Showing that Yang-Mills equations transform homogeneousy

In summary, the differential equation D\star F = 0 transform homogeneously under the adjoint action ##F \mapsto gFg^{-1}## of the lie group ##G##, where ##D## denotes the covariant exterior derivative ##D\alpha = d \alpha + A \wedge \alpha## for some lie algebra valued form ##\alpha##. The third term in the equation is zero if we use that ##gdg^{-1} = - (dg) g^{-1}##, but this is not correct.
  • #1
center o bass
560
2
Hi! I'm trying to show that the differential from equation
$$D \star F = 0$$ transform homogeneously under the adjoint action ##F \mapsto gFg^{-1}## of the lie group ##G##, where ##D## denotes the covariant exterior derivative ##D\alpha = d \alpha + A \wedge \alpha## for some lie algebra valued form ##\alpha##. Since
$$A \mapsto gAg^{-1} + gdg^{-1}$$
we get
$$D\star F \mapsto d(g\star Fg^{-1}) + (A \mapsto gAg^{-1} + gdg^{-1})\wedge \star F = dg \wedge \star F g^{-1} + g d\star F g^{-1} + g \star F \wedge dg^{-1} + g A \wedge \star F g^{-1} + g dg^{-1} \wedge g^{-1} \star F g^{-1}$$
where if we use that ##gdg^{-1} = - (dg) g^{-1}##, then the last term seem to cancel the first. However, if this is correct, then the third term has to be zero.

Is the third term zero, and if so why?

Am I going wrong somewhere else?
 
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  • #2
I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?
 
  • #3
I overlooked this question completely; I'll come back to you asap
 
  • #4
center o bass said:
Hi! I'm trying to show that the differential from equation
$$D \star F = 0$$ transform homogeneously under the adjoint action ##F \mapsto gFg^{-1}## of the lie group ##G##, where ##D## denotes the covariant exterior derivative ##D\alpha = d \alpha + A \wedge \alpha## for some lie algebra valued form ##\alpha##. Since
$$A \mapsto gAg^{-1} + gdg^{-1}$$
we get
$$D\star F \mapsto d(g\star Fg^{-1}) + (A \mapsto gAg^{-1} + gdg^{-1})\wedge \star F = dg \wedge \star F g^{-1} + g d\star F g^{-1} + g \star F \wedge dg^{-1} + g A \wedge \star F g^{-1} + g dg^{-1} \wedge g^{-1} \star F g^{-1}$$
where if we use that ##gdg^{-1} = - (dg) g^{-1}##, then the last term seem to cancel the first. However, if this is correct, then the third term has to be zero.

Is the third term zero, and if so why?

Am I going wrong somewhere else?

Let me try it out :)
[itex]D \star F=0 [/itex]
or
[itex]d \star F + A \wedge \star F = 0 [/itex]

Now make the mappings of [itex]D \star F[/itex]...

[itex] d(g \star Fg^{-1}) + (gAg^{-1} + g dg^{-1}) \wedge \star gFg^{-1} [/itex]

I think here we reach different results... ?
I can't continue because I am not sure what star means... and I am not really familiar with the wedges...
 
  • #5
I'll try to answer using very very old notes; hope the signs and details a correct. I start w/o using forms. I think the problem is not in the exterior derivative but in the su(n) part. In the following F and A are su(N) matrices, g is an SU(N) group element and the commutators [.,.] always refer to the su(N) matrices.

We have the following transformations:

[tex]A_\mu \to A_\mu^\prime = g\,(A_\mu + i\partial_\mu)\,g^\dagger[/tex]
[tex]F_{\mu\nu} \to F_{\mu\nu}^\prime = g\,F_{\mu\nu}\,g^\dagger[/tex]

As far as I remember the covariant derivative of F is defined as

[tex](DF)^\nu = \partial_\mu F^{\mu\nu} - i [A_\mu,F^{\mu\nu}][/tex]

Now we have to check the transformation properties of DF:

[tex](DF)^\nu \to {(DF)^\prime}^\nu = \partial_\mu (g\,F^{\mu\nu}\,g^\dagger) -i[g\,A_\mu\,g^\dagger + ig\,\partial_\mu\,g^\dagger, g\,F^{\mu\nu}\,g^\dagger]
[/tex]

I hope this Ansatz helps ...
 
Last edited:

1. What are Yang-Mills equations and why are they important?

Yang-Mills equations are a set of partial differential equations that describe the behavior of gauge fields, which are fundamental interactions in quantum field theory. They are important because they provide a mathematical framework for understanding the interactions between elementary particles in the Standard Model of particle physics.

2. How do Yang-Mills equations transform homogeneously?

Yang-Mills equations transform homogeneously because they are invariant under local gauge transformations. This means that the equations remain unchanged when the gauge fields are transformed by a differentiable function of the spacetime coordinates.

3. Can you give an example of a homogeneous transformation of Yang-Mills equations?

One example of a homogeneous transformation of Yang-Mills equations is the rotation of a gauge field. This transformation involves rotating the gauge field by a certain angle at each point in spacetime, while keeping the equations themselves unchanged.

4. How does the concept of gauge symmetry relate to the homogeneous transformation of Yang-Mills equations?

The homogeneous transformation of Yang-Mills equations is closely related to the concept of gauge symmetry. Gauge symmetry refers to the idea that the equations remain unchanged under certain transformations of the gauge fields. In the case of Yang-Mills equations, this means that the equations are invariant under local gauge transformations.

5. What are some applications of Yang-Mills equations in physics?

Yang-Mills equations have a wide range of applications in physics, including in the study of quantum chromodynamics (QCD), which describes the strong nuclear force, and in the electroweak theory, which unifies the electromagnetic and weak interactions. They are also used in theories of gravity, such as general relativity, and in models of condensed matter systems.

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