Coulomb obstruction and scalar compatibility

For a classical free scalar,

and direct differentiation gives ${\partial }^{\mu }{T}_{\mu \nu }=(P\varphi ){\partial }_{\nu }\varphi$. Compact smooth scalar data therefore give $T(\varphi )\in \mathcal{Q}$.

At the initial slice, integration of

shows

For $Q=T(\varphi )$ the left side is

unless the data vanish. Hence compactly supported or Schwartz spatial metric data cannot solve the sourced constraint. For the displayed prescriptions,

The derivative energy is finite but these components are not generally in $L^2$.

Nevertheless $S_{t_0}(j)\varphi$ exists and has finite scalar energy on every finite slab. Indeed, $L_j\varphi$ is supported where $\varphi$ is supported; all needed derivatives of the smooth $j$ are bounded on that compact region. Lemma lem:wave applies at every Sobolev order.

The map $\varphi \mapsto T(\varphi )$, and hence $\varphi \mapsto \mathcal{R}T(\varphi )$, is quadratic. The scalar correction $\varphi \mapsto S_{t_0}(\mathcal{R}T(\varphi ))\varphi$ is cubic. Neither this cubic nor a second-order complete scalar observable is quantized in the following section. The classical coefficient construction itself makes no quantum assertion.

Figure fig:Coulomb-and-degree separates the sourced metric tail from the distinct quadratic and cubic constructions.