Why is the following property true for the Schouten tensor? 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.
 A: 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.
