# The basics of Lie algebra cohomology

## Part 3

###### 29th November, 2021

- Lie algebra homology
- Relative Lie algebra (co)homology: some properties
- Relative Lie algebra (co)homology: some definitions

We’ll end this series (for now) by talking about two things that we
really should talk about: Lie algebra *homology*, and
*relative* Lie algebra (co)homology. (We will work over a field
k, but really we are thinking only of
the case k=\mathbb{C}).

# Lie algebra homology

↟Just as Lie algebra cohomology is constructed to be the derived
functor of the functor of invariants, Lie algebra homology is
constructed to be the derived functor of the functor of
*coinvariants*:
\begin{aligned}
\operatorname{Coinv}_\mathfrak{g}(M) = M_\mathfrak{g}&\coloneqq
M/\mathfrak{g}M
\\\operatorname{H}_\bullet(\mathfrak{g},M) &\coloneqq
\operatorname{Tor}_\bullet^{U(\mathfrak{g})}(k,M).
\end{aligned}
We need to justify why this \operatorname{Tor} is indeed the derived
functor of \operatorname{Coinv}, since
the latter doesn’t seem to have anything to do with tensor products a
priori. To do this, we need to make a small definition (which we should
really have made back when we constructed the projective resolution
U(\mathfrak{g})\otimes_k\wedge^p\mathfrak{g}\twoheadrightarrow
k^\mathrm{triv}).

Write i\colon\mathfrak{g}\to
U(\mathfrak{g}) to mean the composite map \mathfrak{g}\hookrightarrow
T(\mathfrak{g})\twoheadrightarrow U(\mathfrak{g}). The k-algebra homomorphism \varepsilon\colon U(\mathfrak{g})\to k
defined by sending i(\mathfrak{g}) to
zero is called the *augmentation map*, and we call its kernel
\mathfrak{J} the *augmentation
ideal*.

The augmentation ideal \mathfrak{J} is generated (as a U(\mathfrak{g})-module) by the image i(\mathfrak{g}). Further, k\cong U(\mathfrak{g})/\mathfrak{J}=U(\mathfrak{g})_\mathfrak{g}.

With this, we can see why the derived functor of the coinvariants functor is calculated by \operatorname{Tor}, since \begin{aligned} k\otimes_{U(\mathfrak{g})}M &\cong U(\mathfrak{g})/\mathfrak{J}_{U(\mathfrak{g})}M \\&\cong M/\mathfrak{J}M \\&\cong M/\mathfrak{g}M = M_\mathfrak{g}. \end{aligned}

As we did for cohomology, we can construct a chain complex that computes the Lie algebra homology: we define \wedge_p = \wedge_p(\mathfrak{g},M) \coloneqq \wedge^p(\mathfrak{g})\otimes_k M (so that \wedge_0=M), with differential \partial\colon\wedge_p\to\wedge_{p-1} given by \begin{aligned} &\partial_p(x_1\wedge\ldots\wedge x_p\otimes m) = \\\,&\sum_{1\leqslant i<j\leqslant p} (-1)^{i+j} [x_i,x_j]\wedge x_1\wedge\ldots\wedge\widehat{x_i}\wedge\ldots\wedge\widehat{x_j}\wedge\ldots\wedge x_p \otimes m \\+&\sum_{i=1}^p (-1)^i x_1\wedge\ldots\wedge\widehat{x_i}\wedge\ldots\wedge x_p \otimes (x_im). \end{aligned} The claim (which we won’t prove) is then that \operatorname{H}_p(\mathfrak{g},M) \cong \operatorname{Z}_p(\mathfrak{g},M)/\operatorname{B}_p(\mathfrak{g},M).

# Relative Lie algebra (co)homology: some properties

↟Again, as is usual for (co)homology theories, we wish to have some
notion of the (co)homology *of a pair*. So let (\mathfrak{g},\mathfrak{h}) be a Lie algebra
pair (i.e. \mathfrak{h} and \mathfrak{g} are both Lie algebras, with
\mathfrak{h}\hookrightarrow\mathfrak{g}
a subalgebra), and M a \mathfrak{g}-module. It should be the case
that, however we define the (co)homology of this pair (with coefficients
in M), we at the very least recover the
non-relative (co)homology of \mathfrak{g} when \mathfrak{h} is zero, i.e. we should
definitely have that
\operatorname{H}^\bullet(\mathfrak{g},0;M) \cong
\operatorname{H}^\bullet(\mathfrak{g};M).
But before giving the definition, it’s maybe also helpful to see
another theorem which will hold, which tells us about the case of the
relative cohomology of a *reductive* subalgebra.

Let (\mathfrak{g},\mathfrak{h}) be a
Lie algebra pair. We say that it is *reductive* if the adjoint
Lie algebra representation of \mathfrak{h} on \mathfrak{g} is completely reducible.

