The Riemann Hypothesis can be reformulated as an infinite sequence of explicit positivity tests. This is not a proof. It turns a statement about complex zeros into inequalities that can be calculated, falsified at finite order, and compared with operator or physical models. The completed zeta function is $$\xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s),$$ with $$\Xi(t)=\xi\left(\frac12+it\right).$$ The function $\Xi$ is even and real on the real axis. A nontrivial zero $\rho=\beta+i\gamma$ corresponds to $$t_\rho=\gamma-i\left(\beta-\frac12\right).$$ Therefore $$t_\rho\in\mathbb R \quad\Longleftrightarrow\quad \beta=\frac12,$$ and hence $$\boxed{\mathrm{RH}\quad\Longleftrightarrow\quad\text{every zero of }\Xi\text{ is real}.}$$ Folding the Zeros Because $\Xi$ is even, write $$\Xi(t)=\sum_{n\ge0}c_nt^{2n}.$$ Now define $$G(z)=\frac{\Xi(i\sqrt z)}{\Xi(0)}.$$ Although this contains $\sqrt z$, the even expansion makes $G$ entire: $$G(z)=...
In a previous post, we showed how, if the ultimate fate of our Universe is space empty of matter...but not quite....of energy (a de Sitter space), then this future cosmic event horizon (CEH) radius of our current, quantum, Universe set a natural maximum length-scale. Amazingly, the future CEH radius also defines the cosmological constant $\Lambda$.
For an observer at O inside the cosmic event horizon (CEH) with radius $l_{\Lambda}$, the universe can be divided into two sub-vacuums, $(A)$ inside the CEH, and $(B)$, outside. The horizon surface $\Sigma$ has entanglement entropy $S_{dS}$ and rest energy $E_H$
What about a Universal minimum length-scale? Quantum localisation plus gravitational collapse gives $\ell_{min}\gtrsim L_p$ (Planck length $L_p$) being a standard minimal length argument. Now, we can find this scale falls out, without needing to be assumed.
Horizon action
Barrow and Gibbons proposed, from pure classical GR, that the maximum tension sustainable by any physical system (e.g. consider a rope spanning a Schwarzschild horizon) is $F_{max}=c^4/4G$.
A Schwarzschild horizon of radius $R$, diameter $\ell=2R$, has energy
$$E=\frac{c^4R}{2G}=F_{\max}\ell.$$ What length does quantum mechanics associate with this force?
Euclidean Schwarzschild time has period
$$T_E=\frac{4\pi R}{c}, \qquad \omega_H=\frac{2\pi}{T_E}=\frac{c}{2R}.$$
Define the horizon action by $dI=dE/\omega_H$. Using $dE=c^4dR/(2G)$ gives
$$\boxed{ I=\frac{c^3A}{8\pi G} =\frac{ER}{c} =\frac{E\ell}{2c} }.$$
Since $S_{\mathrm{BH}}=k_BA/4L_P^2$,
$$\boxed{ \frac{I}{\hbar} = \frac{A}{8\pi L_P^2} = \frac{S_{\mathrm{BH}}}{2\pi k_B} }.$$
Euclidean periodicity supplies an angle $\theta\sim\theta+2\pi$. Introduce canonical coordinates
$$X=\sqrt{2I}\cos\theta, \qquad P=\sqrt{2I}\sin\theta,$$
so that $I=\frac12(X^2+P^2)$. Canonical quantisation, $[X,P]=i\hbar$, then gives
$$\boxed{ I_n=\left(n+\frac12\right)\hbar, \qquad n=0,1,2,\ldots }$$
The half-unit is the minimum expectation value allowed when the action-angle phase space is completed as a regular oscillator plane $$\langle \hat{I} \rangle = \frac{1}{2}\left( \hat{X}^{2} + \hat{P}^{2} \right) \ge \frac{\hbar}{2}$$ Periodicity alone fixes the spacing, but not necessarily this offset.
