# Monday, April 12

## Lie Algebra Homology

:::{.remark}
Last time: Lie algebras.
Fix a cocommutative ring $k$, usually a field, then a Lie algebra $\lieg$ over $k$ is a $k\dash$module with a bilinear product called the bracket such that

- $[xx] = 0$, so $[xy] = -[yx]$
- The Jacobi identity holds.
:::

:::{.definition title="Modules over Lie algebras"}
A left $\lieg\dash$module $M$ is a $k\dash$module with a $k\dash$bilinear product 
\[
\cdot: \lieg \tensor_k M &\to M \\
x\tensor m &\mapsto x\cdot m
\]
which is compatible with the bracket in the following sense:
\[
[xy]m = x(ym) - y(xm) \quad  \forall x,y\in \lieg, m\in M \label{eq:assoc_formula_lie_algebra}
,\]
i.e. there is a Lie algebra morphism $\lieg \to \gl(M) \da \Lie( \Endo_k(M))$, the Lie algebra of the endomorphism algebra.

:::

:::{.example title="Algebra Commutators"}
For $A \in \kalg(\Assoc)$ and $\lieg \in \Lie(A)$, then any $M\in \modsleft{A}$ (so the action is associative) can be made into an $M' \in \modsleft{\lieg}$ by the formula \cref{eq:assoc_formula_lie_algebra}.
:::

:::{.example title="Adjoint Representations"}
Any Lie algebra $\lieg$ is a module over itself by the **adjoint representation**, where $\ad_x(\wait) \da [x, \wait]$.
:::

:::{.example title="Trivial Modules"}
Any $M\in\modsleft{k}$ becomes a trivial $\lieg\dash$module by defining $xm = 0$ for all $x\in \lieg, m\in M$.
Note that this is acting by zero instead of the identity: this is motivated from Lie algebras obtained from Lie groups by taking tangent spaces at the identity.
A trivial group action on the elements would be the identity, but then taking its derivative acting on tangent vectors to curves would be zero.

\

There is a *unique* trivial $\lieg\dash$module, namely $k$ with this trivial action.
:::

:::{.definition title="Morphisms of Lie algebra modules"}
A morphism $M \mapsvia{f} N$ of $\lieg\dash$modules is a morphism of $k\dash$modules commuting with the module action, so $f(xm) = x(fm)$ for $x\in \lieg, m\in M$.
This yields $\Hom_\lieg(M, N) \leq \Hom_k(M, N)$ as a $k\dash$submodule.
:::

:::{.remark}
This yields a category $\modsleft{\lieg} \leq \modsleft{k}$ which is a subcategory of $k\dash$modules, and this is in fact an abelian category.
So we have notions of (co)kernels, injectives and projectives, etc.
There is also a category $\modsright{\lieg}$, but these can be sent to left $\lieg\dash$modules by defining $x\cdot m \da -mx$ which makes $\lieg$ anticommutative.
Thus there is an equivalence of categories
\[
\modsleft{\lieg} \mapsvia{\sim} \modsright{\lieg}
,\]
and so we usually just refer to left modules.
:::

:::{.remark}
We'll want to take homology and cohomology.
There are some relevant functors:

- The trivial module functor:
\[
\Triv: \modsleft{k} \to \modsleft{\lieg}
,\]
which sends $M$ to itself, adding the structure of a trivial $\lieg\dash$action.
- $\lieg\dash$invariants:
\[
(\wait)^\lieg: \liegmod &\to \kmod \\
M &\mapsto M^g \da \ts{ x\in M \st xm = 0 \forall \,\, x\in \lieg}
.\]

    - This yields the largest $\lieg\dash$trivial submodule, and similarly $(\wait)^\lieg$ is right-adjoint to $\Triv$.
  \[
  \adjunction{\Triv}{(\wait)^\lieg}{\kmod}{\liegmod}
  .\]
    - There is an isomorphism 
  \[
  \ev_1: \Hom_\lieg(k, M)
  &\mapsvia{\sim} 
  M^\lieg \\
  f &\mapsto f(1_k)
  .\]
  where $k$ is the trivial $\lieg\dash$module.

- $\lieg\dash$coinvariants:
\[
(\wait)_\lieg: \liegmod &\to \kmod \\
M &\mapsto M/\lieg M
.\]

  - This is the largest $\lieg\dash$trivial *quotient* of $M$, so this is left-adjoint to $\Triv$:
  \[
  \adjunction{(\wait)^\lieg}{\Triv}{\liegmod}{\kmod}
  .\]

> We might expect this is related to some tensor product, but it may not be clear what ring one should tensor over.

