Lemma: Kinetic sign

For ${\gamma }_{0\mu }=0$, ${\partial }_i{\gamma }_{ij}=0$, and ${\delta }^{ij}{\gamma }_{ij}=0$, the quadratic density becomes

Its time-derivative term has the same sign as $\frac{1}{2}{\dot{\varphi }}^2$.

Proof. The trace and divergence terms in (eq:FP) vanish. Raising both spatial indices contributes two minus signs, so ${\gamma }^{ij}={\gamma }_{ij}$. The stated expression follows.