Simple identity for antisymmetric tensor

[paweld]

Is it true that for all antisymmetric tensors $$F^{\mu\nu}$$
the following identity is true:
$$\nabla_\mu \nabla_\nu F^{\mu\nu}=0$$
(I've checked it but I'm not absolutely sure).

 

[tiny-tim]

hi paweld!

Is it true that for all antisymmetric tensors $$F^{\mu\nu}$$
the following identity is true:
$$\nabla_\mu \nabla_\nu F^{\mu\nu}=0$$
(I've checked it but I'm not absolutely sure).

yup, because ∇_µ∇_ρ is symmetric in µ and ρ, so it zeroes anything antisymmetric in µ and ρ

 

[arkajad]

That depends on how you define $$\nabla_\mu$$. For a general affine connection you get, more or less, $$\pm R_{\mu\nu}F^{\mu\nu}$$ (plus or minus depending on which convention is being used in the definition of the Ricci tensor). When there is no torsion, Ricci tensor is symmetric and you get zero. But not so for a general connection.

 

[paweld]

Thanks, I always assume that connection is torsion-free.

 

[arkajad]

BTW:

∇µ∇ρ is not symmetric in µ and ρ. Its antisymmetric part is related to the curvature tensor.

The above holds for u,v commuting vector fields like $$\partial_\mu,\, \partial_\nu$$