Proper-orthochronous Poincaré pullbacks act continuously on Schwartz space

Fix $g=(\mathrm{\Lambda },a)$. The affine chain rule expresses each ${\mathrm{\partial }}^{\beta }(T_{g}f)(x)$ as a finite sum $c_{\beta \gamma }(\mathrm{\Lambda })({\mathrm{\partial }}^{\gamma }f)({\mathrm{\Lambda }}^{-1}(x-a))$, where $\mathrm{\mid }\gamma \mathrm{\mid }=\mathrm{\mid }\beta \mathrm{\mid }$ and the coefficients are polynomials in the entries of ${\mathrm{\Lambda }}^{-1}$. Substituting $y={\mathrm{\Lambda }}^{-1}(x-a)$ gives $x=\mathrm{\Lambda }y+a$; expanding each monomial $x^{\alpha }$ therefore bounds every $p_{\alpha \beta }(T_{g}f)$ by a finite sum of Schwartz seminorms of $f$ with coefficients depending on $g$. Hence $T_{g}f\in S$ and $T_g:S\to S$ is a continuous linear map.

For $g$ near $e$, put ${\epsilon }_g:=\mathrm{\parallel }{\mathrm{\Lambda }}^{-1}-I\mathrm{\parallel }+\mathrm{\mid }{\mathrm{\Lambda }}^{-1}a\mathrm{\mid }$ and $y_t:=(I+t({\mathrm{\Lambda }}^{-1}-I))x-t{\mathrm{\Lambda }}^{-1}a$ for $0\le t\le 1$. Choose the neighborhood so $\mathrm{\parallel }t({\mathrm{\Lambda }}^{-1}-I)\mathrm{\parallel }<1\mathrm{/}2$ for every $t$; the geometric series for the inverse then gives $\mathrm{\parallel }(I+t({\mathrm{\Lambda }}^{-1}-I){)}^{-1}\mathrm{\parallel }\le 2$. Thus $1+\mathrm{\mid }x\mathrm{\mid }\le C(1+\mathrm{\mid }{y}_t\mathrm{\mid })$ uniformly in $x,t,g$. Writing ${\delta }_g(x):=({\mathrm{\Lambda }}^{-1}-I)x-{\mathrm{\Lambda }}^{-1}a$, the fundamental theorem of calculus gives ${\mathrm{\partial }}^{\beta }f(y_1)-{\mathrm{\partial }}^{\beta }f(x)={\int }_0^1{\sum }_{j=1}^4{\mathrm{\partial }}^{\beta +e_j}f(y_t){\delta }_{g,j}(x)dt\mathrm{.}$ Here $\mathrm{\mid }{\delta }_{g,j}(x)\mathrm{\mid }\le {\epsilon }_g(1+\mathrm{\mid }x\mathrm{\mid })$. After multiplying by $x^{\alpha }$, the weight comparison bounds the absolute value by $C_{\alpha \beta }{\epsilon }_g$ times a finite sum of seminorms $p_{\gamma ,\beta +e_j}(f)$ with $\mathrm{\mid }\gamma \mathrm{\mid }\le \mathrm{\mid }\alpha \mathrm{\mid }+1$: expand $(1+\mathrm{\mid }{y}_t\mathrm{\mid }{)}^{\mathrm{\mid }\alpha \mathrm{\mid }+1}$ into finitely many coordinate monomials and use the definition of each seminorm. In step 1.1 the chain-rule coefficients are polynomials in ${\mathrm{\Lambda }}^{-1}$ and tend to their identity coefficients as $g\to e$; their differences are bounded by $C_{\alpha \beta }{\epsilon }_g$, while each remaining weighted derivative is bounded by finitely many Schwartz seminorms using the same comparison. Therefore $p_{\alpha \beta }(T_{g}f-f)\to 0$ for every $\alpha ,\beta$ as $g\to e$. By [F3] this is convergence in the Schwartz topology.

For any $g_0$, the group law gives $T_{g}f=T_{g_0}(T_{g_0^{-1}g}f)$. Step 2.1 and continuity of the fixed map $T_{g_0}$ from step 1.1 show $T_{g}f\to T_{g_0}f$ as $g\to g_0$. Thus the pullback is continuous in the group parameter everywhere. The universal property of algebraic tensor products defines ${\alpha }_g$ on each tensor degree; concatenation, conjugation, reversal, and $T_g{T}_h=T_{gh}$ give the unital star-automorphism and action identities.