BRST cohomology vs Lie algebra cohomology

#### 1` `Relation between the SYM algebra and the susy algebra

The Lie superalgebra is formed by the operators and with the following commutation relations:

The algebra (unlike SYM algebra) is finite-dimensional.
There is a homomorphism from the SYM algebra to :

The kernel of this homomorphism is an ideal in SYM, which we will call . We have:

#### 2` `Cohomology of SYM algebra and pure spinor cohomology

Now we want to relate the cohomology of the operator (6) to the cohomology of the Lie superalgebra SYM.

We observe that the components of pure spinor get contracted with the differential operators
which form a representation of the algebra (11).

Let us therefore consider any representation of . To every such representation corresponds a pure spinor BRST complex, which is constructed as follows.

Let be the space of polynomials of the pure spinor variables of degree . It is assumed that ; in other words is the commutative algebra of polynomials of the degree of the variables modulo the ideal of polynomials divisible by the expression .

where the BRST operator acts as follows (compare to (14)):

The action of on where is a polynomial of degree of and
is equal to . Notice that is a polynomial
of the degree in , so we got an element of .

The cohomology of this complex is related to the Lie algebra cohomology of the SYM algebra.

In order to explain how, we have to notice that , being a representation of , is also a representation of
(because of the homomorphism (12)). Then we have:

This is a very nontrivial theorem. Our goal is to establish a similar theorem in the SUGRA context.

An important special case is when is
the dual space of the universal enveloping algebra of the algebra:
This is the representation in the space of Taylor series of functions on the super-space-time.
In this case the cohomology of the BRST complex computes the solutions of the Maxwell equations.
In this case the operator (17) becomes precisely (14).