BRST cohomology vs Lie algebra cohomology
1 Relation between the SYM algebra and the susy algebra
The Lie superalgebra
is formed by the operators
with the following commutation relations:
(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.
It is convenient to start with a generalization of (6
). Remember that in (6
the BRST operator is:
We observe that the components of pure spinor
get contracted with the differential operators
which form a representation
of the algebra
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 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.