spin.md

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--- title: Spin lead: A design turned about its centre: the ripple identity, the complete spin spectrum, the sponge's opaque diagonal, and a Gaussian Farey. figure: research-spin slug: spin ---

Turn a design about its centre and look at the average. The instrument is exact, the identities are proved, and the census is run.

Proved means a proof is given here; Verified means recomputed by a crate test or a lab study; Conjecture means neither. The generators are mrlynum::spin, the host fixture of mrlydemo, and lab/rs/spin-census, the one pass behind every number of the census below. The spin demo shows the infinite spin, the radial demo the finite ones and the harmonics each keeps.

The identities

  • Write a picture in circular harmonics, f(r, phi) = sum_m f_m(r) e^(i m phi). The average over the q rotations by multiples of 2 pi / q keeps exactly the orders with q | m; the average over all rotations keeps m = 0. Proved: a rotation by 2 pi / q multiplies f_m by e^(2 pi i m / q), and the q-th roots of unity sum to zero unless q | m.
  • A screen turning a design by p/q of a turn per frame, p/q in lowest terms, shows only q orientations, so a long afterglow shows the q-average. A design of rotation order g has f_m = 0 unless g | m, so the lowest surviving order is lcm(q, g): the petal count. Proved; witness the_harmonics_read_the_rotation_order.
  • The infinite spin of a raster is its ring profile F(r), the exact circle mean, and int 2 pi r F(r) dr is the fill. Proved; witness the_mass_of_the_profile_is_the_fill. The carpet's profile is zero to r = side/6, its central hole. Verified.
  • A plane wave averaged over a turn is J0(|k| r), so the infinite spin is the Hankel transform of the radially averaged spectrum, ringing at the lattice norms a^2 + b^2 and a^2 + ab + b^2. Proved.
  • The primes decide the rings. The square lattice rings at sqrt(n), n in A001481, with weight r2(n) = 4 (d1 - d3), A004018: silent exactly where a prime 3 (mod 4) divides n to an odd power, and summing to 4 zeta(s) L(s, chi_4), the zeta of Z[i]. The hexagonal lattice rings at A003136 with weight A004016, 6 zeta(s) L(s, chi_-3), the zeta of Z[omega]: the two L-functions of pi and bases. The mass of a spun lattice is the Gauss circle count, and Hardy's Bessel series for its error is this ring expansion (Hardy 1915). Proved. The gaussian demo paints the primes of both rings and bars the weights.

Known elsewhere

The census

A spin needs a centre, and the raster's centre is the wrong one for half the family. Every filled digit d of a design is the attractor of one map S(x) = (x + d)/base, whose fixed point is p_d = d/(base-1); the design is exactly self-similar about each. Read M(r) = int_0^r 2 pi s F(s) ds about p_d.

