TL;DR: I have something Ricci-flat which is not turning out to be scalar-flat and this is absurd.
Consider $P_I\times_r \Bbb S^2$, where $P_I =\{(t,r) \in \Bbb R^2 \mid r>2m\}$, with $m>0$, and the Lorentzian metric $g^{P_I} =-h(r)\,{\rm d}t^2+h(r)^{-1}\,{\rm d}r^2$, where $h(r) = 1-2mr^{-1}$. The metric on $P_I\times_r \Bbb S^2$ is $$g=-h(r)\,{\rm d}t^2+h(r)^{-1}\,{\rm d}r^2 + r^2\,{\rm d}\Omega^2,$$where ${\rm d}\Omega^2$ is the standard round metric on $\Bbb S^2$. I have computed that the Gaussian curvature of the half-plane $P_I$ is given by $K^{P_I}(t,r) = -h''(r)/2 = 2mr^{-3}$. Hence $${\rm Ric}^{P_I} = -\frac{h''(r)}{2} g^{P_I} \quad \mbox{and}\quad {\rm s}^{P_I} = -h''(r) = 4mr^{-3},$$where ${\rm s}$ stands for scalar curvature. It's not hard to check that $\nabla r = h(r)\partial_r$ and ${\rm Hess}\,r = (h'(r)/2) g^{P_I}$. Great. Now in O'Neill's Semi-Riemannian Geometry book we have that for an arbitrary $B\times_\phi F$, the formulas hold:
- ${\rm Ric}(X,Y) = {\rm Ric}^B(X,Y) - \dfrac{(\dim F)}{\phi}\,{\rm Hess}\,\phi$ for horizontal $X,Y$;
- ${\rm Ric}(X,Y) = 0$ for horizontal $X$ and vertical $Y$.
- ${\rm Ric}(V,W) = {\rm Ric}^F(V,W) - g(V,W) \phi^\#$ for vertical $V$, $W$, where $$\phi^\# = \frac{\triangle \phi}{\phi} + (\dim F-1)\frac{g(\nabla\phi,\nabla\phi)}{\phi^2}.$$
This is corollary 43 in page 211. Now, the Schwarzschild black hole is supposed to be Ricci-flat, right? Sure, take $\phi = r$, so $\triangle r = h'(r)$ and thus $r^\# = r^{-2}$. For horizontal vectors we have $$-\frac{h''(r)}{2}g^{P_I}_{ij} - \frac{2}{r} \frac{h'(r)}{2}g^{P_I}_{ij} = 0$$for all $i,j \in \{t,r\}$, by plugging in $h(r) = 1-2mr^{-1}$. Similarly, $r^\# = r^{-2}$ cancels the warping factor in ${\rm d}\Omega^2$ and since the Gaussian curvature of $\Bbb S^2$ is $1$, ${\rm Ric}$ vanishes on pairs of vertical vectors as well. Awesome, right?
Now look at exercise 13a) in page 214:
$$ {\rm s} = {\rm s}^B + \frac{{\rm s}^F}{\phi^2} - 2\dim F\frac{\triangle \phi}{\phi} - \dim F(\dim F-1)\frac{g(\nabla \phi, \nabla \phi)}{\phi^2}.$$
We have $\dim F = 2$, so: $${\rm s} = -h''(r) + r^{-2} - 4r^{-1} h'(r) - 2r^{-2}h(r).$$But for $h(r) = 1-2mr^{-1}$, we have $h'(r) = 2mr^{-2}$ and $h''(r) = -4mr^{-3}$, leading to: $$4mr^{-3} + r^{-2}-8mr^{-3}-2r^{-2} + 4mr^{-3} = {\color{red}{-r^{-2} \neq 0}}.$$
What gives?
I couldn't sleep thinking of this and I got it! The Gaussian curvature of $\Bbb S^2$ is $1$, so the scalar curvature is $2$, not $1$!
$$4mr^{-3} + {\color{blue}{2r^{-2}}}-8mr^{-3}-2r^{-2} + 4mr^{-3} = {\color{red}{0}}.$$
(I figure registering this here might be better than deleting the post, so...)