The gauge-invariant test space

Define

The denominator lies in the numerator because $K^{†}\mathcal{H}=0$. Further quotienting by a presymplectic radical, if required, is not assumed to have been completed here.

The causal form induced by $E_{\mathcal{B}}$ descends to (eq:quotient). Indeed, on a compact test $a$,

which pairs to zero against a test in $\ker K^{†}$.

A symmetric contraction need not descend merely because its causal part does. Its descent and positivity are additional conditions.