Full arbitrary-algebra and faithful n.s.f. hypotheses are retained.

Entire modular elements and a finite-domain strip criterion

CC0-1.0 to the extent of rights held.

Let $\varphi$ be a faithful normal semifinite weight on an arbitrary von Neumann algebra $M$. All Hilbert inner products are linear in the first variable. Write $N=\mathfrak n_\varphi$, $m=\mathfrak m_\varphi$, $A=N\cap N^*$, and $(H,\pi,\Lambda)$ for its faithful normal GNS representation. Put $S=J\Delta^{1/2}$, $F=S^*=J\Delta^{-1/2}$, and $U_t=\Delta^{it}$.

The exact earlier written proofs are GW1, GW2, GW3, GW4, GW5 for the finite algebra and the increasing finite contraction net; WR3, WR4 for the complete closed involution and left-bounded vectors; MW3, MW4 for covariance and the recovered modular group; KT3, KT4 for the analytic algebra and every spectral graph core; CI1, CI2, CI3 for the polar and adjoint domains; ST2 ultraweak image and inverse and NF5 arbitrary normal-weight GNS continuity; BC1, BC2, BC3, BC4 for every balanced-matrix domain and finite identity; AS1, AS2 for bounded strips and weak-star analytic reconstruction; CP1, CP2, CP3, CP4, CP5, CP6 for concrete vector series; CF8 Hilbert–Riesz; and SF0, SB0, SB1, SB2, SB3, SB4, SB5, SB6, SF4 for the full spectral calculus and scalar holomorphy. ST2 applies to arbitrary faithful normal representations; no faithful-state or separability restriction is imposed here.

The finite-domain criterion is discussed in Hiai's author manuscript, Theorem 8.10 and Lemma 8.12, printed pp.72–75. The converse below first proves a closed intertwining relation, regularizes both endpoints, and uses an explicit rectangle estimate to construct the limiting strip.

MA1. The actual common core

Let $\mathcal T\subset A$ be KT-4's algebra generated by Gaussian regularizations, and put $\mathcal H_0=\Lambda(\mathcal T)$. KT17 gives

$\Delta^r\Lambda(x)=\Lambda(\sigma_{-ir}^\varphi(x)) \quad(x\in\mathcal T,\ r\in\mathbb R), \tag{MA1}$ with full spectral domains, and says that $\mathcal H_0$ is a graph core for every $\Delta^r$. Its vectors have norm-entire $U_t$-orbits. It is preserved by $S$, because $\mathcal T$ is a *-algebra. CI's reciprocal-power identities give $J=\Delta^{1/2}S$ on $D(S)$; hence

$J\Lambda(x)=\Delta^{1/2}\Lambda(x^*) =\Lambda(\sigma_{-i/2}^\varphi(x^*))\in\mathcal H_0. \tag{MA2}$ Thus $J$, $F=J\Delta^{-1/2}$, and every complex power in (MA1) preserve $\mathcal H_0$. In particular $FS=\Delta$, $SF=\Delta^{-1}$, $S^2=F^2=1$, when applied to these vectors. These assertions concern genuine operator domains, rather than formal products of unbounded operators.

MA2. Entire elements act on the full finite ideal on the right

Suppose $a\in M$ has a norm-entire modular orbit $a(z)$, with $a(t)=\sigma_t^\varphi(a)$. Scalar identity and the real group law give $a(z+t)=\sigma_t^\varphi(a(z))$. Write $\sigma_z^\varphi(a)=a(z)$. Each horizontal band has a finite bound on $\|a(z)\|$, since real translations are isometries and the imaginary parameter ranges over a compact interval.

For every real $r$ and $\xi\in\mathcal H_0$,

$\pi(a)\xi\in D(\Delta^r),\qquad \Delta^r\pi(a)\xi =\pi(\sigma_{-ir}^\varphi(a))\Delta^r\xi. \tag{MA3}$ Here is the domain proof. Put $L=\log\Delta$ and $p_k=1_{[-k,k]}(L)$. The real covariance from MW says $p_k\Delta^{it}\pi(a)\xi =p_k\pi(a(t))\Delta^{it}\xi.$ Both sides are entire in $t$ replaced by $z$: on the left $p_k\Delta^{iz}$ is a bounded entire multiplier, and on the right $\Delta^{iz}\xi$ is the entire core orbit. Scalar identity extends the equality to $z=-ir$. Thus $p_k\Delta^r\pi(a)\xi =p_k\pi(a(-ir))\Delta^r\xi.$ The notation on the left is the bounded compact-spectral multiplier. The right sides converge in norm. The spectral domain criterion therefore puts $\pi(a)\xi$ in $D(\Delta^r)$ and proves (MA3). For an arbitrary $\xi\in D(\Delta^{1/2})$, approximate in graph norm by $\mathcal H_0$; boundedness of $\pi(a)$ and $\pi(a(-i/2))$, followed by closedness of $\Delta^{1/2}$, proves (MA3) also on that whole domain when $r=1/2$.

