Associativity of the Killing form

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Discussion Overview

The discussion centers around proving the associativity of the Killing form in the context of Lie algebras, specifically the identity B([X,Y],Z) = B(X,[Y,Z]). Participants explore various approaches to the proof, including the use of traces and commutators.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents an initial attempt at the proof using the trace of the adjoint representation and expresses uncertainty about the next steps.
  • Another participant suggests expanding the last commutator and applying the property of trace that allows rearranging terms.
  • A participant questions whether the product of the adjoint representations can be treated as a commutator, indicating a potential misunderstanding of the definitions in general Lie algebras.
  • Further clarification is provided that the adjoint representation can indeed be expanded as a commutator, reinforcing the linear operator nature of ad(X) and ad(Y).
  • One participant reflects on their exam experience, expressing frustration over not recalling the simpler proof during the test.
  • Another participant shares their experience of needing to prove similar identities multiple times in a different exam context, indicating the practical relevance of the discussion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to the proof, and there are differing views on the treatment of the Lie bracket in general Lie algebras. The discussion remains unresolved regarding the most effective method to demonstrate the associativity of the Killing form.

Contextual Notes

Participants express uncertainty about the definitions and properties of the adjoint representation in the context of general Lie algebras, which may affect their reasoning. There is also mention of varying levels of difficulty between applied and pure exams.

Who May Find This Useful

Students and researchers interested in Lie algebras, the properties of the Killing form, and those preparing for exams in theoretical physics or mathematics may find this discussion relevant.

Cexy
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This came up in an exam on Lie algebras that I had today, and it's been bugging me. How do you prove that

[tex]B([X,Y],Z)=B(X,[Y,Z])[/tex]?

The best I've managed is writing

[tex]B([X,Y],Z)=\mathrm{Tr}(\mathrm{ad}([X,Y])\mathrm{ad}(Z))=\mathrm{Trace}([\mathrm{ad}(X),\mathrm{ad}(Y)]\mathrm{ad}(Z))[/tex]

but I have no idea where to go from there. Hints and/or a complete proof are both appreciated :)
 
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Expand the last commutator and use [itex]\mathrm{Trace} \left( AB \right) = \mathrm{Trace} \left( BA \right)[/itex] on one of the terms.
 
Sure, that's how I would do it if the Lie bracket was defined as a commutator, but it's not in a general Lie algebra.

Oh, hold on... since [tex]\mathrm{ad}:g\to gl(g)[/tex] does that mean that I can expand the product [ad(X),ad(Y)] as a commutator?
 
Cexy said:
Sure, that's how I would do it if the Lie bracket was defined as a commutator, but it's not in a general Lie algebra.

Oh, hold on... since [tex]\mathrm{ad}:g\to gl(g)[/tex] does that mean that I can expand the product [ad(X),ad(Y)] as a commutator?

Yes,

[ad(X) , ad(Y)] = ad(X)ad(Y) - ad(Y)ad(X),

where the associative products are defined, since ad(X) and ad(Y) are both linear operators on g considered as a vector space.
 
It's annoying that it's that simple but I couldn't do it in the exam. :(

Still, I did much more complicated stuff so hopefully the examiners will assume I just had a dim moment - rather than assuming that I'm dim altogether!
 
I thought I'd post to mention that I had to do this proof (well, variants of it) THREE times in a particle physics exam today, so thanks for clearing it up in my mind - you scored me some marks!

Applied exams are much easier than pure ones.
 

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