Modular Analyticity and Chaos Bounds

Complex modular time can bound the growth of a normalized correlator, but only after one proves both an analytic strip and a uniform norm bound in that strip. The resulting rate is dimensionless and belongs to modular flow. It becomes a physical Lyapunov bound only when an independent theorem identifies modular time with physical time and the correlator satisfies the dynamical assumptions used in a chaos diagnostic.

Required background. Modular KMS correlators fix the modular-flow sign, operator order, KMS boundary relation, and analytic domains used here.

Helpful background. Modular spectral measures explain why a finite-mode modular correlator is quasiperiodic and how continuum frequency support changes that conclusion.

The chapter’s differentiable-family map identifies modular-time response as one branch of the problem, while its independent validity gates keep strip analyticity, normalization, regulator control, and physical interpretation separate. Passing the analytic gate does not pass the others.

The strip-to-disk statement

Let $f$ be analytic on the symmetric strip

$$S_a:=\{z\in\mathbb C:\lvert\operatorname{Im}z\rvert<a\}$$

and suppose $\lvert f(z)\rvert\le1$ throughout $S_a$, with the same bound on controlled boundary limits. The conformal map $\zeta=\tanh(\pi z/4a)$ sends $S_a$ to the unit disk. Schwarz–Pick then gives, for real $x$,

$$\frac{\lvert f'(x)\rvert}{1-\lvert f(x)\rvert^2} \le\frac{\pi}{4a}.$$

This is the full-strip version of the complex-analysis step used in the thermal chaos argument Maldacena, Shenker, and Stanford 2016, §4.1, eqs. (17)–(18). Their physical problem is a half-strip and therefore also has a vertical boundary and a transient $\coth$ correction; the constant $2\pi/\beta$ is approached after a few thermal times rather than asserted at the boundary of the half-strip.

If, uniformly in a complex neighborhood of a real interval (so the differentiated remainder is controlled),

$$f(x)=1-\epsilon g(x)+O(\epsilon^2), \qquad g(x)>0,$$

then for increasing $g$,

$$\frac{d}{dx}\log g(x) \le\frac{\pi}{2a}+O(\epsilon).$$

The numerical rate comes entirely from the proven half-width $a$. Analyticity alone is insufficient: an analytic function that leaves the unit disk can have an arbitrarily large fitted exponent.

Thermal time is a special conversion

Use the modular convention

$$\sigma_s(A)=\rho^{is}A\rho^{-is}.$$

For $\rho=Z^{-1}e^{-\beta H}$ and the Heisenberg convention $A(t)=e^{iHt}Ae^{-iHt}$,

$$\sigma_s(A)=A(-\beta s).$$

Thus a dimensionless modular rate $\lambda_{\rm mod}$ converts to the magnitude $\lambda_{\rm phys}=\lambda_{\rm mod}/\beta$; the minus sign reverses the flow orientation, not the positive growth rate. For the quarter-density ordering $y=\rho^{1/4}$, the centered OTOC strip has $a=1/4$, so the conditional modular number $2\pi$ becomes $2\pi/\beta$.

The physical result needs much more than KMS periodicity. The thermal proof normalizes by a disconnected correlator plus a controlled factorization error, establishes bounds on all boundaries of a half-strip, and assumes a window after dissipation but before scrambling Maldacena, Shenker, and Stanford 2016, §4.2, eqs. (19)–(26). A generic subregion modular flow has no universal conversion to laboratory time and no spatial metric from which to infer a butterfly velocity.

Exactly solvable regulated modular OTOC

Consider one fermionic wave-packet mode, so its occupation algebra is $M_2$, with faithful Gaussian state

$$\rho=\begin{pmatrix}p&0\\0&1-p\end{pmatrix}, \qquad 0<p<1,$$

and bounded Majorana insertion $A=B=c+c^\dagger=\sigma_x$. Put $y=\rho^{1/4}$ and

$$F(z):=\operatorname{Tr} \left[yAy\,\sigma_z(B)\,yAy\,\sigma_z(B)\right], \qquad \sigma_z(B)=\rho^{iz}B\rho^{-iz}.$$

With $\omega=\log[p/(1-p)]$, direct matrix multiplication gives

$$F(z)=2\sqrt{p(1-p)}\cos(2\omega z).$$

This finite regulated answer is entire and quasiperiodic on the real modular axis; it has no positive late-time Lyapunov regime. To verify the strip theorem rather than merely inspect the real line, normalize it as

$$f_a(z):= \frac{F(z)}{2\sqrt{p(1-p)}\cosh(2\lvert\omega\rvert a)}.$$

Because $\lvert\cos(u+iv)\rvert\le\cosh\lvert v\rvert$, $\lvert f_a(z)\rvert\le1$ for $\lvert\operatorname{Im}z\rvert\le a$. This proves the boundary norm instead of assuming it.

For $p=1/4$, $a=1/4$, and $x_*=\pi/(4\log3)$, the check is fully numerical:

Quantity	Exact expression	Value
Strip normalization	$\cosh[(\log3)/2]$	$1.15470$
$f_a(x_*)$	$0$	$0$
Measured $\lvert f_a'(x_*)\rvert$	$2\log3/\cosh[(\log3)/2]$	$1.90285$
Schwarz–Pick cap	$\pi/(4a)$	$3.14159$
Cap minus measured derivative	—	$1.23874$

The stated domain is the whole matrix algebra, faithfulness is controlled by $p\in[\delta,1-\delta]$, the observable norm is one, and the analytic target has no truncation error. This verifies a modular strip inequality in a Gaussian mode; it does not demonstrate physical chaos.

