Construction of Lie algebra $g(A)$ in Victor Kac's book

Question

In Kac's book "Infinite Dimensional Lie Algebras" Chapter I, he constructed an infinite Lie algebra $g(A)$ starts from any $n\times n$ complex matrix $A$ as follows:

Let $\mathfrak{h}$ be a vector space and $\mathfrak{h}^\ast$ be its dual space, $\alpha_1^\vee,\cdots,\alpha_n^\vee$ are linearly independent vectors in $\mathfrak{h}$ and $\alpha_1,\cdots,\alpha_n$ are linear independent vectors in $\mathfrak{h}^\ast$ such that $(\alpha_i^\vee,\alpha_j)=a_{ij}$. The pair $\mathfrak{h}$ and $\mathfrak{h}^\ast$ always exists and is called the realization of $A$.

Let $g(A)$ be the lie algebra generated by $e_i,f_i$ and $\mathfrak{h}$ with following restrictions:

1. $[h.h']=0$ for any $h,h'\in\mathfrak{h}$.

2. $[e_i,f_j]=\delta_{ij}\alpha^{\vee}_i$.

3. $[h,e_i]=\alpha_i(h)e_i$.

4. $[h,f_i]=-\alpha_i(h)f_i$.

Kac proved the existence of $g(A)$ by constructed a highest weight representation (with any given highest weight $\lambda\in\mathfrak{h}^\ast$) of the generators $e_i,f_i,\mathfrak{h}$ as follows:

Let $V$ be a $n$ dimensional space with a basis $v_1,\cdots,v_n$, $T(V)$ be the associative tensor algebra of $V$, Let $e_i,f_j,\mathfrak{h}$ act on $T(v)$:

1. $f_i(a)=v_i\otimes a$ for any $a\in T(V)$.

2. $h(1)=\lambda(h)1$ and inductively on the length of $a\in T(V)$ for any $h\in\mathfrak{h}$.

3. $e_i(1)=0$ and inductively on the length of $a\in T(v)$.

Here I have suppressed the concrete expressions because they are tedious but can be easily recovered if one keeps the "highest weight representation" in mind.

Here is the stone that stucked me: Kac asserted that $U(\mathfrak{n}^-)$ is nothing but just $T(V)$. Where $\mathfrak{n^-}$ is the free Lie algebra generated by the $f_i$'s and $U(\mathfrak{n}^-)$ be its universal enveloping algebra. He just mensioned that the mapping $f_i\rightarrow v_i$ gives the associative algebra isomorphism between $U(\mathfrak{n}^-)$ and $T(V)$. I want to know why this is true and how one can deduce from this that $\mathfrak{n}^-$ is a free Lie algebra.

Thanks for any help in advance!

Answer 1

As far as I can tell, most of the text in the body is not relevant to the question you actually want to ask. Let $V$ be a vector space and $L(V)$ the free Lie algebra on $V$. You want to know why $U(L(V))$ is the tensor algebra $T(V)$. (Right?) The reason is that they have the same universal property: recall that, by definition,

• the universal enveloping algebra construction is left adjoint to the forgetful functor from associative algebras to Lie algebras,
• the free Lie algebra construction is left adjoint to the forgetful functor from Lie algebras to vector spaces, and
• the tensor algebra construction is left adjoint to the forgetful functor from associative algebras to vector spaces.

Hence $U(L(V))$, the free associative algebra on the free Lie algebra on $V$, is just the free associative algebra on $V$; that is, adjoints compose. More formally, if $A$ is an algebra, we have natural bijections

$$\text{Hom}_{\text{Alg}}(U(L(V)), A) \cong \text{Hom}_{\text{Lie}}(L(V), A) \cong \text{Hom}_{\text{Vect}}(V, A) \cong \text{Hom}_{\text{Alg}}(T(V), A).$$

Answer 2

From the action of $g(A)$ on $T(V)$, we get an action of $\mathfrak{n}^-$ on $T(V)$ by restriction. Consider the linear map $\mathfrak{n}^- \rightarrow T(V)$ defined by $x \mapsto x(1)$. This maps $f_i$ to $f_i(1)=v_i$ and is in fact a Lie algebra homomorphism. To see this, first observe

$$[f_i, f_j](1)=f_i(f_j(1)) - f_j(f_i(1)) = v_i \otimes v_j - v_j \otimes v_i = [v_i, v_j] = [f_i(1), f_j(1)] $$

Let's say $S \subseteq \mathfrak{n}^-$ satisfies the property $P$ if $[x,y](1)=[x(1), y(1)]$ hold for all $x, y \in S$. By the Jacobi identity, $\{x, y, z\}$ satisfies $P$ implies $\{[x,y], z\}$ satisfies $P$. Furthermore, if $\{x, y, z, w\}$ satisfies $P$, then $\{[x,y], z,w\}$ satisfies $P$, whence $\{[x,y], [z,w]\}$ satisfies $P$. As a result, if a linear subspace $W$ of $\mathfrak{n}^-$ satisfies $P$, then $W + [W, W]$ satisfies $P$. Now put $Y_0=\mbox{span}\{f_1, \cdots, f_n\}$ and $Y_{n+1} = Y_n + [Y_n, Y_n]$. Then $Y=\cup_{n=0}^{\infty}Y_n$ satifies $P$. Note that $Y$ is a Lie subalgebra of $\mathfrak{n}^-$ because if $a \in Y_m$ and $b \in Y_n$, then $[a, b] \in [Y_k, Y_k] \subseteq Y_{k+1}\subseteq Y$ where $k = \max(m, n)$. Since $Y$ contains all $f_i$, $Y$ must equal to whole $\mathfrak{n}^-$. In other words, $\mathfrak{n}^- \ni x \mapsto x(1) \in T(V)$ is a Lie algebra homomorphism.

Note that this fails when $\mathfrak{n}^-$ is replaced by $g(A)$. For example, $[e_i(1), f_j(1)]= [0, v_j] = 0$ whereas $[e_i, f_j](1)=\delta_{ij} \alpha^\vee_i(1)=\delta_{ij}\lambda(\alpha^\vee_i)$.

Using the Lie algebra homomorphism $f_i \mapsto v_i$, we are able to construct an isomorphism between $U(\mathfrak{n}^-)$ and $T(V)$. By the universal property of $U(\mathfrak{n}^-)$, there exists the (unital) algebra homomorphism $U(\mathfrak{n}^-) \to T(V)$ sending $f_i$ to $v_i$. Conversely, the universal property of $T(V)$ yields the algebra homomorphism $T(V)$ to $U(\mathfrak{n}^-)$ sending $v_i$ to $f_i$. These two maps are the inverses of each other.

Under this isomorphism, $\mathfrak{n}^-$ maps onto the Lie subalgebra of $T(V)$ generated by $v_i$. This is the free Lie algebra on $\{v_1, \cdots, v_n\}$. Thus $\mathfrak{n}^-$ is freely generated by $f_i$'s.