Section 3 9 October 2026

Temporal dressing in the physical form dual

Jeremy Rodgers · Independent Researcher
doi:10.5281/zenodo.23259571

Original PDF ↗LaTeX source Section Markdown Full Markdown All contents

Reading position 4 of 16

3 Temporal dressing in the physical form dual

Response estimates must keep the input that actually drives the complement. For an admitted fixed block decomposition $\mathcal H_P\oplus\mathcal H_Q$ , the equation

i\dot q=Aq+Lp

uses the actual $p=P\Psi$ , including its feedback through $q$ , source phases and clock dynamics. An incident population or a frozen electronic amplitude is not a substitute for this vector. Throughout this section energies are in the units of $i\dot q=Aq+f$ ; a physical Hamiltonian must be divided by $\hbar$ if time is measured in seconds.

Let $\mathcal V\subset\mathcal H\subset\mathcal V^*$ be a dense separable Hilbert form triple, with the usual pivot identification. A positive form $K\ge I$ identifies $\mathcal V$ with $\operatorname{Dom}K^{1/2}$ and gives the dual norm $\|f\|_{\mathcal V^*,K}=\|K^{-1/2}f\|_{\mathcal H}$ . Here the square root on a dual vector is its continuous extension, rather than an assertion that $f\in\mathcal H$ . We use the same letter for a self-adjoint operator and its continuous form map $\mathcal V\to\mathcal V^*$ when the domain makes the meaning unambiguous.

Assumption 3.1 (Common physical form and relative work)

On $[0,T]$ , $A(t)$ is associated with a closed Hermitian form $a_t$ on the same $\mathcal V$ . There is a real bounded absolutely continuous scalar $c(t)$ such that

k_t=a_t+c(t)\langle\cdot,\cdot\rangle,\qquad K(t)=A(t)+c(t)\ge I.

The $k_t$ norms are uniformly equivalent to one fixed form norm. For fixed $u,v\in\mathcal V$ , $k_t(u,v)$ is absolutely continuous, with a strongly measurable form derivative and its integral identity, satisfying

|\dot k_t(u,v)| \le \nu(t)\|K(t)^{1/2}u\|\|K(t)^{1/2}v\|, \qquad \nu\ge0,\quad \nu\in L^1(0,T).

(3.1)

Strong measurability here includes that the derivative applied to any fixed form vector is measurable in $\mathcal V^*$ .

The shift is an estimate of positivity, not a spectral gap at an incident energy. Nor does this assumption require an operator-norm derivative of a point-Coulomb force. Classical common-form evolution under smoother form hypotheses goes back to Kisyński [8]. We give the conforming argument for the precise absolutely continuous relative-work assumptions used here.

Lemma 3.2 (Homogeneous common-form propagation)

Under Assumption 3.1 there is a unique unitary propagator $U(t,s)$ on $\mathcal H$ . It preserves $\mathcal V$ and, for $s\le t$ ,

\|K(t)^{1/2}U(t,s)v\| \le \exp\!\left(\frac12\int_s^t\nu(r)\,dr\right) \|K(s)^{1/2}v\|,\qquad v\in\mathcal V.

(3.2)

For $v\in\mathcal V$ its trajectory is continuous in $\mathcal H$ , bounded in $\mathcal V$ , and solves $i\partial_t U(t,s)v=A(t)U(t,s)v$ in $\mathcal V^*$ .

Proof

Fix a reference form $K_*$ and let $P_n=1_{[1,n]}(K_*)$ . These need not have finite rank. They are contractions in its form and dual norms, converge strongly there, and their ranges lie in $\mathcal V$ . On $\mathcal H_n=P_n\mathcal H$ , the restricted forms $K_n(t)$ and $A_n(t)=K_n(t)-c(t)I$ are bounded self-adjoint operators. Uniform equivalence gives their local uniform operator bounds. The integral form identity and (3.1) give absolute continuity in operator norm on each such subspace. The integral equation for the bounded-operator ODE is solved by successive approximations on short intervals and concatenation; its Hermitian generator gives a unitary evolution.

