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\newtheorem{lemma}[thm]{Lemma}
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\title{Enveloping Algebras and The Poincar\'{e}-Birkhoff-Witt Theorem}
\author{Rahbar Virk}
\date{\today}
\begin{document}
\maketitle
\addtocounter{section}{1}
A Lie algebra $\mathfrak{g}$ is not an algebra in the sense that the Lie bracket is not associative. We would like to find an associative algebra $U(\mathfrak{g})$ such that the modules for $\mathfrak{g}$ are the same as those for $U(\mathfrak{g})$. \\
Let $\mathfrak{g}$ be a Lie algebra and let $T$ be the tensor algebra of the vector space $\mathfrak{g}$. Recall that
\[T = T^0 \oplus T^1 \oplus \cdots \oplus T^n \oplus \cdots ,\]
where $T^n = \mathfrak{g} \otimes \mathfrak{g} \otimes \cdots \mathfrak{g}$ ($n$ times); in particular $T^0 = k.1$ and $T^1 = \mathfrak{g}$; the product in $T$ is simply tensor multiplication.
Let $J$ be the two sided ideal of $T$ generated by the tensors
\[ x \otimes y - [x,y],\]
where $x,y \in \mathfrak{g}$. The associative algebra $T/J$ is called the \emph{enveloping algebra} (or sometimes the \emph{universal enveloping algebra}) of $\mathfrak{g}$ and is denoted by $U(\mathfrak{g})$. The composite mapping $\sigma$ of the canonical mappings $\mathfrak{g} \rightarrow T \rightarrow U(\mathfrak{g})$ is termed the canonical mapping of $\mathfrak{g}$ into $U(\mathfrak{g})$; observe that for all $x,y \in \mathfrak{g}$ we have that
\[ \sigma(x)\sigma(y) - \sigma(y)\sigma(x) = \sigma([x,y]) \]
We denote the canonical image in $U(\mathfrak{g})$ of $T^0 + T^1 + \cdots + T^q$ by $U_q(\mathfrak{g})$.
\begin{lemma}Let $\sigma$ be the canonical mapping of $\mathfrak{g}$ into $U(\mathfrak{g})$, let $A$ be an algebra with unity, and let $\tau$ be a linear mapping of $\mathfrak{g}$ into $A$ such that
\[\tau(x)\tau(y) - \tau(y)\tau(x) = \tau([x,y]) \]
for all $x,y \in \mathfrak{g}$. There exists one and only one homomorphism $\tau'$ of $U(\mathfrak{g})$ into $A$ such that $\tau'(1) = 1$ and $\tau' \circ \sigma = \tau$.
\end{lemma}
\begin{proof}Note that $U(\mathfrak{g})$ is generated by $1$ and $\sigma(\mathfrak{g})$ so $\tau'$ must be unique. Now let $\varphi$ be the unique homomorphism of $T$ into $A$ that extends $\tau$ and such that $\varphi(1) = 1$. For $x,y \in \mathfrak{g}$, we have
\[ \varphi(x \otimes y - y \otimes x - [x,y]) = \tau(x)\tau(y) - \tau(y)\tau(x) - \tau([x,y]) = 0, \]
hence $\varphi(J) = 0$ and thus $\varphi$ gives us a homomorphism $\tau'$ of $U(\mathfrak{g})$ into $A$ such that $\tau'(1) = 1$ and $\tau' \circ \sigma = \tau$.
\end{proof}
\begin{lemma}
Let $a_1, \ldots, a_p \in \mathfrak{g}$, $\sigma$ the canonical mapping of $\mathfrak{g}$ into $U(\mathfrak{g})$, and $\pi$ be a permutation of $\{1, \ldots, p \}$. Then
\[\sigma(a_1)\cdots \sigma(a_p) - \sigma(a_{\pi(1)})\cdots \sigma(a_{\pi(p)}) \in U_{p-1}(\mathfrak{g}). \]
\end{lemma}
\begin{proof}It suffices to prove the statement when $\pi$ is the transposition of $j$ and $j+1$. But observe that
\[ \sigma(a_j)\sigma(a_{j+1}) - \sigma(a_{j+1})\sigma(a_j) = \sigma([a_j, a_{j+1}]). \]
The lemma now follows.
\end{proof}
We now assume $\mathfrak{g}$ to a be finite dimensional Lie algebra and fix a basis $(x_1, \ldots, x_n)$ for $\mathfrak{g}$. We denote the canonical image of $x_i$ in $U(\mathfrak{g})$ by $y_i$. For every finite sequence $I=(i_1, \ldots, i_p)$ of integers between $1$ and $n$, we set $y_I = y_{i_1}y_{i_2}\ldots y_{i_p}$.
\begin{lemma}The $y_I$, for all increasing sequences $I$ of length $\leq p$, generate the vector space $U_{p}(\mathfrak{g})$.
