The self-dual width is not flat

The two lattices where a four-dimensional descent stops short searched every four-dimensional lattice of unit covolume for the one with the smallest Gaussian sum $\theta(t) = \sum_v e^{-\pi t|v|^2}$, the generating function of how many vectors of each length a lattice has, read at one width. It found $D_4$ from almost every start, and two exceptions. $A_4$ traps a descent when the Gaussian is narrow and its dual $A_4^*$ when it is wide. Where each trap opens was located by the curvature of the sum at the lattice, measured by finite differences. $A_4$’s curvature along a five-dimensional family of changes of shape changed sign at $t = 0.99998$, and $A_4^*$'s at $1.00002$. The essay read those as a single crossing at the self-dual width $t = 1$, where both lattices would sit on a flat of the sum. It asked why the crossing falls there and nowhere else, and it suggested a path of lattices from $A_4$ to $A_4^*$ along which the sum stays constant as the likeliest reason.

The question had a false premise. Computed exactly, the curvature at $t = 1$ is not zero. It is positive, six parts in a million of the other curvature, and exactly the same for both lattices. The crossings sit at $t = 0.99998110$ and at its reciprocal $1.00001890$, and between them both lattices are genuine minima. What needs explaining is not a zero at one. It is why two lattices have identical curvature there, which duality settles in a line, and why that curvature is so nearly zero, which the lattice’s shells settle by cancelling.

The curvature without a step size #

A change of shape that keeps the covolume is a map $x \mapsto e^{\varepsilon S}x$, with $S$ symmetric and traceless. Under it a lattice vector’s squared length becomes $|v|^2 + 2\varepsilon\, v^\top S v + 2\varepsilon^2\, v^\top S^2 v$ plus terms of higher order. Differentiating the sum twice at $\varepsilon = 0$ gives

$$H(S) = \sum_v e^{-\pi t |v|^2}\left[\,4\pi^2 t^2\,(v^\top S v)^2 - 4\pi t\, v^\top S^2 v\,\right],$$

a sum over lattice vectors with no step size in it. The finite differences of the earlier essay were second differences of the sum itself, each a small number made from nearly equal larger ones, with rounding and truncation errors of their own. They located the crossing at 0.99998, which was right. What they could not do was say whether 0.99998 differed from one by more than their own error, and the essay read it as one.

The lattices are built from $\mathbb{Z}^5$. $A_4$ is the vectors of $\mathbb{Z}^5$ whose coordinates sum to zero, and $A_4^*$ is the projection of $\mathbb{Z}^5$ onto that hyperplane. Both are written in an orthonormal basis of the hyperplane and scaled to unit covolume. That construction makes their symmetry visible: permuting the five coordinates and negating them preserves both lattices, a group of order two hundred and forty.

The magnified figure is the whole correction. At $t = 1$ both curves pass through the same point, $8.59 \times 10^{-6}$, above zero. $A_4$’s rises through zero just below one, and $A_4^*$'s falls through zero just above. The window between the crossings is $3.8 \times 10^{-5}$ wide. Inside it, including at the self-dual width itself, both lattices are strict local minima of the sum. The two traps do not meet at a single width. They overlap, very slightly.

Two curvatures, fixed by symmetry #

The nine directions of change split under the symmetry group into two pieces, one four-dimensional and one five-dimensional. They are irreducible, so by Schur’s lemma the Hessian is a single number on each: $\lambda_4$ on the first piece and $\lambda_5$ on the second. The earlier essay found the same two families by diagonalising a measured matrix. Here each is computed from a single direction. The refusal figure at the end checks irreducibility directly: two orthogonal directions in the five-dimensional piece, and an even mixture of them, give the same curvature to twelve decimal places at three widths.

