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Consider $\mathfrak{g}$ a Lie algebra, its derivated subalgebra $D\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, and the quotient Lie algebra $\mathfrak{g}/D\mathfrak{g}$. To understand why the quotient algebra $\mathfrak{g}/D\mathfrak{g}$ was abelian, that is $[\mathfrak{g}/D\mathfrak{g},\mathfrak{g}/D\mathfrak{g}]=0$, I had the following hand-wavy argument. Every element in $\mathfrak{g}/D\mathfrak{g}$ of the form $ab-ba$ for $a,b\in\mathfrak{g}$ is identified with their commutator $[a,b]$ which is in the $0$ equivalence class in $\mathfrak{g}/D\mathfrak{g}$, and therefore one can write $ab=ba$, that is $\mathfrak{g}/D\mathfrak{g}$ is indeed abelian.

However, I realised this doesn't work; multiplication isn't defined in a Lie algebra, and therefore one can't make sense of an element like $ab$; this could be understood as an element of the universal enveloping algebra $U\mathfrak{g}$ of $\mathfrak{g}$. Therefore, the picture I had of $ab=ba$ doesn't work. This leads me to the following question;

My question: how to think of elements of an abelian Lie algebra if the naive picture of $ab=ba$ for $a,b\in\mathfrak{g}$ doesn't works; does the interpretation of abelian Lie algebra will depend on the Lie bracket chosen?

This question is quite closely related. While it asks about the meaning of the commutator bracket, I am concerned with the notion of abelian Lie algebra, and if its interpretation is related to the Lie bracket.

QuantizedObject
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1 Answers1

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The notion of abelian Lie algebras comes from the notion of abelian Lie groups, like $\Bbb R^n$. But for Lie algebras, it is not the identity $ab=ba$, since a product $ab$ doesn't exist in general on a vector space $V$. The only product, which we have, is the Lie product $x\cdot y$, usually denoted by brackets, i.e. $$ x\cdot y=[x,y]. $$ A Lie algebra $(V,[,])$ with the null product $[x,y]= 0$ for all $x,y\in V$ is called abelian. For a matrix algebra, where we have a natural commutator of matrices $A$ and $B$, it really means $[A,B]=AB-BA=0$ for all $A,B$.

Now we have the result that a connected Lie group is abelian if and only if its Lie algebra is abelian - see here:

If $G$ is an abelian Lie group then the Lie algebra of $G$ is abelian

Commutativity of a Lie algebra $\Leftrightarrow$ the Lie group is abelian

Dietrich Burde
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  • Thank you for the clarifiaction. Thus when thinking of abstract Lie algebras, $ab=ba$ doesn't make sense. But why would it make sense when considering a matrix algebra, i.e. why can we have an element like $AB$. Is this element in the Lie algebra? – QuantizedObject Jun 07 '25 at 11:17
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    In a matrix Lie algebra, only $[A,B]=AB-BA$ is in the Lie algebra, but not $AB$ or $BA$. Consider the matrices with trace zero, i.e $L=\mathfrak{sl}_n(K)$. We do not want that $AB$ is in it, we only want that with $A,B$ in $L$, also the Lie product $A\cdot B:=[A,B]$ is in $L$. Perhaps you are confusing the Lie product with the matrix product. – Dietrich Burde Jun 07 '25 at 11:20
  • Yes, I think you are right; I might have confused the two. Thanks for the answer. – QuantizedObject Jun 07 '25 at 11:23