# Proving the Vacuum Field Equations are Trivial for n=3

Airsteve0
I am trying to prove a concept that was presented in my course but I am having a bit of an issue with understanding the final result. The concept was that in 3 dimension the vanishing of the Riemann tensor leaves only the trivial solution for the vacuum equations. I started by subbing Eq.(1) into Eq.(2) and solving for R (I took the trace in case you were wondering - my result was 6$\Lambda$) and then subbed then this answer back in to solve for $R_{\alpha\beta}$ (answer was 2$\Lambda$$g_{\alpha\beta}$). I then subbed both of these into (3) and solved. However, as you can see my answer is not zero. If anyone could point out a mistake or if I just haven't properly conceptualized the answer I would appreciate the help, thanks.

(1) $G_{\alpha\beta}$=$R_{\alpha\beta}$-$\frac{1}{2}$$g_{\alpha\beta}$R

(2) $G_{\alpha\beta}$+$\Lambda$$g_{\alpha\beta}$=0

(3) $R_{\alpha\beta\gamma\delta}$ = $g_{\alpha\gamma}$$R_{\beta\delta}$+$g_{\beta\delta}$$R_{\alpha\gamma}$-$g_{\alpha\delta}$$R_{\beta\gamma}$-$g_{\beta\gamma}$$R_{\alpha\delta}$-$\frac{R}{2}$($g_{\alpha\gamma}$$g_{\beta\delta}$-$g_{\alpha\delta}$$g_{\beta\gamma}$)

= $\Lambda$($g_{\alpha\gamma}$$g_{\beta\delta}$-$g_{\alpha\delta}$$g_{\beta\gamma}$)

## Answers and Replies

Gold Member
in 3 dimension the vanishing of the Riemann tensor leaves only the trivial solution for the vacuum equations.

I assume you mean "vanishing of the Einstein tensor", not the Riemann tensor.

= $\Lambda$($g_{\alpha\gamma}$$g_{\beta\delta}$-$g_{\alpha\delta}$$g_{\beta\gamma}$)

In the presence of a non-zero cosmological constant, the "trivial solution" is not flat space; it is de Sitter (positive lambda) or anti-de Sitter (negative lambda) space.

Also, I'm not sure the statement you are trying to prove is true in the case of nonzero lambda. The BTZ black hole is not trivial, for example.

Airsteve0
yes, sorry I did mean the Einstein tensor