Positive half-sided groups obey modular scaling and reflection

The data are: Exactly the data, positivity, invariant-vector and half-sided inclusion hypotheses above, and the stated proved modular polar interface.

Put $D={\mathrm{\Delta }}^{1\mathrm{/}2}$. For $a\ge 0$ and $x\in M$, the vector $T(a)x\mathrm{\Omega }=(T(a)xT(-a))\mathrm{\Omega }$ belongs to $M\mathrm{\Omega }$, and $S_{0}T(a)x\mathrm{\Omega }=T(a)x^{\ast }\mathrm{\Omega }=T(a)S_{0}x\mathrm{\Omega }\mathrm{.}$ For $\xi \in \mathrm{Dom}S$, approximate $\xi$ in the graph of $S$ by core vectors $x_n\mathrm{\Omega }$. The boundedness of $T(a)$ and closedness of $S$ imply $T(a)\xi \in \mathrm{Dom}S$ and $ST(a)\xi =T(a)S\xi$. Multiplying this equality on the left by $J$ gives, with $V_a=JT(a)J$, $DT(a)\xi =V_{a}D\xi (\xi \in \mathrm{Dom}D)\mathrm{.}$ Thus $DT(a)\supseteq V_{a}D$. This is a domain statement, not a formal equality obtained by moving unbounded powers through bounded operators.

Taking adjoints of the last inclusion is legitimate for densely defined operators. Since $T(a)$ and $V_a$ are bounded unitaries and $D$ is self-adjoint, it gives $T(-a)D\subseteq D{V}_a^{\ast }\mathrm{.}$ Consequently, for every $\xi \in \mathrm{Dom}D$, one has $V_a^{\ast }\xi \in \mathrm{Dom}D$ and $D{V}_a^{\ast }\xi =T(-a)D\xi$. Injectivity of $D$ then gives the exact identity on this dense domain, $D^{-1}T(-a)D\xi =V_a^{\ast }\xi =JT(-a)J\xi \mathrm{.}$ In particular the operator on the left has bounded unitary closure $JT(-a)J$. The sign $-a$ in this adjoint identity is essential.

Let $A=\log \mathrm{\Delta }$ and $E_R=1_{[-R,R]}(A)$ for $R>0$. These are symmetric spectral cutoffs, and $E_R\to 1$ strongly. Since $JAJ=-A$, $J{E}_{R}J=E_R$. Positive energy gives a contractive holomorphic extension $T(w)=e^{iwP}(\mathrm{Im}w>0),\mathrm{\parallel }T(w)\mathrm{\parallel }\le 1.$ Indeed $\mathrm{\mid }{e}^{iwp}\mathrm{\mid }=e^{-p\mathrm{Im}w}\le 1$ on the spectral support $p\ge 0$; the derivative multipliers $p^n{e}^{-p\mathrm{Im}w}$ are bounded locally away from the real axis, so the extension is norm-holomorphic there. The spectral dominated-convergence theorem gives strong continuity on the closed upper half-plane, including its real boundary. Fix $a>0$ and define on the strip $0<\mathrm{Im}z<1\mathrm{/}2$, $F_R(z)=E_R{\mathrm{\Delta }}^{iz}T(a{e}^{2\pi z}){\mathrm{\Delta }}^{-iz}{E}_R\mathrm{.}$ The exponential $a{e}^{2\pi z}$ belongs to the upper half-plane throughout this strip. The cutoff modular powers are bounded norm-entire functions, so $F_R$ is norm-holomorphic inside, strongly continuous at the horizontal boundaries, and, for $z=t+iy$, $\mathrm{\parallel }{F}_R(t+iy)\mathrm{\parallel }\le e^{2Ry}\le e^R(0\le y\le 1\mathrm{/}2)\mathrm{.}$ This raw bound is uniform in $t$; it is not claimed to be uniform in $R$.

On the lower boundary $\mathrm{\parallel }{F}_R(t)\mathrm{\parallel }\le 1$ because its middle conjugate is unitary. On the upper boundary put $b=a{e}^{2\pi t}>0$. The adjoint-domain identity of step 3.1, applied on the cutoff domain, gives $F_R(t+i\mathrm{/}2)=E_R{\mathrm{\Delta }}^{it}({\mathrm{\Delta }}^{-1\mathrm{/}2}T(-b){\mathrm{\Delta }}^{1\mathrm{/}2}){\mathrm{\Delta }}^{-it}{E}_R=E_R{\mathrm{\Delta }}^{it}JT(-b)J{\mathrm{\Delta }}^{-it}{E}_R\mathrm{.}$ The right side is a compression of a unitary, so it also has norm at most one. Moreover, using $J{E}_{R}J=E_R$ and $J{\mathrm{\Delta }}^{it}J={\mathrm{\Delta }}^{it}$ shows the exact boundary matching $F_R(t+i\mathrm{/}2)=\vartheta (F_R(t)),\vartheta (B)=J{B}^{\ast }J\mathrm{.}$ The map $\vartheta$ is a COMPLEX-LINEAR norm-isometric involution, even though it reverses products: $(\lambda B{)}^{\ast }=\bar{\lambda }{B}^{\ast }$ and antiunitarity of $J$ conjugates $\bar{\lambda }$ back to $\lambda$.

