# Notes on BRST III: Lie Algebra Cohomology

The Invariants Functor

The last posting discussed one of the simplest incarnations of BRST cohomology, in a formalism familiar to physicists. This fits into a much more abstract mathematical context, and that’s what we’ll turn to now.

Given a Lie algebra $\mathfrak g$, we’ll consider Lie algebra representations as modules over $U(\mathfrak g)$. Such modules form a category $\mathcal C_{\mathfrak g}$: what is interesting is not just the objects of the category (the equivalence classes of modules), but also the morphisms between the objects. For two representations $V_1$ and $V_2$ the set of morphisms between them is a linear space denoted $Hom_{U(\mathfrak g)}(V_1,V_2)$. This is just the set of linear maps from $V_1$ to $V_2$ that commute with the action of $\mathfrak g$:

$$Hom_{U(\mathfrak g)}(V_1,V_2)=\{\phi\in Hom_{\mathbf C}(V_1,V_2): \pi(X)\phi=\phi\pi(X)\ \forall X\in \mathfrak g\}$$

Another conventional name for this is the space of intertwining operators between the two representations.

For any representation $V$, its $\mathfrak g$-invariant subspace $V^{\mathfrak g}$ can be identified with the space $Hom_{U(\mathfrak g)}(\mathbf C, V)$, where here $\mathbf C$ is the trivial one-dimensional representation. Having a way to pick out the invariant piece of a representation also allows one to solve the more general problem of picking out the subspace that transforms like a specific irreducible $W$: just find the invariant subspace of $V\otimes W^*$.

The map $V\rightarrow V^{\mathfrak g}$ that takes a representation to its $\mathfrak g$-invariant subspace is a functor: it takes the category $\mathcal C_{\mathfrak g}$ to $\mathcal C_{\mathbf C}$, the category of vector spaces and linear maps ($\mathbf C$ – modules and $\mathbf C$ – homomorphisms). If, instead of taking

$$V\rightarrow V^{\mathfrak g}$$

one takes

$$V\rightarrow V^{\mathfrak h}$$

where $$\mathfrak h$$ is a Lie subalgebra of $$\mathfrak g$$, one again gets a functor. If $$\mathfrak h$$ is an ideal in $$\mathfrak g$$ (so that $$\mathfrak g/\mathfrak h$$ is a Lie algebra), then this functor takes $$\mathcal C_{\mathfrak g}$$ to $$\mathcal C_{\mathfrak g/\mathfrak h}$$. This is a simple version of the situation of interest in the case of gauge theory: if $$V$$ is a state space with $$\mathfrak h$$ acting as a gauge symmetry, then $$V^{\mathfrak h}$$ will be the physical subspace, carrying an action of the algebra of operators $$U(\mathfrak g/\mathfrak h)$$.

Some Homological Algebra

It turns out that when one has a category of modules like $$\mathcal C_{\mathfrak g}$$, these can usefully be studied by considering complexes of modules, and this is the subject of homological algebra. A complex of modules is a sequence of modules and homomorphisms

$$\cdots\stackrel{\partial}\longrightarrow U\stackrel{\partial}\longrightarrow V \stackrel{\partial}\longrightarrow W\stackrel{\partial}\longrightarrow\cdots$$

such that $$\partial\circ\partial =0$$. If the complex satisfies $$im\ \partial=ker\ \partial$$ at each module, the complex is said to be an “exact complex”.

To motivate the notion of exact complex, note that

$$0\longrightarrow V_0\longrightarrow V \longrightarrow 0$$

is exact iff $$V_0$$ is isomorphic to $$V$$, and an exact sequence

$$0\longrightarrow V_1 \longrightarrow V_0 \longrightarrow V \longrightarrow 0$$

represents the module $$V$$ as the quotient $$V_0/V_1$$. Using longer complexes, one gets the notion of a resolution of a module $$V$$ by a sequence of n modules $$V_i$$. This is an exact complex

$$0\longrightarrow V_n \longrightarrow\cdots\longrightarrow V_1 \longrightarrow V_0\longrightarrow V\longrightarrow 0$$

The deviation of a sequence from being exact is measured by its homology $H^*=\frac{ker\ \partial}{im\ \partial}$. Note that if one deletes $$V$$ from its resolution, the sequence

$$0\longrightarrow V_n \longrightarrow\cdots\longrightarrow V_1 \longrightarrow V_0\longrightarrow 0$$

is exact except at $$V_0$$. Indexing the homology in the obvious way, one has $$H^i =0$$ for $$i>0$$, and $$H^0=V$$. A sequence like this whose only homology is $$V$$ at $$H^0$$ is another manifestation of a resolution of $$V$$.

The reason this construction is useful is that, for many purposes, it allows us to replace a module whose structure we may not understand by a sequence of modules whose structure we do understand. In particular, we can replace a $$U(\mathfrak g)$$ module $$V$$ by a sequence of free modules, i.e. modules that are just sums of copies of $$U(\mathfrak g)$$ itself. This is called a free resolution, and more generally one can work with projective modules (direct summands of free modules).

A functor that takes exact complexes to exact complexes is called an exact functor. Homological invariants of modules come about in cases where one has a functor on a category of modules that is not exact. Applying such a functor to a free or projective resolution gives the homological invariants.