Let (\mathfrak{g},\mathfrak{h}) be a Lie algebra pair, and M a \mathfrak{g}-module. Assume that

- everything (i.e. \mathfrak{h}, \mathfrak{g}, and M) is finite dimensional (I’m not actually sure that this is necessary, but I’m just being safe here);
- (\mathfrak{g},\mathfrak{h}) is a reductive pair; and
- we can write \mathfrak{g} as a
semi-direct product \mathfrak{g}=\mathfrak{h}\ltimes\mathfrak{a}
(where \mathfrak{a} is thus, in
particular, a
*Lie ideal*of \mathfrak{g}, i.e. a vector subspace such that [\mathfrak{a},\mathfrak{g}]\subseteq\mathfrak{a}).

Then \operatorname{H}^\bullet(\mathfrak{a};M)^\mathfrak{g}\simeq \operatorname{H}^\bullet(\mathfrak{a};M)^\mathfrak{h}\simeq \operatorname{H}^\bullet(\mathfrak{g},\mathfrak{h};M) (where the superscript denotes the space of invariants).

# Relative Lie algebra (co)homology: some definitions

↟As before, we will first give a complex whose internal cohomology computes relative Lie algebra cohomology, and then construct a projective resolution of k^\mathrm{triv} so that we can show how to recover the derived functor definition.

The *relative Chevalley–Eilenberg complex \operatorname{CE}^\bullet(\mathfrak{g},\mathfrak{h};M)*
is defined by
\operatorname{CE}^p(\mathfrak{g},\mathfrak{h};M) \coloneqq \big(
(\wedge^p(\mathfrak{g}/\mathfrak{h})^*)\otimes_k M \big)^\mathfrak{h}
and the differential is given by the duals of the Lie bracket
and the action of \mathfrak{g} on M, i.e.
d \coloneqq \rho^* + [-,-]^*.

By definition, if \mathfrak{h}=0, then we recover exactly the Chevalley–Eilenberg complex of \mathfrak{g}, which is one thing that we wanted. (To see this, we use the general fact that we have an isomorphism of the form \operatorname{Hom}_k(A,V)\cong A^*\otimes_k V).

As for relative homology, we define the complex \wedge_p = \wedge_p(\mathfrak{g},\mathfrak{h};M) \coloneqq \frac{\wedge^p(\mathfrak{g}/\mathfrak{h})\otimes_k M}{\mathfrak{h}\cdot(\wedge^p(\mathfrak{g}/\mathfrak{h})\otimes_k M)} where \mathfrak{h}\cdot V, for an \mathfrak{h}-module V, denotes the span of elements of the form x\cdot v for x\in\mathfrak{h} and v\in V.

So far, we have not really been making any assumptions about whether or not M is finite dimensional. It turns out that we don’t have to, but we do need to impose at least some sort of condition relating to its dimension, whence the following definition.

We say that a \mathfrak{g}-module is
*finitely semisimple* if it is the sum of its finite-dimensional
irreducible \mathfrak{g}-submodules.

Being finitely semisimple is preserved under quotients, taking submodules, and tensor products.

Let (\mathfrak{g},\mathfrak{h}) be a Lie algebra pair such that \mathfrak{g} is finitely semisimple under the adjoint action of \mathfrak{h}. Write \mathcal{G}\coloneqq U(\mathfrak{g}), and \mathcal{H}\coloneqq U(\mathfrak{h}). Then the complex \ldots \to D_2 \to D_1 \to D_0 is a (\mathcal{G},\mathcal{H})-projective resolution of k^\mathrm{triv}, where D_p \coloneqq \mathcal{G}\otimes_\mathcal{H}\wedge^p(\mathfrak{g}/\mathfrak{h}).

Under the same assumptions, the complex D_\bullet\otimes_k M is a (\mathcal{G},\mathcal{H})-projective resolution of M.

Using this resolution, we can show the following:

Let (\mathfrak{g},\mathfrak{h}) be a Lie algebra pair such that \mathfrak{g} is finitely semisimple under the adjoint action of \mathfrak{h}. Then \operatorname{H}_\bullet(\mathfrak{g},\mathfrak{h};M) \simeq \operatorname{Tor}_\bullet^{(\mathcal{G},\mathcal{H})}(M^t,k) \simeq \operatorname{Tor}_\bullet^{(\mathcal{G},\mathcal{H})}(k,M) (where M^t is the right \mathfrak{g}-module with the same underlying space as M, but with m\cdot x\coloneqq -xm for m\in M and x\in\mathfrak{g}), and \operatorname{H}^\bullet(\mathfrak{g},\mathfrak{h};M) \simeq \operatorname{Ext}_{(\mathcal{G},\mathcal{H})}^\bullet(k,M).

A “fun” exercise is then to prove that, for any Lie algebra pair (\mathfrak{g},\mathfrak{h}) and any \mathfrak{g}-module M, \operatorname{H}^\bullet(\mathfrak{g},\mathfrak{h};M^*) \simeq \operatorname{H}_\bullet(\mathfrak{g},\mathfrak{h};M)^*.

Comments can be written in Markdown. To post, you must either provide a name or tick the box to comment anonymously (or, if you see yourself wanting to comment often, you can login).