Each curvature splits further into two parts, and the split explains which of them can change sign. Averaged over every direction, $(v^\top S v)^2$ and $v^\top S^2 v$ depend only on $|v|$. That average produces an isotropic part, the same for every direction of change:

$$I(t) = \sum_v e^{-\pi t |v|^2}\left[\frac{\pi^2 t^2 |v|^4}{3} - \pi t\,|v|^2\right].$$

What is left over comes from how the lattice’s shells depart from being round at fourth order, their degree-four harmonic content. Call it the harmonic part. The second term of $H(S)$ contributes nothing to it, because every shell of $A_4$ is round at second order: a group acting irreducibly leaves no preferred quadratic form. Only the quartic term does.

The harmonic parts of the two pieces are not independent. Summed over an orthonormal basis of all nine directions, $(v^\top S v)^2$ is a function of $|v|$ alone, so the harmonic parts summed over the nine directions vanish. Four directions with harmonic part $h_4$ and five with $h_5$ must satisfy $4h_4 + 5h_5 = 0$. So

$$\lambda_5 = I + h, \qquad \lambda_4 = I - \tfrac{5}{4}h ,$$

and the measured ratio $h_4/h_5$ is $-1.25$ to nine decimal places at every width tried, for both lattices.

For $A_4$ the harmonic part is negative at every width shown, so it lowers $\lambda_5$ and raises $\lambda_4$ by five-fourths as much. $\lambda_4$ can never be the curvature that fails: it is the isotropic part plus something positive. $\lambda_5$ is the isotropic part minus something nearly as large. The figure shows the two parts as separate curves, close together over the whole range, crossing just below one. At $t = 1$ the isotropic part is $0.61402$ and the harmonic part $-0.61402$, the same to five decimals, and their sum is the $8.6 \times 10^{-6}$ of the window figure.

$D_4$ has no harmonic part at all. Its symmetry group, of order eleven hundred and fifty-two, has no invariant harmonic polynomial of degree four, so every one of its shells is round at fourth order as well as second. Its curvature is the isotropic part alone, the same in every direction of change: $0.66830$ at $t = 1$, checked on six random directions to fourteen decimal places. That is the structural difference between $D_4$ and $A_4$ at the level of second derivatives. $D_4$ has one curvature and it is positive at every width tried. $A_4$ has two, and one of them is a difference of two large numbers.

Why the two Hessians are one matrix at the self-dual width #

Duality is what ties $A_4$’s curvature to $A_4^*$'s, and at $t = 1$ it ties them completely. The sum that turns a lattice into its dual derived the identity from Poisson summation: for a lattice of unit covolume in four dimensions, $\theta_L(t) = t^{-2}\,\theta_{L^*}(1/t)$. The change of shape passes through duality with its sign reversed. The dual of $e^{\varepsilon S}L$ is $e^{-\varepsilon S}L^*$, because $S$ is symmetric. So the sum of $A_4$ changed by $\varepsilon S$ at width $t$ equals $t^{-2}$ times the sum of $A_4^*$ changed by $-\varepsilon S$ at width $1/t$.

Differentiating twice removes the sign. $A_4$’s Hessian at width $t$ is $A_4^*$'s at $1/t$, divided by $t^2$, direction for direction. The exact computation confirms it to $10^{-14}$ at four widths in each piece. Two consequences follow. The crossings of $A_4$ and $A_4^*$ are reciprocal, which the earlier essay observed and explained. And at $t = 1$, where $t$ and $1/t$ are the same width and $t^2$ is one, the two Hessians are the same matrix.

That second consequence is where the earlier reading went wrong. If both lattices had zero five-fold curvature at $t = 1$, it would take a reason. That both have the same five-fold curvature there takes none, because duality alone forces it. What remains is a single number, $\lambda_5$ at $t = 1$, which duality does not constrain at all. It could be anything. It happens to be $8.59 \times 10^{-6}$.

Where the near-zero comes from #

A number that small, made from terms of order one, is a cancellation. The sum defining $\lambda_5$ is a sum over shells, and each shell’s contribution can be read separately.

