Disproof of Nelson-time hypercontractivity for the Poisson semigroup on S⁴

Executive summary

For the Poisson semigroup P_t = e^−t√Δ on the sphere S⁴, the natural “Nelson-time” conjecture — that ‖P_tf‖_q ≤ ‖f‖_p as soon as e^−t√λ₁ ≤ √((p−1)/(q−1)), in direct analogy with Mueller–Weissler’s sharp theorem for the heat semigroup — is false. Conjecture disproved

The witness is the low-degree polynomial f = 1 + (5/8)x₁ + (1/3)(x₁² − 1/5) with (p, q) = (2, 4). At the critical time t* = ln(3)/4 one has ‖P_t*f‖₄ / ‖f‖₂ = 1.00004018842… > 1, and the violation persists on the whole interval t ∈ [t*, t* + 0.0002527…), strictly inside the conjectured region. The gap reduces to an exact rational-coefficient quartic whose positivity is certified by interval arithmetic (and is checkable by hand).

The mechanism is the concavity of k ↦ √(k(k+3)): relative to the heat flow, the Poisson flow under-damps the degree-2 mode at Nelson time, and a fourth-order two-mode perturbation test shows this over-transmission becomes fatal exactly when n ≥ 4. S³ escapes by a margin of about 0.002 in the relevant exponent.

The conjecture

Work on the unit sphere S⁴ ⊂ ℝ⁵ with normalized surface measure σ, and let Δ be the (positive) Laplace–Beltrami operator. Its eigenvalues on degree-k spherical harmonics are

λ_k = k(k + 3), so λ₁ = 4, λ₂ = 10.

The Poisson semigroup is the subordinated flow

P_t = e^−t√Δ,  P_t Y_k = e^{−t√(k(k+3))} Y_k.

For the heat semigroup e^−tΔ on Sⁿ, Mueller and Weissler (1982) proved the sharp hypercontractivity theorem: ‖e^−tΔf‖_q ≤ ‖f‖_p holds precisely when e^−tλ₁ ≤ √((p−1)/(q−1)) — the same “Nelson time” condition, driven by the spectral gap alone, that Nelson established for the Ornstein–Uhlenbeck semigroup in Gauss space. The conjecture under test is the verbatim analogue for the Poisson semigroup:

‖P_tf‖_q ≤ ‖f‖_p whenever e^−t√λ₁ ≤ √((p−1)/(q−1)).

For p = 2, q = 4 on S⁴ we have √λ₁ = 2, so the conjectured region is

e^−2t ≤ 1/√3, i.e. t ≥ t* = ln(3)/4 = 0.274653072167…

The condition is necessary (second-order test)

The threshold cannot be moved earlier. For mean-zero g one has the standard expansion ‖1 + εg‖_p = 1 + ((p−1)/2) ε² E[g²] + O(ε³). Testing f = 1 + εY₁ (so P_tf = 1 + εm₁Y₁ with m₁ = e^−t√λ₁), the inequality at order ε² forces

(q−1) m₁² ≤ (p−1), i.e. m₁ ≤ √((p−1)/(q−1)).

So t* is exactly the first time the inequality can possibly hold, and at t = t* the ε² terms cancel identically — the fate of the conjecture is decided at fourth order, which is where the counterexample lives.

The counterexample

Let x = x₁ be a Cartesian coordinate restricted to S⁴, and set

f = 1 + (5/8) x₁ + (1/3)(x₁² − 1/5) = 1 + a Y₁ + b Y₂,

where Y₁ = x₁ is a degree-1 spherical harmonic, Y₂ = x₁² − 1/5 is a degree-2 spherical harmonic (the restriction of x₁² − |x|²/5), and a = 5/8, b = 1/3. At the critical time t* = ln(3)/4 the Poisson multipliers are

m₁ = e^−2t* = 3^−1/2 ≈ 0.577350,  m₂ = e^{−√10 t*} = 3^−√10/4 ≈ 0.419568,

so P_t*f = 1 + am₁Y₁ + bm₂Y₂ exactly. The verified facts:

Hence ‖P_tf‖₄ > ‖f‖₂ on a whole interval of times satisfying t ≥ t*, and the conjecture is false on S⁴. The violation is small in absolute size (a ratio of ≈ 1 + 4·10⁻⁵) but it is not a numerical artifact: the sign of the gap is an exact statement about rational numbers and the single algebraic quantity m₂ = 3^−√10/4 (see verification).

