Proper-orthochronous Poincaré action and test pullback

Fix coordinates $x=(x^0,\mathbf{x})\in {\mathbb{R}}^4$ and the Minkowski matrix $\eta :=\mathrm{diag}(1,-1,-1,-1)\mathrm{.}$ The proper-orthochronous Lorentz group is $S{O}^{+}(1,3):=\{\mathrm{\Lambda }\in G{L}_4(\mathbb{R}):{\mathrm{\Lambda }}^{\mathsf{T}}\eta \mathrm{\Lambda }=\eta ,\det \mathrm{\Lambda }=1,{\mathrm{\Lambda }}^0{}_0\ge 1\},$ with its subspace topology from real $4\times 4$ matrices. The proper-orthochronous Poincaré group is the semidirect product $P_{+}^{\uparrow }:=S{O}^{+}(1,3)⋉{\mathbb{R}}^4,$ with multiplication $(\mathrm{\Lambda },a)(\mathrm{\Gamma },b)=(\mathrm{\Lambda }\mathrm{\Gamma },a+\mathrm{\Lambda }b)$, identity $e=(I,0)$, and affine action $gx=\mathrm{\Lambda }x+a$. Its topology is the product topology on the matrix and translation coordinates.

For $f\in S:=\mathcal{S}({\mathbb{R}}^4;\mathbb{C})$ define the pointwise pullback formula, viewed initially as a smooth function, $T_{g}f(x):=f(g^{-1}x)=f({\mathrm{\Lambda }}^{-1}(x-a))\mathrm{.}$ On smooth functions this formula obeys $T_g{T}_h=T_{gh}$ and $T_{e}f=f$, by the group law. This definition specifies the formula and conventions; it does not assert that $T_{g}f$ belongs to $S$. Translations have the convention $T_{(I,z)}f(x)=f(x-z)$.

With this signature, $p\cdot z:=p^0{z}^0-\mathbf{p}\cdot \mathbf{z}$, ${\bar{V}}_{+}:=\{p\in {\mathbb{R}}^4:p^0\ge \mathrm{\mid }\mathbf{p}\mathrm{\mid }\}$ is the closed forward cone, and $x-y$ is spacelike exactly when $(x^0-y^0{)}^2-\mathrm{\mid }\mathbf{x}-\mathbf{y}{\mathrm{\mid }}^2<0$.