Action of a symmetry in local coordinates Suppose that a Lie algebra acts on a manifold and leaves invariant a point . Then naturally acts in the tangent space . Consider maps parameterized by a small parameter , satisfying:

As we discussed in Section Geometrical abstraction, there are many such maps. Let us ask the following question: is it possible to construct such a map which would also commute with the action of ? (We are interested in a formal map, i.e. a map specified as an infinite series in ; we will not discuss convergence.)

Let us start by picking some map (not necessarily -invariant) satisfying Eqs. (6), (7) and (8). For each element there is a corresponding vector field on . Let us consider . It is a vector field on :

where is of the form with a polynomial of the degree in .

The vector field has a very straightforward meaning. Our map turns a sufficiently small open neighborhood of into a chart of . In this context, is just the “coordinate representation” of the vector field in that chart. Therefore, our question becomes:

We will see that obstacles are certain cohomology classes.

For two elements and of , we have:

This means that, for :

where , denote the coordinates on . Besides that:

Define the “BRST operator”:

where . We have . This defines the differential in the Lia algebra cohomology complex Knapp of with values in the space of vector fields on having zero of at least second order at the point . (The action of the second term, , is by the commutator of vector fields.)

Let us define as follows ( cf. Eq. (9)):

Eq. (10) implies that satisfies the MC equation:

Suppose that we replace with another function , where is any (nonlinear) function such that and . Then gets replaced with where:

This is the gauge transformation. An infinitesimal gauge transformation is:

The tangent space to the solutions of Eq. (13) at the point is the cohomology of the operator :

The Lie algebra of nonlinear

vector fields has a filtration by the scaling degree. Therefore the cohomology of can be computed by a spectral sequence. The first page of this spectral sequence is:

where .

Monodromy transformation Additional assumption We now have two actions of on : the linearized one, which is given by of Eq. (9), and the nonlinear action given by . Suppose that the linearized one integrates to some action of a group :

Suppose that this group has a non-contractible one-dimensional cycle.

Consider the path ordered exponential over this non-contractible cycle. Without loss of generality, we can start and end the loop at the unit. We define:

where is the action of on . This is the monodromy transformation:

Notice that the derivative of at the point is zero. Therefore we can define its second derivative:

Eq. (17) implies:

Usually the cycle is such that is constant. Then the meaning of the integration in Eq. (18), is that that we pick the resonant terms in the quadratic vector field .

Symmetries The monodromy transformation commutes with the action of , but we have to remember that the action of is given by nonlinear vector fields — see Eq. (9). If it were given just by linearized vector fields, i.e. the of Eq. (9), life would be easier. But this is, generally speaking, not the case. Notice that . Instead of being zero, we have:

But of Eq. (18) does commute with the undeformed action of on ( i.e. with ). This is because are of quadratic and higher order, and the first derivative of vanishes.

Sometimes is zero on some subspace . Then, on this subspace, we can define the third derivative . Suppose that, in addition, the restriction of on is parallel to , i.e.:

Then commutes with the undeformed action of on .

Closed subsectors Suppose that , as a representation of , has an invariant subspace:

It may happen that the restriction of to is tangent to . This, essentially, means that is closed under the action of the symmetry. In particular, the monodromy transformation of Section Monodromy transformation acts within . The sufficient condition for this is: