Undoing a twin shows only one of its two mistakes

A twin leaves an edge that nothing else can rested on one piece of algebra. A crystal twinned at a fraction $\alpha$ records, for each pair of reflections related by the twin law, a mixture of their two intensities that keeps the pair’s sum and multiplies the pair’s difference by exactly $1 - 2\alpha$. That identity turned the pair test into a measurement with a hard edge, and it says something further, about undoing. A multiplication that loses nothing can be reversed, and if the sum is kept and the difference shrunk by a known factor, dividing the difference by that factor gives back the untwinned intensities exactly.

Crystallographers do this, and call it detwinning. Each measured pair $J_1$, $J_2$ is replaced by

$$I_1' = \frac{(1 - a)J_1 - aJ_2}{1 - 2a}, \qquad I_2' = \frac{(1 - a)J_2 - aJ_1}{1 - 2a},$$

with $a$ an estimate of the twin fraction, and the structure is then solved from the $I'$ as though the crystal had never been twinned. Two things can go wrong. The estimate $a$ may not be the true $\alpha$, and the measured intensities carry errors, which the division does not leave alone. Both have exact answers, and they combine in an awkward way. A wrong estimate is visible on one side of the truth and invisible on the other, and noise always pushes the estimate to the invisible side.

The inverse divides the difference #

For intensities without error and the true fraction, the inverse is exact. Over 594 pairs of a 24-atom structure twinned at 0.3 it returns every untwinned intensity to within $4 \times 10^{-16}$ of its own size, the rounding of the arithmetic. In the language of the pair contrast, $H = |I_1 - I_2|/(I_1 + I_2)$, detwinning with a fraction $a$ multiplies each pair’s twinned contrast by $1/(1 - 2a)$. Since the twinned contrast was the untwinned one times $1 - 2\alpha$, the detwinned contrast is

$$H' = H \cdot \frac{1 - 2\alpha}{1 - 2a}.$$

When $a = \alpha$ the factor is one and every pair is restored. When $a$ is larger the factor exceeds one, every contrast is stretched, and a contrast can be stretched past one. A pair whose contrast exceeds one has a negative member, because the difference has become larger than the sum.

A negative intensity cannot exist. It is the square of the modulus of a structure factor, and a computation that returns one has been handed a wrong number. That makes the negatives an instrument. They cannot appear at the true fraction, they must appear above it, and how many appear is fixed exactly by the distribution of untwinned contrasts.

Negatives begin exactly at the truth #

A pair goes negative when $H' > 1$, that is when its untwinned contrast exceeds $(1 - 2a)/(1 - 2\alpha)$. For a structure without a centre of symmetry the untwinned contrast is uniform between zero and one, the fact that gave the pair test its straight lines, and one that holds only after the intensities have been put on a common scale, shell by shell, as the average that knows the atoms does. So the share of pairs that go negative is one minus that threshold:

$$\text{share} = \frac{2(a - \alpha)}{1 - 2\alpha} \quad \text{for } a > \alpha, \qquad 0 \text{ for } a \le \alpha.$$

It is zero up to the true fraction, exactly zero and not merely small, and then a straight line that reaches one at $a = \frac12$, with slope $2/(1 - 2\alpha)$.

This is Britton’s plot, published in 1972, and the reason it works is the uniform distribution read from its far end. The pairs that go negative first are those with the largest untwinned contrast, the ones where one member was nearly absent. Uniformity means there is a fixed density of them all the way to a contrast of one, and so a fixed rate of new negatives per unit of overcorrection. At the three fractions in the figure, fifteen trial fractions each, the measured share departs from the line by at most 0.03, which is the scatter of 594 contrasts about a uniform distribution. Below the truth, at every trial, the count is zero.

An underestimate leaves a twin #

Now take $a$ smaller than $\alpha$. The factor $(1 - 2\alpha)/(1 - 2a)$ is less than one, every contrast is still shrunk, and nothing can go negative. Every intensity is positive and every pair looks plausible. But the result is not the untwinned crystal. Each contrast is the untwinned one times a single factor less than one, and that is precisely what a twin does. An underestimate does not undo the twin: it leaves a smaller twin, of a fraction $\beta$ fixed by

$$1 - 2\beta = \frac{1 - 2\alpha}{1 - 2a}, \qquad \beta = \frac{\alpha - a}{1 - 2a}.$$

The pair test of the earlier essay finds the residual twin to three places: undone with 0.1, a crystal twinned at 0.3 still carries a twin of 0.250, and undone with 0.25 it carries one of 0.100, both on the formula. Not one pair goes negative anywhere on that side. The two instruments divide the line between them. The count of negatives sees only overcorrection, and the pair test sees only undercorrection. At the true fraction both read zero, and that agreement is the evidence that the twin has been undone.

