Lorentzian signature and time orientation of the Kerr exterior datum

Since $0<\sqrt{M^2-a^2}\le M$, with equality in the upper bound exactly when $a=0$, one has $0\le r_{-}<r_{+}$ and $r_{-}=0$ exactly when $a=0$. Thus $\mathrm{\Delta }=(r-r_{+})(r-r_{-})>0$ for $r>r_{+}$. Also $\mathrm{\Sigma }\ge r^2>0$, $A:=r^2+a^2>0$, and $q=\gamma (\xi ,\xi )\ge 0$. The global formula in [F1] is a smooth symmetric tensor because its sphere tensors and scalar functions are smooth, its denominators are nonzero, and its finite sums and products are smooth by [F6]. The stated formula for $N$ is smooth: repeated use of [F5] makes $x\mapsto x^{-1}$ smooth on $(0,\mathrm{\infty })$, [F6] composes this with $\mathrm{\Sigma }\mathrm{\Delta }$, and $T,\mathrm{\Phi }$ are smooth by [F1]; hence $N$ is a smooth vector field in the sense of [F7].

On a polar chart put $q={\sin }^2\theta$. The one-forms $e^0=\sqrt{\mathrm{\Delta }\mathrm{/}\mathrm{\Sigma }}(dt-aqd\varphi )$, $e^1=\sqrt{\mathrm{\Sigma }\mathrm{/}\mathrm{\Delta }}dr$, $e^2=\sqrt{\mathrm{\Sigma }}d\theta$, and $e^3=\sin \theta {\mathrm{\Sigma }}^{-1\mathrm{/}2}(Ad\varphi -adt)$ are smooth: repeated application of [F5] gives every derivative of the positive powers $x^{1\mathrm{/}2}$ and $x^{-1\mathrm{/}2}$ on $(0,\mathrm{\infty })$, and [F6] gives smoothness after composition with the positive smooth functions $\mathrm{\Delta },\mathrm{\Sigma }$. The determinant of the coefficient matrix of $(e^0,e^3)$ relative to $(dt,d\varphi )$ is $\sqrt{\mathrm{\Delta }}\sin \theta$, which is nonzero when $0<\theta <\pi$. Expanding the global formula of [F1] gives $g=-(e^0{)}^2+(e^1{)}^2+(e^2{)}^2+(e^3{)}^2$; hence $g$ has signature $(-,+,+,+)$ at every point off the axis.

At either axis point ${\xi }^{\mathrm{♭}}=0$, $\mathrm{\Sigma }=A$, and the global expression in [F1] restricts to $g=-(\mathrm{\Delta }\mathrm{/}A)d{t}^2+(A\mathrm{/}\mathrm{\Delta })d{r}^2+A\gamma$ on the tangent directions there. The first coefficient is negative and the radial and sphere terms are positive because $A,\mathrm{\Delta }>0$ and $\gamma$ is positive definite. Thus $g$ is nondegenerate with the same signature at the axes. The global definition in [F1] already gives smoothness there, so this proves the asserted signature on all of ${\mathcal{M}}_{\mathrm{K}}$.

Put $D:=A^2-a^2\mathrm{\Delta }q$. Since $\mathrm{\Delta }=A-2Mr$, $0<\mathrm{\Delta }<A$ and $0\le q\le 1$, one has $D\ge A^2-a^2\mathrm{\Delta }\ge A^2-a^{2}A=A{r}^2>0$. Off the axis, direct contraction of the covariant metric in [F1] with the specified vector $N$ gives $g(N,T)=-1$, $g(N,\mathrm{\Phi })=0$, and $g(N,{\mathrm{\partial }}_r)=g(N,{\mathrm{\partial }}_{\theta })=0$; hence $g(N,\cdot )=-dt$ there. At an axis point $q=0$, $\mathrm{\Phi }=0$, and $\mathrm{\Sigma }=A$, so $N=(A\mathrm{/}\mathrm{\Delta })T$ and the metric of step 2.2 gives the same identity on all tangent directions. Thus everywhere $g(N,N)=-dt(N)=-D\mathrm{/}(\mathrm{\Sigma }\mathrm{\Delta })<0$. Step 1.1 and [F4] now make $({\mathcal{M}}_{\mathrm{K}},g,N)$ a smooth time-oriented Lorentzian four-manifold. Only at this point apply [F2]: nondegeneracy and $g(N,\cdot )=-dt$ identify $N=-{\mathrm{♯}}_g(dt)$; [F3] supplies the smooth inverse metric in its stated scope.