Lie Algebras
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7 Homological Methods
Lets expand on homological methods, which were first mentioned in the proof of Weyl’s theorem that a semisimple Lie algebra has semisimple finite dimensional representations.
Usually \(\Ext \) and \(\Tor \) are the main tools of studying representations of algebras but for a Hopf algebra, we can do with a bit less. First observe that the antipode makes left and right modules canonically equivalent, so
we can drop the lefts and rights.

Definition 7.1. \(H_*(\mg ,M)\) is \(\Tor _*^{U(\mg )}(k,M)\), \(H^*(\mg ,M)\) is \(\Ext
^*_{U(\mg )}(k,M)\).
In otherwords, \(H_*\), the homology, is the left derived functors of the coinvariants and \(H^*\), cohomology, is the right derived functors of invariants. The \(\mg \) will be dropped if unambiguous. The homology, cohomology
should be thought of as sheaf cohomology on the associated Lie group (the sheaf being trivial for the module \(k\)). When the module is \(k\) itself, we can also refer to it as cohomology and homology of the Lie algebra. \(M\)
should correspond to a right invariant trivial vector bundle such that differentiating the \(\Ad \) action gives the actiono of \(\mg \) on the fibre. Indeed this construction coincides with the construction of \(\ad \). The reason
these are well equipped to replace \(\Ext \) and \(\Tor \) is the following:

Proposition 7.2. \(\Ext ^*(M,N) = H^*(\Hom (M,N))\), \(\Tor ^*(M,N) = H_*(\Hom (M,N)))\).

Proof. \(\Hom _{\mg }(M,N) \cong \Hom _{\mg }(k,\Hom (M,N))\). Take right derived functors in \(N\). On the right hand side, \(\Hom (M,)\) is exact and sends injectives to
injectives by the tensor hom adjunction since tensor products are exact, so the edge homomorphism in the Grothendieck spectral sequence is an isomorphism.
The same argument works for \(\Tor \) since \(M\otimes _{\mg } N \cong k \otimes _{\mg } (M \otimes N)\). □
Since \(U(\mg ) \otimes _k U(\mg ')\) = \(U(\mg \oplus \mg ')\) we get Kunneth isomorphisms:
\[\bigoplus _{p+q=n} H_p(\mg ,k)\otimes H_p(\mg ',k) = H_n(\mg \oplus \mg ',k)\]
\[\bigoplus _{p+q=n} H^p(\mg ,k)\otimes H^p(\mg ',k) =H^n(\mg \oplus \mg ',k)\]
Using the diagonal \(\mg \to \mg \oplus \mg \) we get a coalgebra structure and an algebra structure on the homology and cohomology of \(\mg \).
\(H_1,H^1\) are easy to interpret. Let \(J\) denote the augementation ideal of \(U(\mg )\).

Proof. \(0 \to J \to U(\mg ) \to k \to 0\), so taking the long exact sequence in homology, we get \(H_1(k) = H_0(J) = \mg /[\mg ,\mg ]\). □

