# Why isn't the Cotton tensor identical zero?

1. Jan 29, 2010

### 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?

2. Jan 29, 2010

### Altabeh

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

3. Jan 29, 2010

### archipatelin

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)?

4. Jan 29, 2010

### Altabeh

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

5. May 19, 2010

### archipatelin

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 $$n>2$$.

Last edited: May 19, 2010
6. May 20, 2010

Seconded.