# Friday, April 16

## The Enveloping Algebra (Continued)

:::{.remark}
Last time: the PBW theorem. 
Let $\lieg \in \liealg$ and free as a $k\dash$module with $k\dash$basis \( \ts{ x_ \alpha }_{\alpha\in A} \). 
Then $\Ug$ has a $k\dash$basis \( \ts{ x_I } \) where \( I = ( \alpha_1, \cdots, \alpha_p) \) is a finite increasing sequence from $A$
:::

:::{.example title="?"}
If $k$ is a field and $\dim_k \lieg$ is finite with basis \( \ts{ x_1, \cdots, x_n } \).
Take $I = (1,\cdots, 1, 2\cdots, 2, n\cdots, n)$ where each $i$ occurs $a_i$ times.
Then a basis for $\Ug$ is \( \ts{ x_1^{a_1} x_2^{a_2} \cdots x_n^{a_n} \st a_i \geq 0 }  \).
:::

:::{.corollary title="?"}
The map $\iota: \lieg \to \Ug$ is injective, so we can identify $\iota(\lieg)$ with $\lieg$.
:::

:::{.proof title="?"}
The elements \( x_{(\alpha)} \da \iota(x_{ \alpha} ) \in \Ug \) are $k\dash$linearly independent.
:::

:::{.corollary title="?"}
If $\lieh \leq \lieg$ is a subalgebra and $k$ is a field, then $\Ug$ is free as a \( \mathcal{U}(\lieh)\dash\)module.
:::

:::{.proof title="?"}
Choose an ordered basis for $\lieh$ first and then extend this to an ordered basis for $\lieg$ -- that one can do this is a fact from linear algebra.
Then the $x_I$ where \( I = ( \alpha_1, \cdots, \alpha_p ) \) is increasing and no $x_{\alpha_i} \in \lieh$ will be a basis for $\Ug$ over $\Uh$.
:::

:::{.example title="?"}
If $\dim_k \lieg < \infty$ and \( \ts{ x_1, \cdots, x_k } \) is a basis for $\lieh$ and  \( \ts{ x_1,\cdots, x_k, x_{k+1} \cdots, x_n } \) is a basis for $\lieg$, then the PBW basis is given by \( \ts{ x_1^{a_1} \cdots x_k^{a_k} x_{k+1}^{a_{k+1}} \cdots x_n^{a_n} \st a_i \geq 0 } \).
Then \( \ts{ x_{k+1]^{a_{k+1}} \cdots x_n ^{a_n} }} \) form a free left $\Uh\dash$module basis for $\Ug$.
:::

:::{.exercise title="?"}
Some suggested exercises:

- 7.3.4
- 7.3.6
- 7.3.7 for working with $\Uh$ as a Hopf algebra.
- 7.3.9 for representations of Lie algebras in characteristic $p$.

:::

## $H^1$ for Lie Algebras (Weibel 7.4)

:::{.remark}
Recall that we have an augmentation ideal $\mathcal{I} \normal \Uh$ and a SES
\[
0 \to\mathcal{I}\to\Ug\to k\to 0
.\]
Applying the functor $\Hom_{\Ug}(\wait, M)$ for a fixed $M\in \liegmod$ yields a LES:

\begin{tikzcd}
	0 && {\Hom_{\Ug}(k, M)} && {\Hom_{\Ug}(\Ug, M)} && {\Hom_{\Ug}(\mathcal{I}, M)} \\
	\\
	&& {\Ext^1_{\Ug}(k, M) = H^1(\lieg; M)} && \textcolor{rgb,255:red,214;green,92;blue,92}{\Ext^1_{\Ug}(\Ug, M)=0} && \cdots
	\arrow["\delta", from=1-7, to=3-3]
	\arrow[from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
	\arrow[from=1-5, to=1-7]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMCwwLCIwIl0sWzIsMCwiXFxIb21fe1xcVWd9KGssIE0pIl0sWzQsMCwiXFxIb21fe1xcVWd9KFxcVWcsIE0pIl0sWzYsMCwiXFxIb21fe1xcVWd9KFxcbWF0aGNhbHtJfSwgTSkiXSxbMiwyLCJcXEV4dF4xX3tcXFVnfShrLCBNKSA9IEheMShcXGxpZWc7IE0pIl0sWzQsMiwiXFxFeHReMV97XFxVZ30oXFxVZywgTSk9MCIsWzAsNjAsNjAsMV1dLFs2LDIsIlxcY2RvdHMiXSxbMyw0LCJcXGRlbHRhIl0sWzAsMV0sWzEsMl0sWzIsM10sWzQsNV0sWzUsNl1d)