Put $c=\sigma_{-i/2}^\varphi(a)$. If $x\in A$, then $\Lambda(x^*)\in D(\Delta^{1/2})$, so $\pi(a)\Lambda(x^*)$ is in that domain. It is also left bounded, with multiplier $\pi(ax^*)$, because $ax^*\in N$. WR-4's exact identity $B_l\cap D(S)=\Lambda(A)$ now shows $ax^*\in A$, and identifies its $S$-image. Applying (MA3) and $S=J\Delta^{1/2}$ yields

$\Lambda(xa^*)=J\pi(c)J\Lambda(x)\qquad(x\in A). \tag{MA4}$

If $h\in M_+$ and $\varphi(h)<\infty$, then $h^{1/2}\in A$, so (MA4) gives

$\varphi(aha^*)\leq \|c\|^2\varphi(h). \tag{MA5}$ For every $h\geq0$, including infinite values, the same inequality holds with any fixed $k\geq\max(1,\|c\|)$ replacing $\|c\|$. In particular $Na^*\subset N$, and right multiplication by $a^*$ induces a bounded operator on the GNS space. Formula (MA4) holds on all of $N$: if $u_i$ are GW's finite positive contractions increasing to $1$, then $u_ix\in m\subset A$, $\Lambda(u_ix)\to\Lambda(x)$, and $\Lambda((u_ix)a^*)=\Lambda(u_i(xa^*))\longrightarrow\Lambda(xa^*).$ Taking limits in (MA4) proves this assertion without presupposing continuity of the unbounded GNS map in an operator topology.

Let $b=\sigma_{-i}^\varphi(a)$. Identity on the real line, followed by holomorphic continuation, gives $\sigma_{-i/2}^\varphi(b^*)=c^*.$ Applying the preceding result to $b^*$ shows that right multiplication by $b$ on $N$ has GNS operator $J\pi(c^*)J=(J\pi(c)J)^*$. Therefore, for $x,y\in N$,

$\begin{split} \varphi_0(ay^*x) &=\langle\Lambda(x),\Lambda(ya^*)\rangle\\ &=\langle(J\pi(c)J)^*\Lambda(x),\Lambda(y)\rangle =\varphi_0(y^*xb). \tag{MA6} \end{split}$ All products in this calculation belong to $m$: $ay^*x=(ya^*)^*x$ and $y^*xb=y^*(xb)$, with all four factors in $N$. Since $m=\operatorname{span}N^*N$, we have proved

$Na^*\subset N,\quad Nb\subset N,\quad \varphi_0(az)=\varphi_0(zb)\quad(z\in m). \tag{MA7}$

MA3. A converse strip from the finite-domain identity

Suppose $a,b\in M$ satisfy (MA7). We construct a bounded weak-star continuous strip function $B$, holomorphic inside, with

$B(t)=\sigma_t^\varphi(a),\qquad B(t-i)=\sigma_t^\varphi(b),\qquad \|B(z)\|\leq C:=\max(\|a\|,\|b\|). \tag{MA8}$

No analytic property of $a$ or $b$ is assumed. First we turn the finite identity into a relation between closed operators. We then regularize that relation, obtaining uniformly bounded entire strip functions whose boundary values converge.

The finite identity determines a closed intertwining relation. On $\Lambda(A)$ define the two linear maps $Q_a\Lambda(x)=\Lambda(xa^*),\qquad Q_b\Lambda(x)=\Lambda(xb).$ They are well defined by faithfulness and the two finite-ideal hypotheses. Both $xa^*$ and $xb$ lie in $A$: their membership in $N$ is assumed, and their adjoints are in $N$ by its left-ideal property. For $x,y\in A$, the finite identity gives $\langle\Lambda(x),Q_a\Lambda(y)\rangle =\varphi_0(ay^*x) =\varphi_0(y^*xb) =\langle Q_b\Lambda(x),\Lambda(y)\rangle .$ This relation does not assume that either $Q_a$ or $Q_b$ is bounded.