For $A_4$, the twenty shortest vectors contribute $-0.2008$ and the next thirty $+0.1967$. The third shell adds $0.0039$ and the fourth $0.0003$. The shortest vectors pull the curvature down, the second shell restores almost all of it, and the outer shells supply the last few thousandths that tip the total just above zero. For $A_4^*$ the pattern is the same with larger numbers. Its ten shortest vectors contribute $-0.8766$ and its next twenty $+0.8562$, followed by $0.0137$, $0.0060$ and smaller terms. The shells of the two lattices have different counts and different lengths, so their contributions differ at every shell. Their totals agree to every digit computed, as duality requires.

The shortest vectors pull downward for the reason one perfect form in space described. At narrow widths the sum is dominated by the shortest vectors, and $A_4$’s are the minimal vectors of a perfect, eutactic form, a local maximum of the packing density. That makes $A_4$ a minimum of the sum when the first shell dominates, which is the narrow-Gaussian trap. As the Gaussian widens the second shell’s weight grows relative to the first. Around $t = 1$ the pull and the push, with the outer shells added, balance to about four parts in a hundred thousand of either, and the balance tips at $0.99998$.

Nothing found here forces the balance to fall that close to one. The duality argument places the two crossings at reciprocal widths, and the shells place $A_4$’s at $0.99998$. They do not say why the shells balance so nearly at the self-dual width. The earlier essay’s question survives in a smaller form: not why the crossing is at one, but why it is within two parts in a hundred thousand of one. The exact computation makes that closeness sharper and does not explain it.

A minimum six millionths wide #

Positive curvature makes a lattice a minimum, but it does not say how large a change of shape the minimum survives. The earlier search probed the traps by perturbing each lattice by a few hundredths and descending. At $t = 1$ every perturbed descent left both lattices, which is what a flat, or a saddle, would do. The window figure says neither lattice is a saddle there. The explanation is a term of the next order.

Along the five-fold direction the sum is not symmetric. Measured at $\varepsilon = \pm 0.002$ its odd part is a cubic term $c\,\varepsilon^3$, with $c = 0.480$ at $t = 1$. The five-dimensional piece of the symmetry admits a cubic invariant, so nothing forbids one. Along the direction where the cubic term is negative, the sum rises as $\lambda_5 \varepsilon^2 / 2$ and falls as $c\,\varepsilon^3$. It turns over at $\varepsilon = \lambda_5 / 3c$. At $t = 1$ that is $6.0 \times 10^{-6}$. A change of shape of six parts in a million leaves the basin.

The radius grows steadily as the width moves away from one: $3.1 \times 10^{-3}$ at $t = 1.01$, $0.029$ at $1.1$, $0.062$ at $1.25$, $0.14$ at $2$. It is the ratio of a curvature that grows from zero to a cubic coefficient that changes slowly. The perturbations of the earlier search were of order a few hundredths, and the figure marks $0.03$ for comparison. The two parametrise a change of shape differently, so the comparison is of orders of magnitude, not of exact thresholds. Only near $t = 2$ is the basin clearly larger than such a kick, and $t = 2$ is where the search found $A_4$ holding its descents. At $t = 1$ the basin is four orders of magnitude too small for any finite search to find.

The same competition can be watched directly. Along the five-fold direction the sum at a change of shape $\varepsilon$ is computed exactly, for every $\varepsilon$ up to a tenth, at a chosen width.

At $t = 1$ the curve through $A_4$ looks like a plain cubic, rising on one side and falling on the other, because the well is six millionths wide and far below the resolution of the drawing. By $t = 1.1$ a shallow lip appears on the falling side, about three hundredths from $A_4$. By $t = 1.3$ the lip has moved out to seven hundredths. The quadratic term is what grows, from almost nothing to 0.07; the cubic coefficient falls only from 0.48 to 0.33. A minimum that exists at every width above one becomes visible only well above it, which is why a search that perturbs by a few hundredths finds $A_4$ holding at $t = 2$ and not at $t = 1$.

So both the finite-difference reading and the descents were right about what they measured and wrong about what it meant. The descents found no trap at $t = 1$ because the trap there is far too small to hold a perturbed start. The finite differences found a crossing indistinguishable from one because the gap is below their error. Together they looked like a flat. The self-dual width holds two minima, each a few millionths wide, and no flat.