Why it works — and why S⁴ is the smallest sphere

Concavity of the subordinated symbol

Write every multiplier in terms of the first: under the heat flow at its Nelson time, m_k = m₁^λ_k/λ₁, whereas under the Poisson flow m_k = m₁^{√(λ_k/λ₁)}. Because k ↦ √(k(k+3)) is concave, the Poisson exponent √(λ₂/λ₁) = √10/2 ≈ 1.5811 is much smaller than the heat exponent λ₂/λ₁ = 5/2. At m₁ = 3^−1/2 this means the degree-2 mode is over-transmitted:

Poisson: m₂ = 3^−√10/4 ≈ 0.41957  vs.  heat: m₂ = 3^−5/4 ≈ 0.25334.

The fourth-order two-mode test

Take f = 1 + aY₁ + bY₂ with a ~ ε, b ~ ε². The relevant moments on S⁴ (computable from the one-dimensional marginal below) are

E[Y₁²] = 1/5, E[Y₂²] = 8/175, E[Y₁²Y₂] = 8/175, E[Y₁⁴] = 3/35.

Expanding G = ‖P_t*f‖₄⁴ − ‖f‖₂⁴ through order ε⁴ with m₁² = 1/3 (the ε² terms cancel exactly, as they must at Nelson time) gives a quadratic in b:

G(b) = (16/175)(3m₂² − 1) b² + (32/175) m₂ a² b − (16/525) a⁴.

Each piece is legible. The b² term compares 4-norm vs 2-norm transmission of the pure degree-2 mode; the constant term −(16/525)a⁴ is the pure degree-1 fourth-order contribution, which is strictly safe (negative); and the linear term is the pump: the positive cross moment E[Y₁²Y₂] = 8/175 lets the squared degree-1 mode feed energy coherently into the over-transmitted degree-2 mode. Maximizing over b (the quadratic opens downward since m₂² < 1/3), the optimum b* = m₂a²/(1 − 3m₂²) yields

max_b G = (16/525) a⁴ · (6m₂² − 1)/(1 − 3m₂²),

which is positive iff m₂² > 1/6, i.e. m₂ > 6^−1/2 ≈ 0.408248. The Poisson value m₂ ≈ 0.419568 clears this threshold (m₂² ≈ 0.17604 > 1/6 ≈ 0.16667); the heat value 3^−5/4 gives m₂² ≈ 0.0642, far below it — consistent with Mueller–Weissler, as it must be. The actual coefficients a = 5/8, b = 1/3 are clean rationals near this perturbative optimum (b* ≈ 0.347 for a = 5/8), with a chosen large enough that the ε⁴ gain survives the neglected higher-order (negative) corrections; the final claim is then verified exactly, not perturbatively.

The dimension count

On Sⁿ the eigenvalues are λ_k = k(k+n−1), so at 2→4 Nelson time m₂ = 3^{−√(λ₂/λ₁)/2} with λ₂/λ₁ = 2(n+1)/n. The failure threshold m₂² > 1/6 becomes

√(2(n+1)/n) < log₃ 6 = 1.630930…

So S⁴ is the smallest sphere on which this two-mode mechanism produces a counterexample — and S³ misses it by only ≈ 0.002 in the exponent (1.63299 vs 1.63093). Note the threshold test rules out counterexamples of this two-mode perturbative type on S² and S³; it does not by itself prove the conjecture true there.

Why the verification is rigorous

Everything is one-dimensional and exact

Since f depends only on x = x₁, all norms are integrals against the marginal density of x₁ on S⁴, which is

(3/4)(1 − x²) dx on [−1, 1].

The script computes every moment symbolically (sympy) and asserts the known values E[x^2k] = 1, 1/5, 3/35, 1/21, 1/33 as a sanity check, along with the orthogonality E[Y₁] = E[Y₂] = E[Y₁Y₂] = 0. The action of P_t on f is exact by definition (it just scales the two harmonics), so no discretization or quadrature enters anywhere.

An exact rational certificate

Expanding ‖P_t*f‖₄⁴ and substituting the exact relation m₁² = 1/3 (only even powers of m₁ occur), the gap becomes a quartic in the single quantity m₂ with exact rational coefficients:

D(m₂) = (128/1299375) m₂⁴ + (128/70875) m₂³ + (611/18900) m₂² + (1/42) m₂ − 9924709/635040000,

where the constant absorbs ‖f‖₂⁴ = 11921800969/10160640000. The whole disproof is now the single claim D(3^−√10/4) > 0.