:::

:::{.remark}
Assume that $\liegmod$ has enough projectives, which we'll see is true in a later section by identifying this with a category $\rmod$ of modules over a ring.
:::

:::{.definition title="Cohomology of Lie algebras"}
Define the **(co)homology of $\lieg$ with coefficients in $M$** as
\[
H_n(\lieg; M) &\da \LL(\wait)_\lieg (M) \\
H^n(\lieg; M) &\da \RR(\wait)^\lieg(M)
.\]
:::

:::{.example title="?"}
If $\lieg = \ts{ 0 }$, then $M^\lieg = M = M_\lieg$ and these functors are exact (and are essentially the identity) and thus their higher derived functors are zero.
So $H^n(0; M) = 0 = H_n(0; M)$.
:::

## The Universal Enveloping Algebra

:::{.remark}
A better name might be the universal *associative* algebra.
This plays an analogous role to the group algebra $\ZZG$ of a group.
We'll assign an associative algebra $\Ug$ to $\lieg$, and there will be an equivalence of categories
\[
\modsleft{\lieg} \mapsvia{\sim} \modsleft{\Ug}
,\]
where we'll know that the latter has enough projectives and injectives, allowing us to compute homology and cohomology with injective and projective resolutions.
:::

:::{.definition title="Tensor Algebra"}
For $k \in \CRing$ and $M\in \kmod$, and **tensor algebra** is defined as
\[
T(M) \da \bigoplus_{i\geq 0}  M^{\tensor_k n} \da k \tensor \bigoplus _{n\geq 1} M^{\tensor_k n}
.\]

:::

:::{.remark}
Note that $T(M) \in \kmod$ by extending the $k\dash$action over sums and tensor products in the obvious way, and in fact $T(M) \in \gr(\kalg)$ where tensors in different degrees are juxtaposed.
Explicitly, for $m\in M^{\tensor n}$ and $m' \in M^{\tensor n'}$, we write $m\tensor m' \in M^{\tensor (n+n')}$, which is what it means to be a *graded* algebra.
:::

:::{.remark}
There is an inclusion map 
\[
M = M^{\tensor 1} \injectsvia{\iota} T(M)_1 \injects T(M)
.\]
where $T(M)_j \da \bigoplus_{n\geq j} M^{\tensor n}$, and in fact $T(M)$ is generated as a $k\dash$algebra by $\iota(M)$.
For example, for $m, m' \in M$, we have $\iota(m) \tensor \iota(m') \in T(M)_2$.
This yields a functor
\[
T: \modsleft{k} \to \kalg(\Assoc, \mathsf{Unital})
,\]
as well as a forgetful functor
\[
\Forget: \kalg \to \kmod
.\]
The pair $(T, i)$ is a **universal** associative algebra in the following sense: if $M\in \kmod$ and $A\in \kalg(\Assoc)$, then there is a $k\dash$module morphism $M\to \Forget(A)$ making the following diagram commute:

\begin{tikzcd}
	\textcolor{rgb,255:red,214;green,92;blue,92}{M} && \textcolor{rgb,255:red,92;green,92;blue,214}{T(M)} \\
	\\
	&& \textcolor{rgb,255:red,92;green,92;blue,214}{A}
	\arrow["f", color={rgb,255:red,214;green,92;blue,92}, from=1-1, to=3-3]
	\arrow["{\exists ! \tilde f \in \kalg}", color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-3, to=3-3]
	\arrow["\iota", from=1-1, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJNIixbMCw2MCw2MCwxXV0sWzIsMiwiQSIsWzI0MCw2MCw2MCwxXV0sWzIsMCwiVChNKSIsWzI0MCw2MCw2MCwxXV0sWzAsMSwiZiIsMCx7ImNvbG91ciI6WzAsNjAsNjBdfSxbMCw2MCw2MCwxXV0sWzIsMSwiXFxleGlzdHMgISBcXHRpbGRlIGYgXFxpbiBcXGthbGciLDAseyJjb2xvdXIiOlsyNDAsNjAsNjBdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19LFsyNDAsNjAsNjAsMV1dLFswLDIsIlxcaW90YSJdXQ==)



Note that the red portion of the diagram happens in $\kmod$, while the blue portion is in $\kalg$, so this allows lifting module morphisms to algebra morphisms.
Commuting here means that 
\[
f(m_1) f(m_2) = \tilde f(m_1 m_2) \da f( \iota(m_1) \tensor \iota(m_2))
.\]
There is thus a natural isomorphism
\[
\Hom_{\kmod}(M, \Forget(A)) \mapsvia{\sim} \Hom_{\kalg}( T(M), A)
.\]

:::