# Wednesday, November 02

:::{.remark}
Clarification from last time: for $L\in \Lie\falg$ over $\FF$ an arbitrary field:

\begin{tikzcd}
	&& {T(L)} & {\trianglerighteq J \da \gens{x\tensor y-y\tensor x -[xy] \st x,y\in L = T^1(L)}} \\
	L \\
	&& {U(L)} \\
	&& { A}
	\arrow["{i_0}", from=2-1, to=1-3]
	\arrow["i"', from=2-1, to=3-3]
	\arrow["\pi", from=1-3, to=3-3]
	\arrow["{\exists \psi}"', dashed, from=3-3, to=4-3]
	\arrow["{\exists ! \phi}"', curve={height=-30pt}, dashed, from=1-3, to=4-3]
	\arrow["j"', from=2-1, to=4-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMCwxLCJMIl0sWzIsMiwiVShMKSJdLFsyLDAsIlQoTCkiXSxbMywwLCJcXHRyaWFuZ2xlcmlnaHRlcSBKIFxcZGEgXFxnZW5ze3hcXHRlbnNvciB5LXlcXHRlbnNvciB4IC1beHldIFxcc3QgeCx5XFxpbiBMID0gVF4xKEwpfSJdLFsyLDMsIiBBIl0sWzAsMiwiaV8wIl0sWzAsMSwiaSIsMl0sWzIsMSwiXFxwaSJdLFsxLDQsIlxcZXhpc3RzIFxccHNpIiwyLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzIsNCwiXFxleGlzdHMgISBcXHBoaSIsMix7ImN1cnZlIjotNSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzAsNCwiaiIsMl1d)

Then $i$ is a Lie algebra morphism since $i([xy]) = i(x)i(y) - i(x)i(y) = [i(x) i(y)]$.
We know
\[
0 
&= \pi(x\tensor y - y\tensor x - [xy]) \\
&= \pi(i_0(x) i_0(y) - i_0(y) i_0(x) - i_0([xy])) \\
&= i(x) i(y) - i(y) i(x) - i([xy])
.\]

:::

:::{.remark}
Recall that we filtered $0 \subseteq U_1 \subseteq \cdots \subseteq U(L)$ and defined the associated graded $G^m = U_m / U_{m-1}$ and $G(L) \da \bigoplus _{m\geq 0} G^m$, and we saw by example that $yhx = hyx = xhy$ in $G^3(\liesl_2)$.
There is a projection map $T^m(L) \to U(L)$ whose image is contained in $U_m$, so there is a composite map
\[
T^m(L) \to U_m \to U_m/U_{m-1} = G^m
.\]
Since $T(L) = \bigoplus _{m\geq 0} T^m(L)$, these can be combined into an algebra morphism $T(L) \to G(L)$.
It's not hard to check that this factors through $S(L) \da T(L)/\gens{x\tensor y - y\tensor x\st x,y\in L}$ since $x\tensor y = y\tensor x + [xy]$ and the $[xy]$ term is in lower degree.
So this induces $w: S(L) \to G(L)$, and the PBW theorem states that this is an isomorphism of graded algebras.
:::

:::{.corollary title="C"}
Let $\ts{x_k}_{k\in K}$ be an ordered basis for $L$, then the collection of monomials $x_{k_1}\cdots x_{k_m}$ for $m\geq 0$ where $k_1\leq \cdots k_m$ is a basis for $U(L)$.
:::

:::{.proof title="?"}
The collection of such monomials of length exactly $n$ forms a basis for $S^m(L)$, and via $w$, a basis for $G^m(L)$.
In particular these monomials form a linearly independent in $U_m/U_{m-1}$, since taking quotients can only increase linear dependencies, and hence these are linearly independent in $U_m$ and $U(L)$.
By induction on $m$, $U_{m-1}$ has a basis consisting of all monomials of length $\leq m-1$.
We can then get a basis of $U_m$ by adjoining to this basis of $U_{m-1}$ any preimages in $U_m$ of basis elements for the quotient $U_m/U_{m-1}$.
So a basis for $U_m$ is all ordered monomials of length $\leq m$.
Since $U(L) = \union_{m\geq 0} U_m$, taking the union of bases over all $m$ yields the result.
:::

:::{.corollary title="B"}
The canonical map $i: L\to U(L)$ is injective.