Here the red term vanishes since $\Ug$ is free and this projective as a $\lieg\dash$module.
Note that for $n\geq 2$, we have the following situation:

\begin{tikzcd}
	\cdots && \textcolor{rgb,255:red,214;green,92;blue,92}{\Ext^{n-1}_{\Ug}(\Ug, M)=0} && {\Ext_{\Ug}^{n-1}(\mathcal{I}, M)} \\
	\\
	{\Ext^1_{\Ug}(k, M) = H^1(\lieg; M)} && \textcolor{rgb,255:red,214;green,92;blue,92}{\Ext^{n}_{\Ug}(\Ug, M)=0} && \cdots
	\arrow["{\delta \quad \cong}", color={rgb,255:red,92;green,92;blue,214}, from=1-5, to=3-1]
	\arrow[from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
	\arrow[from=3-1, to=3-3]
	\arrow[from=3-3, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMCwwLCJcXGNkb3RzIl0sWzIsMCwiXFxFeHRee24tMX1fe1xcVWd9KFxcVWcsIE0pPTAiLFswLDYwLDYwLDFdXSxbNCwwLCJcXEV4dF97XFxVZ31ee24tMX0oXFxtYXRoY2Fse0l9LCBNKSJdLFswLDIsIlxcRXh0XjFfe1xcVWd9KGssIE0pID0gSF4xKFxcbGllZzsgTSkiXSxbMiwyLCJcXEV4dF57bn1fe1xcVWd9KFxcVWcsIE0pPTAiLFswLDYwLDYwLDFdXSxbNCwyLCJcXGNkb3RzIl0sWzIsMywiXFxkZWx0YSBcXHF1YWQgXFxjb25nIiwwLHsiY29sb3VyIjpbMjQwLDYwLDYwXX0sWzI0MCw2MCw2MCwxXV0sWzAsMV0sWzEsMl0sWzMsNF0sWzQsNV1d)

Thus we get a degree shifting isomorphism 
\[
H^n(\lieg; M) \cong \Ext_{\Ug}^{n-1}(\mathcal{I}, M)
.\]
:::

:::{.remark}
We thus have
\[
H^1(\lieg; M) \cong { \Hom_\Ug(\mathcal{I}, M) / \im \qty{ \Hom_\Ug(\Ug, M) \cong M \to \Hom(\mathcal{I}, M)  } }
.\]
Next goal: to more concretely express all of the terms here as $M\dash$valued derivations on $\lieg$.
:::

:::{.definition title="Derivations of an algebra"}
Let $M\in \liegmod$, then a **derivation from $\lieg$ into $M$** is a $k\dash$linear map $D:\lieg\to M$ satisfying the Leibniz rule:
\[
D([xy]) = x\cdot (Dy) - y\cdot (Dx) && x,y\in \lieg
.\]
:::

:::{.remark}
The set of all such maps $\Der(\lieg, M) \leq_{\kmod} \Hom_{\kmod}(\lieg, M)$ is a $k\dash$submodule.
A special case is taking $M \da \lieg$, regarded as a $\lieg\dash$module using the adjoint representation.
In fact, for any $k\dash$algebra (not necessarily associative), we get 
\[
D(ab) = (Da)\cdot b + a\cdot(Db)
.\]
When $A \da \lieg \in \liealg$ with the adjoint action, we obtain
\[
D([xy]) 
&= [x, Dy] + [Dx, y] \\
&= [x, Dy] - [y, Dx] \\
&= x\cdot Dy - y\cdot (Dx)
,\]
recovering the previous definition.
:::

