# How to show that the graded dual of the universal enveloping algebra of a free Lie algebra on a finite set is the shuffle algebra

In the article, the universal enveloping algebra of a free Lie algebra on a set X is defined to be the free associative algebra generated by X.

It is said that the graded dual of the universal enveloping algebra of a free Lie algebra on a finite set is the shuffle algebra.

What is the meaning of "graded dual"? How to show that the graded dual of the universal enveloping algebra of a free Lie algebra on a finite set is the shuffle algebra? Some examples to explain this result will be greatly appreciated!

• The universal enveloping algebra of the free Lie algebra on a set $X$ is (canonically isomorphic to) the free algebra on $X$. Nov 24 '16 at 8:03
• For graded duals, see (e.g.) Section 1.6 in Vic Reiner's and my Hopf algebras in Combinatorics ( cip.ifi.lmu.de/~grinberg/algebra/HopfComb-sols.pdf ). There you will also find the fact that the graded dual of the free algebra is the shuffle algebra; combined with my previous comment, this should answer your question. (The free algebra appears in the guise of the tensor algebra of a free module.) Nov 24 '16 at 8:07
• @darij grinberg, thank you very much. What is the canonical isomorphism from $U(Lie(V))$ to $T(V)$? Jan 18 '17 at 15:34
• There is a canonical vector space homomorphism $V \to T\left(V\right)$. By the universal property of the free Lie algebra, it gives rise to a Lie algebra homomorphism $\operatorname{Lie}\left(V\right) \to T\left(V\right)$. By the universal property of the universal enveloping algebra, the latter gives rise to an algebra homomorphism $U\left(\operatorname{Lie}\left(V\right)\right) \to T\left(V\right)$. This is the isomorphism I was talking about. Its inverse can also be constructed using a universal property (this time, the one of $T\left(V\right)$). Jan 18 '17 at 15:38
• @darij grinberg, thank you very much. Jan 18 '17 at 15:45

All that has been said in the excellent comments by Darij is true for modules (the set of scalars being a ring) except maybe Radford's theorem (see below). I complete here what has been said, too long for a comment though.

The shuffle product is called such because it "shuffles" the letters of two words considered as card decks. If you want to get all shuffles between two words you quickly come to the recursion (with $$a,b$$ being letters of an alphabet $$X$$ and $$u,v\in X^*$$ words with their letters in $$X$$) $$au * bv=a(u * bv)+b(au * v)\quad (R2)$$ which means $$\mbox{ all shuffles}=\mbox{shuffles beginning by "a"}+\mbox{shuffles beginning by "b"}$$ ($$a$$ being seen as the first card of the deck $$au$$ and $$b$$, the first of the deck $$bv$$). If you pursue the recursion, you arrive at the empty word and must initialize the recursion, considering it as a neutral $$u* 1_{X^*}=1_{X^*}* u=u\quad (R1)$$ Of course, the output being a sum, you must linearize the situation and consider it as an algebra law on $$R\langle X\rangle$$ ($$=R[X^*]$$, the $$R$$-algebra of the free monoid $$X^*$$ i.e. the free algebra or the algebra of noncommutative polynomials) defined by $$R1-R2$$. It turns out that $$(R\langle X\rangle,*,1_{X^*})$$ is a $$R$$-CAAU (commutative, associative algebra with unit) in all cases (and an algebra of polynomials, a free CAAU, in case $$R$$ is a $$\mathbb{Q}$$-algebra, which is Radford's theorem).

Now, to come to your question, given a set $$X$$ and a ring $$R$$, one builds

1. the free monoid $$X^*$$
2. the free Lie algebra $$Lie_R\langle X\rangle$$ (in the category of $$R$$-Lie algebras)
3. the free algebra $$R\langle X\rangle$$ (in the category of $$R$$-AAU)

by transitivity of free objects, one gets at once (and for all $$R$$) that $$\mathcal{U}(Lie_R\langle X\rangle)$$ is isomorphic to $$R\langle X\rangle$$.

At this point, it is not granted that $$Lie_R\langle X\rangle$$ be the Lie subalgebra of $$R\langle X\rangle$$ generated by letters. This is due to the magic of Lyndon words [Lothaire, Combinatorics on words] (or Hall sets [Bourbaki, Lie chapter II, Reutenauer, Free Lie Algebras]) saying that, for all $$R$$, $$Lie_R\langle X\rangle$$ is free as a $$R$$-module.

As for the graded dual, one can prove your statement by constructing the enveloping bialgebra $$\mathcal{B}=(R\langle X\rangle,conc,1_{X^*},\Delta,\epsilon)$$ (considering that $$R\langle X\rangle\simeq \mathcal{U}(Lie_R\langle X\rangle)$$ (with, this time, $$Lie_R\langle X\rangle$$ the Lie subalgebra of $$R\langle X\rangle$$ generated by the letters). Then considering, for $$x\in X$$, $$\Delta(x)=x\otimes 1+1\otimes x$$, one gets the very nice expression (for any word $$w\in X^*$$ of length $$n$$) $$\Delta(w)=\sum_{I+J=[1..n]} w[I]\otimes w[J]\quad (3)\ .$$ In order to get the dual, one must restrict the linear forms. The largest dual is Sweedler's one (you have a quick definition here), here your question is about the graded dual. In both cases one gets (with a small abuse of language) $$\mathcal{B}^\vee=(\mathcal{B}^\vee,*,1_{X^*},\Delta_{conc},\epsilon)$$ because the dual of $$\Delta$$ (which is a law of algebra) is precisely the shuffle product.

Now, just a final remark : if the alphabet in finite, you use the grading by the length of the words and if the alphabet is infinite you must use the grading by multidegree (because $$S_X=\sum_{x\in X}\,x$$ is not rational, which means that $$\Delta_{conc}\,S_X$$ cannot be computed in our framework) and then get $$R\langle X\rangle^\vee=R\langle X\rangle$$.

Hope it helps ! (do not hesitate to ask for clarification.)