Let $u_n(s)=P_nv$ . Differentiation of its form energy yields

\frac d{dt}\langle u_n,K_nu_n\rangle =\dot k_t(u_n,u_n).

The two evolution terms cancel because $A_n=K_n-cI$ . Applying (3.1) and Gronwall gives (3.2) for $u_n$ , with $P_nv$ at the entrance. In particular $u_n$ is uniformly bounded in $\mathcal V$ . Its derivative, viewed in the full dual by testing against $P_nw$ , is uniformly bounded in $\mathcal V^*$ .

Weak compactness, followed by a diagonal argument on a countable dense set of form test vectors, gives a limit $u$ weakly continuous in $\mathcal H$ and weakly bounded in $\mathcal V$ , with $u(s)=v$ . In the integrated equation each test $P_nw$ converges strongly in $\mathcal V$ ; uniform boundedness of the forms therefore gives $i\dot u=A(t)u$ in $\mathcal V^*$ . At every fixed time weak lower semicontinuity of $k_t$ gives the asserted energy bound. This passage uses conforming compressions of the same forms.

We recall explicitly the norm fact that makes this weak construction unique. If $w\in L^2(\mathcal V)$ and $\dot w\in L^2(\mathcal V^*)$ , then

\|w(t)\|^2-\|w(s)\|^2 =2\operatorname{Re}\int_s^t\langle\dot w(r),w(r)\rangle\,dr,

(3.3)

and $w$ has a continuous $\mathcal H$ representative. One obtains this by time smoothing: for smooth $\mathcal V$ -valued functions it is the product rule; Cauchy–Schwarz in the dual pairing passes the integral under convergence in $L^2(\mathcal V)$ and $W^{1,2}(\mathcal V^*)$ . The same product rule, integrated from a time whose norm is bounded by its time average, bounds the supremum of the $\mathcal H$ norm by these two space norms. It consequently supplies continuous traces and justifies the endpoint passage as well.

Apply (3.3) to the limit and to differences of solutions. Hermiticity gives $\operatorname{Re}\langle-iA(t)w,w\rangle=0$ , proving norm conservation and uniqueness before any strong-convergence claim. Construction backwards from any terminal form vector gives an inverse evolution. Density extends the isometries to mutually inverse unitaries on $\mathcal H$ . Uniqueness gives their composition rule and strong continuity. The displayed energy bound proves invariance of $\mathcal V$ . It also gives weak measurability there; separability gives the strong measurability needed for the form-valued integrals below.

□

Theorem 3.3 (Ordered inverse dressing for dual forcing)

Under Assumption 3.1, let $q_0\in\mathcal V$ and $f\in W^{1,1}(0,T;\mathcal V^*)$ . Define

\ell(t)=K(t)^{-1/2}f(t),\qquad h(t)=K(t)^{-1/2}\dot f(t),\qquad N(t)=\int_0^t\nu(s)\,ds .

Then $i\dot q=A(t)q+f$ , $q(0)=q_0$ , has a unique solution in $C([0,T];\mathcal H)\cap L^\infty(0,T;\mathcal V)$ . The inverse lift $g=K^{-1}f$ belongs to $W^{1,1}(0,T;\mathcal V)$ and obeys the ordered identity

\dot g=K^{-1}\dot f-K^{-1}\dot K K^{-1}f,\qquad K^{1/2}\dot g=h-C_t\ell,\qquad C_t=K^{-1/2}\dot K K^{-1/2}.

(3.4)

For $z=q+g$ and $\mathcal E_z(t)=\|K(t)^{1/2}z(t)\|$ ,

\begin{align}\mathcal E_z(t) &\le e^{N(t)/2}\left[ \mathcal E_z(0)+ \int_0^t e^{-N(s)/2} \{(|c|+\nu)\|\ell\|+\|h\|\}(s)\,ds\right], \tag{3.5}\\ \|K(t)^{1/2}q(t)\|&\le\mathcal E_z(t)+\|\ell(t)\|. \tag{3.6}\end{align}

The actual entrance term is $\mathcal E_z(0)=\|K(0)^{1/2}(q_0+K(0)^{-1}f(0))\|$ .

