# 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 $$H^*_{deRham}(G)$$ as the cohomology of the complex

$$0\longrightarrow \Omega^0(G)\stackrel{d}\longrightarrow \Omega^1(G)\stackrel{d}\longrightarrow\cdots\stackrel{d}\longrightarrow\Omega^{dim\ G}(G)\longrightarrow 0$$

where $$\Omega^i(G)$$ 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 $$\int_g \Pi (g)$$ 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 $$\Omega^*(G)$$ 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 $$(\Pi(g), \Omega^*(G))$$ 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

$$0\longrightarrow \Omega^0(G)^G\stackrel{d}\longrightarrow \Omega^1(G)^G\stackrel{d}\longrightarrow\cdots\stackrel{d}\longrightarrow\Omega^{dim\ G}(G)^G\longrightarrow 0$$

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 $$C^*(\mathfrak g, \mathbf R)= \Lambda^*(\mathfrak g^*)=\Omega^*(G)^G$$, and the differentials coincide. So, what we have shown is that

$$H^*(\mathfrak g, \mathbf C)= H^*_{de Rham}(G)$$

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 $$\Lambda^*(\mathfrak g)$$, which again commutes with the differential. We end up with a complex
$$0\longrightarrow (\Lambda^0(\mathfrak g^*))^G \longrightarrow (\Lambda^1(\mathfrak g^*))^G\longrightarrow\cdots\longrightarrow (\Lambda^{\dim\ \mathfrak g}(\mathfrak g^*))^G\longrightarrow 0$$

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

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

the adjoint-invariant pieces of the exterior algebra on $$\mathfrak g^*$$. 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 $$H^*(\mathbf R, \mathbf C)=\Lambda^*(\mathbf C)$$, 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

$$H^*(\mathbf R^n, \mathbf C)=\Lambda^*(\mathbf C^n)$$

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

$$H^*(\mathfrak g_{\mathbf C},\mathbf C)= H^*(\mathfrak g,\mathbf C)\otimes \mathbf C$$

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

$$H^*(\mathfrak g_{\mathbf C}, \mathbf C) = (\Lambda^*(\mathfrak g_{\mathbf C}))^{\mathfrak g_\mathbf C}$$

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 $$\omega_3(X_1,X_2,X_3)=$$. For $$G=SU(n)$$, $$\mathfrak g_{\mathbf C}=\mathfrak{sl}(n,\mathbf C})$$, and one gets non-trivial cohomology classes $$\omega_{2i+1}$$ for $$i=1,2,\cdots n$$, such that

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

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

To compute Lie algebra cohomology $$H^*(\mathfrak g, V)$$ 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

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

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 $$H^*(\mathfrak g, \mathbf C)$$. 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.

Last Updated on

This entry was posted in BRST. Bookmark the permalink.

### 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:

Anon,

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 $$\Omega^*(G)^G$$ is $$H^*(G)^G$$

2. But $$H^*(G)$$ is $$G$$-invariant, since it is homotopy-invariant, and $$G$$ is connected.