A metric for poincare half model hyperbolic 3D space time can be written as follows;
\(ds^2=-dt^2+\frac{dx^2}{y^2}+\frac{dy^2}{y^2}\)
Now comparing this  with general spacetime interval,
\(ds^2=g_{_{\alpha\beta}}dx^{\alpha}dy^{\beta}\)
where  \(g_{_{\alpha\beta}}\) is a metric tensor of index 3,[ where \(\alpha,\beta=1,2,3\)]
\(g_{11}=-1\)\(g_{22}=\frac{1}{y^2}\)\(g_{33}=\frac{1}{y^2}\) and rest of all mertrics is zero's
Now we can find all components of christoffel's symbols because christoffel's symbols are partial derivative of metrics given we can use a formulea,
we got,
\(\gamma_{23}^2=\frac{1}{y}\)\(\gamma_{32}^2=-\frac{1}{y}\)\(\gamma_{22}^3=-\frac{1}{y}\)\(\gamma_{33}^3=-\frac{1}{y}\)  Rest of all components in christoffels symbols are automatically zero because all other metric are also zero's
Now we have to find Ricci tensor's components by tedious formulea but i am skipping this and writing ricci components,R
\(R_{\alpha\beta}\)
R11=R12=R13=R21=R23=R31=R32=0, But we have some non-zer components too which is
R22=\(-\frac{1}{y^2}\), R33=\(-\frac{1}{y^2}\)
Now i am finding Reimann tensor component after huge calculation i finally reached there is only one non zero Reimann component which is
R2323=\(-\frac{1}{y^4}\)
Curvature scalar R can be found by following relations :
R= g11R11+g22R22+g33R33
=\(-1\left(0\right)\)+\(y^2\left(-\frac{1}{y^2}\right)\)+\(y^2\left(-\frac{1}{y^2}\right)\)      [ as \(g^{\alpha\beta}=\frac{1}{g_{\alpha\beta}}\)]
= -2
Finally we have reached last step by calculating Einstein's Tensor which is equal to
\(G_{\alpha\beta}=R_{\alpha\beta}-\frac{1}{2}g_{\alpha\beta}R\)    [ where \(\alpha,\beta=1,2,3\)]
\(G_{11}=R_{11}-\frac{1}{2}g_{11}R\)
  = \(0-\frac{1}{2}\left(-1\right)\left(-2\right)\)
=-1
\(G_{22}=R_{22}-\frac{1}{2}g_{22}R\)
       =\(-\frac{1}{y^2}-\frac{1}{2}\left(\frac{1}{y^2}\right)\left(-2\right)\)
     = 0
\(G_{33}=R_{33}-\frac{1}{2}g_{33}R\)
      = \(-\frac{1}{y^2}-\frac{1}{2}\left(\frac{1}{y^2}\right)\left(-2\right)\)
       = 0
rest of all components in Einstein's tensor is zero in matrix form we can wrirte this in
\(\left[-1\ \ \ \ 0\ \ \ \ 0\right]\)
We can calculate Einstein's equation from these data
\(G=kT_{\alpha\beta}\)
\(G\)  Can be calculated by following relation: \(\Sigma_{\alpha,\beta=1}^3G_{\alpha\beta}g^{\alpha\beta}\)
\(G=G_{11}g^{11}+G_{22}g^{22}+G_{33}g^{33}\)
\(=-1\left(-1\right)+0+0\)\(=1\)
final Einstein's field equation is
\(kT_{\alpha\beta}=1\)\cite{Sharan_2009,Ahsan_1995,Padmanabhan_2014,2011,2011a}
Conclusion:
I showed that hyperbolic geometry has constant negtaive curvature  Thus, this metric is a solution for the vacuum Einstein equations. This seems strange because this is a space of constant negative curvature, which would seem to require some sort of matter/energy.