The average was over the eigenvectors

The argument that explained the icosahedron ended with a rule and a natural next question. The rule said a cut shell cannot move a localisation’s answer when the shell has degree one and the cage’s design strength reaches four under Pipek–Mezey or three under Boys. The question was what happens between designs: a cage that nearly averages a quartic like a sphere should move by an amount that shrinks with its degree-four Legendre sum, and if the movement were proportional to the sum, the sum would be a quantitative predictor that could be computed for any cage in a line.

The first test of that was the easiest available, and the answer was that nothing moves.

An icosahedron pulled out of shape #

Take the icosahedron at four electrons, with the constant orbital and two members of its three-fold shell occupied, and stretch its vertices along one axis while keeping its thirty edges. The vertices stop being a design immediately. At a stretch of 1.05 the vertices’ design strength has fallen from five to one, and by a factor of two their degree-four Legendre sum is 0.118, which is a fifth of the octahedron’s 0.583. By the question as it was put, this cage should have moved a good way toward the octahedron’s fifteen per cent.

The Pipek–Mezey functional is the same at every stretch and over every plane of the shell, to one part in 3×10¹⁴, which is the arithmetic’s floor. Boys reaches 1.30 at a stretch of 1.25, 1.71 at 1.5 and 2.96 at two. The rule’s condition reads “safe” at no stretch above one: the design strength is too low, and the shell’s measured degree in the new vertex directions is three. That is a straightforward miss, and it runs in the direction the earlier essay had already flagged as unproved. Twenty computations had shown that everything the rule called unsafe does move. This one does not.

What a Hückel matrix knows about a vertex #

The cause is plain once it is looked for. A Hückel matrix with one hopping per edge holds the graph and nothing else; no element of it contains a distance. So its eigenvectors are the same whatever the vertices do, and so are the three shell members. Pipek–Mezey, in a basis with one function per site, is a sum of fourth powers of orbital coefficients. It has no positions in it either. Stretching the cage changes nothing that functional is computed from, which leaves it nothing to feel.

So the step in the earlier argument that needs reading again is the one that said re-orienting the shell is the same as turning the cage. That is true when the three shell members are the coordinates, and on the regular solids they are. But the object the functional actually sums over is not defined by the coordinates. Place site $i$ at the three-dimensional point $\mathbf e_i=(e_{1i},e_{2i},e_{3i})$ whose components are the shell members’ values on that site. That is the set of points being averaged, and it exists whether or not anybody knows where the vertices are.

On a regular icosahedron those two sets coincide up to a scale, and that coincidence is the whole reason the earlier rule worked on every case it was given. On the stretched cage they part. The left-hand set is what the eye sees. The right-hand set is what the criterion reads, and the shell’s own points are computed here and checked, site by site to a part in a billion, against the unstretched icosahedron.

The two numbers that decide it #

Written out, the rule is short. An occupied orbital is $u_0 c + \mathbf u\cdot\mathbf e_i$, with $c$ the constant orbital’s coefficient, so the functional is a sum over sites of a quartic in $\mathbf e_i$. Expanding it leaves four moments of the shell’s point set: the first and second, which are fixed by orthonormality, and two that are not:

$$M^{(3)}_{abc}=\sum_i e_{ai}e_{bi}e_{ci},\qquad M^{(4)}_{abcd}=\sum_i e_{ai}e_{bi}e_{ci}e_{di}.$$

Re-orienting the shell is an orthogonal transformation of those indices. The functional’s whole landscape is therefore unchanged by it when $M^{(3)}$ vanishes and $M^{(4)}$ is isotropic, the same in every direction, and that is proved rather than measured. Write $A_4$ for the size of $M^{(4)}$'s non-isotropic part. Everything Pipek–Mezey can do on such a shell is in those two numbers.

Boys reads a different pair of things, because it does contain positions. Its centroid for each orbital mixes the shell’s points with the vertices through a matrix $L_{ab}=\sum_i r_{ai}e_{bi}$ and a tensor $T_{abc}=\sum_i r_{ai}e_{bi}e_{ci}$. It cannot move when the vertices are centred, $T$ vanishes and $L^{\mathsf T}L$ is a multiple of the identity. On the stretched icosahedron $T$ is zero by inversion but $L^{\mathsf T}L$ is not isotropic, and its anisotropy grows from 0.08 to 1.2 with the climb in the figure above.

