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We define Lie algebras abstractly as algebras whose multiplication satisfies anti-commutativity and Jacobi's Identity. A particular instance of this is an associative algebra equipped with the commutator bracket: $[a,b]=ab-ba$. However, the notation suggests that this bracket is the one we think about. Additionally, the right adjoint to the functor I just mentioned creates the universal enveloping algebra by quotienting the tensor algebra by the tensor version of this bracket; but we could always start with some arbitrary Lie algebra with some other satisfactory bracket and apply this functor.

My question is

"Why the commutator bracket?"

Is it purely from a historical standpoint (and if so could you explain why)? Or is there a result that says any Lie algebra is essentially one with the commutator bracket (maybe something about the faithfulness of the functor from above)?

I know of (a colleague told me) a proof that the Jacobi identity is also an artifact of the right adjoint to the universal enveloping algebra. He can show that it is the necessary identity for the universal enveloping algebra to be associative (if someone knows of this in the literature I would also appreciate the link to this!)

I hope this question is clear, if not, I can revise and try to make it a bit more specific.

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I would have responded that "any Lie algebra can be obtained as a subalgebra from Lie of an associative algebra," but you've already mentioned the enveloping algebra functor in your answer. Also, I'm pretty sure the enveloping algebra functor is always faithful. I'm not sure what you mean by "the Jacobi identity is the necessary identity for the universal enveloping algebra to be associative." I learned the enveloping alg as a quotient of a tensor algebra, which is always associative. –  Akhil Mathew Jul 22 '10 at 0:51
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I'm also not sure what the Jacobi identity has to do with the enveloping algebra being associative, but it is pivotal in the proof of the Poincare-Birkoff-Witt theorem, which says that the universal algebra is isomorphic, as a vector space, to the symmetric algebra on the underlying vector space of the Lie algebra. –  Eric O. Korman Jul 22 '10 at 1:00
    
Eric apparently can read my mind and through my stupid errors. Sorry for the stupidity of this error. –  BBischof Jul 22 '10 at 1:07
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That's a forgetful functor, so it's faithful too. –  Akhil Mathew Jul 22 '10 at 1:23
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The Jacobi identity is also necessary for the adjoint map to be a Lie algebra representation. –  B R Aug 23 '12 at 15:20

1 Answer 1

up vote 2 down vote accepted

Well Lie algebras naturally arise from the Lie bracket of vector fields and from taking the Lie algebra of a Lie group. If we look at a the Lie algebra of a matrix subgroup, then the Lie bracket is the commutator of matrices.

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Also the Lie bracket of vector fields is really a commutator (when considered as operators on the space of smooth functions). –  Akhil Mathew Jul 22 '10 at 1:03
    
Yes this what I was looking for. Thanks for the quick response, I am unsure how I missed this, it now seems obvious, but still glad I asked. –  BBischof Jul 22 '10 at 1:07

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