Kerr geometry and hidden symmetry — Schwarzschild limit

Setting $a=0$ gives $r_{+}=2M$, $\mathrm{\Sigma }=r^2$, $\mathrm{\Delta }=r(r-2M)$, and $g_{t\varphi }=0$. The remaining components are $g_{tt}=-(1-2M\mathrm{/}r)$, $g_{rr}=(1-2M\mathrm{/}r{)}^{-1}$, $g_{\theta \theta }=r^2$, and $g_{\varphi \varphi }=r^2{\sin }^2\theta$. These are exactly [F2] with $s=2m_{\ast }=2M$ and the same domain $r>2M$.

Since $\mathrm{\Phi }$ is Killing by [F3] and [F8], set $K_{ab}^{\mathrm{\Phi }}:={\mathrm{\Phi }}_a{\mathrm{\Phi }}_b$. The product rule gives ${\mathrm{\nabla }}_{(a}{K}_{bc)}^{\mathrm{\Phi }}=\frac{1}{3}[{\mathrm{\Phi }}_b({\mathrm{\nabla }}_a{\mathrm{\Phi }}_c+{\mathrm{\nabla }}_c{\mathrm{\Phi }}_a)+{\mathrm{\Phi }}_c({\mathrm{\nabla }}_a{\mathrm{\Phi }}_b+{\mathrm{\nabla }}_b{\mathrm{\Phi }}_a)+{\mathrm{\Phi }}_a({\mathrm{\nabla }}_b{\mathrm{\Phi }}_c+{\mathrm{\nabla }}_c{\mathrm{\Phi }}_b)]=0$. Thus $K^{\mathrm{\Phi }}$ is a reducible Killing tensor by [F4].

By [F3], ${\mathrm{\Omega }}_H=a\mathrm{/}(r_{+}^2+a^2)$, so ${\mathrm{\Omega }}_H=0$ at $a=0$. The vacuum statement follows by applying [F6] to this same metric and restricting its domain to $r>2M$.

Along an affine geodesic, $K_{ab}^{\mathrm{\Phi }}{v}^a{v}^b=({\mathrm{\Phi }}_a{v}^a{)}^2$ is constant by [F5]. This is the axial quadratic integral; it is reducible and is not identified with an irreducible Carter tensor.

Setting $a=0$ in [F7] gives $Q=p_{\theta }^2+L_z^2{\cos }^2\theta \mathrm{/}{\sin }^2\theta =p_{\theta }^2+L_z^2{\cot }^2\theta$. Since $1\mathrm{/}{\sin }^2\theta -{\cot }^2\theta =1$, this equals $L_{\mathrm{s}\mathrm{p}\mathrm{h}}^2-L_z^2$ under the definitions in the Example.