Prove that deSitter-Schwarzschild metric with constant Ricci scalar is Einstein

Question

Reference: this paper page 4

Prove that the following metric is Einstein

Given deSitter-Schwarzschild metric with mass $m > 0$ and scalar Ricci $R=2$ by

\begin{align} \bigg( 1-\frac{r^2}{3}-\frac{2m}{r} \bigg)^{-1} dr^2 + r^2 g_{\Bbb S^2} \end{align}

where $g_{\Bbb S^2} = d\theta^2 + \text{sin}^2 \theta\ d\phi^2$.

Proof.

\begin{align} R_{11} =\ & \Lambda g_{11} \nonumber\\ % % \Lambda =\ & -\frac{2m}{r^3} + \frac{2}{3} \tag{1} \end{align}

\begin{align} R_{22} =\ & \Lambda g_{22} \nonumber\\ % % \Lambda =\ & \frac{m}{r^3} + \frac{2}{3} \tag{2} \end{align}

\begin{align} R_{33} =\ & \Lambda g_{33} \nonumber\\ % % \Lambda =\ & \frac{m}{r^3} + \frac{2}{3} \tag{3} \end{align}

Since $\Lambda_{(1)}$ is not the same with $\Lambda_{(2)}$ and $\Lambda_{(3)}$, then the metric is NOT Einstein. On the other hand, the Scalar Ricci is constant, thus the metric is supposed to be Einstein. And now I am confused.

Answer 1

1, First, You should check if Your metric is vacuum solution of Einstein equation with non-zero cosmological constant $\Lambda$. Vanishing $\Lambda$ will give a trivial equality (if the metric is vacuum solution).

Generally, Einstein equation with zero matter tensor can be rewritten as (Einstein metric) $$ R_{\mu \nu}=\frac{2\Lambda}{d-2}g_{\mu\nu} $$

2, Your metric (as I checked) is neither the solution of $d=3$ nor $d=4$ (with a constant time-time component) Einstein equation. So I think Your problem is here.

3, Your statement

the Scalar Ricci is constant, thus the metric is supposed to be Einstein

is incorrect. "Scalar Ricci is constant" is only necessary condition of Einstein metric, not sufficient.