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In wikipedia, free Lie algebras are defined using the universal property. Can anyone give some concrete examples of free Lie algebras?

In mathematics, a free Lie algebra, over a given field $K$, is a Lie algebra generated by a set $X$, without any imposed relations." If we take the field $K$ to be the field $\mathbb{C}$ of complex numbers, what is the free algebra if

  1. $X=\emptyset$?
  2. $X=\{0\}$?
  3. $X=\{0,1,2\}$?
  4. $X=\mathbb{R}$?
  5. $X=\mathbb{C}$?
Jendrik Stelzner
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Zuriel
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  • Are you asking for the construction? Or an example where it comes up in practice? See page 4 of http://www.math.uni-hamburg.de/home/wockel/teaching/data/cohomology_of_lie_algebras_imani.pdf for the construction. – dumbquestiondude Nov 13 '14 at 15:54
  • @dumbquestiondude, I am trying to understand this concept and I thought it might be helpful to look at some examples. – Zuriel Nov 13 '14 at 15:57

2 Answers2

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So if, say, $X=\{1,2\}$, then the free Lie algebra $L(X)$, with $x\mapsto e_x$, consists of all linear combinations of the elements $e_1,e_2$, $[e_1,e_2]$, $[e_1,[e_1,e_2]]$, $[e_2,[e_1,e_2]]$, $[e_1,[e_1,[e_1,e_2]]$, $[e_1,[e_2,[e_1,e_2]]$, $[e_2,[e_1,[e_1,e_2]]$, $[e_2,[e_2,[e_1,e_2]]$, and so forth, such that skew-symmetry and the Jacobi identity holds.

More formally, a free Lie algebra $L(x,y)$ on two letters $x,y$ has its elements in the ring of polynomials $K\langle x,y\rangle$ in the non-commuting variables $x,y$, where $K$ is a field.

Dat Minh Ha
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Dietrich Burde
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As I don't have enough rep to comment under @Dietrich Burde answer, and since an edit to his answer I performed was rejected, here is an interesting comment (in my opinion) on the free Lie algebra $L(X)$ (where $X = \lbrace 1,2\rbrace $) that Dietrich Burde starts to describe in his answer.

As it happens, all the vectors in $ B = \lbrace e_1, e_2, [e_1,e_2], [e_1,[e_1,e_2], [e_2,[e_1,e_2]]\rbrace $ are linearly independent in $L(X)$, i.e. the skew-symmetry and the Jacobi identity do not produce any unexpected relation.

On the other hand, by skew-symmetry and the Jacobi identity, we actually have $[e_1,[e_2,[e_1,e_2]]] = [e_2,[e_1,[e_1,e_2]]]$. If one wishes to go further and add "words of length 4" in our linearly independent set $B$, one can use the so called Lyndon basis to get that the vectors $e_1, e_2, [e_1,e_2], [e_1,[e_1,e_2], [[e_1,e_2],e_2], [e_1,[e_1,[e_1,e_2]]], [[e_1,[e_1,e_2]],e_2]$ and $[[[e_1,e_2],e_2],e_2]$ are linearly independent in $L(X)$.

Going to the next word length would not be too tedious either, since by the Necklace polynomial formula, the number of linearly independent "words of length 5" in $L(X)$ is $\frac{1}{5}(2^5-2) = 6$. Those $6$ linearly independent "words of length 5" in $L(X)$ can be constructed using the 6 Lyndon words of length 5 on two symbols (and the recipe for this construction is given in the Wikipedia article on free Lie algebras).

user293657
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  • Do you have any literature recommendation for Lyndon bases in the context of Lie algebras? – bliipbluup May 05 '20 at 06:35
  • @UnexpectedExpectation Sorry, I don't have recommendations for that. I just blindly trusted the wikipedia article to write this answer. The way I would go would be to start by looking at the references given in this wikipedia article. – user293657 May 18 '20 at 16:40