# Part V: Existence Theorem. Ch. 17: The Universal Enveloping Algebra (Monday, October 31)

## 17.1: The tensor algebra and symmetric algebra

:::{.remark}
Let $\FF$ be an arbitrary field, not necessarily characteristic zero, and let $L\in \Lie\Alg\slice\FF$ be an arbitrary Lie algebra, not necessarily finite-dimensional.
Recall that the tensor algebra $T(V)$ is the $\ZZ_{\geq 0}$ graded unital algebra where $\gr_n T(V) = T^n(V) \da V\tensorpower{\FF}{n}$ where $T^0(V) \da \FF$.
Note $T(V) = \bigoplus _{n\geq 0} T^n(V)$.
If $V$ has a basis $\ts{x_k}_{k\in K}$ then $T(V) \cong \FF\gens{x_k \st k\in K}$, a polynomial ring in the noncommuting variables $x_k$.
Degree $n$ monomials in this correspond to pure tensors with $n$ components in $T(V)$.

There is an $\FF\dash$linear map $V \mapsvia{i} T(V)$, and $T(V)$ satisfies a universal property: given any linear map $\phi\in \mods{\FF}(V, A)$ where $A$ has the structure of an associative algebra, there exists a unique $\psi\in \Assoc\Alg\slice\FF(T(V), A)$ making the diagram commute:

\begin{tikzcd}
	V && {T(V)} \\
	\\
	&& A
	\arrow["i", from=1-1, to=1-3]
	\arrow["{\exists !}", dashed, from=1-3, to=3-3]
	\arrow["\phi"', from=1-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJWIl0sWzIsMCwiVChWKSJdLFsyLDIsIkEiXSxbMCwxLCJpIl0sWzEsMiwiXFxleGlzdHMgISIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFswLDIsIlxccGhpIiwyXV0=)

In fact, one can explicitly write $\psi$ as $\psi(x_{k_1}\tensor \cdots x_{k_n}) = \phi(x_k)\cdots \phi(x_{k_n})$ using the multiplication in $A$.

The **symmetric algebra** and **exterior algebra** are defined as 
\[
S(V) \da T(V)/\gens{x\tensor y -y\tensor x \st x,y\in V}, 
\Extalg(V) \da T(V)/\gens{x\tensor y + y\tensor x \st x,y\in V}
.\]
:::


:::{.definition title="The Universal Enveloping Algebra"}
Let $L\in \Lie\Alg\slice \FF$ with basis $\ts{x_k}_{k\in K}$.
A **universal enveloping algebra** for $L$ is a pair $(U, i)$ where $U$ is a unital associative $\FF\dash$algebra and $i: L \to U_L$ (where $U_L$ is $U$ equipped with the commutator bracket multiplication) is a morphism of Lie algebras, i.e.
\[
i([xy]) = i(x) i(y) - i(y) i(x) = [i(x) i(y) ] \quad \forall x,y\in L
.\]
It satisfies a universal property: for any unital associative algebra $A$ receiving a Lie algebra morphism $j: L\to A_L$, there is a unique $\phi$ in the following:

\begin{tikzcd}
	\Lie\Alg && \Assoc\Alg \\
	L && U \\
	\\
	&& A
	\arrow["i", from=2-1, to=2-3]
	\arrow["{\exists ! \phi}", dashed, from=2-3, to=4-3]
	\arrow["j"', from=2-1, to=4-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMiwxLCJVIl0sWzIsMywiQSJdLFsyLDAsIlxcQXNzb2NcXEFsZyJdLFswLDAsIlxcTGllXFxBbGciXSxbMCwxLCJMIl0sWzQsMCwiaSJdLFswLDEsIlxcZXhpc3RzICEgXFxwaGkiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbNCwxLCJqIiwyXV0=)

:::

:::{.remark}
Uniqueness follows from the usual proof for universal objects.
Existence: let
\[
U(L) \da T(L) / J, \qquad J \da \gens{x\tensor y - y\tensor x - [xy] \st x,y\in L}
.\]
Warning: $J$ is a two-sided ideal, but is not homogeneous!