Let $\xi,\zeta\in\mathcal H_0$, the common core in MA1. The vectors $S\xi,S\zeta$ belong to $\Lambda(A)$. Also $\pi(a)\zeta,\pi(b^*)\xi\in\Lambda(A)$, by the same finite-ideal tests, and $Q_aS\zeta=S\pi(a)\zeta,\qquad Q_bS\xi=S\pi(b^*)\xi .$ Applying the preceding adjoint relation at $S\xi,S\zeta$ gives $\langle S\xi,S\pi(a)\zeta\rangle =\langle S\pi(b^*)\xi,S\zeta\rangle .$ Since $\xi,\zeta\in D(\Delta)=D(S^*S)$, the adjoint definition, with the order reversed for $S$, rewrites this equality as $\langle\pi(a)\zeta,\Delta\xi\rangle =\langle\pi(b)\Delta\zeta,\xi\rangle .$ The core $\mathcal H_0$ is a graph core for $\Delta$. Extending in the test vector $\xi$ to $D(\Delta)$, the definition of $\Delta^*=\Delta$ shows that $\pi(a)\zeta\in D(\Delta)$ and that its image is $\pi(b)\Delta\zeta$. Finally approximate any $\zeta\in D(\Delta)$ in graph norm by $\mathcal H_0$. Boundedness of $\pi(a),\pi(b)$ and closedness of $\Delta$ give the full-domain relation

$\pi(a)D(\Delta)\subseteq D(\Delta),\qquad \Delta\pi(a)\zeta=\pi(b)\Delta\zeta \quad(\zeta\in D(\Delta)). \tag{MA9}$

Regularization preserves both endpoints. Write $p_n(t)=\sqrt{n/\pi}\,e^{-nt^2}$, and set $a_n=\int_{\mathbb R}p_n(t)\sigma_t^\varphi(a)\,dt,\qquad b_n=\int_{\mathbb R}p_n(t)\sigma_t^\varphi(b)\,dt .$ These are the ultraweak integrals of KT-3; $a_n$ is norm entire, $\|a_n\|\leq\|a\|$, and $\|b_n\|\leq\|b\|$. Covariance and the commutation of $U_t$ with $\Delta$ transport MA9 to $\Delta\pi(\sigma_t^\varphi(a))\zeta =\pi(\sigma_t^\varphi(b))\Delta\zeta \quad(\zeta\in D(\Delta)).$ Integrate this equality in the graph of $\Delta$. More explicitly, compact-interval Riemann sums approximate both Hilbert-valued integrands, and their tails are bounded by $\|a\|\|\zeta\|\int p_n$ and $\|b\|\|\Delta\zeta\|\int p_n$ over the omitted region. The real orbits are strongly continuous by MW. The vector integrals in SF-3, followed by closedness of $\Delta$, therefore give $\Delta\pi(a_n)\zeta=\pi(b_n)\Delta\zeta \quad(\zeta\in D(\Delta)).$ On the other hand, MA2's compact-spectral argument for equation MA3 at $r=1$, followed by the same graph approximation, gives $\Delta\pi(a_n)\zeta=\pi(\sigma_{-i}^\varphi(a_n))\Delta\zeta$ on $D(\Delta)$. The range of the injective self-adjoint operator $\Delta$ is dense. Comparing the bounded operators on that range, then using faithfulness of $\pi$, yields

$\sigma_{-i}^\varphi(a_n)=b_n,\qquad B_n(z):=\sigma_z^\varphi(a_n). \tag{MA10}$

The real translation rule gives $B_n(t)=\sigma_t^\varphi(a_n)$ and $B_n(t-i)=\sigma_t^\varphi(b_n)$. For each fixed $n$, norm analyticity and real isometry give a finite bound on the entire closed strip. AS1, applied to every normal functional, then gives the uniform bound $\|B_n(z)\|\leq C\qquad(-1\leq\operatorname{Im}z\leq0).$

Convergence reaches both boundary lines. Fix $f\in M_*$, and put $h_n(z)=f(B_n(z))$. On each bounded real interval the upper-edge values converge uniformly to $f(\sigma_t^\varphi(a))$, and the lower-edge values converge uniformly to $f(\sigma_t^\varphi(b))$. Indeed both are Gaussian averages of a bounded continuous scalar orbit; uniform continuity on a slightly larger compact interval and the Gaussian tail estimate prove these assertions.

