How to calculate the Ricci curvature of the following metric? Does anyone know that how to calculate the Ricci curvature of the following metric of $\Bbb R^3$?
$$g=a(t)^2dt\otimes dt+b(t)^2d\sigma\otimes d\sigma,$$
where $d\sigma\otimes d\sigma$ is the standard round metric on the unit $2$-sphere and $a,b$ are function.
 A: Using the usual (American) spherical coordinates $(\phi,\theta)$, we have the orthonormal coframe
\begin{align*}
\omega_1 &= a(t)\,dt \\
\omega_2 &= b(t)\,d\phi \\
\omega_3 &= b(t)\sin\phi\,d\theta.
\end{align*}
We solve for the connection $1$-forms using $d\omega_i = \sum\limits_j \omega_{ij}\wedge\omega_j$ (and $\omega_{ji} = -\omega_{ij}$) and find that
\begin{align*}
\omega_{12} &= \frac{b'(t)}{a(t)}d\phi \\
\omega_{13} &= \frac{b'(t)\sin\phi}{a(t)}d\theta \\
\omega_{23} &= \cos\phi\,d\theta.
\end{align*} 
We then compute the curvature $2$-forms using $\Omega_{ij} = d\omega_{ij} - \sum\limits_k \omega_{ik}\wedge\omega_{kj}$:
\begin{align*}
\Omega_{12} &= \big(\frac{b'}a\big)'(t)\,dt\wedge d\phi = \big(\frac{b'}a\big)'\frac1{ab}\omega_1\wedge\omega_2 \\
\Omega_{13} &= \big(\frac{b'}a\big)'(t)\sin\phi\,dt\wedge d\theta = \big(\frac{b'}a\big)' \frac1{ab}\omega_1\wedge\omega_3 \\
\Omega_{23} &= \left(\big(\frac{b'(t)}{a(t)}\big)^2 - 1\right)\sin\phi\,d\phi\wedge d\theta = \left(\big(\frac{b'}{a}\big)^2 - 1\right)\frac1{b^2}\omega_2\wedge\omega_3.
\end{align*}
Setting $\Omega_{ij} = \sum\limits_{k<\ell} R_{ijk\ell}\omega_k\wedge\omega_\ell$, we find
$$R_{1212} = R_{1313} = \big(\frac{b'}a\big)'\frac1{ab}, \quad R_{2323} = \left(\big(\frac{b'}a\big)^2-1\right)\frac 1{b^2},$$
with the usual symmetries, and other entries $0$.
Since $\text{Ric}(e_i,e_j) = -\sum\limits_k R_{ikjk}$ [check your conventions!], we have
\begin{align*}
\text{Ric}(e_1,e_1) &= -2\big(\frac{b'}a\big)'\frac1{ab} \\
\text{Ric}(e_1,e_2) &= 0 \\
\text{Ric}(e_1,e_3) &= 0 \\
\text{Ric}(e_2,e_2) &= \big(\frac{b'}a\big)'\frac1{ab}-\left(\big(\frac{b'}a\big)^2-1\right)\frac 1{b^2} \\
\text{Ric}(e_2,e_3) &= 0 \\
\text{Ric}(e_3,e_3) &= \big(\frac{b'}a\big)'\frac1{ab}+\left(\big(\frac{b'}a\big)^2-1\right)\frac 1{b^2}.
\end{align*}
(Here, of course, the $\{e_i\}$ are the orthonormal basis dual to the $\{\omega_i\}$.) I don't vouch for perfection with the algebra and typing.
