Ricci curvature under Ricci flow

In fact ,I got stuck in some compute. Let $B_{ijkl}=g^{pr}g^{qs}R_{piqj}R_{rksl}$, in the below picture,I can't compute out the 1 and 2 equation above red line.

In the first, I know how to get $\Delta R_{ik}$, but the other part ,it's seemly too complex that I can't compute out.

In the second , I think I need some exchange equation of $B_{ijkl}$.

Besides, I always get stuck in compute , when I read paper about Ricci flow. What should I do? And there are so many index that let me dazzle, whether there are some easy way to deal it ?

Thanks for useful hint or answer.Sorry for my poor English.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~  • Well I would say writing this in terms of indices is probably the easiest way to deal with it. On one hand you can use normal coordinate to simplify calculations, on the other hand you there is no such representation (in terms of indices), then it is even harder to compute. You need to be familiar with this. As far as I know, almost all papers/books on Ricci flow starts the calculations right at page 1. – user99914 Oct 21 '15 at 2:31
• @John I also think I need to be familiar with this , But I think it's good to make a soft that help checking the index and giving some possible transformation. I feel assistant software is good for situation of many index. But now ,I want to get the way of computing 1 and 2, do you have any idea? – lanse2pty Oct 21 '15 at 2:52

(1) If we use normal coordinate, then the term in second line is $$-g^{jl}g^{pq} (R_{pjkl} R_{qi} + R_{ipkl} R_{qj} + R_{ijpl} R_{qk} + R_{ijkp} R_{ql} )$$

$$= -g^{pq} ( R_{pk} R_{qi} + R_{ipkl} R_{ql} + R_{ip} R_{qk} + R_{ilkp} R_{ql} )$$

$$=-2g^{pq} R_{pk} R_{qi}\underbrace{ - g^{pq} ( R_{ipkl} R_{ql} + R_{ilkp} R_{ql} )}_{=A}$$

The term in the first line is $$2g^{jl} ( B_{ijkl} - B_{ijlk} - B_{il jk} + B_{ikjl} ) = 2g^{jl} ( B_{ijkl} - 2B_{ijlk} ) + \underbrace{2B_{ikll} }_{=B}$$ so that $$A+ B = - g^{pq} ( R_{ipkl} R_{ql} + R_{ilkp} R_{ql} ) +2 R_{piqk } R_{pq} =0$$

by Bianchi identity

(2) Recall $$B_{ijkl} = R_{piqj} R^{p\ q}_{\ k\ l}$$ so that $$g^{jl} B_{ijkl}= R_{piqj} R^{p\ qj}_{\ k} = R_{piqj} R^{qjp}_k$$ $$g^{jl} B_{ijlk}= R_{piqj} R^{pj q}_{\ \ \ k}$$

The term in third line can be expanded into $$R_{piqj} R^{p\ qj}_{\ k} - R_{pjqi} R^{p\ qj}_{\ k} - R_{piqj} R^{pjq}_{\ \ \ k} + R_{pjqi } R^{pjq}_{\ \ \ k}$$

$$= R_{piqj} R^{qjp}_{\ \ \ k} - R_{pjqi} R^{qjp}_{\ \ \ k} - g^{jl} B_{ijlk} + R_{qjpi } R^{qjp}_{\ \ \ k}$$

$$= g^{jl} B_{ijkl} - g^{jl} B_{ijlk}- g^{jl} B_{ijlk} + g^{jl} B_{ijkl}$$ Hence we complete the proof