For fixed $R$ and unit vectors $\xi ,\eta$, the scalar function $f(z)=⟨F_R(z)\xi ,\eta ⟩$ is holomorphic, continuous at both boundaries, bounded by $e^R$ in the whole strip and bounded by one on its horizontal boundaries. Here is the bounded-strip maximum argument with no uncontrolled limit: multiply $f(z)$ by $e^{-ϵz^2}$ and apply the maximum principle on rectangles $[-L,L]\times [0,1\mathrm{/}2]$. On the horizontal sides the modulus is at most $e^{ϵ\mathrm{/}4}$; on the vertical sides the raw bound is at most $e^R{e}^{-ϵL^2+ϵ\mathrm{/}4}$, tending to zero as $L\to \mathrm{\infty }$. Thus $\mathrm{\mid }f(z)e^{-ϵz^2}\mathrm{\mid }\le e^{ϵ\mathrm{/}4}$, and letting $ϵ\downarrow 0$ gives $\mathrm{\mid }f(z)\mathrm{\mid }\le 1$. Therefore $\mathrm{\parallel }{F}_R(z)\mathrm{\parallel }\le 1$ throughout the closed strip, now INDEPENDENT OF $R$.

Let $H_{\mathrm{c}\mathrm{u}\mathrm{t}}={\bigcup }_R{E}_{R}H$, a dense linear subspace. For $\xi ,\eta \in H_{\mathrm{c}\mathrm{u}\mathrm{t}}$, once $R$ contains both spectral supports, the scalar matrix element $⟨F_R(z)\xi ,\eta ⟩$ is EXACTLY independent of further enlargement of $R$: the external projections can be removed against these vectors and commute with the modular powers. In the linear-first convention its value can equivalently be written $⟨T(a{e}^{2\pi z}){\mathrm{\Delta }}^{-iz}\xi ,{\mathrm{\Delta }}^{-i\bar{z}}\eta ⟩\mathrm{.}$ Step 6.1 bounds this sesquilinear form by $\mathrm{\parallel }\xi \mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }$. It extends uniquely to a bounded operator $F(z)$ on all of $H$, with $\mathrm{\parallel }F(z)\mathrm{\parallel }\le 1$. Matrix elements on cutoff vectors are holomorphic inside and continuous at the boundary; approximating arbitrary vectors by cutoff vectors, using the same uniform bound, proves weak holomorphy and weak boundary continuity for every matrix element. On the lower boundary, $F(t)={\mathrm{\Delta }}^{it}T(a{e}^{2\pi t}){\mathrm{\Delta }}^{-it}\mathrm{.}$ Since $\vartheta$ is weak-operator continuous, the upper boundary satisfies $F(t+i\mathrm{/}2)=\vartheta (F(t))\mathrm{.}$ For clarity, weak continuity and complex linearity follow directly from $⟨\vartheta (B)\xi ,\eta ⟩=⟨BJ\eta ,J\xi ⟩\mathrm{.}$ Thus no weakly convergent subsequence, Banach limit, or interchange of unbounded operators is used in constructing $F$.

Repeat this strip over the entire plane by setting, for each integer $k$ and $0\le \mathrm{Im}z\le 1\mathrm{/}2$, $\tilde{F}(z+ki\mathrm{/}2)={\vartheta }^k(F(z))\mathrm{.}$ Because $\vartheta$ is complex-linear, it preserves weak holomorphy. Since ${\vartheta }^2=1$ and the exact upper/lower boundary relation holds, neighboring definitions agree pointwise at each horizontal seam. All matrix elements of $\tilde{F}$ are therefore continuous on the plane and holomorphic away from the seams, uniformly bounded by $\mathrm{\parallel }\xi \mathrm{\parallel }\mathrm{\parallel }\eta \mathrm{\parallel }$. They are holomorphic across every seam by Morera's theorem: split any triangle meeting a seam into its upper and lower pieces, displace the common edge by a small distance inside the strips, apply Cauchy's theorem separately, and pass to the limit using continuity; the two edge integrals cancel. A compact triangle meets only finitely many seams, so the same argument covers every triangle. Thus each matrix element is a bounded entire function, and Liouville's theorem makes it constant. It follows that the operator $\tilde{F}$ itself is constant. Its value at $z=0$ is $T(a)$.

For real $t$, constancy gives ${\mathrm{\Delta }}^{it}T(a{e}^{2\pi t}){\mathrm{\Delta }}^{-it}=T(a)\mathrm{.}$ Replace $a$ by $b{e}^{-2\pi t}>0$ to obtain ${\mathrm{\Delta }}^{it}T(b){\mathrm{\Delta }}^{-it}=T(e^{-2\pi t}b)(b>0)\mathrm{.}$ The top boundary and the same constancy give $\vartheta (T(a))=T(a)$, equivalently $JT(a)J=T(-a)$. Taking adjoints extends both identities from positive parameters to negative ones; at zero both are immediate. Hence the asserted two commutation relations hold for ALL real $a,t$.