Quasi-Perelman: PT as Asymptotic Expanding Soliton

The picture

Imagine a ball of clay deformed slowly according to a well-defined rule. Perelman proved that under certain conditions, the evolution makes the clay perfectly spherical: it reaches its soliton, its absolute equilibrium form where nothing moves anymore.

PT is not a flow toward a soliton; it is a family of geometries parameterized by $\mu$, with a fixed point at $\mu^* = 15$. At that point, the PT metric satisfies the Ricci-soliton equation up to a residue. This residue is measurable, structured, and nonzero. It crushes as $1/\mu^4$ at large scales, but at $\mu^* = 15$ it remains small yet significant ($\sim 10^{-5}$).

This difference is not an error. It co-occurs with the entropic arrow of PT-time (the $q^-$ branch). A perfect soliton would correspond to a geometry without this dynamical signature. PT avoids this fate by not quite closing its soliton equation.

A perfect soliton is a frozen ball. PT is a ball that never quite manages to freeze. This residual gap is time.

The slow-roll analogy

In cosmology, we know a similar situation: inflation. The primordial universe was nearly in equilibrium — nearly, but not exactly. This “nearly” — a tiny parameter called slow-roll — is precisely what allowed the universe to expand. Exact equilibrium would have been sterile: nothing would have happened.

PT works the same way, but the slow-roll parameter is forced by arithmetic: it comes directly from the structure of the prime number sieve.

Two branches of the sieve

PT explicitly distinguishes two readings at $\mu^* = 15$:

• $q^+$: the static branch, which measures forces (vertex, couplings, $\alpha_{\rm EM}$).
• $q^-$: the dynamic branch, which measures flow (propagation, dissipation, arrow of time).

If PT were a steady soliton ($\lambda \equiv 0$), the anisotropy between the three Hubble rates $H_3, H_5, H_7$ would have to vanish. The entropic arrow on $q^-$ (BT13) remains independently true, but it co-occurs with the residue: the two say, from two different angles, that PT is dynamically non-trivial.

Why this matters

If the PT–Perelman correspondence were strictly exact, PT would be a particular case of a theory that has already earned a Fields medal. That would be encouraging but not distinctive. The fact that there is a residual gap — with a precise, computable mathematical form — makes PT a theory that dialogues with Perelman while retaining its arithmetic specificity.

Why almost, and why it matters

What an exact soliton would do

If PT were a Ricci soliton in the strict sense, its geometry at $\mu^* = 15$ would be in absolute equilibrium. No evolution, no fluctuation, no possible trajectory. Mathematically attractive; physically sterile.

Yet PT explicitly possesses two natural branches at $\mu^* = 15$:

• $q^+ = 1 - 2/\mu$: the static branch (couplings, $\alpha_{\rm EM}$, PMNS). The frozen.
• $q^- = e^{-1/\mu}$: the dynamic branch (propagator, dissipation, arrow of time). The flow.

Theorem BT13 (EML closure) states strict monotonicity of the Shannon entropy $H$ on $q^-$:

This property is independently true on the $q^-$ branch (it does not depend on the Perelman analysis). But it co-occurs with the non-triviality of the Perelman residue: both properties are jointly true in PT, and together they express the dynamical non-triviality of the sieve.

The computation

We write Perelman’s equation for PT:

with $g_{\rm PT}$ the Fisher-Bianchi metric and $f = -\ln\alpha$ the potential. For a perfect soliton, $\lambda \equiv 0$. For PT, numerically:

a tiny but nonzero value. And crucially, when $\mu$ grows large:

The coefficient $-1$ is exact. Verified at $\mu = 30{,}000$ in 60-digit arithmetic: $\lambda \cdot \mu^4 = -1.00011$.

The conceptual inversion

The initial reading would be: “PT misses the perfect soliton, too bad.” The correct reading is opposite:

The fact that $\lambda \neq 0$ IS the geometric condition that makes PT time possible. An exact soliton would correspond to a frozen universe, with no temporal arrow, no dynamics.

Proposition (proved in Appendix U and in the transversal chapter on spectral geometry): for the PT Fisher–Bianchi metric, the following two properties are equivalent:

1. Effective anisotropy of Hubble rates: there exist two active primes $p_1 \neq p_2$ with $H_{p_1}(\mu) \neq H_{p_2}(\mu)$.
2. Unique determination of $\lambda$: the system of $(pp)$ equations for $p \in \{3,5,7\}$ admits a unique solution $\lambda$.

