# Unravelling the Mystery of the (0,3) Symmetric Tensor

• christodouloum
In summary, this statement holds for every pair of indices not generally, but there is no way to generalize it.f

#### christodouloum

I am a bit confused by this observation.
Every tensor is it's symmetric plus antisymmetric part.

Thus for the components of a (0,3) tensor

$$F_{\lambda\mu\nu}=F_{[\lambda\mu\nu]}+F_{\{\lambda\mu\nu\}}$$

and if I write this down explicitly I end up that for the components of ANY (0,3) tensor

$$F_{\lambda\mu\nu}=(1/3)(F_{\lambda\mu\nu} +F_{\mu\nu\lambda}+F_{\nu\lambda\mu} )$$

Huh? Does this indeed hold?

I am replying to my self since I searched around a bit and the statement
Every tensor is it's symmetric plus antisymmetric part

holds for every pair of indices not generally. So now I know that the last equality I wrote does not hold but still, is there any way to generalize this idea? I mean, please correct me if I am wrong but we have

$$F_{\lambda\mu}=F_{[\lambda\mu]}+F_{\{\lambda\mu\}}$$
$$F_{\lambda\mu\nu}=F_{[\lambda\mu]\nu}+F_{\{\lambda\mu\}\nu}$$

how about a relation between $$F_{\lambda\mu\nu}$$ ,$$F_{\{\lambda\mu\nu\}}$$ and $$F_{[\lambda\mu\nu]}$$??

It is a useful idea, because symmetric and antisymmetric tensors each have useful properties, but the decomposition is basically trivial arithmetic, i.e.

a = (a+b)/2 + (a-b)/2
b = (a+b)/2 - (a-b)/2

I don't think there is much to generalize about that.

I know how to work it out . It is surely not a trivial idea, the symmetric and antisymmetric parts of the exponential function in the reals are the hyberbolic functions cosh and sinh which make the splitting idea quite important. Also by the property that a totally symmetric tensor contracted with a totally antisymmetric one gives nul I am just wondering, is there any way to split an arbitrary tensor in two additive parts the symmetric and antisymmetric one? I