Horizon Spectrum
Substitution gives $$\boxed{ \begin{aligned} A_n &=4\pi(2n+1)L_P^2 \\ R_n &=\sqrt{2n+1}L_P \\ E_n &=\frac{\sqrt{2n+1}}{2} m_Pc^2 \\ S_n &=\pi(2n+1)k_B. \end{aligned} }$$ Adjacent states therefore satisfy $$\boxed{ \Delta I=\hbar \qquad \Delta A=8\pi L_P^2 \qquad \Delta S=2\pi k_B }$$The First Horizon
For $n=0$,
$$\boxed{ R_0=L_P, \qquad \ell_0=2L_P, \qquad E_0=\frac12m_Pc^2, \qquad I_0=\frac{\hbar}{2} }.$$
Thus the Planck radius need not be assumed separately. It follows from the zero-point horizon action. Since $E=F_{\max}\ell$,
$$\boxed{ E_n\ell_n = F_{\max}\ell_n^2 = (2n+1)\hbar c }.$$
For the first horizon,
$$\boxed{ F_{\max}\ell_{\min}^2 = E_0\ell_{\min} = \hbar c },$$
and hence $\boxed{\ell_{\min}=2L_P}$. The same scale follows from quantum localisation, $\ell\gtrsim\hbar c/E$, and gravitational collapse, $\ell\gtrsim 4GE/c^4$. Their intersection gives
$$\boxed{ \ell_{\min}=2L_P, \qquad E_{\mathrm{gap}}=\frac12m_Pc^2, \qquad E_{\mathrm{gap}}\ell_{\min}=\hbar c }$$
This is not a minimum particle mass: there is no localised particle. It is better interpreted as a proposed horizon energy gap between no-horizon geometry and the first horizon state.
Result
The complete spectrum can be written
$$\boxed{ \frac{2I_n}{\hbar} = \frac{A_n}{4\pi L_P^2} = \frac{R_n^2}{L_P^2} = \frac{S_n}{\pi k_B} = \frac{E_n\ell_n}{\hbar c} = \frac{F_{\max}\ell_n^2}{\hbar c} = 2n+1 }$$
Flat space is the separate no-horizon configuration $I=0$. The essential assumption is that the Euclidean Schwarzschild action-angle phase space can be canonically quantised. Under this assumption, the Planck-radius ground state and the $8\pi L_P^2$ area spacing follow together.
Maximum Acceleration and Gravitational Schwinger Effect
For mass $m$, the Schwinger critical acceleration is
$$a_S=\frac{mc^3}{\hbar}, \qquad \frac{c^2}{a_S}=\frac{\hbar}{mc}.$$
For the smallest proposed horizon,
$$m_0=\frac{M_P}{2},\qquad \ell_0=2L_P,$$
so
$$\frac{\hbar}{m_0c}=2L_P=\ell_0, \qquad a_0=\frac{m_0c^3}{\hbar}=\frac{c^2}{2L_P}.$$
Its critical force is
$$m_0a_0=\frac{c^4}{4G}=F_{\max}.$$
The associated Unruh temperature equals its Hawking temperature because $a_0$ equals the horizon surface gravity:
$$k_BT_U=k_BT_H=\frac{\hbar a_0}{2\pi c} =\frac{m_0c^2}{2\pi}.$$
Finally, the Schwinger exponent is
$$B_0=\frac{\pi m_0^2c^3}{\hbar F_{\max}} =\pi=\frac{S_0}{k_B}.$$
Thus
$$\boxed{ \frac{\hbar}{m_0c} =\frac{c^2}{a_0} =\ell_0=2L_P,\quad m_0a_0=F_{\max},\quad T_U=T_H,\quad B_0=\pi }$$
with candidate tunnelling weight $e^{-\pi}$. A physical production rate still requires a gravitational tunnelling solution.
Maximum Force from Quantum Localisation
Quantum localisation and gravitational backreaction give$$\Delta x\gtrsim \frac{\hbar}{2\Delta p} +\frac{2G\Delta p}{c^3}.$$
This is minimised at
$$\Delta p_*=\frac12M_Pc,$$
giving
$$\Delta x_{\min}=2L_P, \qquad \Delta E = c\Delta p=\frac12M_Pc^2$$
Therefore,
$$\boxed{ F_{\max} =\frac{\Delta E_*}{\Delta x_{\min}} =\frac{c^4}{4G}. }$$
Thus quantum localisation and gravitational backreaction recover the minimum length, its corresponding energy, and the maximum-force scale.