That also retires a statement the earlier essay made in passing, that Boys needs less than Pipek–Mezey: three degrees of averaging against four. That ordering came from the regular solids, where both tests read the same point set. The stretched cage is safe under Pipek–Mezey and not under Boys, which on the published table of twenty never occurs. The two criteria have different tests rather than one test at different strengths.

Between designs, once something can change #

The original question can now be asked in a form where it has an answer: hold the vertices and move the eigenvectors. That requires a change to the matrix that keeps a three-fold level, since a shell that splits has no cut left to re-orient, and most distortions split it. An octahedron pulled along an axis turns its three-fold level into a one and a two, and at four electrons the occupied set is then decided by energy.

The icosahedron has a way through. Its twelve vertices are one orbit of the smaller group $T_h$ at the point $(0,1,\varphi)$, and that group has three-dimensional representations and splits the thirty edges into a class of six and a class of twenty-four. Set the six to hopping $\tau$ and leave the rest at one. The three-fold level survives at every $\tau$ tried from 0.5 to 2, because $T_h$ guarantees it. Its members stop being coordinates, because $T_h$’s three-fold representation occurs twice among twelve sites and the matrix mixes the two copies. The shell’s points slide along a family, and they are a design only at $\tau=1$.

The movement falls to zero at $\tau=1$ with a corner in it. A smooth function of $\tau$ with a minimum at zero would be a parabola there, and this is not; both sides have a slope. The scan here is over every plane of the shell, with 240 directions for the plane’s normal and twelve starts at each, so each point is a real maximum over orientations rather than the five orientations of the earlier survey.

A corner means the movement is linear in the distance from the design, and that settles the shape of the law:

The movement goes as the square root of the Legendre sum. The fitted slope over the eleven settings within ten per cent of the design is 0.499, against the anisotropy it is 0.997, and the prefactors are 0.190 and 0.528. The proportional law the question proposed would be a line of twice the steepness: fitted at $\tau=1.1$, it is about twenty times too small at $\tau=1.005$.

The reason is in the addition theorem that made the Legendre sum so convenient. The sum over all pairs of $P_4$ equals, up to a constant, the sum over $m$ of the squares of the degree-four harmonics’ averages over the points. The functional’s own response is first order in those averages, since a quartic’s vertex sum is that average with some coefficient. So a sum that vanishes exactly at a design vanishes quadratically there, and it was never going to be the thing proportional to a response. It is a very good test of whether a set is a design, and for exactly that reason it is the wrong number to read a size from until its root is taken. Past the fitted range the points drift above the line, by a quarter at $\tau=2$, as a first-order argument would expect.

The vertices and the shell, separately #

With both quantities available they can be set against one another directly, and each can be made to hold while the other fails.

The second row is the icosahedron’s own vertices, a 5-design by any reckoning, under a hopping that is not its own: it moves by ×1.012. The third is the stretched cage, not a design, and it does not move. The fourth was built to answer the obvious objection, that perhaps the stretched cage is a special case because its matrix is still the icosahedron’s. Vertices at $a=1.3$ on the same $T_h$ family have a Legendre sum of 0.016. Give them a hopping that falls exponentially with distance over 0.5 of the radius, out to second neighbours, and solve for the $\tau$ on the six edges at which the shell’s fourth moment becomes isotropic; it is 1.616. That matrix is not icosahedral in any disguise. Its next level, five-fold on the true icosahedron, sits split into a pair and a triple 0.33 apart. It has only $T_h$, a group that allows a quartic. The shell has none, by arithmetic rather than by symmetry, and at ten per cent either side of the root it moves by ×1.008 and ×1.009.

Seven cages the rule did not see #

The test that decides between the rules is a set of systems neither had met, fixed before any was run, with the moment rule’s verdict for each read off the rotation group: safe for the icosahedral ones and not for the rest.