Adversarial edge-bound failure

Take the same half-width $a=1/4$ and the analytic trial function

$$g_\epsilon(z)=1-\epsilon e^{3\pi z}, \qquad \epsilon=10^{-4}.$$

On $0\le x\le0.2$, its real values lie between $0.9999$ and $0.99934$, and an exact fit of $1-g_\epsilon(x)$ returns the apparently clean rate $3\pi>2\pi$. Yet on the upper strip edge, with $\delta_x=\epsilon e^{3\pi x}>0$,

$$\lvert g_\epsilon(x+ia)\rvert^2 =1+\delta_x^2-2\delta_x\cos(3\pi/4)>1.$$

The unit-disk hypothesis fails precisely where a real-axis fit cannot see it. Schwarz–Pick therefore licenses no growth bound for this trial function. This adversarial example also shows why dividing by a convenient real-axis value is not a substitute for proving a boundary norm.

Modular growth is not automatically chaos

An $e^{\pm2\pi s}$ component can be meaningful as a modular-response mode when its operator domain, zero-mode quotient, strip, and norm control have all been supplied. In holographic examples it can additionally encode bulk causal structure de Boer and Lamprou 2020, §§2–5. Neither fact promotes a generic modular slope to a many-body Lyapunov exponent.

A physical chaos claim must identify the physical observable, a parametric factorization error, a dissipation-to-scrambling window, the conversion between $s$ and $t$, and a regulator limit that preserves the strip. A butterfly velocity needs an additional notion of spatial support and a locality estimate. The comparison of information geometries is also a reminder that an analytic response coefficient is not automatically a Fisher or fidelity susceptibility.

Common pitfalls

Checking only the real axis. Real boundedness does not imply a strip-to-disk map. Test both horizontal edges and every additional boundary of a half-strip.

Calling $2\pi$ a physical exponent without units. It is a dimensionless modular number for $a=1/4$. A physical exponent needs a justified time conversion.

Inferring propagation from analyticity. A complex strip contains no distance, causal cone, or Lieb–Robinson input, so it cannot determine a butterfly velocity.

Exercises

1. Derive the strip derivative bound

For $\zeta(z)=\tanh(\pi z/4a)$, show that

$$\frac{\lvert\zeta'(x)\rvert}{1-\lvert\zeta(x)\rvert^2}=\frac{\pi}{4a}$$

on the real axis, and use Schwarz–Pick to obtain the bound for $f$.

Solution

Write $u=\pi x/(4a)$. Since $\zeta(x)=\tanh u$ and $\zeta'(x)=(\pi/4a)\operatorname{sech}^2u$,

$$1-\lvert\zeta(x)\rvert^2 =1-\tanh^2u =\operatorname{sech}^2u.$$

Their ratio is $\pi/(4a)$. The map $F=f\circ\zeta^{-1}$ sends the disk to itself. Schwarz–Pick says

$$\frac{\lvert F'(\zeta)\rvert}{1-\lvert F(\zeta)\rvert^2} \le\frac{1}{1-\lvert\zeta\rvert^2}.$$

Using $f'=F'(\zeta)\zeta'$ and evaluating at real $x$ gives $\lvert f'\rvert/(1-\lvert f\rvert^2)\le\pi/(4a)$.

2. Compute the Gaussian-mode correlator

Derive $F(z)=2\sqrt{p(1-p)}\cos(2\omega z)$ from the matrices above and explain why no late-time exponential can be extracted from the exact real-axis answer.

Solution

Let $q=1-p$. Then

$$\sigma_z(B)= \begin{pmatrix} 0&e^{i\omega z}\\ e^{-i\omega z}&0 \end{pmatrix}, \qquad yAy=(pq)^{1/4}\sigma_x.$$

Therefore

$$yAy\,\sigma_z(B) =(pq)^{1/4} \begin{pmatrix} e^{-i\omega z}&0\\0&e^{i\omega z} \end{pmatrix}.$$

Squaring and taking the trace gives $2\sqrt{pq}\cos(2\omega z)$. For real $z=s$, this is bounded and periodic when $\omega\ne0$, while it is constant when $\omega=0$. Neither case has an asymptotic interval with positive exponential growth.

3. Test an unlicensed exponent above 2π

For $g_\epsilon(z)=1-\epsilon e^{\lambda z}$, find a condition on $\lambda a$ that makes $\lvert g_\epsilon(x+ia)\rvert>1$ for every real $x$. Apply it to $\lambda=3\pi$ and $a=1/4$. What, if anything, can the fitted exponent establish?

Solution

With $\delta_x=\epsilon e^{\lambda x}>0$,

$$\lvert g_\epsilon(x+ia)\rvert^2 =1+\delta_x^2-2\delta_x\cos(\lambda a).$$

If $\cos(\lambda a)\le0$, the right side is strictly larger than one. For $\lambda a=3\pi/4$, this condition holds. The fitted rate is an algebraic property of the chosen real-axis function, but the strip theorem is inapplicable; it establishes neither a modular growth bound nor physical chaos. If a thermal identification were separately valid, only a licensed modular rate would convert by $\lambda_{\rm phys}=\lambda_{\rm mod}/\beta$.

References

• de Boer, Jan, and Lampros Lamprou. “Holographic Order from Modular Chaos.” Journal of High Energy Physics 2020, no. 6 (2020): 024. DOI; Open PDF.
• Maldacena, Juan, Stephen H. Shenker, and Douglas Stanford. “A Bound on Chaos.” Journal of High Energy Physics 2016, no. 8 (2016): 106. DOI; Open PDF.