Interval arithmetic closes the loop

The script evaluates D by Horner’s rule in mpmath’s interval arithmetic at 40 digits, starting from a rigorous enclosure of m₂ = 3^−√10/4. The output enclosure is

D ∈ [0.000188628768632176447771679061152975685933…, …936…],

whose left endpoint is positive; the assert D_iv.a > 0 line makes the run fail loudly otherwise. Every rounding error is contained in the interval, so D > 0 is a theorem, not a floating-point observation.

A hand-checkable route

No computer is strictly necessary: all four non-constant coefficients of D are positive, so D is strictly increasing on (0, ∞). It therefore suffices to check two elementary facts: (i) m₂ = 3^−√10/4 > 0.4195, which unwinds to the integer inequality 10 < (4 ln(1/0.4195)/ln 3)² ≈ 10.0024; and (ii) D(0.4195) > 0, a finite rational computation (it evaluates to ≈ +1.85·10⁻⁴). Monotonicity then gives D(m₂) > D(0.4195) > 0.

Sanity checks that the mechanism is real

• Heat control. With the heat multiplier m₂ = 3^−5/4 at Nelson time, scanning the general two-mode family 1 + AY₁ + BY₂ over a 121×121 grid of (A, B) ∈ [−3, 3]² gives max gap = 0.0 — never positive, exactly as Mueller–Weissler requires. The same grid under the Poisson multipliers has max gap 0.00018889512 > 0.
• Dimension threshold. The test √(2(n+1)/n) < log₃ 6 correctly separates S², S³ (pass) from S⁴, S⁵ (fail), matching the perturbative analysis.

Reproducing the result

The verification script is /Users/vikvang/Projects/poisson_hc_counterexample.py (no hosted repository; the full source is embedded below). It needs Python with sympy and mpmath; a virtualenv that has both is at /tmp/hcenv:

/tmp/hcenv/bin/python /Users/vikvang/Projects/poisson_hc_counterexample.py

The script has four sections:

1. Exact norms. Builds the 1-D marginal (3/4)(1−x²), asserts the moment table and orthogonality, and computes ‖f‖₂⁴ exactly and ‖P_tf‖₄⁴ as a polynomial in (m₁, m₂) via sympy.
2. Certificate at t*. Substitutes m₁² = 1/3, extracts the rational quartic D(m₂), asserts all coefficients are rational, and certifies D > 0 by Horner evaluation in mpmath interval arithmetic; then prints the norm ratio.
3. Failure window. Treats the gap as a function of t and root-finds the endpoint of the interval on which it stays positive.
4. Sanity checks. The heat-semigroup grid scan (must be ≤ 0), the Poisson grid scan (> 0), and the per-dimension threshold test.

Actual output of the run captured for this readout (2026-07-20, /tmp/hcenv/bin/python):

||f||_2^4 = 11921800969/10160640000 = 1.1733316965269904
||P f||_4^4 = 375*m1**4/28672 + m1**2*m2**2/180 + m1**2*m2/14 + 15*m1**2/32 + 128*m2**4/1299375 + 128*m2**3/70875 + 16*m2**2/525 + 1

Gap polynomial  D(m2) = ||P_{t*} f||_4^4 - ||f||_2^4 :
   D(m2) = 128*m2**4/1299375 + 128*m2**3/70875 + 611*m2**2/18900 + m2/42 - 9924709/635040000

Interval enclosure of m2 = 3^(-sqrt(10)/4): [0.419568165337302392115621311240859743127729348, 0.419568165337302392115621311240859743127758047]
Interval enclosure of D(m2): [0.000188628768632176447771679061152975685933764079, 0.000188628768632176447771679061152975685936275206]
=> D > 0 RIGOROUSLY:  ||P_{t*} f||_4 > ||f||_2.   CONJECTURE FALSE.

||P_{t*} f||_4 / ||f||_2 = 1.0000401884221086894  ( > 1 )

t* = ln(3)/4 = 0.274653072167
gap(t*)      = 0.0001886287686  (>0)
failure window: t in [ t*,  0.274905807072 )  i.e. t* +  0.000252735

heat check, max gap over (a,b) grid [-3,3]^2: 0.0  (<= 0: heat is hypercontractive, Mueller-Weissler)
Poisson, same grid: max gap = 0.00018889512  (> 0)

threshold test  sqrt(2(n+1)/n) < log_3(6) = 1.63093:
  S^2: exponent = 1.73205  -> test passes
  S^3: exponent = 1.63299  -> test passes
  S^4: exponent = 1.58114  -> FAILS (counterexample exists)
  S^5: exponent = 1.54919  -> FAILS (counterexample exists)

"""
Counterexample to Nelson-time hypercontractivity of the Poisson semigroup on S^4.
==================================================================================

Setting.  On the unit sphere S^4 in R^5 with normalized surface measure sigma,
let Delta be the (positive) Laplace-Beltrami operator, with eigenvalues
lambda_k = k(k+3) on degree-k spherical harmonics.  The Poisson semigroup is

    P_t = e^{-t sqrt(Delta)},   P_t Y_k = e^{-t sqrt(k(k+3))} Y_k .

