Original complete reconstruction by GPT-6 Astra (OpenAI), Ultra; reader integration by GPT-6.1 Sol (OpenAI), Ultra. Self-checked by the writing AI. Full arbitrary-algebra and faithful n.s.f. hypotheses are retained.

Entire modular elements and a finite-domain strip criterion

Fresh local reconstruction, GPT-6 Astra (OpenAI), Ultra, 2026-10-04. 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 free development source is Hiai's author manuscript, Theorem 8.10 and Lemma 8.12, printed pp.72–75. We prove the part needed for entire elements directly, with the full domains. We do not import the manuscript's general analytic-generator results.

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

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

$B(t)=\sigma_t^\varphi(a),\qquad B(t-i)=\sigma_t^\varphi(b),\qquad \|B(z)\|\leq\max(\|a\|,\|b\|). \tag{MA8}$ No prior analytic property of $a$ or $b$ is assumed in this direction.

Take $\xi,\eta\in\mathcal H_0$, and choose $x,y\in\mathcal T$ with $\Lambda(x)=S\xi,\quad \Lambda(y)=F\eta.$ MA1 gives $\Delta\xi=F\Lambda(x),\quad \xi=S\Lambda(x),\qquad \Delta^{-1}\eta=S\Lambda(y),\quad \eta=F\Lambda(y).$ The elements $ya^*$ and $xb$ belong to $A$. Indeed, membership in $N$ is a hypothesis, while their adjoints $ay^*$ and $b^*x^*$ are in the left ideal $N$. The conjugate-linear adjoint convention is $\langle Su,v\rangle=\langle Fv,u\rangle$. Applying it twice, using $S^2=F^2=1$ on their domains and (MA7), gives

$\begin{split} \langle\pi(a)\Delta^{-1}\eta,\Delta\xi\rangle &=\langle S\Lambda(ya^*),F\Lambda(x)\rangle\\ &=\langle\Lambda(x),\Lambda(ya^*)\rangle =\varphi_0(ay^*x)\\ &=\varphi_0(y^*xb) =\langle\Lambda(xb),\Lambda(y)\rangle\\ &=\langle F\Lambda(y),S\Lambda(xb)\rangle\\ &=\langle F\Lambda(y),\pi(b^*)S\Lambda(x)\rangle =\langle\pi(b)\eta,\xi\rangle. \tag{MA9} \end{split}$ Every application of $S$ or $F$ is justified above or in MA1.

Define, on the closed lower unit strip,

$G_{\xi,\eta}(z) =\langle\pi(a)\Delta^{-iz}\eta,\Delta^{-i\bar z}\xi\rangle. \tag{MA10}$ This is holomorphic: the first vector is entire, the second is anti-entire, and the inner product is linear first. It is bounded on the entire strip for these fixed core vectors, since $|G_{\xi,\eta}(t+iv)| \leq\|a\|\,\|\Delta^v\eta\|\,\|\Delta^{-v}\xi\|, \qquad -1\leq v\leq0.$ The right side has a finite maximum independent of $t$. On the upper boundary, covariance gives $G_{\xi,\eta}(t)=\langle\pi(\sigma_t^\varphi(a))\eta,\xi\rangle.$ On the lower boundary, apply (MA9) to $U_{-t}\xi,U_{-t}\eta\in\mathcal H_0$; it gives $G_{\xi,\eta}(t-i)=\langle\pi(\sigma_t^\varphi(b))\eta,\xi\rangle.$ AS1 therefore bounds (MA10) by $\max(\|a\|,\|b\|)\|\xi\|\|\eta\|$. Bounded sesquilinear representation by the Hilbert Riesz theorem produces an operator $D(z)\in B(H)$ with these coefficients and the same norm bound.

Approximate arbitrary vectors by $\mathcal H_0$. The uniform operator bound makes their coefficients uniform limits on the entire strip of the core coefficients. They are continuous on the closed strip and holomorphic inside by the scalar Cauchy formula. Any concrete ultraweak functional is a summable vector series by CP; Cauchy–Schwarz and the uniform norm bound make the series tails uniform in $z$. Hence $D$ is ultraweakly continuous on the closed strip and ultraweakly holomorphic inside.

For $d'\in\pi(M)'$, every coefficient of $D(z)d'-d'D(z)$ is such a continuous strip function and vanishes on its real edge. SF4 zero-edge uniqueness makes it zero throughout the strip. Thus $D(z)\in\pi(M)''=\pi(M)$; the last equality is the actual weakly closed image assertion of ST-2. That same theorem makes $\pi^{-1}$ ultraweakly continuous. Set $B(z)=\pi^{-1}(D(z))$. It has exactly (MA8), the claimed topology, and the required analyticity.

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.