The Koszul Resolution and Lie Algebra Cohomology

There are many possible choices of a free resolution of a module. For the case of $$U(\mathfrak g)$$ modules, one convenient choice is known as the Koszul (or Chevalley-Eilenberg) resolution. To construct a resolution of the trivial module $$\mathbf C$$, one uses the exterior algebra on $$\mathfrak g$$ to make free modules

$$Y_k=U(\mathfrak g)\otimes_{\mathbf C}\Lambda^k(\mathfrak g)$$

and get a resolution of $$\mathbf C$$

$$0\longrightarrow Y_{dim\ \mathfrak g}\stackrel{\partial_{dim\ \mathfrak g -1}}\longrightarrow\cdots\stackrel{\partial_1}\longrightarrow Y_1\stackrel{\partial_0}\longrightarrow Y_0\stackrel{\epsilon}\longrightarrow \mathbf C\longrightarrow 0$$

The maps are given by
$$\epsilon : u\in Y_0=U(\mathfrak g) \rightarrow \epsilon (u) = const.\ term\ of\ u$$

and
$$\partial_{k-1} (u\otimes X_1\wedge\cdots\wedge X_k)=$$
$$\sum_{i=1}^k(-1)^{i+1}(uX_i\otimes X_1\wedge\cdots\wedge\hat X_i\wedge\cdots \wedge X_k)$$
$$+\sum_{i<j} (-1)^{i+j}(u\otimes[X_i,X_j]\wedge X_1\wedge\cdots\wedge \hat X_i\wedge\cdots\wedge \hat X_j\wedge\cdots\wedge X_k)$$

To get Lie algebra cohomology, we apply the invariants functor

$$V\longrightarrow V^{\mathfrak g}=Hom_{U(\mathfrak g)}(\mathbf C, V)$$

replacing the trivial representation by its Koszul resolution. This gives us a complex with terms

$$C^k(\mathfrak g, V)=Hom_{U(\mathfrak g)}(Y_k,V)= Hom_{U(\mathfrak g)}(U(\mathfrak g)\otimes \Lambda^k(\mathfrak g),V)$$
$$=Hom_{U(\mathfrak g)}(U(\mathfrak g),Hom_{\mathbf C}(\Lambda^k(\mathfrak g),V))$$
$$=Hom_{\mathbf C}(\Lambda^k(\mathfrak g),V) =V\otimes\Lambda^k(\mathfrak g^*)$$

and induced maps $d_i$

$$0\longrightarrow C^0(\mathfrak g, V)\stackrel{d_0}\longrightarrow C^1(\mathfrak g, V)\cdots\stackrel{d_{dim\ \mathfrak g -1}}\longrightarrow C^{dim\ \mathfrak g}(\mathfrak g, V)\longrightarrow 0$$

The Lie algebra cohomology $H^*(\mathfrak g, V)$ is just the cohomology of this complex, i.e.

$$H^i(\mathfrak g, V)=\frac{ker\ d_i}{im\ d_{i-1}}|_{C^i(\mathfrak g, V)}$$

This is exactly the same definition as that of the BRST cohomology defined in physicist’s formalism in the last posting with $\mathcal H =C^*(\mathfrak g, V)$.

One has $H^0(\mathfrak g, V)=V^{\mathfrak g}$ and so gets the $\mathfrak g$-invariants as expected, but in general the cohomology will be non-zero also in other degrees.

This is all rather abstract, so in the next posting some examples will be worked out, as well as the relationship of all this to the de Rham cohomology of the group. Anthony Knapp’s book Lie Groups, Lie Algebras, and Cohomology is an excellent reference for details on Lie algebra cohomology.

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### 6 Responses to Notes on BRST III: Lie Algebra Cohomology

1. Serifo says:

Hi professor Peter , I think you forgot to index the homomorphisms of the complex in your first definition of complex . It may be convenient also , to specify over which ring R is the complex defined.

2. Serifo says:

Ok, you are considering the complex to be defined over the ring of complex numbers C ( which is a field ). Would it be relevant to your research if you consider complexes defined over another type of rings, for example the ring Z of integers ?

3. Peter Woit says:

Serifo,

I’m trying to avoid over-generality here, just using what I need, and trying to get the basic ideas across. So, unless otherwise specified, everything is defined over C.

I’m also trying to avoid explicitly dealing with some points that would require adding a lot of details that aren’t so relevant, until the point at which they become relevant. An example would be the indexing question.

In general, if someone hasn’t seen any of this before, I hope they’ll use what I write to get a general idea of what is going on, then go to a textbook like Knapp to see the details, and to see more general versions of the story.

4. a.k. says:

..sorry if I am posing a stupid question, but: the above cohomology groups should be expressable as some sort of Ext-groups, that is as appropriate n-fold extensions of V by C, that is, they should be more or less the same as (an equivariant version of) Ext*(C,V), where C are the complex numbers. Is this true and if yes, is there a physical meaning of this correspondance?

5. woit says:

a.k.

What I gave was basically the definition of Lie algebra cohomology as
$$Ext_{U(\mathfrak g)}(\mathbf C, V)$$

This is an Ext for modules over a non-commutative algebra, I don’t know of any physical meaning for such objects, curious if anyone else does…

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