  • The spin mass is exactly log-periodic about a fixed point. Proved. S(F) = F ∩ cell_d for F the design, and the equal-weight self-similar measure divides by the fill under one map, so M(r/base) = mu(S(B(p_d, r) ∩ F)) = M(r)/fill for every r with r/base below the distance from p_d to the other filled cells. Hence M(r) = r^(log(fill)/log(base)) G(log_base r) with G periodic of period exactly log base - the ripple's period is an identity, not a fit. The corner digit gives the widest window, r <= side; the centre digit only r <= side/2. Witness the_spin_mass_scales_by_the_fill_about_a_filled_corner; the instrument is mrlynum::spin::mass_within.
  • About the raster centre the spin dimension is undefined for half the designs. Proved. An empty centre digit removes the open square of side side/3 about the raster centre, hence its inscribed disc of radius side/6, so M(r) = 0 for every r <= side/6: the centre-spun mass carries neither power law nor ripple over a whole factor of the base, and the raster centre is not a fixed point at all. The bound is attained, in exact integer arithmetic on doubled coordinates rather than on cell centres, which would return hole + 1/2 whatever the hole: the squared distance from the centre to the nearest filled cell is (side/3)^2 = 6561 at level 5 for both bang dim 2, base 3, code 239 and the carpet 495, that is side/6 = 40.5 exactly, while 79 empties more, out to 56.572962. Reading the spin dimension at the raster centre separates 127 from 239 only by which of them has a centre at all.
  • The slope is the dimension. Verified (lab/rs/spin-census). Corner fixed point at level 6, window 27 <= r <= 729, three whole periods: slopes 1.465054, 1.649432, 1.783588, 1.761814, 1.897854, 1.879522, 2.000100 for codes 79, 95, 127, 239, 255, 495, 511 against the exact log(fill)/log 3 = 1.464974, 1.630930, 1.771244, 1.771244, 1.892789, 1.892789, 2 - every gap at or below 1.9e-2, and the solid square reads 2.000100. The discretisation of the identity, max |M(3r)/(fill M(r)) - 1| over the window, runs 5.3e-3 to 1.8e-2 and is the size of the error the slopes carry.
  • The ripple separates the equal-mass pairs. Verified (lab/rs/spin-census). Fold ln M(r) - (log(fill)/log 3) ln r with that exponent taken exactly, never fitted, into 24 bins of log_3 r over whole periods; the drift bar is the same fold on the first half of the window against the second. At level 7 about the corner, 127 against 239 gives ripple gap 0.11984 on bar 0.04126, and 255 against the carpet 495 gives 0.12042 on bar 0.01461 - both separated, where dimension and every density reading are identical. The solid square is the control and has no ripple: swing 0.00277 under its own bar 0.00578. The pipeline control is exact: a code and its mirror return the same ripple to the last bit, worst gap 0.00e0. About the corner the instrument is invariant under the one transposition fixing that corner and not under the whole square group, so the corner ripple is a function of the transpose class, and a design's full spin fingerprint is the family of ripples over all its filled digits.
  • The ripple is not a complete invariant of the transpose class. Proved. A design that is a solid segment has M(r) = c r exactly about a fixed point on it, so ln M(r) - D ln r is constant and the ripple vanishes identically. Code 7 draws the solid row and code 273 the solid diagonal; both are of dimension exactly 1, they lie in different transpose classes, and both carry the zero ripple. No bar is needed for the conclusion; the census reads them at swings 0.01114 and 0.01217 and at mutual gap 0.01371, all of it discretisation.
  • Near-degeneracies outside the segment case exist, and their count is not stable. Verified as a phenomenon (lab/rs/spin-census). De-duplicated to transpose classes, the equal-fill class pairs whose ripples sit inside their own drift bar number 13 at level 6 and 6 at level 7, at gap-to-bar ratios 0.71 to 0.95, the tightest being 287 against 315 at fill 6, gap 0.03574 on bar 0.04543 - two designs that differ by moving one cell from (0,2) to (1,2). Six survive both levels: 287-315, 63-123, 123-187, 31-59, 437-485, 37-261. The count moves with the level and with the estimator, so this is a list of near-coincidences and not a census of them.
  • No two designs outside one symmetry orbit are spin-isospectral. Verified (lab/rs/spin-census). P_m = int |f_m(r)|^2 2 pi r dr for m = 0..12, read at levels 1 and 2 over all 511 nonempty base-3 plane codes, splits them into exactly 101 spectra - the count of nonempty orbits of the square group, whose largest orbit has 8 members and whose largest spectral bucket has 8. Not one pair outside one orbit agrees to 1e-9. Within the family the spin spectrum is a complete invariant of the dihedral class, and the isospectral witness the question asks for does not exist here.
  • The spin spectrum is a quadratic form, so it reads a pair census and nothing else. Proved. Write the render at level level as f = sum_j x_j 1_(cell_j) over its cells. Every ring coefficient c_m(r) is linear in the indicators, so P_m = sum_(j,k) x_j x_k Q_m[j,k] with Q_m[j,k] = int Re(g_(m,j) conj(g_(m,k))) 2 pi r dr and g_(m,j) the m-th coefficient of one cell. Turning a pair of cells about the raster centre by theta multiplies both g by e^(-i m theta) and leaves their product fixed; reflecting conjugates both and leaves the real part fixed. So Q_m is constant on the orbits of the raster's symmetry group acting on pairs of cells, and P_m is a linear functional of the pair census Phi_level: the number of filled cell pairs in each such orbit, an exact integer vector computed with no transform. Two designs sharing a pair census share P_m identically, at every order, every ring count and every truncation. The hypotheses are that the render is a constant-valued 0/1 indicator on a raster of side base^level, that the ring radii are data-independent, and nothing more; at dim 3 the covariant object is the degree-l power summed over its 2l + 1 orders, not a single (l, m). The base-3 plane has 11 pair classes at level 1 and 461 at level 2; solving the level-1 coefficients from 11 independent censuses reproduces mrlynum::spin::harmonics, read at 1024 rings and m = 0..12, on all 511 codes at worst relative residual 1.14e-14.
