The Boyer–Lindquist spatial end and differentiated expansion

The slice metric is diagonal in the polar chart: $h_{rr}=\mathrm{\Sigma }\mathrm{/}\mathrm{\Delta }=1+2M\mathrm{/}r+O(r^{-2})$, $h_{\theta \theta }=r^2+a^2{\cos }^2\theta$, and $h_{\varphi \varphi }=(r^2+a^2+2M{a}^{2}r{\sin }^2\theta \mathrm{/}\mathrm{\Sigma }){\sin }^2\theta$. Relative to the smooth sphere metric and its rotation one-form, the last two correction terms are bounded smooth sphere tensors. Transforming their covariant components to Cartesian coordinates introduces $r^{-2}$, whereas $dr=n_{i}d{x}^i$. This gives the displayed Cartesian expansion. Smooth rational functions of $1\mathrm{/}r$ and smooth angular tensors have the indicated derivative bounds; all denominators are bounded away from zero for sufficiently large $r$. Moreover, globally on the sphere the induced metric is $h=(\mathrm{\Sigma }\mathrm{/}\mathrm{\Delta })d{r}^2+\mathrm{\Sigma }\gamma +a^2(1+2Mr\mathrm{/}\mathrm{\Sigma })\alpha \otimes \alpha$, where $\gamma$ is the round metric and $\alpha ={\sin }^2\theta d\varphi$ is its smooth rotation one-form. Since $\mathrm{\Delta },\mathrm{\Sigma },r,M>0$, this is positive definite everywhere on each exterior slice, including the axes. Thus the slices are spacelike and [F2] applies.

Set $D=(r^2+a^2{)}^2-a^2\mathrm{\Delta }{\sin }^2\theta =(r^2+a^2)\mathrm{\Sigma }+2M{a}^{2}r{\sin }^2\theta >0$. Inverting the exterior $t,\varphi$ block gives $g^{tt}=-D\mathrm{/}(\mathrm{\Sigma }\mathrm{\Delta })<0$ away from the axes; this smooth scalar formula extends across them. For the unit normal in [F2], the covector $g(\nu ,\cdot )$ vanishes on the slice tangents and takes the value $-N$ on $T$, so $g(\nu ,\cdot )=-Ndt$. Taking its inverse-metric squared norm gives $-1=N^2{g}^{tt}$. Also $1=dt(T)=Ndt(\nu )$, since $X$ is tangent, so the chosen orientation gives $N>0$. Consequently $N=(-g^{tt}{)}^{-1\mathrm{/}2}=(\mathrm{\Sigma }\mathrm{\Delta }\mathrm{/}D{)}^{1\mathrm{/}2}$. This is a smooth function of $1\mathrm{/}r$ and smooth angular data near $1\mathrm{/}r=0$, with value $1$ there and denominators bounded away from zero, hence $N=1+O_2(r^{-1})$ in the convention of [F1].

For every tangent vector $U$, orthogonality gives $\beta (U)=h(X,U)=g(T,U)$, so in the adapted coordinates ${\beta }_i=g_{ti}$, as claimed in the Statement. The exterior tensor therefore gives $\beta =-2Mar{\sin }^2\theta d\varphi \mathrm{/}\mathrm{\Sigma }$. Since ${\sin }^2\theta d\varphi =(-x^{2}d{x}^1+x^{1}d{x}^2)\mathrm{/}{r}^2$, its Cartesian coefficients are $O_2(r^{-2})$, smoothly across the axes. The smooth rational coefficients give the differentiated bounds by the same fixed-parameter argument as in step 1.1.