Causal propagator induces the free scalar symplectic quotient

Since $P{E}_{+}f=P{E}_{-}f=f$, $P(Ef)=0$. Its support is contained in $J^{+}(\mathrm{supp}f)\cup J^{-}(\mathrm{supp}f)$ by [F1], so $Ef\in {\mathrm{Sol}}_{\mathrm{s}\mathrm{c}}(M)$. Also $E(Pf)=E_{-}Pf-E_{+}Pf=f-f=0$. Hence $P{\mathcal{D}}_g(M)\subseteq \ker E$.

Let $f,h\in {\mathcal{D}}_g(M)$, $u=E_{-}f$ and $v=E_{+}h$. Their supports meet in the closed set $C:=\mathrm{supp}u\cap \mathrm{supp}v$, contained in $J^{-}(\mathrm{supp}f)\cap J^{+}(\mathrm{supp}h)$ and compact by [F2]. If $C=\mathrm{\varnothing }$, both $fv$ and $uh$ vanish; otherwise choose by [F3] a compactly supported smooth cutoff $\eta$ equal to $1$ on a neighborhood of $C$. Since $f=Pu$ and $h=Pv$, formal self-adjointness applied to the compactly supported functions $\eta u$ and $\eta v$ gives ${\int }_{M}fv{\mu }_g={\int }_{M}uh{\mu }_g$: after expanding $P(\eta u)$ and $P(\eta v)$, each commutator term is supported where a derivative of $\eta$ meets $\mathrm{supp}u\cap \mathrm{supp}v\subseteq C$, so it vanishes because $\eta =1$ near $C$; the remaining integrands $fv$ and $uh$ are supported in $C$. Applying the same calculation with $u=E_{+}f$ and $v=E_{-}h$ gives ${\int }_{M}f(E_{-}h){\mu }_g={\int }_M(E_{+}f)h{\mu }_g$. Subtracting these two Green adjoint identities yields $E^{\ast }=-E$ under the compact test-density pairing.

Conversely suppose $Ef=0$, so $u:=E_{-}f=E_{+}f$. Then $Pu=f$ by [F1] and $\mathrm{supp}u\subseteq J^{-}(K)\cap J^{+}(K)$, where $K=\mathrm{supp}f$ is compact. Its closed support is compact by [F2], so the smooth function $u$ is compactly supported. Thus $u\in {\mathcal{D}}_g(M)$ and $f=Pu\in P{\mathcal{D}}_g(M)$. Together with step 1.1, $\ker E=P{\mathcal{D}}_g(M)$.

If $f$ is replaced by $f+Pk$, then ${\int }_M(Pk)(Eh){\mu }_g={\int }_{M}kP(Eh){\mu }_g=0$ by the formal adjoint identity in [F1] and step 1.1; if $h$ is replaced by $h+Pk$, then $E(Pk)=0$ by step 1.1. Thus ${\sigma }_M$ is well-defined on $V(M)$. Bilinearity follows from linearity of $E$ and the integral. From step 1.2, ${\sigma }_M([h],[f])=\int hEf{\mu }_g=-\int fEh{\mu }_g=-{\sigma }_M([f],[h])$, so it is alternating. If ${\sigma }_M([f],[h])=0$ for every $[h]$, step 1.2 gives ${\int }_M(Ef)h{\mu }_g=0$ for every compact smooth real $h$. If $Ef$ were nonzero at a point, continuity would give a relatively compact coordinate neighborhood on which it has one strict sign; choose a smaller compact coordinate ball there and use [F3] to obtain a nonnegative test bump equal to $1$ on that ball. Its integral against $Ef{\mu }_g$ is then nonzero because ${\mu }_g$ is a positive smooth density and the ball has positive volume, a contradiction. Hence $Ef=0$, and step 2.1 gives $[f]=0$. This proves nondegeneracy and completes the proof.