Free Lie algebras and Witt's formula

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Prove that there exists a universal object for the category LAn(K)LA_n(K) of Lie algebras with nn distinguished generators over the field KK. The corresponding object is called the free Lie algebra with nn generators.

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Remark.
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An object AA of a category is called a universal repelling object (or, briefly, a universal object) if for every object BB there is exactly one morphism from AA to BB.

Let G\mathfrak G be a Lie algebra over a field KK of characteristic zero. We consider the category K(G)K(\mathfrak G) whose objects are those linear mappings φ\varphi from G\mathfrak G into associative algebras (depending on the object) having the property that

φ([x,y])=φ(x)φ(y)−φ(y)φ(x). \varphi ([x,y])=\varphi (x)\varphi (y)-\varphi (y)\varphi (x).

A morphism of an object φ ⁣:G→A\varphi \colon \mathfrak G\to A to an object ψ ⁣:G→B\psi \colon \mathfrak G\to B is defined to be a homomorphism χ ⁣:A→B\chi \colon A\to B which makes the following diagram commute:

Prove that K(G)K(\mathfrak G) has a universal object φ0 ⁣:G→U(G)\varphi_0\colon \mathfrak G\to U(\mathfrak G). The algebra U(G)U(\mathfrak G) is called the associative hull or enveloping algebra of G\mathfrak G.

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Remark.
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An object AA of a category is called a universal repelling object (or, briefly, a universal object) if for every object BB there is exactly one morphism from AA to BB.

Prove that the associative hull (see Problem 51) of the free Lie algebra with nn generators is isomorphic to the tensor algebra over an nn-dimensional space.

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Remark.
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An object AA of a category is called a universal repelling object (or, briefly, a universal object) if for every object BB there is exactly one morphism from AA to BB.

Let A(q)=Ass⁡F(X1,…,Xq)A(q)=\operatorname {Ass}_{F}\left(X_{1}, \ldots , X_{q}\right) be the free associative algebra over a field FF, generated by qq free generators — noncommuting variables X1,…,XqX_{1}, \ldots , X_{q}. Setting

Am(q)=⟨Xi1Xi2…Xim∣1⩽ij⩽q⟩F,dim⁡Am(q)=qm A_{m}(q)= \left\langle X_{i_{1}} X_{i_{2}} \ldots X_{i_{m}} \mid 1 \leqslant i_{j} \leqslant q\right\rangle _{F}, \quad \operatorname {dim} A_{m}(q)=q^{m}

we note that A(q)A(q) is a graded algebra:

A(q)=F⋅1⊕A1(q)⊕A2(q)⊕A3(q)⊕… A(q)=F \cdot 1 \oplus A_{1}(q) \oplus A_{2}(q) \oplus A_{3}(q) \oplus \ldots

A(q)A(q) contains the free Lie algebra L(q)=Lie⁡(X1,…,Xq)L(q)=\operatorname {Lie}\left(X_{1}, \ldots , X_{q}\right) with the same free generators and the commutator operation [UV]=UV−VU[U V]=U V-V U. The algebra L(q)L(q) is also graded:

L(q)=L1(q)⊕L2(q)⊕L3(q)⊕… L(q)=L_{1}(q) \oplus L_{2}(q) \oplus L_{3}(q) \oplus \ldots

where

L1(q)=⟨X1,…,Xq⟩F,L2(q)=⟨[Xi,Xj]∣i<j⟩F,… L_{1}(q)= \left\langle X_{1}, \ldots , X_{q}\right\rangle _{F}, \quad L_{2}(q)= \left\langle \left[X_{i}, X_{j}\right] \mid i<j\right\rangle _{F}, \quad \ldots

Using the Jacobi identity, verify that

L3(q)=⟨[[Xi,Xj],Xk]⟩,i<j,k⩽j;dim⁡L3(q)=13(q3−q) L_{3}(q)= \left\langle \left[\left[X_{i}, X_{j}\right], X_{k}\right]\right\rangle , \quad i<j, k \leqslant j ; \quad \operatorname {dim} L_{3}(q)=\frac{1}{3}\left(q^{3}-q\right)

In fact, the general Witt formula holds

dim⁡Lm(q)=Ψm(q)=1m∑d∣mμ(md)qd \operatorname {dim} L_{m}(q)=\Psi _{m}(q)=\frac{1}{m} \sum _{d \mid m} \mu \left(\frac{m}{d}\right) q^{d}

coinciding exactly with formula (12). The essential difference is only that in (12') qq is an arbitrary natural number, while in (12) it is a power of a prime number.

The number of necklaces from Problem 2 in the preamble to Ch. 3 is expressed by the same formula.

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References

[KiGv88] Kirillov A.A., Gvishiani A.D. Теоремы и задачи функционального анализа. 2-е изд., Наука, 1988.
[Кос04] Кострикин А.И. Основные структуры алгебры. ФИЗМАТЛИТ, 2004.