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Why is the following property true for the Schouten tensor $P_{ij}$:

$$\nabla^k\nabla_iP_{kj}=\nabla^k\nabla_jP_{ki}$$

The Schouten tensor is defined as $$P_{ij}=\frac{1}{n-2}\left(\operatorname{Ric}_{ij}-\frac{1}{2(n-1)}Rg_{ij}\right)$$

I need this fact to prove that the Bach tensor is symmetric. But I don't know why the above fact should be true. I know that Bianchi identities imply that $\nabla^kP_{kj}=\nabla_jP^k_k$. But this fact is also not helping me prove the above identity.

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  • $\begingroup$ You need to use the formula for $\nabla_a\nabla_bP_{ij} - \nabla_b\nabla_aP_{ij}$ $\endgroup$
    – Deane
    Commented May 23, 2022 at 0:55
  • $\begingroup$ @Deane- What is the formula for that? This is just $R_{ab}P_{ij}$ right? I think that the answer to this will be able to solve my problem. $\endgroup$ Commented May 23, 2022 at 1:04
  • $\begingroup$ @Deane I don't think this will work... the LHS is $\nabla^k\nabla_i$ whereas the right hand side is $\nabla^k\nabla_j$. $\endgroup$
    – K.defaoite
    Commented May 23, 2022 at 1:06
  • $\begingroup$ @Deane Sorry, I see your idea now. My fault. $\endgroup$
    – K.defaoite
    Commented May 23, 2022 at 1:09
  • $\begingroup$ @K.defaoite- What is the idea? $\endgroup$ Commented May 23, 2022 at 1:11

1 Answer 1

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Remember the Ricci Identity, $$[\nabla_a,\nabla_b]A_{c_1\dots c_n}=A_{d~c_2\dots c_n}R^d{}_{c_1ab}+A_{c_1d~c_3\dots c_n}R^d{}_{c_2ab}+\dots A_{c_1\dots c_{n-1}d}R^d{}_{c_nab}$$ And so,

$$[\nabla^k,\nabla_i]P_{kj}=P_{lj}R^l{}_{k}{}^k{}_{i}+P_{kl}R^l{}_{j}{}^k{}_{i}$$

Now, note that $R^l{}_k{}^k{}_{i}=R^{lk}{}_{ki}=-R^{kl}{}_{ki}=-R^l{}_{i}$

So,

$$[\nabla^k,\nabla_i]P_{kj}=-P_{lj}R^l{}_i+P_{kl}R^l{}_{j}{}^k{}_{i}$$

And so, expanding the commutator,

$$\nabla^k\nabla_iP_{kj}=\nabla_i\nabla^kP_{kj}-P_{lj}R^l{}_i+P_{kl}R^l{}_{j}{}^k{}_{i} \\ =\nabla_i\nabla_jP^k{}_{k}-P_{lj}R^l{}_i+P_{kl}R^l{}_{j}{}^k{}_{i}$$

And of course, swapping the roles of $i,j$ this gives

$$\nabla^k\nabla_jP_{ki} =\nabla_j\nabla_iP^k{}_{k}-P_{li}R^l{}_j+P_{kl}R^l{}_{i}{}^k{}_{j}$$

Combining, $$\nabla^k\nabla_iP_{kj}-\nabla^k\nabla_jP_{ki}=[\nabla_i,\nabla_j]P^k{}_k+(P_{li}R^l{}_j-P_{lj}R^l{}_i)+(P_{kl}R^l{}_{j}{}^k{}_{i}-P_{kl}R^l{}_{i}{}^k{}_{j}) \tag{*}$$

First term: Since $P^k{}_k$ is a scalar, $[\nabla_i,\nabla_j]P^k{}_k=0$.

Second term: Observe

$$P_{li}R^l{}_j=\frac{1}{n-2}R^l{}_j\left(R_{li}-\frac{1}{2(n-1)}g_{li}\right) \\ = \frac{1}{n-2}R^l{}_j R_{li}-\frac{1}{2(n-1)(n-2)}R_{ij}$$

Which is very obviously symmetric w.r.t $i,j$.

Last term: We can exploit the Riemann tensor identities here.

$$P_{kl}R^l{}_j{}^k{}_i=P^{kl}R_{lj~ki} \\ =P^{kl}R_{ki~lj}$$

Now using the symmetry of $P$,

$$P^{kl}R_{ki~lj}=P^{lk}R_{ki~lj}$$

$l,k$ are dummy indices so we can swap them:

$$P^{lk}R_{ki~lj}=P^{kl}R_{li~kj} \\ = P_{kl}R^l{}_i{}^k{}_j$$

So the third term of $(*)$ vanishes as well. Thus finally,

$$\boxed{\nabla^k\nabla_iP_{kj}=\nabla^k\nabla_jP_{ki}}$$

As desired.

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  • $\begingroup$ Please note that when I expanded the commutator I used your identity that you say follow from Bianchi as given. $\endgroup$
    – K.defaoite
    Commented May 23, 2022 at 1:52
  • $\begingroup$ Thanks so much for the detailed and complete answer! $\endgroup$ Commented May 23, 2022 at 2:21
  • $\begingroup$ Also worth noting this can be summarized neatly as $P_{k[j;i]}{}^{;k}=0$. $\endgroup$
    – K.defaoite
    Commented May 23, 2022 at 14:02
  • $\begingroup$ Can I also ask what is the correct way to interpret $R_i^pv_{pk}$? Is it just $\nabla_i\nabla_pv^{p}_k-\nabla_p\nabla_i v^p_k=R_{ip}{}^{pl}v_{lk}+R_{ipk}{}^l v^p{}_l$? $\endgroup$ Commented May 23, 2022 at 14:09
  • $\begingroup$ I just contracted the first and third indices of Riemann to get the Ricci tensor. $\endgroup$
    – K.defaoite
    Commented May 23, 2022 at 14:13

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