Strong Constraint DFT Scalar and Action Gauge Invariance

From [F1], ${\delta }_{\xi }{e}^{-2d}=-2e^{-2d}{\delta }_{\xi }d={\mathrm{\partial }}_M({\xi }^M{e}^{-2d})$. For any expression $W$, write ${\delta }_{\xi }W={\widehat{\mathcal{L}}}_{\xi }W+{\mathrm{\Delta }}_{\xi }W$, where ${\mathrm{\Delta }}_{\xi }$ is the failure of its ordinary component variation to be a generalized tensor variation. The product rule makes ${\mathrm{\Delta }}_{\xi }$ a derivation, while ${\mathrm{\Delta }}_{\xi }\eta =0$ because $\eta$ is constant.

Differentiating the transformation in [F1] and raising indices with the invariant $\eta$ gives ${\mathrm{\Delta }}_{\xi }({\mathrm{\partial }}_M{\mathcal{H}}_{KL})=-2{\mathrm{\partial }}_M{\mathrm{\partial }}_P{\xi }_{(K}{\mathcal{H}}_{L)}{}^P+2{\mathrm{\partial }}_M{\mathrm{\partial }}_{(K}{\xi }^P{\mathcal{H}}_{L)P}$, ${\mathrm{\Delta }}_{\xi }({\mathrm{\partial }}_M{\mathrm{\partial }}_{N}d)={\mathrm{\partial }}_M{\mathrm{\partial }}_N{\xi }^P{\mathrm{\partial }}_{P}d-\frac{1}{2}{\mathrm{\partial }}_M{\mathrm{\partial }}_N({\mathrm{\partial }}_P{\xi }^P)$, and ${\mathrm{\Delta }}_{\xi }({\mathrm{\partial }}_M{\mathrm{\partial }}_N{\mathcal{H}}^{MN})=-2{\mathrm{\partial }}_M({\mathrm{\partial }}_P{\xi }^P){\mathrm{\partial }}_N{\mathcal{H}}^{MN}-2{\mathrm{\partial }}_M{\mathrm{\partial }}_N({\mathrm{\partial }}_P{\xi }^P){\mathcal{H}}^{MN}-{\mathrm{\partial }}_M{\mathrm{\partial }}_N{\xi }^P{\mathrm{\partial }}_P{\mathcal{H}}^{MN}$. Parentheses symmetrize the enclosed metric indices. Terms containing ${\eta }^{MN}{\mathrm{\partial }}_{M}A{\mathrm{\partial }}_{N}B$ vanish by the strong constraint.

Substituting step 2.1 into the first two terms of $\mathcal{R}$ gives ${\mathrm{\Delta }}_{\xi }(4{\mathcal{H}}^{MN}{\mathrm{\partial }}_M{\mathrm{\partial }}_{N}d-{\mathrm{\partial }}_M{\mathrm{\partial }}_N{\mathcal{H}}^{MN})=4{\mathcal{H}}^{MN}{\mathrm{\partial }}_M{\mathrm{\partial }}_N{\xi }^P{\mathrm{\partial }}_{P}d+2{\mathrm{\partial }}_M({\mathrm{\partial }}_P{\xi }^P){\mathrm{\partial }}_N{\mathcal{H}}^{MN}+{\mathrm{\partial }}_M{\mathrm{\partial }}_N{\xi }^P{\mathrm{\partial }}_P{\mathcal{H}}^{MN}$. Substitution into the next two terms gives ${\mathrm{\Delta }}_{\xi }(4{\mathrm{\partial }}_M{\mathcal{H}}^{MN}{\mathrm{\partial }}_{N}d-4{\mathcal{H}}^{MN}{\mathrm{\partial }}_{M}d{\mathrm{\partial }}_{N}d)=-4{\mathcal{H}}^{MN}{\mathrm{\partial }}_M{\mathrm{\partial }}_N{\xi }^P{\mathrm{\partial }}_{P}d-2{\mathrm{\partial }}_M({\mathrm{\partial }}_P{\xi }^P){\mathrm{\partial }}_N{\mathcal{H}}^{MN}$. Adding these four variations leaves only $+{\mathrm{\partial }}_M{\mathrm{\partial }}_N{\xi }^P{\mathrm{\partial }}_P{\mathcal{H}}^{MN}$.