The formula for $\beta$ is a composition law, and it says something about what a twin fraction is. Every twin of a given law multiplies pair differences by its own factor $1 - 2\alpha$, and doing two such mixings in succession multiplies the factors. Twin fractions do not add. Their factors $1 - 2\alpha$ multiply, so the natural measure of how twinned a crystal is would be the logarithm of that factor, which does add. On that scale a twin of 0.45 is not one and a half times a twin of 0.3. Its logarithm is 2.30 against 0.92, two and a half times as much twinning. That is why the last few hundredths before a half cost so much more than the first.

Noise moves the corner, always the same way #

Real intensities carry measurement errors, and an error enters the count of negatives from both directions. Weak reflections can be measured as negative even in an untwinned crystal, because a measured intensity is a count with the background subtracted. And near the true fraction, pairs that should stay just inside the quadrant are pushed across an axis by their errors. The corner of Britton’s plot rounds off.

The standard way to read a fraction from a rounded curve is to fit the straight part and extrapolate it down to where noise alone would leave the count. Doing that gives 0.303 for exact data, 0.296 with 2 per cent error, 0.289 with 5 and 0.269 with 10. Every estimate is low, and lower the larger the error. The reason is in the shape. Noise adds negatives below the truth, where there should be none, and the rising part is pulled down and to the left with them, while far above the truth the share is near one whatever the noise and the curve is pinned. The fitted line pivots about its top and its foot moves left.

The plot carries a second reading of the fraction, which is a check on the first. The line’s slope is $2/(1 - 2\alpha)$, so the slope alone gives $\alpha = \frac12 - 1/\text{slope}$, with no reference to where the line meets the floor. For exact data the fitted slope is 4.99 and reads 0.2995. With error the slope flattens, to 4.68, 4.44 and 3.81, and reads 0.286, 0.275 and 0.238, low again and further off than the intercept. On exact data the two readings agree to the third decimal; on noisy data they part, and the size of the disagreement is a measure of how much the noise has bent the plot. A fraction read from a Britton plot without that comparison carries a bias it gives no sign of.

A low estimate is the invisible kind of mistake. A crystal twinned at 0.3, undone with the 10 per cent estimate of 0.269, keeps a twin of $(0.3 - 0.269)/(1 - 0.538)$, about 0.07, and produces no excess negatives to say so. The remedy the previous section points to is to run the pair test on the detwinned data, which is the one instrument that sees that side. The same errors that biased the estimate blur the pair test’s edge, though, and a residual of seven per cent is near what it can resolve on a few hundred pairs. Noise does not merely make the estimate uncertain. It makes it uncertain in the direction its own check cannot confirm.

Using both instruments, and why they still fall short #

The composition law suggests a repair. Estimate the fraction from the negatives, undo the twin with that estimate, run the pair test on what is left to measure the residual twin, and compose the two: $1 - 2\alpha = (1 - 2a)(1 - 2\beta)$. If the pair test read the residual exactly, the composed fraction would be exact, whatever the bias of the first estimate, because the composition law is algebra rather than statistics.

It does not read the residual exactly, and for the same reason the first estimate was low. A measured error adds a random difference to every pair, so contrasts come out larger than they are, and a larger mean contrast reports a smaller twin. Both instruments read low under noise, and the second cannot fully correct the first. On the crystal twinned at 0.3, composing the two moves the estimate from 0.289 to 0.295 with 5 per cent error, and from 0.269 to 0.280 with 10. The gap to the truth is roughly halved and does not close. The pair test run on the raw twinned data with 10 per cent error reads 0.26, lower than either.

The lesson is not that the methods fail; at a few per cent error they are good to a hundredth. It is that the error in the estimate has a sign, known in advance, and that the sign points at the side no internal check can see. A detwinned data set with no excess negatives and a small residual pair contrast is consistent with a correct fraction and also with a fraction a few hundredths low. The difference shows only in the refinement, where an undercorrected twin leaves the model’s intensities systematically too far apart within each pair. A translation that is nearly there met the same shape of problem, a bias that pushes a statistic toward a wrong but self-consistent answer, and how many reflections a centre needs the cure that does not work here: more reflections reduce the scatter of an estimate, not its bias.

What the division costs #

The division by $1 - 2a$ applies to errors as well as to differences. If each twinned intensity carries an independent error $\sigma$, the detwinned intensity $((1 - a)J_1 - aJ_2)/(1 - 2a)$ carries

$$\sigma' = \sigma\,\frac{\sqrt{(1 - a)^2 + a^2}}{1 - 2a}.$$

At a fraction of 0.1 the factor is 1.13. At 0.3 it is 1.90, at 0.4 it is 3.61, and at 0.45 it is 7.11. As the fraction approaches a half it grows without bound, and at a half it is infinite: the difference was multiplied by zero and nothing divides it back.

