In three Euclidean dimensions, the curvature 2-forms vanish. Thus, \begin{equation} 0 = \Omega^i{}_j = d\omega^i{}_j + \omega^i{}_k\wedge\omega^k{}_j \end{equation} and in particular \begin{equation} 0 = \Omega^1{}_2 = d\omega^1{}_2 + \omega^1{}_k\wedge\omega^k{}_2 = d\omega^1{}_2 + \omega^1{}_3\wedge\omega^3{}_2 \label{3dcurv} \end{equation} If we were instead to look only at the two-dimensional surface with $x^3=\hbox{constant}$, we would compute \begin{equation} \tilde\Omega^1{}_2 = d\omega^1{}_2 + \omega^1{}_k\wedge\omega^k{}_2 = d\omega^1{}_2 \ne 0 \label{2dcurv} \end{equation} since now $k=1,2$; we have written $\tilde\Omega^i{}_j$ for the two-dimensional curvature 2-forms to distinguish them from the three-dimensional curvature 2-forms $\Omega^i{}_j$.

However, it is important to realize that no tilde is needed on $\omega^1{}_2$. To see this, recall the structure equation \begin{equation} d\sigma^i + \omega^i{}_j\wedge\sigma^j = 0 \end{equation} The two-dimensional version of this equation is obtained from the three-dimensional version simply by setting $x^3=\hbox{constant}$ and $\sigma^3=0$. Thus, the same is true for the solutions, and the two-dimensional $\omega^1{}_2$ can be similarly obtained from the three-dimensional $\omega^1{}_2$. 1)

But we already know $\omega^1{}_2$, since on the surface we must have \begin{align} d\omega^1{}_2 &= - \omega^1{}_3\wedge\omega^3{}_2 \nonumber\\ &= \omega^3{}_1\wedge\omega^3{}_2 \nonumber\\ &= \Gamma^3{}_{1k}\,\sigma^k \wedge \Gamma^3{}_{2l}\,\sigma^l \nonumber\\ &= K\,\sigma^1\wedge\sigma^2 \end{align} where \begin{equation} K = \det(-S) = \det(S) \end{equation} is the Gaussian curvature. Thus, for a two-dimensional surface, we have \begin{equation} \Omega^1{}_2 = d\omega^1{}_2 = K\,\omega \end{equation} where we have dropped the tilde.

This remarkable result is due to Gauss, and is known as the Theorema Egregium (“Outrageous Theorem”): The Gaussian curvature, originally defined extrinsically, is in fact an intrinsic property of the surface.