Varying the $-\frac{1}{2}$ kinetic term in $\mathcal{R}$ and using [F1] gives

where $T:={\mathrm{\partial }}_K{\mathcal{H}}^{MN}{\mathcal{H}}^{KP}{\mathcal{H}}_{NL}{\mathrm{\partial }}_M{\mathrm{\partial }}_P{\xi }^L$. For the remaining $+\frac{1}{8}$ kinetic term, the same per-index variation gives

The two terms produced by the parentheses are equal after using symmetry of $\mathcal{H}$, ${\mathcal{H}}^{MN}{\mathcal{H}}_{NL}={\delta }^M{}_L$, and commuting partial derivatives. Thus this variation equals ${\mathrm{\partial }}_M{\mathcal{H}}^{KL}{\mathcal{H}}^{MN}{\mathcal{H}}_{LP}{\mathrm{\partial }}_N{\mathrm{\partial }}_K{\xi }^P=T$, after relabeling contracted indices. The $T$ terms cancel, while the first term cancels the residual in step 3.1. Therefore ${\mathrm{\Delta }}_{\xi }\mathcal{R}=0$ and ${\delta }_{\xi }\mathcal{R}={\widehat{\mathcal{L}}}_{\xi }\mathcal{R}={\xi }^M{\mathrm{\partial }}_M\mathcal{R}$.

The product rule gives

where symmetry of ${\mathcal{H}}^{MN}$ and interchange of dummy indices identify ${\mathrm{\partial }}_{M}d{\mathrm{\partial }}_N{\mathcal{H}}^{MN}={\mathrm{\partial }}_M{\mathcal{H}}^{MN}{\mathrm{\partial }}_{N}d$. Adding $e^{-2d}\mathcal{R}$ cancels the $\mathrm{\partial }\mathrm{\partial }\mathcal{H}$ and $\mathcal{H}\mathrm{\partial }\mathrm{\partial }d$ terms, changes the coefficient $-6$ of this contraction to $-2$, and leaves $4e^{-2d}{\mathcal{H}}^{MN}{\mathrm{\partial }}_{M}d{\mathrm{\partial }}_{N}d$ together with the two generalized-metric kinetic terms. Relabeling dummy indices in the second kinetic term gives exactly ${\mathcal{L}}_{\mathrm{D}\mathrm{F}\mathrm{T}}$ in [F1]. Thus ${\mathcal{L}}_{\mathrm{D}\mathrm{F}\mathrm{T}}=e^{-2d}\mathcal{R}+{\mathrm{\partial }}_M{J}^M$, where $J^M:=e^{-2d}({\mathrm{\partial }}_N{\mathcal{H}}^{MN}-4{\mathcal{H}}^{MN}{\mathrm{\partial }}_{N}d)$. By step 4.1 and step 1.1, ${\delta }_{\xi }(e^{-2d}\mathcal{R})={\mathrm{\partial }}_M({\xi }^M{e}^{-2d}\mathcal{R})$. Since ${\delta }_{\xi }$ commutes with ${\mathrm{\partial }}_M$, the full density variation is ${\delta }_{\xi }{\mathcal{L}}_{\mathrm{D}\mathrm{F}\mathrm{T}}={\mathrm{\partial }}_M({\xi }^M{e}^{-2d}\mathcal{R}+{\delta }_{\xi }{J}^M)$. Every term in ${\delta }_{\xi }{J}^M$ contains $\xi$ or one of its derivatives, so compact support of $\xi$ makes the gauge-variation flux vanish; otherwise impose the zero-flux condition in the Given clause. The integrated DFT action is invariant under this hypothesis, without requiring $J^M$ itself to have zero boundary flux.