What is the difference between lie product and commutation relation?

mak_294
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Hi guys

I'm quit confuse here I want to know what's the difference between lie product and commutation relation? I've been told that every commutator is lie product but not the way around, but I can't see the difference?
 
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The moral of the story is: the commutator is an example of a Lie product, but not every Lie Product is given by the commutator.

The commutator is defined as:

[A,B] = AB - BA

It satisfies three basic properties, namely:
-Linearity
-Antisymmetry
-Jacobi Identity

The Lie product is defined to be an operation which satisfies these three properties (including closure of the Lie product). Since the commutator satisfies these three identities, it is therefore a Lie product. On top of that, in a lot of practical cases it turns out that the Lie Product coincides with the operation of the ordinary commutator. A famous example are the Lie algebras of matrix groups, such as su(n) and so(n). The operators A and B are then matrices and the commutator is given in the usual way.

But still, not every Lie Product is given by the commutator.

A famous example is the Poisson bracket. Given x and p as canonical coordinates, the Poisson bracket of two functions f and g of these coordinates is given by the following operation

\{f,g\} = \partial_x f \partial_p g - \partial_g f \partial_x g

You can check that this operation is indeed a Lie product since it satisfies all the identites given above. However, it is clear that it is not a commutator.

For mathematicians the Lie product is an abstract product, that doesn't even need a practical realisation - it just satisfies the identities given above. To make things worse, mathematicians often use the same notation [A,B] to denote the Lie product, even though they do not mean the ordinary commutator. This often leads to confusion. The reason why it is an abstract product is that some algebraic spaces have the notion of a Lie product [A,B], but not of the ordinary product AB. So in these spaces you are only allowed to multiply elements using the Lie Product [A,B], but not the ordinary multiplication of AB -- one reason is that sometimes the algebraic space is closed uner the Lie product [A,B], but not under the ordinary multiplication AB meaning multiplication takes maps to something outside the algebraic space.
 
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Xepma, many thanks for your answer. It helped me a lot. Even though I've read the Poisson bracket before in Jones' book, you make it much clearer. However, and excuse my naive question, is the difference because of the existence of the derivation?
 
Yes that's exactly the reason! Another example of a Lie product is the wedge/cross product by the way.

(there's a type in my definition of the Poisson bracket by the way).
 
Xepma, many thanks. that helped me a lot. And yes I got Poisson bracket mistake.
 
mak_294 said:
Xepma, many thanks for your answer. It helped me a lot. Even though I've read the Poisson bracket before in Jones' book, you make it much clearer. However, and excuse my naive question, is the difference because of the existence of the derivation?

The main difference is that the Poisson bracket is used to describe classical physics, and the commutator (after multiplication with i/hbar) is its quantum analogue.

See http://de.arxiv.org/abs/0810.1019 for more.
 
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