Another way to show all this
The central mathematical structure is expressed as$$(E,t);\longrightarrow;(I,\theta);\longrightarrow;(X,P)$$
with each transformation canonical. The uncertainty relation then gives
$$\langle \hat{I} \rangle \geq \frac{\hbar}{2}$$
without requiring the whole oscillator spectrum. Schwarzschild thermodynamics converts that directly into
$$R_{\min}=L_P,\qquad \ell_{\min}=2L_P,\qquad A_{\min}=4\pi L_P^2,\qquad E_{\min}=\frac{1}{2}M_Pc^2,\qquad S_{\min}=\pi k_B$$
The independent localisation-backreaction calculation gives precisely the same point and simultaneously recovers
$$F_{\max}=\frac{c^4}{4G}$$
De Sitter Limits and the First Causal Cell
The evolution of our Universe can be considered as two asymptotic de Sitter epochs connected by a transition phase parameterised by the brief moment of matter-radiation equality. We are presently living in the second epoch - our accelerating Universe is quasi-de Sitter.
Perhaps then, dS space is a natural limiting state with positive vacuum energy. In fact, drawing on Sean Carroll, then inflation need not be the original state. Inflation may begin as a later transition from an earlier, lower-energy dS-like phase. By logical extension (i.e. we exist!) it also implies that dS space is highly unstable, giving common ground with Swampland, i.e. in string theory it is basically impossible to construct a meta-stable de Sitter theory. Rajaraman also showed dS space to be unstable once quantum gravity is included.
This implies that the Universe before inflation would be the actual initial past boundary condition, with the dS-like start of inflation being a later phase change. A key takeaway here is that during and up during and up to the end of inflation, the proper distance $l_{\Lambda}$ to the CEH stays constant (i.e. a dS-like state), while the proper distance between two points increases exponentially. I know, this is a little difficult to get your head around. What it means is that inflation is not superluminal expansion.
As Lineweaver explains, during inflation (a period of around 60 e-folds) all the energy is in the inflaton which has very few degrees of freedom and low entropy. Inflation ends with a period of reheating, during which the inflation's energy is transferred into a relativistic fluid. This is also known as the Hot Big Bang. After reheating (in $\Lambda$CDM) during radiation domination, the CEH is approximately constant (the CEH proper radius increases as $l_{\Lambda} \propto a$ ), and in the DE dominated future, the CEH is a constant proper radius. Here $a$ the cosmic scale factor (refer Figure below, also from Lineweaver).
Alright then, but what about the minimal length-scale? Well, there is an argument (derived from the holographic principle), that the entire universe must be contained within the past horizon of a so-called eternal observer.
Now, there are multiple lines of evidence that at Planckian scales spacetime behaves effectively two-dimensional, so the natural "cell" for information is a 2-sphere of radius $L_p$ (linear size $l_{UV} = 2L_p$, i.e., its diameter). The unit of information is then the area of that sphere in Planck units:
CosmIn counts the modes crossing the Hubble radius between the early and late accelerating phases and proposes$$I_c=4\pi.$$
But what fixes this value? The previous horizon calculation found a first nonzero horizon state with
$$R_0=L_P, \qquad \ell_0=2L_P, \qquad A_0=4\pi L_P^2.$$
Its area in Planck units is therefore
$$\boxed{ I_c=\frac{A_0}{L_P^2}=4\pi. }$$
Thus $4\pi$ is not obtained by arbitrarily choosing a Planck-radius sphere. It follows from the minimum horizon action,
$$I_0=\frac{\hbar}{2},$$
which fixes the first horizon radius to $L_P$. This also connects with Lineweaver and Patel's proposal that the intersection of quantum localisation and gravitational collapse identifies the smallest possible object and a candidate initial condition for the Universe. Here that intersection is
$$\boxed{ R_0=L_P, \qquad \ell_0=2L_P, \qquad E_0=\frac{1}{2}M_Pc^2. }$$
It is better understood not as a particle or an ordinary black hole, but as a first causal cell: the smallest nonzero region bounded by a quantum horizon. The proposed sequence is therefore
$$\boxed{ \text{first causal cell} \longrightarrow \text{pre-inflationary transition} \longrightarrow \text{inflation} \longrightarrow \text{Hot Big Bang} \longrightarrow \text{late de Sitter limit}. }$$
The entire cosmic history accessible to an eternal observer may thus originate within a Planck-radius causal boundary whose area is $4\pi$ in Planck units, matching the proposed cosmic-information count.
CosmIn can therefore be interpreted geometrically as the area of the unit two-sphere belonging to the first allowed horizon radius.
What remains conjectural is the statement that the cosmological mode count is physically identical to this geometric area count. But the direct de Sitter derivation makes that identification substantially more plausible than a purely numerical comparison with a Schwarzschild state.
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