The Robertson-Walker line element is \begin{align} ds^2 &= -dt^2 + a(t)^2 \Biggl( \frac{dr^2}{1-kr^2} \Biggr.\nonumber\\ &\qquad\Biggl.{} + r^2 \left( d\theta^2 + \sin^2\theta\,d\phi^2 \right) \Biggr) \end{align} It now follows immediately that \begin{align} d\rr &= dt\,\That + \frac{a(t)\,dr}{\sqrt{1-\kappa r^2}}\,\rhat + a(t)\,r\,d\theta\,\that + a(t)\,r\,\sin\theta\,d\phi\,\phat \end{align} and the basis 1-forms are \begin{align} \sigma^t &= dt \\ \sigma^r &= \frac{a(t)\,dr}{\sqrt{1-\kappa r^2}} \\ \sigma^\theta &= a(t)\,r\,d\theta \\ \sigma^\phi &= a(t)\,r\,\sin\theta\,d\phi \end{align} The structure equations therefore become \begin{align} d\sigma^t &= 0 = -\omega^t{}_m\wedge\sigma^m \\ d\sigma^r &= \frac{\dot{a}\,dt\wedge dr}{\sqrt{1-\kappa r^2}} = -\omega^r{}_m\wedge\sigma^m \\ d\sigma^\theta &= \dot{a}r\,dt\wedge d\theta + a\,dr\wedge d\theta = -\omega^\theta{}_m\wedge\sigma^m \\ d\sigma^\phi &= \dot{a}r\,\sin\theta\,dt\wedge d\phi + a\,\sin\theta\,dr\wedge d\phi + ar\,\cos\theta\,d\theta\wedge d\phi \nonumber\\ &= -\omega^\phi{}_m\wedge\sigma^m \end{align} These equations suggest that \begin{align} \omega^r{}_t &= \frac{\dot{a}\,dr}{\sqrt{1-\kappa r^2}} = \frac{\dot{a}}{a}\,\sigma^r \\ \omega^\theta{}_t &= r\dot{a}\,d\theta = \frac{\dot{a}}{a}\,\sigma^\theta \\ \omega^\theta{}_r &= \sqrt{1-\kappa r^2}\,d\theta = \frac{\sqrt{1-\kappa r^2}}{ar}\,\sigma^\theta \\ \omega^\phi{}_t &= r\dot{a}\,\sin\theta\,d\phi = \frac{\dot{a}}{a}\,\sigma^\phi \\ \omega^\phi{}_r &= \sqrt{1-\kappa r^2}\,\sin\theta\,d\phi = \frac{\sqrt{1-\kappa r^2}}{ar}\,\sigma^\phi \\ \omega^\phi{}_\theta &= \cos\theta\,d\phi = \frac{\cot\theta}{ar}\,\sigma^\phi \end{align} and it is easy to check that these educated guesses actually do satisfy the structure equations, remembering that \begin{align} \omega^r{}_t &= \omega^t{}_r, & \omega^\theta{}_t &= \omega^t{}_\theta, & \omega^\phi{}_t &= \omega^t{}_\phi, \nonumber\\ \omega^r{}_\theta &= -\omega^\theta{}_r, & \omega^r{}_\phi &= -\omega^\phi{}_r, & \omega^\theta{}_\phi &= -\omega^\phi{}_\theta \end{align} Since we are guaranteed a unique solution, we have found our connection 1-forms.

The curvature 2-forms are now given by \begin{align} \Omega^t{}_r &= d\omega^t{}_r + \omega^t{}_m\wedge\omega^m{}_r = \frac{\ddot{a}\,dt\wedge dr}{\sqrt{1-\kappa r^2}} = \frac{\ddot{a}}{a}\sigma^t\wedge \sigma^r \\ \Omega^t{}_\theta &= d\omega^t{}_\theta + \omega^t{}_m\wedge\omega^m{}_\theta = {\ddot{a}r\,dt\wedge d\theta} = \frac{\ddot{a}}{a}\sigma^t\wedge \sigma^\theta \\ \Omega^t{}_\phi &= d\omega^t{}_\phi + \omega^t{}_m\wedge\omega^m{}_\phi = {\ddot{a}r\,\sin\theta\,dt\wedge d\phi} = \frac{\ddot{a}}{a}\sigma^t\wedge \sigma^\phi \\ \Omega^r{}_\theta &= d\omega^r{}_\theta + \omega^r{}_m\wedge\omega^m{}_\theta = \frac{(\dot{a}^2+\kappa)r\,dr\wedge d\theta}{\sqrt{1-\kappa r^2}} = \frac{\dot{a}^2+\kappa}{a^2}\sigma^r\wedge\sigma^\theta \\ \Omega^r{}_\phi &= d\omega^r{}_\phi + \omega^r{}_m\wedge\omega^m{}_\phi = \frac{(\dot{a}^2+\kappa)r\,\sin\theta\,dr\wedge d\phi}{\sqrt{1-\kappa r^2}} \nonumber\\ &= \frac{\dot{a}^2+\kappa}{a^2}\sigma^r\wedge\sigma^\phi \\ \Omega^\theta{}_\phi &= d\omega^\phi{}_\theta + \omega^\phi{}_m\wedge\omega^m{}_\theta = (\dot{a}^2+\kappa)r^2\,\sin\theta\,d\theta\wedge d\phi \nonumber\\ &= \frac{\dot{a}^2+\kappa}{a^2}\sigma^\theta\wedge\sigma^\phi \end{align}

Again we have \begin{align} \Omega^r{}_t &= \Omega^t{}_r, & \Omega^\theta{}_t &= \Omega^t{}_\theta, & \Omega^\phi{}_t &= \Omega^t{}_\phi \nonumber\\ \Omega^r{}_\theta &= -\Omega^\theta{}_r, & \Omega^r{}_\phi &= -\Omega^\phi{}_r, & \Omega^\theta{}_\phi &= -\Omega^\phi{}_\theta \end{align}