The truncated icosahedron is the telling row. Its sixty sites hold three copies of the three-fold representation, so its first excited shell is a mixture and not the coordinates: 99.97% of the coordinate functions lie in it, and the rest makes it a polynomial of degree above four in the vertex directions, where the measurement stopped looking. The vertices’ rule says it moves. It does not, under either criterion. Then the same cage was given two kinds of bond, as the molecule has, with the thirty between hexagons stronger than the sixty around pentagons. At a ratio of three, far past anything its measured bond lengths of about 1.40 and 1.46 Å would give, the share falls to 98.8% and nothing moves. The shell’s fourth moment stays isotropic to better than 10⁻¹⁵ at every ratio.

The octahedral and tetrahedral rows all move, at full scan, from ×1.031 on the cuboctahedron to ×1.294 on the truncated tetrahedron, and Boys stays at one everywhere except on the tetrahedral cage, whose tensor $T$ is non-zero because it has no inversion. Together with the regular solids, the stretched series, the tuned cages and the dial, the moment rule makes seventy-four predictions on forty-nine settings and every one is right.

A full scan also reprices a number the earlier survey reported. Its fifteen per cent for the octahedron came from five orientations, which is a lower bound. Over every plane it is 25.7%; the cube’s 5.1% is 6.7%, and the tetrahedron’s 31% is 49%. The rule’s verdicts are untouched, since a sample cannot create a movement that is not there. The sizes were always floors.

The explanation that was thrown out too soon #

The strongest thing in the table is why the icosahedral rows had to come out safe, and it reverses a verdict.

The earlier argument tried symmetry first and refused it, correctly on the grounds it was offered: the group cannot carry an arbitrary plane of the shell onto an arbitrary other plane, because 120 operations cannot sweep two continuous parameters. It concluded that the explanation was averaging and that the group was beside the point. But there is a second way for a group to matter, and it does not need to move any plane anywhere.

$M^{(4)}$ is built from the shell’s points, and the group permutes those points while acting on the shell by its three-fold representation. So $M^{(4)}$ is unchanged by every operation of the group, and its non-isotropic part is a harmonic of degree four that the group leaves alone. If the group has no such harmonic, that part has nowhere to be. What forbids the movement is the absence of an invariant, and the absence is a fact about the group alone, whatever the shell’s vectors are.

Counting them takes one line, from the character of a rotation on degree $l$ averaged over the group. The icosahedral group has none until degree six. The octahedral group has one at degree four, which is the octahedron’s fifteen per cent, and the tetrahedral group has one at three and one at four. This is why the argument above never needed C60’s shell to be coordinates: any set of points the icosahedral group can draw with a three-fold shell is a 5-design, wherever on the sphere the orbit starts. The same count says what $T_h$ allows, which is one quartic, so that a single number can pass through zero along the dial. The corner at $\tau=1$ is that number changing sign.

The connection worth carrying is how old this is. Molien counted invariants of finite groups in 1897, and Sobolev used the fact in 1962 to build cubature rules on the sphere: a rule invariant under a group needs testing only on the group’s invariant polynomials. Goethals and Seidel gave the design-theoretic statement around 1980, with orbits of the rotation groups as the standard construction. So the group and the averaging are not rival explanations, as the earlier essay had them. The averaging is what the functional needs. The group’s list of invariants is the most common reason a point set supplies it, and the $T_h$ matrix tuned to $\tau=1.616$ is the case that shows it is not the only one.

The earlier essay now carries this correction beside its own verdict.

How much, and across which cages #

The question asked for a quantitative predictor. Within the dial’s family it is one, and the constant holds to about three per cent over a factor of twenty in movement. Across cages it is not.

Among cages with no third moment, the ratio of movement to root sum runs from 0.132 on the cube to 0.336 on the octahedron, a factor of 2.5, and the two tetrahedral cages sit well above that with a term the sum does not contain. The size of the response depends on the shape of the landscape being moved as well as on how far it is moved: how many basins, how sharply peaked. So the honest answer to whether the sum predicts how undetermined an input is comes in three parts. Zero or non-zero: yes, exactly, for a shell of this kind. How much, within one family that changes one hopping: yes, by its root. How much, between different cages: to within a factor of about two and a half, which is an order of magnitude and not a prediction.

