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  SUSY Dirac operator

+ 2 like - 0 dislike
84 views
For a supermanifold, can we define a supersymmetric Dirac operator ?
asked Apr 4 in Mathematics by Antoine Balan (60 points) [ no revision ]

1 Answer

+ 3 like - 0 dislike

Yes. In physics, we think of a Dirac operator \(D\) as a first-order differential operator such that \(D^2=\Delta\) is the Laplacian. (Mathematically, its an operator acting on the sections of a spinor bundle, see here for details). In the case of supersymmetry, the interesting question to ask is about the possible supersymmetric decompositions of the Dirac operator, i.e., decompositions that satisfy an appropriate SUSY algebra.

For the sake of pedagogy, consider \(d=4\), where we have 2 SUSY decompositions of \(D\), one of which is called chiral SUSY, and the other complex SUSY. Here I will consider only the first, since it's well-known and a little easier to construct. For the complex case cf. the ref. below. 

Let \((\gamma^m)\) be our Clifford algebra generators (i.e., \(\{\gamma^m,\gamma^n\}=2g^{mn}1\)), \(D_m=\partial_m-iA_m\) the covariant derivative for a vector potential \(A_m\), and \({\not}D=\gamma^m D_m\). Then Dirac operator can be decomposed as \[{\not}D = Q_+ + Q_-,\]

where \[Q_+=\frac{1}{2}(1+\gamma^5)\gamma^m D_m, \quad Q_-=\frac{1}{2}(1-\gamma^5)\gamma^m D_m.\]

As an easy exercise, you can see that \(Q_+\) and \(Q_-\) are nilpotent operators, and letting \(H=-(\gamma^m D_m)^2\), the usual \(\mathcal{N}=1\) SUSY algebra is satisfied: \[H = \{Q_+,Q_-\}, \quad [Q_+, H] = [Q_-,H]=0,\]

as you learn in a SUSY QM course.

For details and a discussion of the complex SUSY case, cf.

answered Apr 4 by Igor Mol (500 points) [ no revision ]

You are right, but I believe that it is not the object of my question. I didn't ask if the Dirac operator is supersymmetric; but I would like to have a Dirac operator for a supermanifold and not only for a spin-manifold. I think that we would have to study the spin supersymmetric Lie groups and the supersymmetric connections.

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