The measured factors, from five independent draws of noise at each fraction, agree with the formula to within two per cent: 1.93 against 1.90 at 0.3, 7.22 against 7.11 at 0.45. Since a counting error falls only as the square root of the counting time, matching the precision of untwinned data takes the square of the factor in time. That is 3.6 times as long at 0.3, 13 times at 0.4, and about fifty times at 0.45. A twin near a half is not a nuisance that a computation removes. It is a fifty-fold loss of measuring time, paid in the only currency a diffraction experiment has. The redundancy of a data set, the number of times every reflection is measured over, is where that time would have to come from.

Detwinning is not the only way to use twinned data. A refinement can compare a twinned model with the twinned measurements directly, mixing the model’s intensities by the same fraction rather than unmixing the data, and that is now the usual practice. It never divides, so it never produces a negative intensity and never inflates an error bar. It does not escape the cost. The information about each pair’s difference reaches the detector at a fraction $1 - 2\alpha$ of its strength whichever way the data are handled, and a refinement that never divides simply pays the same price as a wider uncertainty on the parameters that depend on those differences.

Correctly undone, and still misleading #

Suppose the fraction is known exactly and the twin undone with it. The detwinned intensities are then unbiased, but each carries an amplified error, and an amplified error is extra variance. Extra variance widens the intensity distribution, which is what a centre of symmetry does to it. Whether there is a centre is a statistic reads the centre from exactly that widening.

With 10 per cent error the second moment of correctly detwinned data stays near 1.97 up to a fraction of 0.3, then climbs to 2.06 at 0.4 and 2.42 at 0.45, a third of the way from the acentric value of two to the centric value of three. Nineteen per cent of the intensities are negative, against four per cent for the same error without a twin. A detwinned data set from a crystal twinned near a half therefore argues, through the same statistic that a twin hides in, for a centre of symmetry the crystal does not have. The twin narrowed the distribution, detwinning restored it, and the amplified error then widened it past where it started. Each step is correct arithmetic, and the last statistic is wrong.

What the account has to refuse #

The refused claim is the limit of everything above: that a twin of exactly one half can be undone. Its mixing multiplies every pair’s difference by zero, the two members of every pair are recorded as equal, and the information about which was the stronger is gone from the data. No division recovers it and no refinement does either. The remaining tests check each statement of the essay against an independent computation. The inverse is checked against the untwinned intensities, the negative share and the residual twin against their formulae, the measured error against the propagated one, and the noisy estimates against the truth they miss.

What the figures cannot show #

The intensities come from a two-dimensional model of 24 equal atoms without a centre of symmetry, and the errors are Gaussian with one standard deviation for every reflection, a fixed fraction of the mean. Real errors are larger for strong reflections in absolute terms and for weak ones in relative terms, and they are not Gaussian for the weakest. The shares of negatives and the bias of the noisy estimate depend on that error model, and the figures give their direction and rough size rather than numbers for any instrument. The exact results do not depend on it: the line from the true fraction, the composition of twin factors and the propagation formula are algebra.

The account also assumes the law is known and the twin has two domains. For a centrosymmetric structure the untwinned contrast is not uniform, and Britton’s plot is curved rather than straight, though its corner is still at the true fraction. A twin with three or more domains mixes each reflection with several partners, its mixing matrix has more than two eigenvalues, and its cost is set by the smallest of them.

Who worked it out #

Britton described the count of negative intensities as an estimate of the twin fraction in 1972. Yeates set out the pair statistics and their use in detwinning in a review of 1997, and refinement against twinned data, which mixes the model rather than unmixing the data, was made routine by Herbst-Irmer and Sheldrick in 1998. That twinning acts on a pair as a matrix with eigenvalues one and $1 - 2\alpha$ is the algebra under all of it. The exact straight line from the true fraction, the residual twin and its composition law, the direction of the noise bias and the centric drift of correctly detwinned data are computed here.

Still open: three domains, and the eigenvalue that sets the price #

A crystal can be twinned by a three-fold law, with three domains in fractions that sum to one, and then each reflection is mixed with two partners. The mixing matrix is three by three. For equal fractions it has one eigenvalue of one, for the sum, and two equal ones of zero, and the twin cannot be undone at all. For unequal fractions the two smaller eigenvalues are complex conjugates in general, and the cost of undoing the twin is set by their modulus. Which combinations of fractions make that modulus small, whether there is a Britton plot with a corner for each domain, and whether the pair test’s edge has an analogue for triples are all finite questions on the same arithmetic. Twinning by merohedry counts which crystal classes allow such laws, and a twin is a symmetry the crystal lacks is where the coset arithmetic behind that count is set out. None of it has been computed here.