Cohomology of vs Lie algebra cohomology
In this Section we will show that the Modified de Rham complex of is the same as the Lie algebra cohomology complex of with coefficients in .
1 Definition of the Lie algebra cohomology
The space is a representation of ; the action of is slightly easier
to write down at the level of the corresponding action of the Lie group ;
acts on as follows:
Therefore:
To follow the Faddeev-Popov notations, we introduce
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Beware that is not just the Faddeev-Popov ghost; it is the product of the Faddeev-Popov ghost with .
where is the structure constants of :
The subgroup preserves and therefore :
This implies:
2 Proof that is the same as
This is similar to the statement that for any Lie group , the de Rham subcomplex of right-invariant forms on
is the same as the Lie cohomology complex of with coefficients in the trivial representation:
Our case is a variation on this theme:
As we explained,
consists of functions of and . To obtain the corresponding element
of , we replace with
with (as in Eq. (15)), and with .
Under this identification becomes .