Let's say we have a principal bundle $(P,B,\pi;G)$ and associated bundle $E=P \times_{(G,\rho)}V$and $Ad(P)=P\times_{(G,Ad)} \mathfrak{g}$ the adjoint bundle. The Yang-Mills-Higgs action (without potential) is

\begin{equation} \int_M(- \frac{1}{2}\langle F^A, F^A \rangle_{\operatorname{Ad}(P)} +\langle d_A \phi, d_A \phi \rangle_E - m^2 \langle \phi, \phi \rangle_E )d\nu_g \end{equation} with $\phi \in \Gamma(E)$. If we vary the equation, i.e. add $A+t \omega$ and $\phi+t \alpha$ and then take $d/dt$ the equations of motion are

\begin{equation} \delta_A F^A=j \quad \delta_A d_A \phi=0 \end{equation} with the codifferential $\delta_A$ and the implicit defined current

\begin{equation} \langle j, \alpha \rangle = -\langle d_A \phi,\rho_*(\alpha) \phi \rangle \end{equation} In Differential Geometry And Mathematical Physics 2 by Gerd Rudolph and Matthias Schmidt it says that if we take the associated bundle, i.e. $E$ to be the adjoint bundle, the first equation of motion becomes

\begin{equation} \delta_A F^A=[d_A \phi,\phi] \end{equation}

I have two equations:

  1. How can I derive $j=[d_A \phi,\phi]$ form the general equation for the current. And how do I express it in local coordinates to get something similar to the currents in physics.
  2. Is there some Binachi-identity for the curvature form with the codifferential, i.e. $\delta_A \delta_A F^A=0$. This somehow should be the case, since the current should be conserved, i.e. $\delta_A j=0$
  • 1
    $\begingroup$ Who are the authors of the textbook in question? $\endgroup$ Jun 13, 2021 at 11:49
  • $\begingroup$ Sorry, I included the names of the author in my question $\endgroup$
    – NicAG
    Jun 13, 2021 at 12:42
  • $\begingroup$ One can show something like ${\text{d}_A}^2=F^A\wedge(\cdot)$, so your "Bianchi identity" reads $0=\delta_A\delta_AF^A=*\text{d}_A\text{d}_A*F^A=*(F^A\wedge *F^A)$ and since $*$ is an iso you know that $F^A\wedge *F^A=0$, and the dynamic Lagrangian of $A$ vanishes. I don't think this is true. Are you sure that $j$ should be conserved w.r.t $\delta_A$? $\endgroup$
    – nicrot000
    Jun 13, 2021 at 14:58
  • $\begingroup$ @nicrot000 At least in Bleeckers book gauge theory and variational principles $\delta^{\omega}j=0$ is written down. But probably the current is defined in another way. $\endgroup$
    – NicAG
    Jun 13, 2021 at 16:01
  • $\begingroup$ I don't know that reference. But i looked that up, on any associated bundle $E$ of $P$ you can define a wedge of $\text{Ad}(P)$- and of $E$-valued forms, and $\text{d}_A^2$ in $E$ is just this wedge product with $F^A\in\Omega^2(M;\text{Ad}(P))$. So if $E=\text{Ad}(P)$, then this wedge product should be the same as the one you write the Yang-Mills Lagrangian with. Hence, the identity $\delta_A^2F^A$ should render this Lagrangian, which is what I wrote above up to taking the trace, trivial. $\endgroup$
    – nicrot000
    Jun 13, 2021 at 20:52

1 Answer 1

  1. The first question has been already answered in a comment by NicAG, but let us repeat the 3-line long argument here for completeness:

For the adjoint bundle , ρ∗=ad(⋅)(⋅)=[⋅,⋅]. Hence $⟨j,α⟩=−⟨d_Aϕ,[α,ϕ]⟩$. Because of ad-invariance, the commutator can be 'moved': $⟨j,α⟩=⟨[d_Aϕ,ϕ],α⟩$. The rest its then just non degeneracy.

Edit: changed the sign in the latter formula because so does 'moving' the commutator.

Remark. Possible confusion (see the comments) may come from another common formula for the current borrowed here from a different context: $$j_μ=i(ϕ^†D_μϕ−(D_μϕ^†)ϕ).$$ In fact the latter formula is not applicable here because in the extremely abstract setting considered in the book, there is simply no complex structure, hence no complex conjugation. But if we come down to Earth and recall that G is a compact Lie group (this is assumed in the 3rd line of Section 7.2 `Yang–Mills–Higgs Systems' in the book in question), hence, ``more or less'', subset of $U(n)$, hence any element $\phi$ of the Lie algebra satisfies $ϕ^†=-ϕ$, then we come to the same expression (up to constant factor). Then, plugging in the expressions $$ D_μϕ=(d_Aϕ)_\mu=\frac{\partial\phi}{\partial x^\mu}+A_\mu\phi-\phi A_\mu, $$ one gets the expression for the current in local coordinates.

  1. The identities $δ_Aj=0$ and $δ_Aδ_AF_A=0$ indeed hold with a one-line proof: for each element $\alpha$ of the Lie algebra we have $$ \langle δ_Aδ_AF_A,\alpha\rangle=\langle F_A,d_Ad_A\alpha\rangle=\langle F_A,[\alpha,F_A]\rangle=\langle \alpha,[F_A,F_A]\rangle=0. $$ However this is not called ``the Bianchi identity'', the name being already reserved for the identity $d_AF_A=0$ (Eq. 7.2.6 in the book in question).

Remark. One should not be confused (see the comments) by another common formula $d_Ad_A\phi=F\wedge \phi$, which is valid for a 'vector' $\phi$ but not for a 'matrix' $\phi$ (more formally, for the fundamental representation but not for the adjoint representation).

Remark. In fact the material of the book is not properly exposed in the question. For instance, in the right-hand side of the equation of motion $δ_Ad_Aϕ=0$ there should be $mϕ$ or $-mϕ$ instead. Notice that here the Higgs potential $V(\phi)=m^2|\phi|^2$ does not vanish. But in the absence of higher degree terms in the potential, the action should not be called Yang-Mills-Higgs action: it is rather action for a Klein-Gordon field coupled with a Yang-Mills field.

Remark. Maybe the deep source of all those confusions is an extremely abstract style of most literature on the subject. Let me try to advertise a paper with an elementary introduction of gauge theory. See Sections 2.4-2.5 there. It deals with a lattice formulation, thus to get a deep understanding, one even does not need to know what a derivative is.


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