\end{lemma}
\begin{proof}
Clearly the vector space $U_{p}(\mathfrak{g})$ is generated by $y_I$, for \emph{all} sequences $I$ of length $\leq p$. But now the required statement follows by applying the previous lemma.
\end{proof}
Let $P$ be the algebra $k[z_1, \ldots, z_n]$ of polynomials in $n$ indeterminates $z_1, \ldots, z_n$. For every $i \in \mathbb{Z}_{\geq 0}$, let $P_i$ be the set of elements of $P$ of degree $\leq i$. If $I = (i_1, \ldots,i_p)$ is a sequence of integers between $1$ and $n$, we set $z_I = z_{i_1}z_{i_2}\cdots z_{i_p}$.
\begin{lemma}For every integer $p \geq 0$, there exists a unique linear mapping $f_p$ of the vector space $\mathfrak{g} \otimes P_p$ into $P$ which satisfies the following conditions:
\begin{enumerate}
\item[$(A_p)$] $f_p(x_i \otimes z_I) = z_iz_I$ for $i \leq I$, $z_I \in P_p$;
\item[$(B_p)$] $f_p(x_i \otimes z_I) - z_iz_I \in P_q$ for $z_I \in P_q$, $q \leq p$;
\item[$(C_p)$] $f_p(x_i \otimes f_p(x_j \otimes z_J)) = f_p(x_j \otimes f_p(x_i \otimes z_J)) + f_p([x_i, x_j] \otimes z_J)$ for $z_J \in P_{p-1}$. $[$The terms in $(C_p)$ are meaningful by virtue of $(B_p)]$.
\end{enumerate}
Moreover, the restriction of $f_p$ to $\mathfrak{g} \otimes P_{p-1}$ is $f_{p-1}$.
\end{lemma}
\begin{proof}
For $p=0$ the mapping given by $f_0(x_i \otimes 1) = z_i \otimes 1$, satisfies the conditions $(A_0), (B_0)$ and $(C_0)$, furthemore it is clear that the condition $(A_0)$ forces this to be our linear mapping. Proceeding by induction, assume the existence and uniqueness of $f_{p-1}$. If $f_p$ exists then $f_p$ restricted to $\mathfrak{g} \otimes P_{p-1}$ satisfies $(A_{p-1}), (B_{p-1}), (C_{p-1})$ and is hence equal to $f_{p-1}$. Thus, to prove our claim it suffices to show that $f_{p-1}$ has a unique linear extension $f_p$ to $\mathfrak{g} \otimes P_p$ which satisfies $(A_p), (B_p), (C_p)$.
Note that $P_p$ is generated by $z_I$ for an increasing sequence $I$ of $p$ elements. Thus, we must define $f_p(x_i \otimes z_I)$ for such a sequence $I$. If $i \leq I$, then the choice is dictated by $(A_p)$. Otherwise we can write $I$ as $(j, J)$ where $j < i$ and $j \leq J$. Then we must have that
\begin{align*}
f_p(x_i \otimes z_I) &= f_p(x_i \otimes f_{p-1}(z_j \otimes z_J)) \qquad &\mbox{from }& (A_{p-1}) \\
&= f_p(x_j \otimes f_{p-1}(x_i \otimes z_J)) + f_{p-1}([x_i,x_j] \otimes z_J) \qquad &\mbox{from }& (C_p).
\end{align*}
Now $f_{p-1}(x_i \otimes z_J) = z_iz_J + w$, with $w \in P_{p-1}$ (from $(B_{p-1})$). Hence
\begin{align*}
f_p(x_j \otimes f_{p-1}(x_i \otimes z_J)) &= z_jz_iz_J + f_{p-1}(x_j \otimes w) \qquad \mbox{from } (A_p) \\
&= z_jz_I + f_{p-1}(x_j \otimes w).
\end{align*}
The above defines a unique linear extension $f_p$ of $f_{p-1}$ to $\mathfrak{g} \otimes P_p$, and by construction this extension satisfies $(A_p)$ and $(B_p)$. All that remains to show is that $f_p$ when defined this way satisfies $(C_p)$. \\
Observe that if $j \leq i$ and $j \leq J$ then $(C_p)$ is satisfied by construction. Since $[x_j, x_i] = - [x_i, x_j]$, it is also satisfied if $i < j$ and $i \leq J$. Since $(C_p)$ is trivially satisfied if $i=j$, we see that $(C_p)$ is satisfied if $i \leq J$ or $j \leq J$. Otherwise, $J = (k, K)$, where $k \leq K$, $k < i$ and $k < j$. For the sake of brevity we will write $f_p(x \otimes z) = xz$ for $x \in \mathfrak{g}$ and $z \in P_p$. Then from the induction hypothesis we have that
\begin{equation*}
x_iz_J = x_j(x_kz_K) = x_k(x_jz_K) + [x_j,x_k]z_K \qquad (*)
\end{equation*}
Now $x_jz_K$ is of the form $z_jz_K + w$, where $w \in P_{p-2}$. As $k \leq K$ and $k < j$ we can apply $(C_p)$ to $x_i(x_k(z_jz_K))$, and to $x_i(x_kw)$ from the induction hypothesis and hence also to $x_i(x_k(x_jz_K))$; using $(*)$ this then gives us that
\begin{equation*}
x_i(x_jz_J) = x_k(x_i(x_jz_K)) + [x_i, x_k](x_jz_K) + [x_j, x_k](x_iz_K) + [x_i, [x_j,x_k]]z_k.
