Commutator between covariant derivative, field strength

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SUMMARY

The discussion centers on proving the equation ∂μFμν + ig[Aμ, Fμν] = [Dμ, Fμν], where Dμ = ∂μ + igAμ. The user encounters difficulty with the term Fμν∂μ and attempts to demonstrate its nilpotency. The conclusion reached is that the term can be considered nil under the assumption that the test function \Psi vanishes at infinity, which is acceptable for physics applications despite requiring mathematical rigor in a formal context.

PREREQUISITES
  • Understanding of covariant derivatives in gauge theory
  • Familiarity with field strength tensors in theoretical physics
  • Knowledge of the properties of smooth functions and their behavior at infinity
  • Basic concepts of commutators in the context of Lie algebras
NEXT STEPS
  • Study the properties of covariant derivatives in gauge theories
  • Learn about the mathematical foundations of field strength tensors
  • Explore the implications of nilpotent operators in quantum field theory
  • Investigate the role of test functions in distribution theory
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The discussion is beneficial for theoretical physicists, particularly those working in gauge theory and quantum field theory, as well as mathematicians interested in the applications of commutators and differential operators in physics.

oliveriandrea
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Hello,
i try to prove that
μFμ[itex]\nu[/itex] + ig[Aμ, Fμ[itex]\nu[/itex]] = [Dμ,Fμ[itex]\nu[/itex]]
with the Dμ = ∂μ + igAμ

but i have a problem with the term Fμ[itex]\nu[/itex]μ ...
i try to demonstrate that is nil, but i don't know if it's right...

Fμ[itex]\nu[/itex]μ [itex]\Psi[/itex] = [itex]\int[/itex] (∂[itex]\nu[/itex]Fμ[itex]\nu[/itex]) (∂μ[itex]\Psi[/itex]) + [itex]\int[/itex] Fμ[itex]\nu[/itex]μ[itex]\nu[/itex] [itex]\Psi[/itex] = (∂[itex]\nu[/itex]Fμ[itex]\nu[/itex]) [[itex]\Psi[/itex] ] - [itex]\int[/itex][itex]\Psi[/itex]∂μ[itex]\nu[/itex]Fμ[itex]\nu[/itex] = 0

with [itex]\Psi[/itex] a smooth function, nil at infinity

if it's wrong please do you post the right answers? and why it is wrong...
thank you
 
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This is right. Of course, if you need it mathematically accurate, you have to think about, how fast you test function has to go to 0 at infinity, but if it's a physics question, what you did should be sufficient.
 
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Thank you (yes it's physics question)
 

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