Why isn't the Cotton tensor identical zero?

[archipatelin]

Cotton tensor C_{\mu\varkappa\lambda} is define as:

\nabla_{\sigma}W^{\sigma}_{\phantom{M}\mu\varkappa\lambda}=-\frac{n-3}{n-2}C_{\mu\varkappa\lambda}​

where W^{\sigma}_{\phantom{M}\mu\varkappa\lambda} is Weyl tensor and n is dimension of space.

Weyl tensor obey II. Bianchi identity (and all symetries of Rieamann tensor):

\nabla_{\nu}W^{\sigma}_{\phantom{M}\mu\varkappa\lambda} + \nabla_{\varkappa}W^{\sigma}_{\phantom{M}\mu\lambda\nu}+\nabla_{\lambda}W^{\sigma}_{\phantom{M}\mu\nu\varkappa}=0​

and extra it is traceless:

W^{\sigma}_{\phantom{M}\mu\sigma\varkappa}=0​

For a divergence of Weyl tensor can write:

\nabla_{\sigma}W^{\sigma}_{\phantom{M}\mu\varkappa\lambda}=g^{\rho\sigma}\nabla_{\sigma}W_{\rho\mu\varkappa\lambda}=\left<\mbox{II. Bianchi identity}\right>=-g^{\rho\sigma}\left(\nabla_{\varkappa}W_{\rho\mu\lambda\sigma}+\nabla_{\lambda}W_{\rho\mu\sigma\varkappa}\right)=
=\nabla_{\varkappa}W^{\sigma}_{\phantom{M}\mu\sigma\lambda}-\nabla_{\lambda}W^{\sigma}_{\phantom{M}\mu\sigma\varkappa}=0​

Becose Weyl tensor is traceless therefor Cotton tensor must be identical zero!
Is this correct?

 

[Altabeh]

Cotton tensor C_{\mu\varkappa\lambda} is define as:

\nabla_{\sigma}W^{\sigma}_{\phantom{M}\mu\varkappa\lambda}=-\frac{n-3}{n-2}C_{\mu\varkappa\lambda}​

where W^{\sigma}_{\phantom{M}\mu\varkappa\lambda} is Weyl tensor and n is dimension of space.

Weyl tensor obey II. Bianchi identity (and all symetries of Rieamann tensor):

\nabla_{\nu}W^{\sigma}_{\phantom{M}\mu\varkappa\lambda} + \nabla_{\varkappa}W^{\sigma}_{\phantom{M}\mu\lambda\nu}+\nabla_{\lambda}W^{\sigma}_{\phantom{M}\mu\nu\varkappa}=0​

and extra it is traceless:

W^{\sigma}_{\phantom{M}\mu\sigma\varkappa}=0​

For a divergence of Weyl tensor can write:

\nabla_{\sigma}W^{\sigma}_{\phantom{M}\mu\varkappa\lambda}=g^{\rho\sigma}\nabla_{\sigma}W_{\rho\mu\varkappa\lambda}=\left<\mbox{II. Bianchi identity}\right>=-g^{\rho\sigma}\left(\nabla_{\varkappa}W_{\rho\mu\lambda\sigma}+\nabla_{\lambda}W_{\rho\mu\sigma\varkappa}\right)=
=\nabla_{\varkappa}W^{\sigma}_{\phantom{M}\mu\sigma\lambda}-\nabla_{\lambda}W^{\sigma}_{\phantom{M}\mu\sigma\varkappa}=0​

Becose Weyl tensor is traceless therefor Cotton tensor must be identical zero!
Is this correct?

Let's do clean the equations a little bit and see what's wrong:

From the second Bianchi Identity we have

\nabla_{\nu}W^{\sigma}_{\mu \varkappa \lambda}+\nabla_{\varkappa }W^{\sigma}_{\mu \lambda \nu}+\nabla_{\lambda }W^{\sigma}_{\mu \nu\varkappa}=0

Contracting \nu with \sigma gives

\nabla_{\sigma}W^{\sigma}_{\mu \varkappa \lambda}+\nabla_{\varkappa }W^{\sigma}_{\mu \lambda \sigma}+\nabla_{\lambda }W^{\sigma}_{\mu \sigma\varkappa}=0\Rightarrow

\nabla_{\sigma}W^{\sigma}_{\mu \varkappa \lambda}+\nabla_{\varkappa }W^{\sigma}_{\mu \lambda \sigma}-\nabla_{\lambda }W^{\sigma}_{\mu \varkappa\sigma}=0\Rightarrow

\nabla_{\sigma}W^{\sigma}_{\mu \varkappa \lambda}+g^{\rho \sigma}\nabla_{\varkappa }W_{\rho \mu \lambda \sigma}-g^{\rho \sigma}\nabla_{\lambda }W_{\rho \mu \varkappa\sigma}=0. (1)

Now I think by the property of tracelessness of Weyl tensor, or,

g^{\rho \sigma}W_{\rho \mu \varkappa\sigma}=0,

you conclude from (1)

\nabla_{\sigma}W^{\sigma}_{\mu \varkappa \lambda}=0.

So you put this into the equation

\nabla_{\sigma}W^{\sigma}_{\phantom{M}\mu\varkappa \lambda}=\frac{n-3}{n-2}C_{\mu\varkappa\lambda}

and finally claim that the Cotton tensor vanishes. But unfortunately you did make a big mistake. I let you think about where this flaw arises in the above calculations! You can have a look at the fact that in a geodesic coordinates, for example, the second derivatives of metric tensors wrt coordinates do not vanish whereas their first derevatives do.

AB

 

[archipatelin]

But unfortunately you did make a big mistake.

If do you think case when n=3. Yes, I did a big mistake. But I suppose n\geq4.
Or Cotton tensor exist only for n=3 space (for n=2 it is also zero)?

 

[Altabeh]

If do you think case when n=3. Yes, I did a big mistake. But I suppose n\geq4.
Or Cotton tensor exist only for n=3 space (for n=2 it is also zero)?

For n=3, the Cotton tensor does not vanish in general. But for n\geq4, if the weyl tensor vanishes, then the Cotton tensor always vanishes. For n less than 3 the Cotton tensor is not defined.

AB

 

[archipatelin]

For n=3, the Cotton tensor does not vanish in general. But for n\geq4, if the weyl tensor vanishes, then the Cotton tensor always vanishes. For n less than 3 the Cotton tensor is not defined.

AB

This is correct, because I made another big mistake. Weyl tensor isn't governed of analog II. Bianchi identity. Therefore cotton tensor is non-zero generaly for n>2.

 

[Altabeh]

This is correct, because I made another big mistake. Weyl tensor isn't governed of analog II. Bianchi identity. Therefore cotton tensor is non-zero generaly for n>2.

Seconded.