This follows from taking $m=1$ in the previous corollary.
:::

:::{.corollary title="D"}
Let $H\leq L$ be a Lie subalgebra and extend an ordered basis for $H$, say $\ts{h_1, h_2,\cdots}$, and extend it to an ordered basis $\ts{h_1,h_2,\cdots, x_1,x_2,\cdots}$.
Then the injection $H\injects L$ induces an injective morphism $U(H) \injects U(L)$.
Moreover, $U(L)\in \modsleft{U(H)}^\free$ with a basis of monomials $x_{k_1}x_{k_2}\cdots x_{k_m}$ for $m\geq 0$.

This follows directly from corollary C.
:::

:::{.remark}
We'll skip 17.4, which proves the PBW theorem.
The hard part: linear independence, which is done by constructing a representation of $U(L)$ in another algebra.
:::

## 17.5: Free Lie algebras

:::{.definition title="Free Lie algebras"}
Let $L \in \Lie\falg$ which is generated as a Lie algebras (so allowing commutators) by a subset $X \subseteq L$.[^sl2_free_gen]
We say $L$ is **free on $X$** and write $L = L(X)$ if for any set map $\phi: X\to M$ with $M\in \Lie\falg$ there exists an extension:

\begin{tikzcd}
	& {} && \Lie\falg \\
	\Set & X && L \\
	\\
	&&& M
	\arrow[hook, from=2-2, to=2-4]
	\arrow["{\exists !\psi}", dashed, from=2-4, to=4-4]
	\arrow["\phi"', from=2-2, to=4-4]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMSwxLCJYIl0sWzMsMSwiTCJdLFszLDMsIk0iXSxbMywwLCJcXExpZVxcZmFsZyJdLFsxLDBdLFswLDEsIlxcU2V0Il0sWzAsMSwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMSwyLCJcXGV4aXN0cyAhXFxwc2kiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMCwyLCJcXHBoaSIsMl1d)

[^sl2_free_gen]: 
Note that $\liesl_2$ has a basis $\ts{x,h,y}$ but is freely generated by $x,y$ since $h=[xy]$.

:::

:::{.remark}
Existence:

- Let $V = V(X)$ be the free $\fmod$ on $X$.
- Let $L(X) \leq T(V)_L$[^notationsubL] be the Lie subalgebra generated by $X$ (or equivalently by $V$), which has elements like
\[
x,y,z\in X,\qquad x\tensor y-y\tensor x, \qquad z(x\tensor y-y\tensor x) - (x\tensor y - y\tensor x)z,\,\cdots
.\]
- Check that $L(X)$ is free on $X$ by letting $\phi: X\to M$ for $M$ a Lie algebra, then by the universal property of $V(X)$ we get a unique linear map $\tilde \phi: V(X) \to M$ extending $\phi$:

\begin{tikzcd}
	{V(X)} && {T(V)} \\
	\\
	M \\
	\\
	{U(M)}
	\arrow["\phi", from=1-1, to=3-1]
	\arrow[hook, from=3-1, to=5-1]
	\arrow[hook, from=1-1, to=1-3]
	\arrow["{\exists ! \tilde \phi \text{ (an algebra morphism)}}", dashed, from=1-3, to=5-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJWKFgpIl0sWzIsMCwiVChWKSJdLFswLDIsIk0iXSxbMCw0LCJVKE0pIl0sWzAsMiwiXFxwaGkiXSxbMiwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFswLDEsIiIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzEsMywiXFxleGlzdHMgISBcXHRpbGRlIFxccGhpIFxcdGV4dHsgKGFuIGFsZ2VicmEgbW9ycGhpc20pfSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==)

One checks that $\tilde \phi$ restricts to a Lie algebra morphisms $\tilde \phi: L(X) \to U(M)$ whose image is the Lie subalgebra of $U(M)$ generated by $M$ -- but this subalgebra is precisely $M$, since e.g. $U(M)\ni x\tensor y - y\tensor x = [xy]\in M$.
Thus we can view $\tilde \phi$ as a map $\tilde \phi: L(X)\to M$.

[^notationsubL]: 
$W_L$ is $W$ made into a Lie algebra via $[xy] = xy-yx$.

:::

:::{.remark}
One can check that $U(L(X)) = T(V(X))$.
:::