Averaging threw the answer away before the model ever saw it

Take a metal sheet, punch a repeating pattern of holes in it, and ask how stiff the result is. Answering properly means solving a small physics problem on one repeating tile, which is slow enough that nobody wants to do it inside a design loop. So you train a model to guess the answer from a picture of the tile. It works beautifully on the hole shapes you trained it on and collapses on the ones you did not.

I have written about that collapse twice. Both times I was, without noticing, assuming the same thing everyone else in this literature assumes: that the failure lives in the model, or in how few shapes it has been shown. Fix the architecture, or buy more families of shape.

This note is about the other possibility. The number the model is trained to reproduce is computed by taking an average. Averages lose things. If what they lose is the very thing that distinguishes one family of shapes from another, then no model and no quantity of data can recover it, because it was gone before any learning started.

That turns out to be measurably true, and once you measure the missing part you can hand it to the model as three numbers, which then beat all 2,304 pixels of the picture by a distance.

The field names, for anyone searching onward: mechanical metamaterials for the sheets, numerical homogenization for the slow calculation, surrogate for the model that replaces it, out-of-distribution generalization for the way these models fail. I will not lean on those words again.

What an average is allowed to keep

The stiffness of the patterned sheet comes out of the tile calculation by averaging the internal stresses over the whole tile. That average is the answer. It is not a summary of the answer computed afterwards; it is the definition.

Averaging is a projection, which is a formal way of saying it is a doorway with a size limit. Some of what the tile can do passes through and the rest does not, and what does not pass is not arbitrary. The average keeps the two ways the tile can slide bodily as a rigid object. Every other way the material inside the tile can move against itself is projected out.

Now hold that next to the learning problem. The model is fitted to reproduce the average. If two tiles differ only in a respect the average does not record, the model is never once shown that they differ. More data does not help, because the extra data carries the same blind spot. A better architecture does not help, because it is the target that is blind.

This is a claim about identifiability rather than about difficulty. It does not say the model is badly built. It says its answer key is missing a column.

Does the discarded part exist?

An argument like that is only worth anything if the discarded content is real, and if it separates tiles that the stiffness cannot separate at all. So I went and measured it.

Same tile, same equations, same periodic boundaries, but with the mass put back in. Instead of asking how the tile responds to being squeezed, ask what it does when it rings: which frequencies, and what shape the motion takes at each one. Two of those shapes are the rigid slides the average keeps. Everything above them is precisely what the average throws out.

For any one of those shapes the motion splits, exactly and point by point, into three parts: how much of it is stretching, how much is shearing, how much is rotating. Weight each part by where the metal actually is, normalise the three to sum to one, and you have three numbers describing the character of that mode.

The extractor is checked against the one case with an exact answer. On a uniform sheet the modes are plane waves whose characters are known in closed form, and it returns them to six decimal places. The rigid modes come back at 1e-13 instead of zero. Refining the mesh drops the error by a factor of 16.0, where exact second-order convergence predicts 16. Rotating the solution arbitrarily inside a degenerate pair moves the answer in the fourteenth decimal place. Ten tests, all passing, and the homogenization solver underneath has its own six.

Then the evidence. Four of the twenty-two families in the benchmark are exactly isotropic in plane: circular holes, square holes, cross-shaped holes, six-pointed star holes. Their symmetry group forces the two stiffnesses to be equal, and they come out equal to machine precision. A model trained on those four has never observed directional stiffness in its target, because there is none to observe.

Identical in the target, to machine precision, and spanning 0.19 in the thing the target discards, with per-tile distributions that barely overlap. Across the whole pool the rotational fraction runs from about 0.64 to 0.95 and is essentially uncorrelated with the stiffness ratio, which is exactly what you would expect of a quantity the stiffness tensor does not encode.

So the discarded information exists, and it is directional. That is the same kind of information whose absence broke the model in the first study.

three numbers against 2,304

A nice argument. The test of it is whether the discarded content is usable.

So: take the softest discarded mode, take its three character numbers, add the tile's solid fraction, which is free and says how much metal is left, and give those four numbers to the same model in place of the picture. Train on four families, score on the eighteen it has never seen, and repeat over 120 paired runs with independently drawn family subsets and model seeds, so every arm sees identical subsets and only the features differ.

Cost is in units of one physics solve, the thing being predicted. Score is on the scale where zero is what you get by ignoring the input and always answering with the average, so the third row is a model that has been handed a complete description of the tile and is doing worse than a constant.

Four numbers, three of which come from a calculation costing a quarter of the answer it replaces, beat the free control on 97 of every 100 paired runs.

The same contrast in one pair of numbers: on tiles drawn from its own training families the picture model scores 0.994, which is the accuracy this literature reports and the reason anyone uses these models at all. On the eighteen unseen families that same model reaches -0.306, and the four-number model reaches +0.508.

It is not the learner. Both of those are the same learner.

not any spectrum, and not symmetry

Two objections had to be closed before I would believe my own result.

