Notes on BRST IV: Lie Algebra Cohomology for Semi-simple Lie Algebras

In this posting I’ll work out some examples of Lie algebra cohomology, still for finite dimensional Lie algebras and representations.

If $G$ is a compact, connected Lie group, it can be thought of as a compact manifold, and as such one can define its de Rham cohomology $H_{deRham}^{\ast }(G)$ as the cohomology of the complex

$0⟶{\mathrm{\Omega }}^0(G)\stackrel{d}{⟶}{\mathrm{\Omega }}^1(G)\stackrel{d}{⟶}\cdots \stackrel{d}{⟶}{\mathrm{\Omega }}^{dimG}(G)⟶0$

where ${\mathrm{\Omega }}^i(G)$ are the differential i-forms on $G$ (note, we’ll use complex-valued forms), and $d$ is the deRham differential.

For a compact group, one has a bi-invariant Haar measure ${\int }_G$, and can use this to “average” over an action of the group on a space. For a representation $(\pi ,V)$, we get a projection operator ${\int }_g\mathrm{\Pi }(g)$ onto the invariant subspace $V^G$. This projection operator gives explicitly the invariants functor on ${\mathcal{C}}_{\mathfrak{g}}$. It is an exact functor, taking exact sequences to exact sequences.

The differential forms ${\mathrm{\Omega }}^{\ast }(G)$ give a representation of $G$ in two ways, taking the induced action on forms by pullback, using either left or right translation on the group. If $(\mathrm{\Pi }(g),{\mathrm{\Omega }}^{\ast }(G))$ is the representation by left translations, we can use this to apply our “averaging over $G$” projection operator to the de Rham complex. This action commutes with the de Rham differential, so we get a sub-complex of left-invariant forms

$0⟶{\mathrm{\Omega }}^0(G{)}^G\stackrel{d}{⟶}{\mathrm{\Omega }}^1(G{)}^G\stackrel{d}{⟶}\cdots \stackrel{d}{⟶}{\mathrm{\Omega }}^{dimG}(G{)}^G⟶0$

Since elements of the Lie algebra $\mathfrak{g}$ are precisely left-invariant 1-forms, it turns out that this complex is nothing but the Chevalley-Eilenberg complex considered last time to represent Lie algebra cohomology, for the case of the trivial representation. This means we have $C^{\ast }(\mathfrak{g},\mathbf{R})={\mathrm{\Lambda }}^{\ast }({\mathfrak{g}}^{\ast })={\mathrm{\Omega }}^{\ast }(G{)}^G$, and the differentials coincide. So, what we have shown is that

$H^{\ast }(\mathfrak{g},\mathbf{C})=H_{deRham}^{\ast }(G)$

If one knows the cohomology of $G$, the Lie algebra cohomology is thus known, but this identity is normally used in the other direction, to find the cohomology of $G$ from that of the Lie algebra. To compute the Lie-algebra cohomology, we can exploit the right-action of G on the group, averaging over the induced action on the left-invariant forms ${\mathrm{\Lambda }}^{\ast }(\mathfrak{g})$, which again commutes with the differential. We end up with a complex
$0⟶({\mathrm{\Lambda }}^0({\mathfrak{g}}^{\ast }){)}^G⟶({\mathrm{\Lambda }}^1({\mathfrak{g}}^{\ast }){)}^G⟶\cdots ⟶({\mathrm{\Lambda }}^{\dim \mathfrak{g}}({\mathfrak{g}}^{\ast }){)}^G⟶0$

where all the differentials are zero, so the cohomology is given by

$H^{\ast }(\mathfrak{g},\mathbf{C})=({\mathrm{\Lambda }}^{\ast }({\mathfrak{g}}^{\ast }){)}^G=({\mathrm{\Lambda }}^{\ast }({\mathfrak{g}}^{\ast }){)}^{\mathfrak{g}}$

the adjoint-invariant pieces of the exterior algebra on ${\mathfrak{g}}^{\ast }$. Finding the cohomology has now been turned into a purely algebraic problem in invariant theory. For $G=U(1)$, $\mathfrak{g}=\mathbf{R}$, and we have shown that $H^{\ast }(\mathbf{R},\mathbf{C})={\mathrm{\Lambda }}^{\ast }(\mathbf{C})$, this is $\mathbf{C}$ in degrees 0, and 1, as expected for the de Rham cohomology of the circle $U(1)=S^1$. For $G=U(1{)}^n$, we get

$H^{\ast }({\mathbf{R}}^n,\mathbf{C})={\mathrm{\Lambda }}^{\ast }({\mathbf{C}}^n)$

Note that complexifying the Lie algebra and working with ${\mathfrak{g}}_{\mathbf{C}}=\mathfrak{g}\otimes \mathbf{C}$ commutes with taking cohomology, so we get

$H^{\ast }({\mathfrak{g}}_{\mathbf{C}},\mathbf{C})=H^{\ast }(\mathfrak{g},\mathbf{C})\otimes \mathbf{C}$

Complexifying the Lie algebra of a compact semi-simple Lie group gives a complex semi-simple Lie algebra, and we have now computed the cohomology of these as

$H^{\ast }({\mathfrak{g}}_{\mathbf{C}},\mathbf{C})=({\mathrm{\Lambda }}^{\ast }({\mathfrak{g}}_{\mathbf{C}}){)}^{{\mathfrak{g}}_{\mathbf{C}}}$

Besides $H^0$, one always gets a non-trivial $H^3$, since one can use the Killing form $<\cdot ,\cdot >$ to produce an adjoint-invariant 3-form [tex]\omega_3(X_1,X_2,X_3)=[/tex]. For $G=SU(n)$, $\text{Extra close brace or missing open brace}$, and one gets non-trivial cohomology classes ${\omega }_{2i+1}$ for $i=1,2,\cdots n$, such that

$H^{\ast }(\mathfrak{sl}(n,\mathbf{C}))={\mathrm{\Lambda }}^{\ast }({\omega }_3,{\omega }_5,\cdots ,{\omega }_{2n+1})$

the exterior algebra generated by the ${\omega }_{2i+1}$.

To compute Lie algebra cohomology $H^{\ast }(\mathfrak{g},V)$ with coefficients in a representation $V$, we can go through the same procedure as above, starting with differential forms on $G$ taking values in $V$, or we can just use exactness of the averaging functor that takes $V$ to $V^G$. Either way, we end up with the result

$H^{\ast }(\mathfrak{g},V)=H^{\ast }(\mathfrak{g},\mathbf{C})\otimes V^{\mathfrak{g}}$

The $H^0$ piece of this is just the $V^{\mathfrak{g}}$ that we want when we are doing BRST, but we also get quite a bit else: $dim{V}^{\mathfrak{g}}$ copies of the higher degree pieces of the Lie algebra cohomology $H^{\ast }(\mathfrak{g},\mathbf{C})$. The Lie algebra cohomology here is quite non-trivial, but doesn’t interact in a non-trivial way with the process of identifying the invariants $V^{\mathfrak{g}}$ in $V$.

In the next posting I’ll turn to an example where Lie algebra cohomology interacts in a much more interesting way with the representation theory, this will be the highest-weight theory of representations, in a cohomological interpretation first studied by Bott and Kostant.