Proof intuition: with two independent equations $(p_1 p_1)$ and $(p_2 p_2)$ in two unknowns $(\dot f, \lambda)$, Cramer’s method gives a unique solution if and only if the determinant $H_{p_2} - H_{p_1}$ is nonzero. If all $H_p$ are equal to a common $H$, the system degenerates into a single equation in two unknowns: $\lambda$ becomes a free degree of freedom, undetermined.

Asymptotic corollary: as $\mu \to \infty$, the anisotropy $H_{p_1} - H_{p_2} = (p_2 - p_1)/\mu^3 + O(1/\mu^4) \neq 0$ is automatic (active primes are distinct in $\Z$), so the unique value of $\lambda$ satisfies $\lambda(\mu) = -1/\mu^4 + O(1/\mu^5)$, nonzero.

Co-occurrence with BT13. A third property is jointly verified in PT, but it is independent of the previous two:

• (BT13) $dH/d\mu > 0$ strictly on the $q^-$ branch: the entropic temporal arrow is carried by $q^-$ independently of the anisotropy of the $\gamma_p$.

The three properties (1)–(2)–(BT13) are thus jointly true in PT, but their strict equivalence is not proved: (1) ⟺ (2) holds, while (BT13) holds for its own reason. It is this co-occurrence that legitimizes the reading “the residue is the geometric signature of the $q^-$ branch”, without making it a formal equivalence.

The basin of stability

We work inside the reduced stable basin of the odd sieve: $(\mu_7, \mu_{11}) \approx (11.63,\, 17.98)$ which contains the fixed point $\mu^* = 15$. In this interval, the three active primes are exactly $\{3, 5, 7\}$, and the reduced map sends all integers $\{12, 13, 14, 15, 16, 17\}$ to $15$ (theorems BT17 and BT18 of the bridge program). All numerical results are systematically verified on this basin.

Three complementary readings

The deviation from a steady soliton can be read in three complementary ways. Only one (the Perelman reading) is a strict mathematical description; the other two are analogies (qualified below).

Perelman reading (strict mathematical). Perelman studies flows that converge toward a soliton. PT is not a flow: it is a family of geometries parameterized by $\mu$, with a fixed point at $\mu^* = 15$. The metric $g_{\rm PT}(\mu^*)$ satisfies the soliton equation with a constant $\lambda \neq 0$ that vanishes as $1/\mu^4$ at large scales. PT is therefore an expanding quasi-soliton ($\lambda < 0$ in the Perelman convention) whose constant tends to zero asymptotically.

Cosmological analogy (slow-roll). In inflation, an exact de Sitter universe would be sterile: one needs a small parameter $\epsilon_{\rm sr} \neq 0$ that generates the dynamics. PT’s residue $|\lambda| \sim 1/\mu^4$ plays qualitatively the same structural role as a slow-roll parameter: without residue, no dynamics. The analogy is purely structural, not a formal identification (the two parameters do not match term-for-term).

Speculative analogy (Connes, NCG). In noncommutative geometry, time emerges from the modular flow of a KMS state. No modular flow $\Leftrightarrow$ no time. PT’s residue plays a qualitatively comparable role: without residue, no proper dynamics. This analogy remains speculative and is not formalized in the present work.

Closure of the temporal triptych

Three independent results, one mechanism:

Technical proof

Metric, gauge, and method

We work in the Fisher–Bianchi metric in Lorentzian signature $(-,+,+,+)$ — the component $g_{\mu\mu}$ becomes negative for $\mu > \mu_c \approx 6.97$, the Lorentz frontier determined by the vanishing of $d^2(\ln\alpha)/d\mu^2 = 0$ (signature theorem 12.4 of Chapter 13):

where $a_p(\mu) := \gamma_p(\mu) / \mu$ and $\gamma_p$ is the anomalous dimension at prime $p$. We choose the gauge $N(\mu) = \mu$ (technically favorable), giving proper time $\tau = \mu^2/2$. The basin $(\mu_7, \mu_{11}) \approx (11.63,\, 17.98)$ is strictly above $\mu_c$, hence in the Lorentzian regime.