  • Level 1 alone is not complete, and the four witnesses are exact. Proved. The level-1 pair census takes exactly 97 values on the 101 orbits. Codes 45 and 105 at fill 4, 61 and 121 at fill 5, 78 and 102 at fill 4, 94 and 118 at fill 5 are four homometric pairs: different square-group orbits with all 11 class counts equal. By the reduction each pair's whole level-1 spectrum coincides identically, for every m and every resolution, so the 101 is a level-2 fact and never a level-1 one. The instrument agrees rather than being asked to: P_m read at level 1 alone, bucketed greedily by first match at tolerance 1e-9, gives 97 spectra, the four pairs at gaps 1.30e-16, 1.03e-17, 6.51e-17 and 1.64e-16, and level 2 separates the same four at 0.151, 0.0689, 0.253 and 0.105.
  • The 13 orders see exactly 9 of the 11 census directions: the cap is 9 overall, 6 even and 3 odd. Proved for the caps, Verified for their attainment (lab/rs/spin-census). The half turn rho lies in the square group, so it acts trivially on classes and says nothing; the involution that bites half-turns one member only, tau: {j, k} -> {rho j, k}, which is well defined on classes exactly because rho is central in D4, giving {j, rho k} = rho . {rho j, k} and {rho g j, g k} = g . {rho j, k}. Since g_(m, rho j) = (-1)^m g_(m, j), Q_m . tau = (-1)^m Q_m, so at odd m the coefficient vector is tau-antisymmetric. tau fixes 5 of the 11 classes - the two corner-corner and edge-edge classes it closes and the three touching the centre - so the antisymmetric part has dimension (11 - 5)/2 = 3 and no number of odd orders can exceed rank 3. The six odd orders reach exactly 3 and the seven even orders exactly 6, the proved caps, for 9 of 11 in all: the level-1 spectrum is strictly coarser than the census it factors through, and splits it into the same 97 classes anyway.
  • The completeness is not about base 3. Verified (lab/rs/spin-census), as a statement about the census and not yet about the truncated spectrum. At base 5 the level-1 pair census takes 3993511 values on the 4211743 nonempty square-group orbits of the 5 x 5 plane, 204856 of them shared by 423088 orbits with the largest tie holding 8. At base 3 with dim 3, under the order-48 cube group, it takes 1461693 values on the 2852287 nonempty orbits, 757066 shared by 2147660 orbits with the largest tie holding 32. Every one of those ties breaks at level 2: the generator's weight window is budget-capped but did not bind, covering every group, all 204856 of them out to weight 21 and all 757066 out to weight 24, against level-2 censuses of 24805 and 6325 classes. So Phi_1 with Phi_2 is injective on orbits at base 3, at base 5 and at dim 3 alike, and the census collision that would make the theorem a fact about base 3 does not exist. The canonical-form orbit counts match the Burnside averages 4211744 and 2852288 the same pass computes from the cycle index.
  • Census injectivity is necessary, not sufficient. Conjecture. The 26 numbers P_m at levels 1 and 2 are a projection of rank at most 26 of a census of dimension 11 + 461 at base 3, and 55 + 24805 at base 5; at level 1 that projection already loses 2 of 11 directions. So the base-5 and dim-3 results remove the only obstruction that transfers - a shared pair census - without establishing that the truncated spectrum itself separates there, which no run of the transform at those sizes has been made to decide.
  • The powder falls as -log(fill)/log 3. Conjecture. Arithmetic ring average of |F(k)|^2 in 240 logarithmic bins over the band 3 pad/side to pad/8, at level 7 with pad 4096: slopes -1.37986, -1.51762, -1.73012, -1.83071, -1.97886, -2.00433 for 79, 95, 127, 239, 255, 495 against -log(fill)/log 3 = -1.464974, -1.630930, -1.771244, -1.771244, -1.892789, -1.892789, every gap at or below 0.24 over both pads. The agreement is inside the instrument's own spread and so establishes nothing. Sliding a three-period fit window a quarter period at a time inside the same band moves the slope by 0.16 to 0.45, two to four times the gap it would have to establish; doubling the pad to 8192 moves every slope it is run on, 127 from -1.73012 to -1.81607, 255 from -1.97886 to -2.03225, the carpet from -2.00433 to -2.12289. A band-split bar is not an error estimate here - the slide and the pad are - and the readings do not move monotonically toward -log(fill)/log 3 as the level or the band grows.
  • The powder is not Porod. Verified (lab/rs/spin-census). Every sliding-window slope of every fractal code, at both pads, stays above -2.28, hence at least 0.72 from the -3 a sharp interface would give. The solid square is the control and behaves the other way: it slides from -2.75781 to -2.35759, coming within 0.25 of -3 and never near its own -log(fill)/log 3 = -2. Whatever the exponent is, the mass-fractal side and the surface side are cleanly apart.
  • The powder's log-periodic ripple is not resolved. Conjecture. Folding the band's residual at period log 3 swings by 1.5 to 4.4 in ln power, larger than the fit it is a residual of: the ring average of a lattice point set is spiked on the norms of A001481, and 4 periods of band do not average those spikes away. The log-periodicity of (Cherny, Anitas, Osipov and Kuklin 2011) is neither confirmed nor denied here.