Here is an explicit estimate that propagates this convergence across the closed strip. If $h$ is continuous on the rectangle $[-T,T]\times[-1,0]$, holomorphic inside, has $|h|\leq M_0$ everywhere, and has $|h|\leq\varepsilon$ on the two horizontal edges, then for $z=t+iv$, $|h(t+iv)| \leq\varepsilon+ \frac{M_0\cosh t\,\cos(v+1/2)}{\cosh T\,\cos(1/2)} .$ To prove it, apply the scalar maximum principle to the real part of $e^{i\theta}h(z)-\varepsilon -\frac{M_0\cosh(z+i/2)}{\cosh T\,\cos(1/2)}$ for each real $\theta$. Its real part is nonpositive on the horizontal sides. On the vertical sides the last term has real part at least $M_0$, because $\cos(v+1/2)\geq\cos(1/2)>0$. The maximum principle for real parts follows from SF4's maximum-modulus theorem by exponentiation. Choosing $\theta$ at the point under consideration gives the estimate.

Apply this to $h_n-h_m$, with $M_0=2C\|f\|$. For a fixed $L$, first take $T>L$ large enough that $M_0\cosh L/(\cosh T\cos(1/2))$ is small, then take $n,m$ large enough for both boundary differences on $[-T,T]$ to be small. Thus $h_n$ is uniformly Cauchy on $[-L,L]\times[-1,0]$. Its limit $h_f$ is continuous there and holomorphic inside, by the scalar Cauchy formula.

For each $z$, the map $f\mapsto h_f(z)$ is linear and bounded by $C\|f\|$. The proved duality $M=(M_*)^*$ in CP6 therefore defines a unique $B(z)\in M$. Every scalar pairing is the continuous, holomorphic limit just constructed. Consequently $B$ is weak-star continuous on the closed strip, weak-star holomorphic inside, and bounded by $C$. Its two edges are the asserted limits, proving MA8. This construction takes its values directly in $M$; it requires neither an extension of the finite weight to undefined products nor a prior analytic orbit for $a$.

MA4. The mixed-weight form with every domain specified

Let $\varphi,\psi$ be faithful normal semifinite weights on the same $M$. Put $Z_{\varphi,\psi} =\operatorname{span}(\mathfrak n_\varphi^*\mathfrak n_\psi), \qquad \sigma_t^{\Theta(\varphi,\psi)}(xE_{21}) =\beta_t^{\psi,\varphi}(x)E_{21}.$ BC-1–3 prove the full balanced domains and the isometry group $\beta$. Suppose $a$ has a norm-entire $\beta$-orbit, and put $b=\beta_{-i}^{\psi,\varphi}(a)$, $c=\beta_{-i/2}^{\psi,\varphi}(a)$. Apply MA2 to $aE_{21}$ in $M_2(M)$. For $k\geq\max(1,\|c\|)$ the resulting conclusions are

$\psi(aha^*)\leq k^2\varphi(h),\qquad \varphi(b^*hb)\leq k^2\psi(h)\quad(h\in M_+), \tag{MA11}$

$\mathfrak n_\varphi a^*\subset\mathfrak n_\psi,\qquad \mathfrak n_\psi b\subset\mathfrak n_\varphi, \tag{MA12}$

$\psi_0(az)=\varphi_0(zb)\qquad(z\in Z_{\varphi,\psi}). \tag{MA13}$ To check the indices, $aE_{21}$ sends the first diagonal positive entry $h$ to the second one $aha^*$; $(bE_{21})^*$ sends the second to $b^*hb$ in the first. The half-shift of the latter is $c^*E_{12}$, so the same $k$ works. In MA7 for the balanced weight, the two possibly nonzero diagonal products for a matrix $X$ are $a x_{12}$ and $x_{12}b$; BC3 says exactly $x_{12}\in Z_{\varphi,\psi}$. Also (MA12) makes these products elements of the respective finite algebras, so their finite weight values in (MA13) are defined.

Conversely, if $a,b\in M$ satisfy (MA12)–(MA13), the balanced matrices $aE_{21},bE_{21}$ satisfy (MA7) on the entire balanced finite algebra. The domain statement follows directly from BC3's column description of $N_\Theta$, and the scalar identity follows from BC4 and the preceding two diagonal products. MA3 supplies a balanced strip. Its entries other than $(2,1)$ vanish, by zero-edge uniqueness applied to their coefficients. The $(2,1)$ entry is consequently a bounded ultraweakly continuous strip $B$, holomorphic inside, satisfying

$B(t)=\beta_t^{\psi,\varphi}(a),\qquad B(t-i)=\beta_t^{\psi,\varphi}(b). \tag{MA14}$ This is the exact converse used later. In particular no a priori boundedness of a right GNS multiplier is needed in the converse beyond the domains and finite identities explicitly stated.