Stationary horizon Noether-charge entropy variation

By [F4], the selected charge has the stated $W^c{\xi }_c$, $X^{cd}{\mathrm{\nabla }}_{[c}{\xi }_{d]}$, $\mathbf{Y}({\mathcal{L}}_{\xi }\varphi )$, and exact-form decomposition. On $B$, $\xi =0$ and ${\mathcal{L}}_{\xi }\varphi =0$. The $W^c{\xi }_c$ and $\mathbf{Y}$ terms and their field variations vanish there except for $\mathbf{Y}(\varphi ;{\mathcal{L}}_{\xi }\delta \varphi )$. Covariance identifies this term with ${\mathcal{L}}_{\xi }\mathbf{Y}(\varphi ;\delta \varphi )$; Cartan's formula [F2] writes it as ${\iota }_{\xi }d\mathbf{Y}+d({\iota }_{\xi }\mathbf{Y})$, whose pullback integrates to zero because $\xi =0$ on $B$ and $B$ has no boundary. The $dZ$ variation integrates to zero by Stokes [F3]. The Lagrangian ambiguity ${\iota }_{\xi }\mu$ vanishes on $B$, and the exact charge ambiguity also integrates to zero. Therefore $\delta {\int }_B{Q}_{\xi }=\delta {\int }_B{X}^{cd}{\mathrm{\nabla }}_{[c}{\xi }_{d]}$.

Define $w_{cd}:={\mathrm{\nabla }}_{[c}{\xi }_{d]}-\kappa {ϵ}_{cd}$ along $B$. It vanishes on the background. Since $\xi$ and $B$ are fixed and $\xi =0$ on $B$, $\delta ({\mathrm{\nabla }}_c{\xi }^a)=\delta {\mathrm{\Gamma }}_{ce}^a{\xi }^e=0$ there. Lowering the index and using ${\mathrm{\nabla }}_c{\xi }^a=\kappa {ϵ}_c{}^a$ gives $\delta w_{cd}=-\kappa g_{a[d}\delta {ϵ}_{c]}{}^a$. The binormal annihilates tangent vectors, so its variation has no tangent-tangent block; its fixed normalization ${ϵ}^a{}_b{ϵ}^b{}_a=2$ removes the normal-normal antisymmetric block in $g_{a[d}\delta {ϵ}_{c]}{}^a$. Thus $\delta w_{cd}$ has only normal-tangent components. The coefficient $X^{cd}$, including its form slots, is local and covariant in the fields, so stationarity gives ${\mathcal{L}}_{\xi }X=0$. At $p\in B$, define $A^a{}_b:={\mathrm{\nabla }}_b{\xi }^a{\mathrm{\mid }}_p$. It annihilates tangent vectors and is a Lorentz-boost generator on the normal plane with opposite nonzero eigenvalues $+\kappa$ and $-\kappa$. Choose null normal eigenvectors $l,k$ with $g(l,k)=-1$ and a tangent basis $e_A$. After choosing the sign of the flow parameter $s$, the induced linear action sends $l\mapsto e^{\kappa s}l$, $k\mapsto e^{-\kappa s}k$, and fixes every $e_A$. Since $\xi (p)=0$, ${\mathcal{L}}_{\xi }X=0$ is invariance of each component under this pointwise action. In the pullback of $X$ to $B$, all form slots are tangent and have boost weight zero. A component with exactly one of the indices $c,d$ normal and the other tangent has nonzero weight $+\kappa$ or $-\kappa$, so invariance for every $s$ forces that component to vanish. These are precisely the components that could contract with the normal-tangent $\delta w_{cd}$. Hence the pullback of $X^{cd}\delta w_{cd}$ to $B$ vanishes.

On the background, ${\mathrm{\nabla }}_a{\xi }_b=\kappa {ϵ}_{ab}$ and [F4] gives $X^{cd}{ϵ}_{cd}=-P^{abcd}{ϵ}_{ab}{ϵ}_{cd}d{A}_B$. Write the charge term as $X^{cd}(\kappa {ϵ}_{cd}+w_{cd})$. Since $w=0$ on the background, its variation is $\kappa \delta (X^{cd}{ϵ}_{cd})+X^{cd}\delta w_{cd}$. The second term integrates to zero by step 1.2; the first includes the variations of $P$, the binormal, and the induced area form. Hence $\delta {\int }_B{Q}_{\xi }=\kappa \delta {\int }_B{X}^{cd}{ϵ}_{cd}=-\kappa \delta {\int }_B{P}^{abcd}{ϵ}_{ab}{ϵ}_{cd}d{A}_B=(\mathrm{\hslash }\kappa \mathrm{/}(2\pi k_B))\delta S_W$. The coefficient $\kappa$ is the fixed background surface gravity, so no $\delta \kappa$ term appears.