Proof

Uniform equivalence and coercivity make each $K(t)$ a bounded isomorphism $\mathcal V\to\mathcal V^*$ . The exact inverse difference identity is

K(t)^{-1}-K(s)^{-1} =-K(t)^{-1}[K(t)-K(s)]K(s)^{-1}.

The relative-work bound makes its norm difference bounded by a constant times $\int_s^t\nu$ . For a fixed dual vector, the resulting inverse path is absolutely continuous into the Hilbert space $\mathcal V$ . Strong differentiation of the form identity on fixed vectors, the inverse difference identity, and uniform boundedness therefore give $(K^{-1})'f=-K^{-1}\dot K K^{-1}f$ on fixed vectors. For completeness, one can choose a common full-measure set first on a countable dense set of form vectors, then use the integrable relative bound and Lebesgue differentiation to extend the difference quotient to each vector. Approximating the absolutely continuous path $f$ by simple derivatives proves the product rule for $K^{-1}f$ . This establishes (3.4) in $\mathcal V$ . Its second equality is a form-Riesz identification; no derivative of $K^{-1/2}$ has been taken.

Since $A=K-cI$ , direct substitution gives

i\dot z=A(t)z+r,\qquad r=cg+i\dot g\in L^1(0,T;\mathcal V).

(3.7)

Indeed

\|K^{1/2}r\|\le (|c|+\nu)\|\ell\|+\|h\|.

These terms are integrable: $f$ is bounded in the fixed dual norm, $\dot f$ is integrable, and the form norms are uniformly equivalent.

Use Lemma 3.2 to define the form-valued Duhamel integral

z(t)=U(t,0)(q_0+g(0)) -i\int_0^tU(t,s)r(s)\,ds.

Its form norm is bounded by the right side of (3.5). Fubini in the form-dual weak equation verifies (3.7); the usual Hilbert-space Duhamel integral is the same vector, so $z$ is continuous in $\mathcal H$ . Subtracting $g$ gives the asserted solution $q$ . The difference of any two such solutions has derivative $-iA(t)w$ in $L^\infty(\mathcal V^*)$ ; the norm chain rule makes it zero when its entrance is zero. This proves uniqueness and both inequalities.

□

The estimate trades temporal regularity in the form dual for spatial first-form control. It does not require $f\in\mathcal V$ or $Af\in\mathcal H$ . For an actual coupling $f=L(t)p(t)$ , sufficient explicit input contracts are

\begin{align}\|K^{-1/2}LF^{-1}\|&\le b(t),& \|K^{-1/2}\dot L F^{-1}\|&\le b_1(t),\notag\\ \|\ell\|&\le b\|Fp\|,& \|h\|&\le b_1\|Fp\|+b\|F\dot p\|. \tag{3.8}\end{align}

Here $F\ge I$ is a fixed input graph and the products and derivatives must hold on their stated domains. The last quantity is a property of the coupled input, not of a frozen occupation. A changing $F$ adds its own derivative or commutator.

Corollary 3.4 (Retained internal right clock)

For a Hilbert–Schmidt factor satisfying $i\dot q=Aq-qD+f$ , with fixed bounded Hermitian right clock $D$ , the same argument applies with

r=K^{-1}\{i\dot f+cf+fD-i\dot K K^{-1}f\}.

In (3.5) its inhomogeneous norm can be bounded by $\|K^{-1/2}(i\dot f+cf+fD)\|_{\rm HS} +\nu\|\ell\|_{\rm HS}$ .

Proof

With $z=q+g$ , substitution gives $i\dot z=Az-zD+cg+gD+i\dot g$ . Right multiplication by the clock unitary removes $-zD$ and preserves Hilbert–Schmidt norms, including the physical left form norm. Apply the preceding Duhamel estimate and substitute (3.4).

