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In 2D the Einstein tensor is always zero, and we can easily get solution with non-zero Ricci tensor but zero Einstein tensor. But is it possible in 4D? Can we get a space-time with zero Einstein tensor, zero cosmological constant but non-zero Ricci tensor and Ricci scalar? If not, why?

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Can we get a space-time with zero Einstein tensor, zero cosmological constant but non-zero Ricci tensor and scalar?

The answer is negative: If $$R_{ab}-\frac{1}{2}g_{ab}R=0 \tag{1}$$ then $$g^{ab}R_{ab}-\frac{1}{2}g^{ab}g_{ab}R=0\:,$$ namely (in 4D), $$R- 2R =0\:,$$ that is $$R=0\:.$$ Inserting in (1), you have $$R_{ab}=0\:.$$ In summary $\Lambda =0$ and $G_{ab}=0$ entail $R_{ab}=0$ and $R=0$ contrarily to the hypothesis.

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    $\begingroup$ This is correct +1 (and also applies to all $D >2$) $\endgroup$
    – Eletie
    Commented Nov 4, 2023 at 16:45
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If I assume that Einstein tensor is zero for a generic (1,$d$) dimensional spacetime then $$ G_{ab} = R_{ab} - \frac{1}{2}R g_{ab} = 0$$ This means that $$ R_{ab} = \frac{1}{2}R g_{ab}$$ Contracting with $ g^{ab}$

$$ g^{ab}R_{ab} = \frac{1}{2}R g^{ab} g_{ab}$$

gives $R =\frac{1}{2}R g^{ab} g_{ab} $, This implies $$g^{ab}g_{ab} = 2.$$ As I understand, this only holds if the dimension of spacetime is (1,1). So $G_{ab}$ cannot be zero for a generic dimensional spacetime.

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  • $\begingroup$ This equation is incorrect because of your repeated indices. You should have $R g_{ab} = g^{cd} R_{cd} g_{ab}$. If what you wrote were true, the Einstein tensor would simply be $G_{ab} = (1-\frac{n}{2}) R_{ab}$, or $-R_{ab}$ in $D=4$, which is clearly not the case. $\endgroup$
    – Eletie
    Commented Nov 4, 2023 at 16:34
  • $\begingroup$ Thanks for the catch. I have updated my response now $\endgroup$
    – S.G
    Commented Nov 4, 2023 at 18:05

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