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I am reading the paper The Nagata Automorphism is Wild , where the authors state:

Let $L$ be a free Lie algebra with free generators $x_1, x_2, \dots, x_n$.

I understand the categorical (universal property) definition of ``free'' in this context. However, I also know an alternative description:

Categorical Definition: Let $X$ be a set and $i: X \to L$ be morphism of sets from $X$ into a Lie algebra $L$. Then $L$ is called $\textbf{free}$ on $X$ if for every Lie algebra $A$ and every map $f: X \to A$, there exists a unique Lie algebra homomorphism $\tilde{f} : L \to A$ such that $\tilde{f} \circ i = f$.

Presentation Definition: A free Lie algebra over a field $K$ is a Lie algebra generated by the set $X$ without any imposed relations other than those required by the Lie algebra axioms, namely, the properties of alternating $K$-bilinearity (which implies antisymmetry) and the Jacobi identity.

While the first definition is clear to me, I have the following questions regarding the second one:

  1. How should one understand the second definition in more elementary terms?

  2. Is there a simple example that illustrates the construction of a free Lie algebra?

  3. How can one prove that these two definitions are equivalent?

Any insights, examples, or references on proving this equivalence rigorously would be appreciated.

Thank you in advance for your help!

KReiser
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Santa-clause
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    Usually one gives a concrete model for an object satisfying your second definition, and then you prove that it satisfies the universal property. Think about how the tensor product of modules is usually presented. – Randall Apr 10 '25 at 11:32
  • See my example here. perhaps it is helpful. – Dietrich Burde Apr 10 '25 at 11:49
  • For the structure of free Lie algebra, there is a basis called Lyndon basis, with which you can found how a basis looks like. – Peter Wu Apr 10 '25 at 12:54
  • @DietrichBurde Please confirm my understanding: in your example, any element of the free Lie group generated by ${x_1, x_2}$ is one of the elements you explicitly listed, and does not include expressions like $x_1 \cdot x_2$ or $x_1 + x_2$, where $\cdot$ and $+$ denote scalar multiplication and addition as defined in the Lie group. – Santa-clause Apr 10 '25 at 13:32
  • Indeed, $a\cdot b=[a,b]$ in a Lie algebra, so the (bilinear) multiplication in a Lie algebra is denoted by the brackets. So we have products in $x_1$ and $x_2$ here. Of course, a Lie algebra is a vector space $V$. So you can add $x_1+x_2$, and all the vectors in $V$. – Dietrich Burde Apr 10 '25 at 14:50
  • A more abstract version of the presentation definition would be: start with the set of all abstract syntax trees where the leaves are either real scalars or elements of $X$ or the formal symbol $0$ as appropriate, and the internal nodes are either $+$, binary $-$, unary $-$, $\cdot$ (for scalar multiplication), or the Lie bracket. Then, you can form an equivalence relation of which abstract syntax trees always have to evaluate to equal elements in any Lie algebra for any assignment function from $X$ to the algebra, and take the quotient by that equivalence relation. – Daniel Schepler Apr 10 '25 at 15:40
  • Alternatively to @Randall 's comment, you can also define objects by their universal property, and then show existence using the concrete model. Of course, you still have to do exactly the same work, so it's essentially a matter of preference. – aeae Apr 10 '25 at 16:57
  • Please use markdown to format non-mathematical text instead of MathJax. – KReiser Apr 11 '25 at 01:46

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A Lie algebra is a $k$-vector space. And for free Lie algebra the space is spanned by elements like $[\dots[\dots,x_i],[x_j,\dots]\dots]$, namely all possible Lie bracket formed by generators, one might imagine that those are represented by rooted binary trees, with leaf nodes colored by $X$.

Here we've described a linear space by "generators and relations". To find a basis one can look at the Lyndon basis.

Peter Wu
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