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I have already tried to understand this good explanation: Basis of free Lie algebras

But still I have a confusion in the following case: Let $L(a,b)$ be a free Lie algebra of rank 2. How is it possible to say that the elements $a, [b,a], [b,[b,a]], [b,b,[b,a]]], ...$ are a basis for a free Lie subalgebra of $L(a,b)$? Because if we do the following calculation between two elements of this basis and use Jacobi Identity we have

[a,[b,a]]=-[a,[a,b]]-[b,[a,a]]= -[a,[a,b]]

So, the bracket of two elements of the basis can not be written as the bases element! Can we identify that Lie subalgebra with the above-mentioned basis?

Nil
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  • If we assume that b>a then [b,a] is a Lyndon term and ba is Lyndon word. But unfortunately I can not understand how it can be a basis – Nil Jun 12 '23 at 07:37

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