Conjecture (Nelson time, true for the HEAT semigroup by Mueller-Weissler 1982):

    ||P_t f||_q <= ||f||_p    whenever   e^{-t sqrt(lambda_1)} <= sqrt((p-1)/(q-1)).

For p = 2, q = 4:  sqrt(lambda_1) = 2, so the conjectured region is
    e^{-2t} <= 1/sqrt(3),  i.e.  t >= t* = ln(3)/4.
(The condition is NECESSARY: testing f = 1 + eps*Y_1 forces it at order eps^2.)

Counterexample.  With x = x_1 (a Cartesian coordinate restricted to S^4), take

    f(x) = 1 + (5/8) x + (1/3)(x^2 - 1/5)        [ = 1 + a Y_1 + b Y_2 ]

Then AT the critical time t = t* = ln(3)/4, where the multipliers are
    m_1 = 3^{-1/2},   m_2 = 3^{-sqrt(10)/4},
we verify (exactly + by interval arithmetic) that

    ||P_{t*} f||_4  >  ||f||_2 ,

and the failure persists on a whole interval t in [t*, t* + 0.000252...),
strictly inside the conjectured region.  Hence the conjecture is FALSE on S^4.

Why it works: k -> sqrt(k(k+3)) is concave, so at Nelson time the degree-2
multiplier m_2 = 3^{-sqrt(10)/4} = 0.4195... exceeds the threshold 6^{-1/2} =
0.4082... tolerated by a 4th-order two-mode perturbation test (the cross moment
E[Y_1^2 Y_2] > 0 pumps the over-transmitted degree-2 mode).  The heat semigroup
at Nelson time has m_2 = 3^{-5/4} = 0.2533 < 6^{-1/2}: safe, as it must be.

Run:  python3 poisson_hc_counterexample.py     (needs sympy + mpmath)
"""
from fractions import Fraction
from sympy import (symbols, integrate, expand, Rational, sqrt, Poly,
                   lambdify, nsimplify, log)
import mpmath as mp

# ----------------------------------------------------------------------------
# 1. Exact norms.  Everything depends only on x = x_1, whose marginal density
#    on S^4 is (3/4)(1 - x^2) on [-1, 1].  All integrals are exact (sympy).
# ----------------------------------------------------------------------------
x, m1, m2 = symbols('x m1 m2', real=True)
w = Rational(3, 4) * (1 - x**2)
E = lambda expr: integrate(expand(expr) * w, (x, -1, 1))

# sanity: moments of x1 on S^4
assert [E(x**k) for k in range(9)] == \
    [1, 0, Rational(1,5), 0, Rational(3,35), 0, Rational(1,21), 0, Rational(1,33)]

Y1 = x                       # degree-1 spherical harmonic
Y2 = x**2 - Rational(1, 5)   # degree-2 spherical harmonic (x1^2 - |x|^2/5 on S^4)
assert E(Y1) == 0 and E(Y2) == 0 and E(Y1*Y2) == 0        # orthogonality

a, b = Rational(5, 8), Rational(1, 3)                     # the counterexample
f  = 1 + a*Y1 + b*Y2
Pf = 1 + a*m1*Y1 + b*m2*Y2                                # P_t f (exact action)

R = E(f**2)**2               # ||f||_2^4   (exact rational)
L = expand(E(Pf**4))         # ||P_t f||_4^4  as a polynomial in m1, m2
print("||f||_2^4 =", R, "=", float(R))
print("||P f||_4^4 =", L)

# ----------------------------------------------------------------------------
# 2. Exact certificate at the critical time t* = ln(3)/4.
#    m1^2 = 1/3 (only even powers of m1 occur), so the gap
#    D(m2) = ||P f||_4^4 - ||f||_2^4  is a rational-coefficient quartic in m2.
# ----------------------------------------------------------------------------
D = expand(L.subs(m1**2, Rational(1, 3)) - R)
Dpoly = Poly(D, m2)
coeffs = [nsimplify(c) for c in Dpoly.all_coeffs()]       # exact rationals
print("\nGap polynomial  D(m2) = ||P_{t*} f||_4^4 - ||f||_2^4 :")
print("   D(m2) =", Dpoly.as_expr())
assert all(c.is_rational for c in coeffs), "certificate must be rational"