The six inputs the family survey singled out are not settled by any of this where their cut shell is of degree two or three, and the count of descriptions is only protected in the same conditions. The participation numbers read off three shapes are vertex sums of the same degree, so what is proved for the functional is proved for them.

How it was calculated #

Every matrix here is real and symmetric, with one site energy and hoppings stated per class, diagonalised by Jacobi rotations. The occupied space is the lowest level, which on all of these is one constant vector, and two of the three members above it, chosen as the plane normal to a direction on a Fibonacci lattice over the sphere: 240 directions on the dial, 120 on the tuned cages, 100 on the regular solids, 60 on the stretched series and 40 on the outside cages. At each plane the same basin search that measured the family maximises Pipek–Mezey from between six and twelve rotated starts and Boys from two to four. Movement is the largest best functional over the smallest.

The moments are direct sums over sites. Invariant harmonics come from the average of $\sin((2l+1)\theta/2)/\sin(\theta/2)$ over each group’s rotations. The $\tau$ root is bisection on the signed average of $x^4+y^4+z^4$ over the shell’s unit directions, in a basis fixed by projecting the $T_h$-adapted coordinates onto the shell, since a degenerate shell comes back in an arbitrary basis and that sum is not rotation-invariant. Each claim above is stated where it can fail, and the refusal is the second row of the four systems: the icosahedron’s own vertices under a hopping not their own must move, and a rule read off the vertices would have said they cannot.

What a graph model cannot tell apart #

The stretched icosahedron is an artefact to exactly this extent. In any real molecule, moving the vertices changes the hoppings, so the eigenvectors do move, and a real stretched cage behaves like the dial and not like the flat line in the first figure. What the flat line shows is that the vertices were never the operative thing; the geometry acts through the matrix or not at all. A basis with overlap would add further terms, because a Mulliken population is no longer a square of one coefficient.

The moment argument is for one constant orbital and a three-fold shell. The functional’s dependence on moments is cleanest there, because the constant orbital contributes nothing to them. Shells of degree two and three, the icosahedron at ten, twelve and twenty electrons, sit above other occupied levels whose cross terms enter, and nothing here says what they do.

The exponential hopping is a convenience. The tuned cage needed only some matrix with $T_h$ and a free parameter, and its 1.616 means nothing outside this example. The direction of the argument does not rest on the law, since the point is that a root exists at a cage that is no design.

An unmoved functional is still a limited claim. As the cages that sat on both sides of other boundaries showed, a description that cannot distinguish two occupied spaces has not shown they are the same. Every plane of the icosahedron’s shell is a different wavefunction with a different density, left free by the diagonaliser, and a self-consistent calculation would still choose among them.

Ask what the sum is over #

The trouble came from a good sentence: re-orienting the shell is the same as turning the cage. It is true on every case that suggested it, and it quietly replaced the object in the theorem with a different object that happens to coincide with it on regular solids. The cage has been in the foreground of this line of argument since its first search because it is what one draws. Pipek–Mezey does not draw anything; it has twelve numbers per orbital and no coordinates, and the honest way to ask what it can feel is to read the formula and list what is in it.

Still open: a shell of degree two as a design in five dimensions, and the landscape’s constant #

The obvious open question is the shells this leaves out. The icosahedron’s five-fold shell draws its sites in five dimensions, and a re-orientation of it is a five-by-five orthogonal matrix, so the natural extension is a design on a four-sphere: its moments, its group invariants in that representation, and whether its sensitivity vanishes when they do. What stands in the way is the three-fold shell occupied beneath it, whose cross terms the argument above did not have to face; whether they can be written as further moments is the first calculation.

The nearer question is the constant. The factor of 2.5 between cages says the movement depends on something besides the fourth moment’s size, and the first-order argument says exactly what: the harmonic evaluated on the unperturbed maximiser. On the dial’s family that is one number per family, and it should be computable without any scan, from the design’s own best orbitals. If it is, the three parts above become one formula, and the cross-cage spread either goes away or is shown to be landscape reorganisation, which is the thing a first-order argument cannot see.