\end{equation*}
Interchanging $i$ and $j$, this gives us that
\begin{align*}
x_i(x_jz_J) - x_j(x_iz_J) &= x_k(x_i(x_jz_K) - x_j(x_iz_K) + [x_i,[x_j,x_k]]z_K - [x_j,[x_i,x_k]]z_K \\
&= x_k([x_i,x_j]z_K) + (x_i,[x_j,x_k]]z_K + [x_j, [x_k,x_i]]z_K \\
&= [x_i,x_j]x_kz_K + [x_k, [x_i,x_j]]z_K + [x_i,[x_j,x_k]]z_K + [x_j,[x_k,x_i]]z_K \\
&= [x_i,x_j]x_kz_K \\
&=[x_i,x_j]z_J,
\end{align*}
as required.
\end{proof}
\begin{lemma}
The $y_I$, for every increasing sequence $I$, form a basis for the vector space $U(\mathfrak{g})$.
\end{lemma}
\begin{proof}
By the previous lemma (whose notation we will also use), there exists a bilinear mapping $f$ of $\mathfrak{g} \times P$ into $P$ such that $f(x_i, z_I) = z_iz_I$ for $i \leq I$ and
\[ f(x_i,f(x_i,z_J)) = f(x_j,f(x_i,z_J)) + f([x_i,x_j],z_J), \]
for all $i,j,J$. Thus, we have a representation $\varrho$ of $\mathfrak{g}$ in $P$ such that $\varrho(x_i)z_I = z_iz_I$ for $i \leq I$. From lemma 1.1 there exists a homomorphism $\varphi$ of $U(\mathfrak{g}$ into $End((P)$ such that $\varphi(y_i)z_I = z_iz_I$ for $i \leq I$. We thus obtain that, if $i_1 \leq i_2 \leq \cdots \leq i_p$, then
\[\varphi(y_{i_1}y_{i_2}\cdots y_{i_p})\cdot 1 = z_{i_1}z_{i_2}\cdots z_{i_p}.\]
Thus, the $y_I$ are linearly independent, for $I$ increasing. From lemma 1.3 they generate $U(\mathfrak{g})$ and hence form a basis.
\end{proof}
\begin{prop}The canonical mapping of $\mathfrak{g}$ into $U(\mathfrak{g})$ is injective.
\end{prop}
\begin{proof}This is immediate from the previous lemma.\end{proof}
We thus have that $\mathfrak{g}$ is embedded in $U(\mathfrak{g})$, so we will identify every element of $\mathfrak{g}$ with its canonical image in $U(\mathfrak{g})$
\begin{thm}[Poincare\'{e}-Birkhoff-Witt]
Let $(x_1, \ldots, x_n)$ be a basis for the vector space $\mathfrak{g}$. Then the $x_1^{\lambda_1}x_2^{\lambda_2}\cdots x_n^{\lambda_n}$, where $\lambda_1, \ldots, \lambda_n \in \mathbb{N}$, form a basis for $U(\mathfrak{g})$.
\end{thm}
\begin{proof}This is again immediate from the previous lemma.\end{proof}
Taking the Poincar\'{e}-Birkhoff-Witt Theorem into account Lemma 1.1 can be restated as
\begin{lemma}Let $A$ be an algebra with unity, $\tau$ a linear mapping of $\mathfrak{g}$ into $A$ such that $\tau(x)\tau(y) - \tau(y)\tau(x) = \tau([x,y])$ for all $x,y \in \mathfrak{g}$. Then $\tau$ can be uniquely extended to a homomorphism of $U(\mathfrak{g})$ into $A$ which transforms $1$ into $1$.
\end{lemma}
\begin{cor}Let $V$ be a vector space, and $\mathcal{R}$ and $\mathcal{R}'$ the sets of representations of $\mathfrak{g}$ and $U(\mathfrak{g})$ in $V$ respectively. For all $\varrho \in \mathcal{R}$, there exists one and only one $\varrho' \in \mathcal{R}'$ which extends $\varrho$, and the mapping $\varrho \rightarrow \varrho'$ is a bijection of $\mathcal{R}$ onto $\mathcal{R}'$.
\end{cor}
\end{document}