The shadow

  • The sponge blocks every lattice line down its space diagonal. Proved; recomputed by lab/rs/spin-census. Count lattice lines in direction v meeting the sponge at level level, as classes of filled cells under x -> x cross v; the shadow obeys S_(level+1) = union_d (3 S_level + proj d) over the digits, so it is decided by the digit projections alone. Along (1,1,1) the 27 cube digits project onto 19 classes and the 20 sponge digits, being a subset, project onto 19 too - hence onto the same 19, and the induction gives sponge shadow equal to cube shadow at every level. The count is 3^(2 level+1) - 3^(level+1) + 1, the centred hexagonal number A003215 at 3^level - 1 and A220978 at level: 19, 217, 2107, 19441, 176419 for level = 1..5, matched by both objects. No lattice line down the space diagonal passes through the sponge without meeting a cell.
  • The sponge's axis shadow is exactly the Sierpinski carpet. Proved; recomputed by lab/rs/spin-census. The 20 sponge digits project along an axis onto the 8 carpet digits, disjoint modulo 3, so the shadow is the carpet at level level: 8, 64, 512, 4096, 32768 against the cube's 9^level, a share (8/9)^level falling to 0.62430 at level = 4, dimension log 8 / log 3 = 1.892789 against the cube's 2.
  • No other direction in the searched window is deficient. Verified (lab/rs/spin-census). Over the 13 directions with 0 <= a <= b <= c <= 3, read to level 4 against the cube and level 5 against itself, the axis is the only one whose share of the cube's lines falls with the level; every other share rises, (1,1,2) reaching 0.98568 and (0,1,2) 0.97090 at level = 4. Rational exceptional directions exist, as (Falconer, Fraser and Jin 2015) allows; whether the axes are the only ones is not settled by a 13-direction search.

A Gaussian Farey

Scale n draws the lattice (1/n) Z^2 and its radii sqrt(k)/n, k in A001481; read them inside the disc of radius sqrt 2, the spun form of the Farey window of farey.

