FRW metric spacelike slice In Steven Winberg's Cosmology book (p. 6) he says that the spatial part (co-moving part) of the FRW metric
can be written as 
$$
 \tilde g_{ij} = \delta_{ij}+K \frac{x^i x^j}{1- K \mathbf x ^2} 
$$
These qoordinates are "quasi-cartesian". I can get back 
$ \tilde g _{11} = \frac{1}{1-K \mathbf x^2}$ by using $x^1=r$. But, for the other coordinates I can't find the right components using this relations. If I use $x^2=\theta$ then I can't get anything. Any help will be highly appreciated.
 A: Your equation gives the components of the spatial metric in cartesian coordinates: i.e. $x^1 \equiv x$, $x^2 \equiv y$, and $x^3 \equiv z$. You cannot use $x^1=r$.
If you want to get the metric in a different coordinate system (like spherical coordinates) you just need to be able to write $x$, $y$, and $z$ in terms of the new coordinates (i.e. $r$, $\theta$, and $\phi$) and use the normal way tensor components transform:
$$
g'_{ij} = g_{kl} \frac{\partial x^k}{\partial x'^i}\frac{\partial x^l}{\partial x'^j},
$$
where the primed coordinates are your new coordinates and there as an implicit sum over indices $k$ and $l$.
In your case it's probably easier (but totally equivalent) to write out the proper length $ds^2$ in terms of $dx$, $dy$, and $dz$ and then write the $dx$, $dy$, and $dz$ in terms of $dr$, $d\theta$, and $d\phi$. Then you'll end up with $ds^2$ in terms of $dr$, $d\theta$, and $d\phi$ and can read off $\tilde{g}_{rr}$, $\tilde{g}_{\theta\theta}$, $\tilde{g}_{\phi\phi}$, $\tilde{g}_{r\theta}$, etc.
