An Identity in SUSY sigma model

[Claret]

I am struggleing in an identity, i.e. $$\nabla_m R_{ikjl}(\overline{\epsilon}\psi^m)(\overline{\psi^i}\psi^j)(\overline{\psi^k}\psi^l)=0$$ ,

where $$i,j,k,l,m$$ are dummy indices, $$\nabla_m$$ is covariant derivative, $$R_{ikjl}$$ is Riemann-Christoffel curvature tensor, and it is known that, for any two arbitary spinors $$\psi^i,\psi^j$$, $$\overline{\psi^i}\psi^j=\overline{\psi^j}\psi^i$$.

I think one could use Bianchi Identity to prove this, but I failed...who can do me a favor? Thanks!

 

[LightHero]

The Bianchi identity for the Riemann-Christoffel curvature tensor is as follows: \nabla_{[m}R_{ikjl]} = 0.Using this identity, we can rewrite the given equation as: \nabla_m R_{ikjl}(\overline{\epsilon}\psi^m)(\overline{\psi^i}\psi^j)(\overline{\psi^k}\psi^l) = 0.Since the brackets are zero, the two terms inside the brackets must be zero, so we have:\nabla_m R_{ikjl} = 0 \quad and \quad (\overline{\epsilon}\psi^m)(\overline{\psi^i}\psi^j)(\overline{\psi^k}\psi^l) = 0.Thus, the identity holds.

 

[sir-pinski]

It seems like you are struggling with an identity in the context of the SUSY sigma model. This can be a challenging topic, but don't worry, there are some useful tools you can use to help you understand and solve this problem.

Firstly, let's break down the different elements in this identity. The Riemann-Christoffel curvature tensor, denoted by R_{ikjl}, is a fundamental object in differential geometry that describes the curvature of a space. The covariant derivative, \nabla_m, is a way of measuring the change of a vector or tensor as one moves along a curve in a curved space. And the spinors, \psi^i and \psi^j, are mathematical objects used in the description of supersymmetry.

Now, the identity you are struggling with involves the covariant derivative acting on the product of the Riemann tensor and two spinors. This may seem daunting, but as you mentioned, the Bianchi Identity can be a helpful tool in solving this problem.

The Bianchi Identity is a mathematical relation that connects the covariant derivative of the Riemann tensor with the Riemann tensor itself. It states that \nabla_m R_{ikjl} + \nabla_i R_{jklm} + \nabla_j R_{klim} = 0. This is a powerful identity that can be used to simplify and prove various equations involving the Riemann tensor.

Now, let's see how we can use the Bianchi Identity to prove the identity you are struggling with. First, let's expand the covariant derivative of the product of the Riemann tensor and two spinors:

\nabla_m R_{ikjl}(\overline{\epsilon}\psi^m)(\overline{\psi^i}\psi^j)(\overline{\psi^k}\psi^l) = (\nabla_m R_{ikjl}) (\overline{\epsilon}\psi^m)(\overline{\psi^i}\psi^j)(\overline{\psi^k}\psi^l) + R_{ikjl} (\nabla_m (\overline{\epsilon}\psi^m)) (\overline{\psi^i}\psi^j)(\overline{\psi^k}\psi^l) + R_{ikjl} (\overline{\epsilon}\psi^m) (\nabla_m (\