# The divergence of the Weyl tensor

First, the Weyl tensor is given by $$W_{ijkl}=R_{ijkl}-\frac{1}{n-2}(g_{ik}A_{jl}-g_{il}A_{jk}-g_{jk}A_{il}+g_{jl}A_{ik})$$ where, $A_{ij}$ is the Schouten tensor, given by $$A_{ij}=R_{ij}-\frac{R}{2(n-1)}g_{ij}$$ And the Cotton tensor is given by $$C_{ijk}=\nabla_{i}A_{jk}-\nabla_{j}A_{ik}$$ I've to show that the divergence of the Weyl tensor and the Cotton tensor satisfy: $$C_{ijk}=-\frac{n-2}{n-3}\nabla_{l}W_{ijkl}\qquad(*)$$

I know that the divergence of the Weyl tensor is $g^{ls}\nabla_sW_{ijkl}$ and, by the once contracted second Bianchi identity is $g^{ls}\nabla_sR_{ijkl}=-\nabla_iR_{jk}+\nabla_jR_{ik}$.

Someone can help me to show the identity $(*)$?

• Can someone help you do what? Commented Aug 9, 2013 at 18:16
• The identity $(*)$. Now, can you able to help me? Commented Aug 9, 2013 at 18:30

First note that

$$\nabla^l A_{jl}=\nabla^l R_{jl}-\nabla^l \frac{R}{2(n-1)}g_{jl}=\nabla^l R_{jl}- \frac{1}{2(n-1)}\nabla_j R=\frac{1}{2}\nabla_j R- \frac{1}{2(n-1)}\nabla_j R=\frac{n-2}{2(n-1)}\nabla_j R \qquad(1)$$

the last equality is obtained by twice contracted second Bianchi identity ( that is $2\nabla^jR_{jk}=\nabla_k R$)

Now we calculate the divergence of the Weyl tensor:

$$g^{ls}\nabla _sW_{ijkl}=g^{ls}\nabla_s R_{ijkl}-\frac{1}{n-2}(g_{ik}g^{ls}\nabla_s A_{jl}-g_{il}g^{ls}\nabla_s A_{jk}-g_{jk}g^{ls}\nabla_s A_{il}+g_{jl}g^{ls}\nabla_s A_{ik})$$

$$=\nabla_iR_{jk}-\nabla_jR_{ik}-\frac{1}{n-2}(g_{ik}\nabla^l A_{jl}-g_{il}\nabla^l A_{jk}-g_{jk}\nabla^l A_{il}+g_{jl}\nabla^l A_{ik})$$

$$=\nabla_iR_{jk}-\nabla_jR_{ik}-\frac{1}{n-2}(g_{ik}\nabla^l A_{jl}-\nabla_i A_{jk}-g_{jk}\nabla^l A_{il}+\nabla_j A_{ik})$$

We know $C_{ijk}=\nabla_{i}A_{jk}-\nabla_{j}A_{ik}$, then

$$=\nabla_iR_{jk}-\nabla_jR_{ik}-\frac{1}{n-2}(g_{ik}\nabla^l A_{jl}-g_{jk}\nabla^l A_{il}-C_{ijk})$$

Now by considering (1):

$$=\nabla_iR_{jk}-\nabla_jR_{ik}-\frac{1}{n-2}(g_{ik}\frac{n-2}{2(n-1)}\nabla_j R-g_{jk}\frac{n-2}{2(n-1)}\nabla_i R-C_{ijk})$$

$$=\nabla_iR_{jk}-\nabla_jR_{ik}-\frac{1}{2(n-1)}g_{ik}\nabla_j R+\frac{1}{2(n-1)}g_{jk}\nabla_i R+\frac{1}{n-2}C_{ijk}$$

$$=\nabla_i(R_{jk}+\frac{R}{2(n-1)}g_{jk})-\nabla_j(R_{ik}+\frac{R}{2(n-1)}g_{ik})+\frac{1}{n-2}C_{ijk}$$

By using $\frac{R}{2(n-1)}g_{ij}=R_{ij}-A_{ij}$:

$$=\nabla_i(2R_{jk}-A_{jk})-\nabla_j(2R_{ik}-A_{ik})+\frac{1}{n-2}C_{ijk}$$

$$=2\nabla_i R_{jk}-2\nabla_j R_{ik}-\nabla_iA_{ik}+\nabla_jA_{ik}+\frac{1}{n-2}C_{ijk}=-C_{ijk}+\frac{1}{n-2}C_{ijk}=-\frac{n-3}{n-2}C_{ijk}$$

Note that $2\nabla_i R_{jk}-2\nabla_j R_{ik}=0$. So we obtain

$$C_{ijk}=-\frac{n-2}{n-3}g^{ls}\nabla _sW_{ijkl}=-\frac{n-2}{n-3} \nabla^l W_{ijkl}$$

• You answer is totally correct! Thanks very much! Commented Aug 9, 2013 at 22:26
• @DiegoMath Be careful I changed my answer a little. gook luck! Commented Aug 9, 2013 at 22:32
• The last line is definitely wrong: $2\nabla _i R_{jk}-2\nabla _j R_{ik}$ is not zero! That is nonsense. You made a mistake with the sign at the beginning with the once contracted Bianchi identity. If you change that sign, then you get at the end: $\nabla _i R_{jk} - \nabla _i R_{jk}$ and $\nabla _j R_{ik} - \nabla _j R_{ik}$, which is certainly zero.
– user303766
Commented Jan 9, 2016 at 8:34