  • A radius is new at n exactly when k is free of the squares of the primes dividing n. Proved. sqrt(k)/n = sqrt(k')/m needs k m^2 = k' n^2, and k' <= 2m^2 follows from k <= 2n^2; the sum-of-two-squares condition on k' is automatic, since v_p(k') = v_p(k) + 2 v_p(m) - 2 v_p(n) has the parity of v_p(k), which is even at every p = 3 (mod 4), so only integrality binds. The least m with n^2 | k m^2 is n itself exactly when min(v_p(k), 2 v_p(n)) <= 1 for every p | n, that is when no prime of n has its square dividing k.
  • The count is a Mobius sum over the radical. Proved; recomputed by lab/rs/spin-census. k lies in the sum-of-two-squares set and d^2 | k if and only if k/d^2 lies in it too, so new(n) = sum_(d | rad n) mu(d) B(2n^2/d^2) with B the counting function of A001481. Both routes and a direct union over reduced squared radii agree at every n to 64: 2, 3, 9, 11, 22, 18, 40, 38, 55, 52, 91, 64, 123, 97, 128, 126, 199, 136, 243, 180.
  • The spun phi(n) is the Jordan totient. Proved. The local factor of the Mobius sum is prod_(p | n) (1 - 1/p^2) = J_2(n)/n^2, so the new radii are that fraction of all radii at scale n, exactly as the Farey new fractions are phi(n)/n of all - the square lattice raises the exponent, it does not change the shape. The approach is slow because B(X) ~ K X / sqrt(ln X) (A064533) makes B(X/d^2)/B(X) exceed 1/d^2 by a 1/ln X margin: along the radical-6 family the ratio climbs 0.56250, 0.59813, 0.61126, 0.62594, 0.63276, 0.63801 at n = 6, 12, 24, 48, 96, 192 toward 2/3.
  • The new radii are not counted by primitive representations in Z[i]. Refuted. The norms below 2n^2 with a primitive representation run 2, 3, 6, 9, 13, 17, 23, 29, 35, 44 against new(n) = 2, 3, 9, 11, 22, 18, 40, 38, 55, 52, agreeing only at n = 1, 2. Primitivity is the wrong condition - (3,4) is primitive and 25 is a square, so sqrt(25)/5 = 1 is old at 5, while (4,0) is imprimitive and sqrt(16)/3 is new at 3.
  • The disc and the box read different Farey windows. Verified (lab/rs/spin-census). Restricting to the box 0 <= a, b <= n instead of the disc breaks the criterion at n = 3, witness the radius 4/3: 16 is free of 9 and 4/3 < sqrt 2, but 16 = 4^2 + 0^2 needs a coordinate above 3. The box counts run 2, 3, 7, 9, 17, 14, 31, 27, 41, 38 and the two readings part company from n = 3 on. The disc is the reading the spin sees; neither sequence is in the OEIS, so both are candidates.

What is left

  • The coprime law does not spin. Flat layers at coprime odd scales are uncorrelated, the stack's prime detector; their ring profiles correlate at +0.38 for (3, 5), -0.33 for (5, 7), +0.38 for (9, 13), no better than gcd pairs. The cancellation is separable in x and y, and the spin discards the angle. Refuted; witness the_coprime_law_dies_under_the_spin.
  • What the corner ripple does not see beyond the segments: the six class pairs that survive both levels are not explained. Conjecture: the ripple's Fourier coefficient at frequency 2 pi / log 3 is a linear functional of the digit set, and its kernel is what collides.
  • Whether the powder exponent is -log(fill)/log 3 at all. The slide and the pad move it by more than the gap, so the band would have to reach level 9 or so, and the estimator would have to average the lattice spikes, before the question is even asked cleanly. Conjecture.
  • Whether the axes are the sponge's only deficient directions. The window searched is |v| <= 3. Conjecture.
  • The exact growth of new(n): the Mobius sum is proved, but B itself has no closed form, so the Gaussian Farey has a Landau-Ramanujan constant where the Farey has none. Conjecture that new(n) sqrt(ln n) / n^2 converges, to sqrt 2 K prod_(p | n) (1 - 1/p^2) along each radical class.