□

Only a bounded physical-coordinate-independent internal clock is covered by this corollary. An unbounded right source clock needs its own graph contract. A positional purifier cannot be changed into an internal index just because the Hilbert-space expressions have the same size. The scalar $c$ is not free: it appears in the full combination $i\dot f+cf+fD$ , while $\dot c$ already belongs to $\dot K$ . Phases may cancel a term only through that complete physical combination.

3.1 Two sharp boundaries of the temporal argument

Example 3.5 (Continuous dual forcing need not give a wave)

On $\ell^2(n\ge2)$ let $A_n=n^2$ , $K_n=n^2+1$ , and $\ell_n(t)=n^{-1}e^{-in^2t}$ . Dominated convergence makes $\ell(t)$ continuous in $\ell^2$ ; hence $f=K^{1/2}\ell$ is continuous in $\mathcal V^*$ . Its zero-initial causal solution has components

q_n(t)=-it\,\frac{\sqrt{n^2+1}}n e^{-in^2t}.

For every $t>0$ these are not square summable. Thus a bounded form-dual coupling and continuous input alone do not give a physical Hilbert-space response. The time-derivative hypothesis in Theorem 3.3 fails.

Proof

The dual group is diagonal, so substitution in the causal integral gives the displayed formula. Its component magnitude tends to $t$ . On the other hand the derivative of $\ell_n$ has magnitude $n$ , so it cannot supply the required integrable dual derivative of $f$ .

□

Proposition 3.6 (Why an inverse square root is not the lift)

The relative form-speed condition does not give a dimension-independent bound on $K^{1/2}(K^{-1/2})'$ . It does give an ordered bounded inverse derivative as in (3.4).

Proof

In a finite eigenbasis, write $\dot K=K^{1/2}CK^{1/2}$ and $K_{ii}=\lambda_i>0$ . Differentiate $(K^{1/2})^2=K$ and then its inverse; the Sylvester equation gives

\begin{aligned}(K^{-1/2})'_{ij} &=-\frac{C_{ij}}{\sqrt{\lambda_i}+\sqrt{\lambda_j}},\\ [K^{1/2}(K^{-1/2})']_{ij} &=-\frac{\sqrt{\lambda_i}C_{ij}} {\sqrt{\lambda_i}+\sqrt{\lambda_j}}. \end{aligned}

The first kernel equals $-\int_0^\infty e^{-sK^{1/2}}C e^{-sK^{1/2}}\,ds$ , so its norm is at most $\|C\|/(2\sqrt{\lambda_{\min}})$ . The warning concerns the second, weighted derivative.

Take $\lambda_j=4^j$ and the Hermitian matrix $C_{ij}=i/[\pi(i-j)]$ for $i\ne j$ , $C_{ii}=0$ , on indices $1,\ldots,N$ . Its norm is at most one: it is a compression of the convolution operator on $\ell^2(\mathbb Z)$ whose Fourier multiplier is, up to its sign, $(\theta-\pi)/\pi$ on $(0,2\pi)$ . This follows by integrating the Fourier series of the sawtooth, or by Abel summing $\sum_{d\ne0}e^{id\theta}/d=i(\pi-\theta)$ . For the unit vector $v=N^{-1/2}(1,\ldots,1)$ , pairing the two off-diagonal entries at separation $d$ gives

\left|\left\langle v, K^{1/2}(K^{-1/2})'v\right\rangle\right| =\frac1{\pi N}\sum_{d=1}^{N-1} \frac{(N-d)\tanh(d\log2/2)}d .

For $2\le d\le N/2$ , the hyperbolic tangent is at least $3/5$ and $(N-d)/N\ge1/2$ . The harmonic sum diverges, proving the assertion. These are actual positive paths $K(s)=K(0)^{1/2}(I+sC)K(0)^{1/2}$ , $|s|\leq1/2$ . Since $K(0)\geq4I$ , these obey $K(s)\geq2I$ ; their relative speed is bounded by $(1-|s|)^{-1}\leq2$ . The ordered full inverse in the theorem avoids this weighted-square-root obstruction.

□