:::{.definition title="Inner Derivations"}
For $M\in \liegmod$, fix an $m\in M$.
We then define
\[
D_m: \lieg &\to M \\
x &\mapsto x\cdot m
.\]
Any derivation of this form is said to be an **inner derivation**, and this yields a $k\dash$submodule
\[
\Inn(\lieg, M) \leq_{\kmod} \Der(\lieg, M)
.\]
:::

:::{.remark}
Note that this is indeed a derivation:
\[
D_m([xy]) ] [xy]\cdot m = x\cdot(y\cdot m) - y\cdot(x\cdot m) = x\cdot (D_m y) - y\cdot (D_m x)
.\]
It also turns out that any inner derivation is of this form, bracketing against a fixed element.
:::


:::{.proposition title="?"}
\[
\Hom_\liegmod(\mathcal{I}, M) \cong \Der(\lieg, M)
.\]
:::


:::{.proof title="?"}


:::{.claim}
There exists such a map.
:::


Say \( \varphi\in \Hom_\lieg(\mathcal{I}, M) \) and set
\[
D_{\varphi}:\lieg &\to M \\
x &\mapsto \phi(x)
.\]
Then $D_{ \varphi}$ is a derivation, so we have
\[
D_{ \varphi}([xy]) 
&\da \phi([xy]) \\
&= \phi(xy-yx) \\
&= x \varphi(y) - y \varphi(x) && \text{since $\phi$ is $\lieg\dash$linear} \\
&= x D_{\varphi}(y) - y D_{\varphi}(x)
.\]


:::{.claim}
This map is a natural isomorphism, in the sense that it doesn't depend on any choices:
\[
\Hom_\lieg(\mathcal{I}, M) &\to \Der(\lieg, M) \\
\varphi&\mapsto D_{ \varphi}
.\]
:::


:::{.proof title="of surjectivity"}
Recall that we can write \( \mathcal{I} = \Ug\,\lieg \), so the following product map is a surjection:
\[
\theta: \Ug \tensor_k \lieg &\surjects \Ug\,\lieg = \mathcal{I} \\
x\tensor y &\mapsto xy
.\]
One checks that the kernel is given by
\[
\ker(\theta) = \ts{ u \tensor [xy] - \qty{ux\tensor y - uy\tensor x} \st x,y\in \lieg, u\in \Ug } 
.\]
Now given $D \in \Der(\lieg, M)$, consider the map
\[
f: \Ug \tensor_k \lieg &\to M \\
f(u\tensor x) &= u\cdot Dx
.\]

One can compute the following, using that $D$ is a derivation:
\[
f\qty{ u\tensor [xy] - ux\tensor y - uy\tensor x }
&= u D([xy]) - (ux)\cdot D(y) + (uy) \cdot D(x) \\
&= u (x\cdot Dy - y\cdot Dx) - u\cdot(x\cdot Dy) + u\cdot(y\cdot Dx) \\
&= 0
.\]
So $f$ induces a well-defined morphism of $\lieg\dash$modules, and descends to a map
\[
\phi: \Ug\,\lieg = \mathcal{I} &\to M
,\]
which is clearly also a morphism of $\lieg\dash$modules.
So \( \varphi\in \Hom_\lieg(\mathcal{I}, M) \) and \( D_{\varphi}(x) = \varphi(x) = \varphi(1\cdot x) = f(1\cdot x) = 1\cdot Dx = Dx \), and so $D = D_{ \varphi}$.
:::

\todo[inline]{Might as well find-and-replace "map" with "morphism"!}

:::{.proof title="of injectivity"}
Suppose over $D$ that we have $D_{\psi}$ for some $\psi \in \Hom_\lieg(\mathcal{I}, M)$.
We then have
\[
\phi(ux) = uD(x) = u \Psi(x) = \Psi(ux) && \forall u\in \Ug, x\in \lieg
.\]
Since \( \mathcal{I} = \Ug\, \lieg \) and \( \phi = \Psi \), yielding a 1-to-1 map.
:::


:::