One can form the required map:

\begin{tikzcd}
	&& {T(L)} \\
	L \\
	&& {U(L)} \\
	\\
	&& A
	\arrow["{i_*}"', from=2-1, to=1-3]
	\arrow["\pi", from=1-3, to=3-3]
	\arrow["{\exists ! \psi}", dashed, from=3-3, to=5-3]
	\arrow["{j\in \Lie\Alg(L, A)}"', from=2-1, to=5-3]
	\arrow["i"', from=2-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwxLCJMIl0sWzIsMCwiVChMKSJdLFsyLDIsIlUoTCkiXSxbMiw0LCJBIl0sWzAsMSwiaV8qIiwyXSxbMSwyLCJcXHBpIl0sWzIsMywiXFxleGlzdHMgISBcXHBzaSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFswLDMsImpcXGluIFxcTGllXFxBbGcoTCwgQSkiLDJdLFswLDIsImkiLDJdXQ==)

This satisfies $\psi(x\tensor y - y\tensor x - [xy]) = j(x) j(y) - j(y)j(x) - j([xy]) = 0$ using the properties of $j$.
$\phi$ is unique because $U(L)$ is generated by 1 and $\im i$, since $T(L)$ is generated by 1 and the image of $L = T^1(L)$.
:::


:::{.remark}
If $L$ is abelian, $U(L) = S(L)$ is the symmetric algebra.

Note that $J \subseteq \bigoplus _{n\geq 1} T^n(L)$ so $\FF = T^0(L)$ maps isomorphically into $U(L)$ under $\pi$.
So $\FF\injects U(L)$, meaning $U(L) \neq 0$, although we don't yet know if $L$ injects into $U(L)$.
:::


:::{.theorem title="Poincaré-Birkhoff-Witt (PBW) Theorem"}
Let $L$ be a Lie algebra with basis $\ts{x_k}_{k\in K}$, and filter $T(L)$ by $T_m \da \bigoplus _{i\leq m} T^i (L)$.
Then $T_m$ is the span of words of length at most $m$ in the basis elements $x_k$.
Note $T_m \cdot T_n \subseteq T_{m+n}$, and the projection $\pi: T(L) \surjects U(L)$ induces an increasing filtration $U_0 \subseteq U_1 \subseteq \cdots$ of $U(L)$.
Let $G^m \da U_m / U_{m-1}$ be the $m$th graded piece. 
The product on $U(L)$ induces a well-defined product $G^m \times G^n \to G^{m+n}$ since $U_{m-1}\times U_{n-1} \subseteq U_{m+n-2} \subseteq U_{m+n-1}$.
Extending this bilinearly to \( \bigoplus _{m\geq 0} G^m \) to form the associated graded algebra of $U(L)$. 

> Note that this construction generally works for any filtered algebra where the multiplication is compatible with the filtration.

:::


:::{.example title="?"}
Let $L \da \liesl_2(\CC)$ with ordered basis $\ts{x,h,y}$. 
Then $y\tensor h\tensor x\in T^3(L)$ -- denote the image of this monomial in $U(L)$ by $yhx \in U_3$.
We can reorder this:
\[
yhx 
&= hyx + [yh]x
&= hyx + 2yx \qquad \in U_3 + U_2 \\
,\]
so in $G^3$ we have $yhx=hyx$.
This is a general feature: reordering introduces error terms of lower degree, which are quotiented out.
Continuing,
\[
hyx + 2yx
&= hxy + h[yx] + 2xy + 2[yx] \\
&= hxy - h^2  + 2xy - 2h \\
&= xhy + [hx]y - h^2  + 2xy - 2h \\
&= xhy + 2xy - h^2  + 2xy - 2h \\
&= xhy + 4xy - h^2 - 2h \qquad \in U_3 + U_2 + U_2 + U_2
.\]



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