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 [tex]G[/tex] 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 [tex]H^*_{deRham}(G)[/tex] as the cohomology of the complex

[tex]0\longrightarrow \Omega^0(G)\stackrel{d}\longrightarrow \Omega^1(G)\stackrel{d}\longrightarrow\cdots\stackrel{d}\longrightarrow\Omega^{dim\ G}(G)\longrightarrow 0[/tex]

where [tex]\Omega^i(G)[/tex] are the differential i-forms on [tex]G[/tex] (note, we’ll use complex-valued forms), and [tex]d[/tex] is the deRham differential.

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

The differential forms [tex]\Omega^*(G)[/tex] give a representation of [tex]G[/tex] in two ways, taking the induced action on forms by pullback, using either left or right translation on the group. If [tex](\Pi(g), \Omega^*(G))[/tex] is the representation by left translations, we can use this to apply our “averaging over [tex]G[/tex]” 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

[tex]0\longrightarrow \Omega^0(G)^G\stackrel{d}\longrightarrow \Omega^1(G)^G\stackrel{d}\longrightarrow\cdots\stackrel{d}\longrightarrow\Omega^{dim\ G}(G)^G\longrightarrow 0[/tex]

Since elements of the Lie algebra [tex]\mathfrak g[/tex] 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 [tex]C^*(\mathfrak g, \mathbf R)= \Lambda^*(\mathfrak g^*)=\Omega^*(G)^G[/tex], and the differentials coincide. So, what we have shown is that

[tex]H^*(\mathfrak g, \mathbf C)= H^*_{de Rham}(G)[/tex]

If one knows the cohomology of [tex]G[/tex], the Lie algebra cohomology is thus known, but this identity is normally used in the other direction, to find the cohomology of [tex]G[/tex] 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 [tex]\Lambda^*(\mathfrak g)[/tex], which again commutes with the differential. We end up with a complex
[tex]0\longrightarrow (\Lambda^0(\mathfrak g^*))^G \longrightarrow (\Lambda^1(\mathfrak g^*))^G\longrightarrow\cdots\longrightarrow (\Lambda^{\dim\ \mathfrak g}(\mathfrak g^*))^G\longrightarrow 0[/tex]

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

[tex]H^*(\mathfrak g,\mathbf C)=(\Lambda^*(\mathfrak g^*))^G=(\Lambda^*(\mathfrak g^*))^{\mathfrak g}[/tex]

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

[tex]H^*(\mathbf R^n, \mathbf C)=\Lambda^*(\mathbf C^n)[/tex]

Note that complexifying the Lie algebra and working with [tex]\mathfrak g_{\mathbf C}=\mathfrak g\otimes \mathbf C[/tex] commutes with taking cohomology, so we get

[tex]H^*(\mathfrak g_{\mathbf C},\mathbf C)= H^*(\mathfrak g,\mathbf C)\otimes \mathbf C[/tex]

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

[tex]H^*(\mathfrak g_{\mathbf C}, \mathbf C) = (\Lambda^*(\mathfrak g_{\mathbf C}))^{\mathfrak g_\mathbf C}[/tex]

Besides [tex]H^0[/tex], one always gets a non-trivial [tex]H^3[/tex], since one can use the Killing form [tex]< \cdot,\cdot>[/tex] to produce an adjoint-invariant 3-form [tex]\omega_3(X_1,X_2,X_3)=[/tex]. For [tex]G=SU(n)[/tex], [tex]\mathfrak g_{\mathbf C}=\mathfrak{sl}(n,\mathbf C})[/tex], and one gets non-trivial cohomology classes [tex]\omega_{2i+1}[/tex] for [tex]i=1,2,\cdots n[/tex], such that

[tex]H^*(\mathfrak{sl}(n,\mathbf C))=\Lambda^*(\omega_3, \omega_5,\cdots,\omega_{2n+1})[/tex]

the exterior algebra generated by the [tex]\omega_{2i+1}[/tex].

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

[tex]H^*(\mathfrak g, V)=H^*(\mathfrak g, \mathbf C)\otimes V^{\mathfrak g}[/tex]

The [tex]H^0[/tex] piece of this is just the [tex]V^{\mathfrak g}[/tex] that we want when we are doing BRST, but we also get quite a bit else: [tex]dim\ V^{\mathfrak g}[/tex] copies of the higher degree pieces of the Lie algebra cohomology [tex]H^*(\mathfrak g, \mathbf C)[/tex]. 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 [tex]V^{\mathfrak g}[/tex] in [tex]V[/tex].

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.

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

  1. Anon says:

    I’m a bit confused by the step where the De Rham complex of differential forms on the group G is found to have the same cohomology as the complex of left-invariant differential forms. Did I miss the step where this was explained above?

  2. Peter Woit says:


    I guess maybe I’m skipping some steps there…

    The argument is that:

    1. Because the averaging functor is exact, the cohomology of the complex with terms [tex]\Omega^*(G)^G$[/tex] is [tex]H^*(G)^G[/tex]

    2. But [tex]H^*(G)[/tex] is [tex]G[/tex]-invariant, since it is homotopy-invariant, and [tex]G[/tex] is connected.

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