Let and denote the projectors:

For any vector , we will denote by the difference of its and components:

The projector was defined in Bedoya:2010qz as follows:

where is adjusted to satisfy (70). In fact is the projection to the tangent space along the space which is orthogonal to with respect to the metric defined by :

In other words, for generic and we have an exact sequence:

In Section Explicit formula for the projector we will give an explicit formula for following Berkovits:2010zz.

It turns out that computations can often be streamlined by thinking about elements of literally as -matrices. In fact is a factorspace of modulo a subspace generated by the unit matrix. Therefore, when talking about a matrix corresponding to an element of , we have to explain every time how we choose a representative. The grading of can be extended to ; the unit matrix has grade two. Therefore, the ambiguity of adding a unit matrix only arises for representing elements of . To deal with this problem, we introduce some notations. Given a matrix of grade two, we denote by the corresponding traceless matrix:

(The subscript “TL” is an abbreviation for “traceless”.) Also, it is often useful to consider -matrices with nonzero supertrace. Such matrices do not correspond to any elements of . For a -matrix we define:

In particular:

We also define, for any even matrix :

This definition agrees with Eq. (68).

In fact is given by the following expression:

Notice that is actually both super-traceless and traceless; it is the same as (with the overline extending over “”). We have to prove that the defined this way satisfies (70). Indeed, we have:

and we have to prove Eq. (70). We have:

Both and have -grading two. Let us use:

For all grade matrices and such that the following identity holds:

Therefore:

(where “” means “modulo the center of ”, i.e. up to a multiple of the unit matrix). This proves (70).

The central part of is generally speaking nonzero:

In -matrix notations, is and is .

Let us define ( cp. Eq. 71):

so that:

It follows from the definition, that for any we have

Let us verify this explicitly using the definition (69) with the explicit expression for given by (71). We have:

Consider the expression :

Let us consider the first expression on the RHS of (78). Using (73) we rewrite:

This cancels with the first term on the RHS of (77). And the second expression on the RHS of (78) is zero:

Consider the decomposition:

Here is a 4-dimensional subspace -orthogonal to and commuting with , and is -orthogonal to and commuting with .

Similarly we can refine and :