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I’m reading Stillwell’s Naive Lie Theory, where he defined Lie bracket on page 80 as

$$[x, y]=xy-yx$$

However, on page 82 he said

In general , a Lie algebra is a vector space with a bilinear operation $[,]$ , satisfying

$$[x,y]=-[y,x]$$ $$[x,[y,z]] + [y,[z,x]] + [z, [x,y]] =0$$

I’m a bit confused here — is Lie bracket always in the form of $[x, y]=xy-yx$ ? Or actually this is just the most common form, there are other forms, while the basic requirements are only the anti-symmetric and Jacobi identity ?

Sorry for such a naive question , I tried check on wiki but the pages are a bit too advanced for me.

athos
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    I think the first definition is the Lie bracket on vector fields (which creates a Lie algebra) and the other definition is that of an (abstract) Lie algebra in which you have, I believe, no multiplication in general, so $xy-yx$ does not make sense. – fweth Oct 12 '22 at 08:27
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    On any associative algebra $A$, the first expression defines a Lie bracket making it into a Lie algebra. Another, less interesting one, is $[x,y]=0$, which is not the same when $A$ is non-commutative – doetoe Oct 12 '22 at 08:56
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    On $\mathbb{R}^3$ the cross product also defines a Lie bracket. The so called commutator bracket you mentioned is just an example for a Lie bracket. It does however, come up very often when studying lie algebras because it is the Lie bracket of the linear algbera $\mathfrak{gl}(V)$, when $V$ is a vector space. – QED Oct 12 '22 at 09:25

2 Answers2

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No, a Lie bracket $[x,y]$ on a vector space $V$ need not be a commutator $xy-yx$, since there may not be a product $xy$ of vectors defined. Of course, for matrix algebras there is always a product $xy$, so then it is true, see the reference below.

A famous theorem by Ado says that every finite-dimensional Lie algebra $L$ over a field $K$ has a faithful finite-dimenional representation $L\hookrightarrow \mathfrak{gl}_n(K)$. Then we may identify $L$ with the image and the Lie bracket there is given by the commutator. Indeed, $\mathfrak{gl}_n(K)$ is the Lie algebra of the matrix algebra $M_n(K)$, together with the Lie bracket $[A,B]=AB-BA$.

Is the Lie bracket of a Lie Algebra of a Matrix Lie Group always the commutator?

Edit: And for a very similar question see also here:

Does every Lie algebra have commutator as its Lie bracket up to isomorphism?

Dietrich Burde
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You have to separate the Lie bracket like an "abstract" operation, to define a Lie algebra, and the Lie bracket of vector fields.

Something like taking apart the operation "sum" of real number, and the abstract operation sum in the definition of a ring. The former is a particular case of the latter.

In the case of the Lie bracket, apart from the abstract case, the more common realization is the case of vector fields. Given two vector fields $X,Y$ on a manifold $M$, the Lie bracket is a new vector field which acts on a smooth function $f$ in this way: $$ [X,Y](f):=X(Y(f))-Y(X(f)) $$

If the manifold is a Lie group $G$, the Lie bracket of left-invariant vector fields is again a left invariant vector field. This property induces a Lie bracket (in the abstract sense) in the vector space $T_eG$ (the tangent space at the identity). Then you have a Lie algebra associated to the Lie group.

In case the Lie group were a subgroup of $GL(n)$ (a matrix group) then its associated Lie algebra is "made of matrices" and the induced Lie bracket correspond to $$ [A,B]=AB-BA $$

A. J. Pan-Collantes
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