Proof. Again look at the long exact sequence for \(0 \to J \to U(\mg ) \to k \to 0\) after taking \(\Hom \) into \(M\). We get \(0 \to \Hom (k,M) \to \Hom (U(\mg ),M) \to
\Hom (J,M) \to H^1(M) \to 0\). \(\Hom (J,M)\) is the derivations, and the image of \(\Hom (U(\mg )),M\) is the inner derivations. □
If \(M\) is trivial, then \(\Ider (M)\) is trivial, and \(\Der (M)\) is \(\Hom _{\Lie }(\mg ,M)\). Suppose we have a short exact sequence \(0 \to \mh \to \mg \to \mg /\mh \to 0\). The forgerful functor from \(\mg /\mh
\) modules to \(\mg \) modules has left adjoint \(\mh \)coinvariants and right adjoint \(\mh \)invariants. Taking \(\mg \)invariants is the composite of taking \(\mh \)invariants and then taking \(\mg /\mh \)invariants, and
similarly for coinvariants so we get Grothendieck spectral sequences:
\[E_2^{p,q} = H^p(\mg /\mh ,H^q(\mh ,M)) \Rightarrow H^{p+q}(\mg , M) \]
\[E_{p,q}^2 = H_p(\mg /\mh ,H_q(\mh ,M)) \Rightarrow H_{p+q}(\mg , M) \]
These are also called the HochschildSerre spectral sequences.
\(H^2(M)\) has a simple interpretation: it is extensions of \(\mg \) by the abelian Lie algebra \(M\).
Given an extension \(0 \to M \to \me \to \mg \to 0\), we can look at low dimenisonal exact sequence of the spectral sequence in cohomology:
\[0 \to H^1(\mg ,M) \to H^1(\me ,M) \to H^1(M,M)^{\mg } \to H^2(\mg ,M) \to H^2(\me ,M)\]
The image of the identity in \(H^1(M,M)^{\mg } = \Hom _{\mg }(M,M)\) is the class corresponding to the extension. This gives a map \(\Ext (\mg ,M) \xrightarrow {d^2} H^2(\mg ,M)\). To go in the other direction, we can
use a presentation of \(\mg \).
Now consider a presentation \(0 \to \ma \to \mff \to \mg \to 0\), where \(\mff \) is free, and mod out by the ideal generated by \([\ma ,\ma ]\) so that it is an extension of \(\mg \) by \(\mg _{ab}\) which is abelian.
This is actually the unversal extension in that \(\Hom (\ma _{ab}, M) = H^2(\mg ,M)\) classifies extensions by \(M\).
Given any extension \(\me \), we can form the diagram below since \(\mff \) is free and \(M\) is abelian:
Now the map producing the class in \(H^2\) corresponding to \(\me \) factors as
\[\Hom _{\mg }(M,M) \to \Hom _{\mg }(\ma _{ab}, M) = H^1(\ma _{ab},M)^{\mg } \xrightarrow {d^2} H^2(\mg ,M)\]
By naturality the extension class in \(H^2(\mg ,\ma _{ab})\) gets sent to the corresponding class in \(H^2(\mg ,M)\). Since \(H^2(\mff ,M)\) vanishes, the restriction of any class in \(H^2(\mg ,M)\) to \(\mff \) vanishes.
From the lowdimensional exact sequence, this means that there is an element of \(H^1(\ma ,M)^\mg = \Hom _{\mg }(\ma _{ab},M)\) lifting \(H^2(\mg ,M)\). One can then from the semidirect product \(M\ltimes \mff \)
over \(\ma \) to get an extension of \(\mg \) by \(M\) realizing the class.
It suffices to show that the cohomology class determines the extension. But suppose there were two extensions \(\me _1,\me _2\) coming from the same class. They are classified by some maps \(\phi _1,\phi _2 \in H^1(\ma
_{ab},M)^{\mg }\). By assumption their image via \(d^2\) is the same, so there is some element in \(H^1(\mff /[a,a],M)=\Der (\mff ,M)\) whose image is \(\phi _1\phi _2\). WLOG, the map \(\mff \to \me _1,\me
_2\) is surjective, and it is then easy to see that \(\phi _2+(\phi _1\phi _2)\) descends to an isomorphism of extensions from \(\me _1\) to \(\me _2\).

Proof. Choose a presentation \(\mg =\mff /\ma \) that is minimal in the sense that \(\ma \subset [\mff ,\mff ]\). Mod out by \([\mff ,\ma ]\) to see that \(\mff /[\mff ,\ma
] = \mg \oplus \ma /[\mff ,\ma ]\) since extensions by abelian things are trivial. But \(f\) is minimal, so taking the commutator of that equation, we see \(\ma /[\mff ,\ma ]=0\), but any subalgebra of a free algebra
satisfying that is trivial. □
There is a canonical and efficient complex called the ChevalleyEilenberg complex that computes Lie algebra homology and cohomology. It is a free resolution of \(k\), and is essentially the de Rham complex.
Consider the chain complex where \(V_n(\mg ) = U\mg \otimes _k \Lambda ^n\mg \), and \(d(g \otimes w_1 \dots \wedge w_n)= \sum _{i}(1)^{i+1}ux_i \otimes w_1 \dots \wedge \hat {w_i} \dots \wedge
w_n\) \(+ \sum _{i <j}(1)^{i+j}u\otimes [w_i,w_j]\otimes w_1 \dots \wedge \hat {w_i} \dots \wedge \hat {w_j} \dots \wedge w_n\). There is an obvious augmentation \(V_*(\mg ) \to k\).

Proof. There is a filtration on \(V_*\) coming from the tensor products of the filtrations on \(U(\mg ),\Lambda ^*(\mg )\). The associated graded by PBW is \(k[\mg ]\otimes
_k\Lambda ^*(\mg )\), and differential is the Koszul differential. Thus the spectral sequence for the filtration degenerates at \(E^1\) to the expected cohomology.
□

Corollary 7.9. For an \(\mg \)module \(M\),
\[H_*(M) = H_*(M\otimes \Lambda ^*(\mg )), H^*(M) = H^*(\Hom (\Lambda ^*(\mg ),M)\]
This complex can give explicit proofs of the interpretations of \(H^1,H^2\).
The ChevalleyEilenberg complex for the Lie algebra of a Lie group can be used to show that cohomology and homology of the Lie algebra agree with homology and cohomology of a compact connected Lie group (over \(k\)).
Namely, one can choose a biinvariant metric, and use Hodge theory to show the harmonic forms are invariant and correspond to the cohomology of the ChevalleyEilenberg complex. Using this we can see why we should expect
\(H^3(\msl _2)=k\), since \(\msl _2\CC \) is the complexification of the lie algebra of \(\su (2)\), and \(\SU (2)\) is a \(3\)sphere. Indeed, the ChevalleyEilenberg complex verifies this.
We already showed \(H^1(\mg ,M)\) vanishes for a semisimple Lie algebra, but the same is true for \(H^2\). First we can upgrade Weyl’s theorem.