What this changes about the search #

The earlier essay drew a phase diagram for the Gaussian core model in four dimensions, particles repelling with a Gaussian of fixed width on a lattice. Its energy per particle is a lattice sum. Unlike the Madelung sum, whose answer depends on the shape of the region it is added over, a Gaussian sum converges absolutely, so the energy belongs to the lattice alone. $D_4$ is the stable crystal at every density tested, $A_4$ metastable at low density and $A_4^*$ at high. Its reading was that the two metastable regions meet at the self-dual density and do not overlap. The exact curvature says they overlap in a window four parts in a hundred thousand wide. Near its edges the metastability is too weak to trap anything with the slightest disorder, and for any physical system the difference is invisible. For the mathematics it matters: the self-dual width is not a boundary between two regimes. It is inside both.

The question that motivated the search is untouched: whether $D_4$ has the smallest sum of every four-dimensional lattice at every width. The window adds two more local minima at $t = 1$ to compare against $D_4$, and $D_4$ beats both there by 7.2 per cent, the tie the earlier essay measured. In eight dimensions one lattice wins at every width described the analogous question for $E_8$, which is proved, and no lattice in space wins at every width the three-dimensional case, where duality forbids a single winner. Four dimensions sits between: duality neither forbids a winner nor proves one. The traps are where a proof would have to work, and they are now known to overlap at the width where the earlier essay thought the problem was simplest.

What the calculation leaves out #

The exact Hessian is exact only as a sum truncated where its terms fall below $10^{-18}$. Every conclusion depends on differences far larger than that. The smallest quantity used, $\lambda_5$ at $t = 1$, is $8.6 \times 10^{-6}$, and it comes out identical for $A_4$ and $A_4^*$, which are built from different vectors and summed separately, to every digit printed. That agreement is the strongest check on the truncation available, because duality predicts it and nothing in the computation enforces it.

The cubic coefficient is measured along one direction of the five-dimensional piece, the one built from a transposition of two coordinates. A cubic invariant varies over the piece, and along other directions the turning point will differ. So the radius of $6.0 \times 10^{-6}$ is the basin’s extent in one direction, not its smallest. The conclusion that the basin is far below any finite search’s step does not depend on which direction is taken.

The search is still only over lattices. The lengths do not name the lattice showed that the whole theta series does not determine a lattice, and nothing here constrains non-lattice arrangements, where the conjecture that $D_4$ is universally optimal is posed.

What the calculation has to refuse #

The tests check the structure before the numbers: the five-dimensional piece irreducible; the ratio $-5 : 4$ of the harmonic parts that Schur’s lemma and a vanishing trace require; the duality relation between the two lattices’ Hessians; the two Hessians equal at $t = 1$ with a positive five-fold curvature; the reciprocal crossings, neither at one; $D_4$’s curvature the same in every direction; and the cubic term overtaking the quadratic within $10^{-5}$ of a change of shape. The refusal is the earlier essay’s candidate explanation. A path from $A_4$ on which the sum at $t = 1$ stays constant would have to leave along a direction of zero curvature. Along the five-fold direction the sum moves by $4.9 \times 10^{-10}$ at $\varepsilon = 10^{-3}$, mostly through the cubic term. There is no such path.

Still open: why the shells balance at one #

The crossing at $0.99998$ is a balance between the first shell and the rest that happens to fall within two parts in a hundred thousand of the self-dual width. The harmonic part is a theta series with a degree-four harmonic coefficient, and such series obey their own duality identity, with $t^{-6}$ in place of $t^{-2}$. The isotropic part is built from the first two derivatives of the ordinary theta series. Writing both in terms of the modular forms that $A_4$’s theta series belongs to might show whether $\lambda_5(1)$ is a small number with a closed form or a coincidence of size $10^{-5}$. Either answer is a finite computation on known series, and it is the question the earlier essay asked, now in the form the exact curvature leaves.