint bundle on both sides. Without it, we cannot get rid of $$\omega$$. If $$\rho_*(T_a)=T_a^r$$ and $$T_a$$ is a basis for the Lie algebra, $$\omega_{\mu}=\omega_{\mu}^a T_a^r$$. A section in the adjoint or any associated bundle is of the form $$[s(x),\varphi(x)]$$ where $$s$$ is a section in the principal bundle and $$\varphi:U \rightarrow V$$ is a function. When the vector space of the associated bundle is multidimensional, it also has several components, i.e. $$\varphi=(\varphi_1,...,\varphi_n)$$. I suppress that.

Accordingly, take equations above are $$$$j^{\mu}_aT^{a,r}\omega_{\mu}^bT_b^r=-\left( ((\partial^{\mu}+A^{\mu}_a T_a^r)\varphi )^{t} \omega_{\mu}^a T_a^r \varphi^*+ ((\partial^{\mu}+A^{\mu}_a T_a^r)\varphi)^{\dagger} \omega_{\mu}^a T_a^r \varphi \right)$$$$

Since $$\varphi^*$$ is not Lie algebra valued, I guess $$$$(\varphi^*(\partial^{\mu}+A^{\mu}_a T_a^r)\varphi )^{t} \omega_{\mu}^a T_a^r \varphi^* = \varphi^* (\partial^{\mu}+A^{\mu}_a T_a^r)\varphi )^{t} \omega_{\mu}^a T_a^r$$$$ But also $$(\partial^{\mu}+A^{\mu}_a T_a^r)\varphi )^{t}$$ is just a vector so

$$$$\varphi^* (\partial^{\mu}+A^{\mu}_a T_a^r)\varphi )^{t} \omega_{\mu}^a T_a^r=\varphi^{\dagger} (\partial^{\mu}+A^{\mu}_a T_a^r)\varphi ) \omega_{\mu}^a T_a^r$$$$ Do you think that's right?

$$\mathbf{Edit2:}$$ The last formula from yesterday was wrong. On the l.h.s. the result is a number because we take an ad-invariant scalar product. However writing the matrix multiplication in components, we can drop the transpose.

$$\begin{equation*} j_{a}^{\mu}\omega_{\mu}^{b} \langle T^{a, ad}, T_{b}^{ad} \rangle_{\operatorname{Ad}(P)}=-\left(\left(\partial^{\mu}\varphi_i+A_{a}^{\mu} (T_{a}^{r}\varphi)_i\right) \omega_{\mu}^{b} (T_{b}^{r})^{ik} \varphi^{*}_k+\left(\partial^{\mu}\varphi_i+A_{a}^{\mu} (T_{a}^{r}\varphi)_i \right)^{*} \omega_{\mu}^{b} (T_{b}^{r})^{ik} \varphi_k\right) \end{equation*}$$

Rearranging everything gives $$\begin{equation*} j_{a}^{\mu}\omega_{\mu}^{b} \langle T^{a, ad}, T_{b}^{ad} \rangle_{\operatorname{Ad}(P)}=-\left( (T_{b}^{r} \varphi^{*})_i \left(\partial^{\mu}\varphi_i+A_{a}^{\mu} (T_{a}^{r}\varphi)_i\right) +\left(\partial^{\mu}\varphi_i+A_{a}^{\mu} (T_{a}^{r}\varphi)_i \right)^{*} (T_{b}^{r} \varphi)_i\right)\omega_{\mu}^{b} \end{equation*}$$ Now, we can identify the current

$$\begin{equation*} (j^{\mu})_i=-\left( (T_{b}^{r} \varphi^{*})_i \left(D_{\mu} \varphi \right)_i +\left(D^{\mu}\varphi\right)_i^{*} (T_{b}^{r} \varphi)_i\right) \end{equation*}$$

• For the last part, how does Hamilton handle the "$i$-convention". So confusion with real or imaginary part may stem from there. Moreover, I'm not sure whether you screwed up the ${}^\dagger$ and ${}^t$ or not, there may also come an additional sign. On pulling the $\omega_\mu$ outside: My first impression is that there is one basis expansion missing, namely with a local frame of $E$. Then, is this basis also $\phi$ should have an index, and, if I remember correctly, $\rho_*(\omega)$ is $\text{End}(E)$-valued, so should have two indices. Jul 5, 2021 at 15:23
• And since the covariant derivative also gets a section and returns a section, in this space I guess it should have also two indices. Watch out, Greek indices by physicists usually refer to the indices of space-time (i.e. $M$) coordinates, where the covariant derivative has only one index. Jul 5, 2021 at 15:25
• @nicrot000 I edited the post for more detail. The $i$ convention is handled by multiplying every Lie-algebra valued thing by i. I have not done this above. But I think I took the real part right. I.e. $Re(z)=1/2(z^*+z)$ and $z^{\dagger *}=z^t$ Jul 5, 2021 at 16:02
• Sure, that's how this convention works, but if some author takes an imaginary part in this convention, some other takes a $\pm$ real part in the convention without $i$. Jul 5, 2021 at 19:11
• I see thanks. Due to the dagger we get a relative sign change. Do you think, I can commute $\varphi$ with $T_a^r$ and $\omega_{\mu}^a$. Then, I think the problem is solved, if the last equation above holds. Jul 5, 2021 at 19:31