Local coordinate expression for the equations of motion in gauge theory

Question

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 \begin{equation} \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} \end{equation} The variation $A\mapsto A+\omega$ gives the equations of motion \begin{equation} \delta_{A} F^{A}=j \end{equation} \begin{equation} \delta_{A} d_{A} \phi + m^{2} \phi=0 \end{equation} with the codifferential $\delta_A$ and $j \in \Omega^{1}(M , \operatorname{Ad}(P))$ implicitly defined by \begin{equation} \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. \end{equation} In physics, the current is defined by \begin{equation} 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) \end{equation} 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?

Answer 1

$\omega$ 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 \begin{equation} 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) \end{equation}

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

\begin{equation} \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 \end{equation} 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*}

Answer 2

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.