First: perhaps any spectral description of a shape transfers, and elasticity has nothing to do with it. There is published prior art to test that against. The Laplace-Beltrami spectrum, used as shape-DNA, is a rotation-invariant spectral descriptor for exactly this kind of design work, and it is purely geometric. Computed on the same tiles in the same harness it reaches +0.106, and its scale-free ratios reach +0.038. The standard hand-crafted geometric bundle, seven Hu moment invariants plus the radially averaged two-point correlation function, reaches +0.173. The elastic spectrum reaches +0.356. It is not that any spectrum transfers. It is that this one does.

Second: perhaps this is a symmetry problem, and the answer is an architecture that knows about symmetry. I cannot reimplement one of those, but I can impose the tile's exact symmetry group by brute force: train on all eight mirrored and rotated copies of every tile, average each prediction over the eight copies of the test tile, and transform the target along with it. At eight times the training and inference cost, the median moves by -0.010. The spread narrows and the middle does not move.

That is a lower bound on what a properly equivariant model would do and not a test of one, and I claim to have outperformed nothing. What it supports is the weaker statement I need: imposing the symmetry does not recover the information, which is what you expect if the information is missing from the target rather than mishandled by the model.

The sharpest version is against a convolutional network, which is the fair geometric baseline. The network on the image reaches +0.098. Concatenating the four numbers into its head lifts it to +0.179. The four numbers alone, on a plain multilayer perceptron with no image at all, reach +0.362, beating the network on 83 percent of subsets.

Throwing the image away and keeping the physics is worth more than adding the physics to the image.

which of the four numbers does the work

Worth knowing, because this is the part that says what the finding means rather than that it happened.

Three numbers beat twenty-four. Adding the other twenty-one makes transfer worse, which is the signature of a model absorbing information specific to the families it was trained on.

The six raw frequencies are nearly worthless and their five ratios are strong. So the transferable quantity is the shape of the spectrum and not its scale: how stiff a tile is in absolute terms is a fact about its family and does not survive leaving it.

Inside the three, the content is rotational. Rotation alone is worth +0.146 and stretching alone +0.057; rotation and shear together +0.349; all three +0.388. The thing that travels is a ratio, how the tile's softest internal motion divides between rotating and shearing.

That has a name in mechanics. It is the micropolar, or Cosserat, degree of freedom: the ability of the material inside a cell to spin relative to the cell. The ordinary first-order theory of a continuous material has no slot for it, which is precisely why the averaged stiffness cannot record it.

And it helps the component it ought to help. On the stiffness component whose transfer collapsed in the first study the three numbers are worth +0.409. On the component that did not collapse they are worth -0.088.

what this doesn't fix

Absolute accuracy on unseen shapes is still poor. A score of +0.36 is a large relative improvement on a very low base, not a replacement for the physics solve. The honest description of the four numbers is a cheap diagnostic and a prior.

Two families are consistently made worse: chevrons and a two-hole cell. The honeycomb that was mispredicted about sixfold in the first study improves from -66 to -16, a fourfold reduction in squared error and still wrong. Sixteen of twenty-two families improve under a forest and fourteen of twenty-two under a neural network, and the two most extreme families are improved without being repaired.

The clean split between the two stiffness components only holds at full resolution. Everything here is two-dimensional, linear-elastic, one void phase, one stiffness contrast. Nothing shows this is the right descriptor in three dimensions or at large strain.

And the obvious follow-up failed. If these numbers carry the transferable content, they ought to tell you which families to buy, which is the question left open by the previous study. They do not. The range of rotational fraction across a candidate training set correlates with the outcome at about +0.25 to +0.32, which is weak, and the free control, the range of solid fraction, beats it at +0.40. Knowing which families hold the content is not the same as the model being able to use it.

The comparison this work does not contain is against a trained equivariant graph network. That is the missing piece, and I would rather say so than let the orbit-averaging experiment stand in for it.

the part that isn't about metal

Strip the engineering out and the claim is about labels.

If your label is a summary statistic, everything the summary discarded is invisible to anything trained on it. Written down, that is obvious. In practice it is not, because the label is the part of the pipeline nobody treats as a variable. It arrives, it is correct, and all the attention goes to the model and the data.

Three consequences I would actually act on:

• A benchmark for generalization should state which degrees of freedom its training families hold fixed. A training set that is uniformly symmetric cannot teach a model what asymmetry looks like, however large it is.
• More data inside a design space cannot repair a blindness in the target. In the first study more data and a finer mesh made cross-family transfer worse, which is the opposite of what a coverage explanation predicts.
• A richer input is not automatically a safer one. The full picture let the model fit its training families almost exactly, and it was that fit which failed to travel. Four physically chosen numbers could not fit as well, and went further.

Everything is open: the extractor and its validation suite, the descriptors at every resolution and coarsening setting, every result file behind the numbers above, and the preprint, at https://doi.org/10.5281/zenodo.22137052