Proof. Over \(\bar {k}\), \(L\), the Casimir element, acts by a scalar on \(M\) and by \(0\) on \(k\), hence it acts on the homology and cohomology, and the two actions coming from
\(M\) and \(k\) coincide since \(L\) is central. Thus a nonzero scalar is equal to zero, so the cohomology and homology vanish. □
Thus all the interesting (finitedimensional) cohomological information is in the trivial sheaf.

Proof. By the previous proposition, it suffices to show \(H^2(\mg )\) vanishes, which is equivalent to any saying any extension \(k \to \me \to \mg \) splits. To see that is true, simply
observe that the \([\me ,\me ]\) gives a splitting since \(k\) commutes with everything and \([\mg ,\mg ]= \mg \). □
This is analogous to (and partially proves) a fact in Lie groups, that \(\pi _2\) vanishes for any Lie group. Perhaps one complete the proof for compact Lie groups by proving that \(H_2\) is the kernel of the universal central
extension, so there can’t be any torsion as it would give a nontrivial cover.

Proof. Note that \(\ma \) is really the radical, so we want to show that \(\rad (\mg ) \to \mg \to \mg /\rad (\mg )\) splits. We can do this by induction on the length of the
derived series. Mod out by \([\rad (\mg ),\rad (\mg )]\) to get an extension by an abelian Lie algebra, for which there is a splitting by the second Whitehead lemma. The preimage of this splitting is an ideal \(\mh \) such
that there is an extension \([\ma ,\ma ] \to \mh \to \mg \), at which point we can induct. □

Proof. Consider the low term exact sequence from the homology spectral sequence for \(0 \to \ma \to \mff \to \mg \to 0\) and the trivial module:
\[0 \to H_2(\mg ,k) \to \ma /[\ma ,\ma ]_\mg \to \mff /[\mff ,\mff ] \to \mg /[\mg ,\mg ] \to 0\]
The second term is \(\ma /[\mff ,\ma ]\), giving the result. □

Proof. Suppose \(\mg \) has a universal central extension \(\me \), it must have a unique map to the trivial central extension, which implies that \(\mg \) has to be perfect. Conversely
let \(\mg \) be perfect. Choose a presentation \(\mg = \mff /\ma \), for \(\mff \) free. Then \(\mh = [\mff ,\mff ]/[\mff ,\ma ]\) will be shown to be the universal central extension. By the previous lemma, it is an
extension by \(H_2(\mg )\). Moreover, one can see it is perfect. Since \(\mg \) is perfect, any \(x \in \mff \) decomposes as \(x'+r\) where \(x' \in \mff , r \in \ma \). Then choosing such a decomposition for
\(x =x'+r,y =x'+r\) Then \([x,y] = [x',y']\) modulo \([\mff ,\ma ]\). This gives uniqueness of a map to any extension.
Now let \(\me \) be any other central extension. There is a map from \(\mff \) to \(\me \) lifting the projection to \(\mg \), and since \(\me \) is central, it induces a map from \(\mh \). □
Here is a recognition criterion for universal central extensions.

Proof. Consider \(\me '\), the universal central extension of the universal central extension, and let \(x\) be in its kernel to \(\mg \). Since \(\me '\) is perfect, for any
\(a\), there is some \(a',a''\) so that \([x,a] = [x,[a',a']]\). But by the Jacobi identity this is zero since \(\me ',\me \) are central extensions. Thus \(\me '\) is a central
extension so it splits, and must be \(\me \).
\((1) \implies (2)\): Any central extension \(\me '\) of \(\me \) splits. \(\me '\) is thus also a central extension of \(\mg \) and since \(\me \) is perfect, this splitting is unique by Lemma 7.15.
\((2) \implies (3)\): Since the a trivial \(k\) extension splits in a unique way, we see \(H_1(\me )=0\). Moreover since any extension splits, the universal central extension splits, so it must be trivial, showing \(H_2(\me )=0\).
\((3) \implies (1)\): Given any central extension of \(\mg \), we can pull it back to a central extension of \(\me \) which splits, giving a map to the central extension. Uniqueness follows from perfectness. □