A tetrad field that is not the coordinate frame of any chart

By [L1], $E$ is a smooth future-directed tetrad field on the nonempty open subset $W$ of flat Lorentzian four-space, and $[E_0,E_1]=-{\gamma }^{\mathrm{\prime }}{\mathrm{\partial }}_0-(\gamma \mathrm{\Omega }{)}^{\mathrm{\prime }}{\mathrm{\partial }}_2$ with ${\gamma }^{\mathrm{\prime }}={\gamma }^3{\lambda }^2{x}^1$. At every point of $W$ with $x^1\mathrm{\ne }0$ we have ${\gamma }^{\mathrm{\prime }}\mathrm{\ne }0$, so the coefficient of ${\mathrm{\partial }}_0$ in $[E_0,E_1]$ is nonzero and $[E_0,E_1]\mathrm{\ne }0$ there; such points exist in $W$.

Suppose, for contradiction, that $E$ were the coordinate frame of a smooth chart $(V,v)$ on some open subset of $W$: that is, $E_a={\mathrm{\partial }}_{v^a}$ on the chart domain for $a=0,1,2,3$. Then by [L2] we would have $[E_a,E_b]=[{\mathrm{\partial }}_{v^a},{\mathrm{\partial }}_{v^b}]=0$ for all $a,b$, contradicting step 1.1: every nonempty open chart domain $V\subseteq W$ contains a point with $x^1\mathrm{\ne }0$, since the hyperplane $\{x^1=0\}$ has empty interior in ${\mathbb{R}}^4$, and the bracket is nonzero at that point. Hence no such chart exists, and the displayed general statement is refuted.