Killing tensor of Friedman-Robertson-Walker metric (not assuming flatness) There exists an earlier question about how to verify that, if $U^\mu = (1,0,0,0)^\mu$ is the four-velocity of comoving observers, then
$$K_{\mu\nu} = a^2(g_{\mu\nu} + U_\mu U_\nu)\tag{1}$$
is a Killing tensor, $\nabla_{(\sigma} K_{\mu\nu)} = 0$, for the FRW metric
$$ds^2 = -dt^2 + a(t)^2 \left[\frac{dr^2}{1 - \kappa r^2} + r^2 d\Omega^2 \right], \quad d\Omega^2 = d\theta^2 + \sin^2\theta \, d\phi^2.\tag{2}$$
However, they only considered the flat case, $\kappa = 0$, but Sean Carroll claims that this is a Killing tensor without assuming flatness (see section 8.5, page 344, of his textbook, or chapter 8, page 228, in his lecure notes). I have tried for a while now to verify this claim, but I have not managed to do so and would therefore appreciate your help.
Partial results
By using $\nabla_\sigma g_{\mu\nu} = 0$, $\nabla_\sigma a^2 = \partial_\sigma a^2 = \delta_\sigma^0 2 a \dot a$, and the definition of $U^\mu$, I have found that
$$\nabla_\sigma K_{\mu\nu} = (g_{\mu\nu} + \delta_\mu^0 \delta_\nu^0)\delta_\sigma^0 2 a \dot a,\tag{3}$$
where $\dot a \equiv \partial_0 a$. This is diagonal in $\mu$ and $\nu$, and zero for $\sigma \ne 0$. I tried to use this to show that $\nabla_{(\sigma} K_{\mu\nu)} = 0$, but to no avail. Of course, I might have made a computation error somewhere along the way.
 A: I worked this problem in detail, starting with the ansatz:
$u_{a} = (A, 0, 0, 0)$
$K_{ab} = \alpha g_{ab} + \beta u_{a}u_{b}$
With some laborious algebra, you find that the $ttt$ component of $\nabla_{c}K_{ab}$ vanishes if $\alpha = A^{2}\beta$, and for simplicity, assuming that there is not $r$ dependence in any of our unsolved functions.  Then, letting $q_{ab} = {\rm diag}(0, 0, 1, \sin^{2}\theta)$
$$\nabla_{c}K_{ab} = \delta_{a}^{r}\delta_{b}^{r}\delta_{c}^{t}\frac{a^{2}{\dot \alpha}}{1-kr^{2}} + a^{2}r^{2}{\dot \alpha}\delta_{c}^{t}q_{ab} - 2\frac{a{\dot a}\alpha}{1-kr^{2}}\delta_{(a}^{t}\delta_{b)}^{r}\delta_{c}^{r} - 2a {\dot a}r^{2}\alpha\delta_{(a}^{t}q_{b)c}$$
The RHS is manifestly symmetric in a and b, so, if permuting a and c is antisymmetric, then symmetrizing over abc will give zero.  Solving for this, you get the condition $2\frac{\dot a}{a} = \frac{\dot \alpha}{\alpha}$, which, choosing $A = -1$ (future-pointing lowered index), and $\alpha = a^{2}$ we get:
$u_{a} = (-1, 0, 0, 0)$
$K_{ab} = a^{2}(g_{ab} + u_{a}u_{b})$