# Rigorous sign check with interval arithmetic: m2 = 3^(-sqrt(10)/4)
iv = mp.iv
iv.dps = 40
m2_iv = iv.exp(-iv.sqrt(10) / 4 * iv.log(3))
D_iv = iv.mpf(0)
for c in Dpoly.all_coeffs():                              # Horner, intervals
    fr = Fraction(int(c.p), int(c.q))
    D_iv = D_iv * m2_iv + iv.mpf(fr.numerator) / iv.mpf(fr.denominator)
print("\nInterval enclosure of m2 = 3^(-sqrt(10)/4):", m2_iv)
print("Interval enclosure of D(m2):", D_iv)
assert D_iv.a > 0, "certificate failed!"
print("=> D > 0 RIGOROUSLY:  ||P_{t*} f||_4 > ||f||_2.   CONJECTURE FALSE.")

mp.mp.dps = 30
m1n, m2n = mp.mpf(3)**mp.mpf('-0.5'), mp.exp(-mp.sqrt(10)/4*mp.log(3))
Lfun = lambdify((m1, m2), L, 'mpmath')
Rnum = mp.mpf(int(R.p)) / mp.mpf(int(R.q))
ratio = Lfun(m1n, m2n)**mp.mpf('0.25') / Rnum**mp.mpf('0.25')
print("\n||P_{t*} f||_4 / ||f||_2 =", mp.nstr(ratio, 20), " ( > 1 )")

# ----------------------------------------------------------------------------
# 3. The failure persists for an interval of t >= t* (strictly inside the
#    conjectured region).  Find the endpoint.
# ----------------------------------------------------------------------------
Dfun = lambdify((m1, m2), expand(L - R), 'mpmath')
gap_t = lambda t: Dfun(mp.exp(-2*t), mp.exp(-mp.sqrt(10)*t))
tstar = mp.log(3) / 4
t_end = mp.findroot(gap_t, tstar + mp.mpf('0.005'))
print("\nt* = ln(3)/4 =", mp.nstr(tstar, 12))
print("gap(t*)      =", mp.nstr(gap_t(tstar), 10), " (>0)")
print("failure window: t in [ t*, ", mp.nstr(t_end, 12), ")  i.e. t* + ",
      mp.nstr(t_end - tstar, 6))

# ----------------------------------------------------------------------------
# 4. Sanity checks (the mechanism, not an artifact):
#    (i)  heat semigroup at Nelson time (m2 = 3^{-5/4}) must NOT fail: scan a
#         grid of (a,b) with general test function 1 + A*Y1 + B*Y2;
#    (ii) the 4th-order threshold: a counterexample of this two-mode type at
#         2->4 Nelson time exists iff m2 > 6^{-1/2}, i.e. iff
#         sqrt(lambda_2/lambda_1) = sqrt(2(n+1)/n) < log_3(6) <=> n >= 4.
# ----------------------------------------------------------------------------
A, B = symbols('A B', real=True)
Ggen = expand(E((1 + A*m1*Y1 + B*m2*Y2)**4) - E((1 + A*Y1 + B*Y2)**2)**2)
Gfun = lambdify((A, B, m1, m2), Ggen, 'mpmath')
m2_heat = mp.mpf(3)**(mp.mpf(-5)/4)      # heat multiplier on degree 2 at Nelson time
grid = [mp.mpf(i)/20 for i in range(-60, 61)]
max_heat = max(Gfun(aa, bb, m1n, m2_heat) for aa in grid for bb in grid)
print("\nheat check, max gap over (a,b) grid [-3,3]^2:", mp.nstr(max_heat, 8),
      " (<= 0: heat is hypercontractive, Mueller-Weissler)")
max_poisson = max(Gfun(aa, bb, m1n, m2n) for aa in grid for bb in grid)
print("Poisson, same grid: max gap =", mp.nstr(max_poisson, 8), " (> 0)")

print("\nthreshold test  sqrt(2(n+1)/n) < log_3(6) = 1.63093:")
for n in (2, 3, 4, 5):
    expo = mp.sqrt(mp.mpf(2)*(n+1)/n)
    print(f"  S^{n}: exponent = {mp.nstr(expo, 6)}  ->",
          "FAILS (counterexample exists)" if expo < mp.log(6)/mp.log(3) else "test passes")