The Komar angular-momentum flux of the Kerr end

Write $b={\sin }^2\theta$. Expanding, inverting and radially differentiating the exact tensor in [F2] gives the exterior coefficients used below. Since ${\eta }^{\mathrm{♭}}=g_{t\varphi }dt+g_{\varphi \varphi }d\varphi$ and the metric has no $r$ or $\theta$ mixed terms, the pulled-back dual on a $t,r$ sphere has coefficient $(\star d{\eta }^{\mathrm{♭}}{)}_{\theta \varphi }=-\sqrt{-\det g}{g}^{rr}(g^{tt}{\mathrm{\partial }}_r{g}_{t\varphi }+g^{t\varphi }{\mathrm{\partial }}_r{g}_{\varphi \varphi })$. The sign follows from ${ϵ}_{tr\theta \varphi }>0$ and $d{\eta }_{tr}^{\mathrm{♭}}=-{\mathrm{\partial }}_r{g}_{t\varphi }$. The exterior coefficients give $g^{tt}=-1+O(r^{-1})$, $g^{t\varphi }=-2Ma\mathrm{/}{r}^3+O(r^{-4})$, $g^{rr}=1+O(r^{-1})$, ${\mathrm{\partial }}_r{g}_{t\varphi }=2Mab\mathrm{/}{r}^2+O(r^{-4})$, ${\mathrm{\partial }}_r{g}_{\varphi \varphi }=2rb+O(r^{-2})$, and $\sqrt{-\det g}=\mathrm{\Sigma }\sin \theta$. Thus the coefficient is $6Ma{\sin }^3\theta +O(r^{-1})\sin \theta$, with a bounded smooth angular remainder.

Both inverse-metric terms contribute: their leading sum before the overall minus sign is $-6Mab\mathrm{/}{r}^2$, so ignoring the $g^{t\varphi }$ term would give an incorrect factor. Since ${\int }_0^{2\pi }d\varphi {\int }_0^{\pi }{\sin }^3\theta d\theta =8\pi \mathrm{/}3$, the limiting flux is $16\pi Ma$. The Komar normalization therefore gives $J_{\mathrm{K}}=Ma$, including negative $a$, zero $a$ and extremal $\mathrm{\mid }a\mathrm{\mid }=M$.