Pair-compatibility method: the soliton equation $\mathrm{Ric} + \mathrm{Hess}(f) = \lambda g$ has 4 diagonal components (1 temporal $(\tau\tau)$ + 3 spatial $(pp)$ for $p \in \{3,5,7\}$), for 2 unknowns $(\lambda, \dot f)$. The system is overdetermined. The $(pp)$ components read

with $A_p := \dot H_p + H_p \sum_k H_k$ and $H_p := \mu^{-1} d\ln a_p/d\mu$. We extract $\lambda$ via the 3 pairs $(3,5), (5,7), (3,7)$ by elimination of $\dot f$. If all three yield the same value, we have a soliton; otherwise, the spread measures the defect.

Asymptotic expansion of $\gamma_p$

For large $\mu$, $q^+ = 1 - 2/\mu \to 1$. The expansion proceeds in four steps:

1. Binomial: $q^p = 1 - 2p/\mu + 2p(p-1)/\mu^2 + O(1/\mu^3)$.
2. Deficit: $\delta_p := (1-q^p)/p = 2/\mu - 2(p-1)/\mu^2 + O(1/\mu^3)$.
3. Holonomy T6: $\sin^2\theta_p = \delta_p(2-\delta_p) = 4/\mu - 4p/\mu^2 + O(1/\mu^3)$.
4. Logarithm and derivation: $\ln\sin^2\theta_p = \ln 4 - \ln\mu - p/\mu + (p^2-4)/(6\mu^2) + O(1/\mu^3)$, then $\gamma_p = -\mu\, d(\ln\sin^2\theta_p)/d\mu$:

Asymptotically, $\gamma_p \to 1$ for every active prime: the $p$-dependence disappears in the limit.

Kinematics in gauge $N = \mu$

From $\ln a_p = -\ln\mu + \ln\gamma_p$ we derive (gauge $\tau$, $d/d\tau = (1/\mu)\,d/d\mu$):

And the useful combination:

Key step: the coefficient -1

For a pair $(p_1, p_2)$, the compatibility of the $(pp)$ equation gives:

We compute the numerator, keeping all terms up to $1/\mu^7$:

The $1/\mu^6$ terms cancel (the factor $-5/\mu^6$ appears in both). Remaining terms in $1/\mu^7$:

And the denominator:

The quotient thus gives:

The coefficient $-1$ is independent of the choice of pair among the three active primes $\{3, 5, 7\}$. This is what makes the value consistent — without this independence, the pair decomposition would not be well-defined.

The correction term (numerical conjecture)

Numerically (mpmath 60 digits),

with the factor $-10/3$ observed at $\mu = 10{,}000$ and $30{,}000$. This is a numerical conjecture not yet analytically proved: the analytic derivation requires extending the $\gamma_p$ expansion to order $1/\mu^3$, a heavy but elementary calculation.

At the fixed point $\mu^* = 15$

At $\mu = 15$ we are far from the asymptotic regime. The approximation $-1/\mu^4$ gives $-1.975 \times 10^{-5}$, whereas the exact value is $-2.5215 \times 10^{-5}$. The gap is $22\%$ relative to the exact value (or $28\%$ relative to the approximation $-1/\mu^4$). The $1/\mu^{k\geq 5}$ terms exactly account for this gap.

Temptation: a local fit gives $\lambda \approx -(12/5)\, \mu^{-127/30}$ with rel_RMS of 0.07 % on the basin. Rejected as an exact identity: the mpmath calculation shows that the logarithmic derivative $d \ln|\lambda| / d \ln\mu$ varies from $-4.244$ to $-4.211$ across the basin (not constant). The fit is a local artifact of the expansion $\lambda = -\mu^{-4}(1 + 10/(3\mu) + \ldots)$, which looks like an effective power law in a limited $\mu$ range but is not one.

What remains open

• Exact closed form of $\lambda(\mu)$ at finite $\mu$: the function is rational (all PT operations are) but its symbolic expressions explode in size. CRT decomposition modulo $3 \cdot 5 \cdot 7$ or factorization in $\sin^2\theta_p$ factors.

• Analytic confirmation of $-10/3$ in the correction term.

• Global bound $|\lambda(\mu)| \leq C/\mu^4$ on the entire basin with explicit $C$.

• Mirzakhani link: the non-power-law form of $\lambda$ suggests a multi-parameter recursion, which is exactly the territory of Eynard-Orantin topological recursion.