Three rotations of order twelve

Where five-fold becomes legal stated the crystallographic restriction for every dimension in one line: an $n$-fold rotation of a lattice exists in $d$ dimensions exactly when Euler’s totient $\varphi(n)$ is at most $d$. The argument was that a rotation of order $n$ satisfies the $n$-th cyclotomic polynomial, whose degree is $\varphi(n)$, and a matrix of size $d$ cannot satisfy a polynomial of higher degree than $d$ as its minimal polynomial. It gives the familiar five orders in the plane and in space and admits five-, eight-, ten- and twelve-fold in four dimensions.

The degrees that name the restriction noticed the flaw and did not follow it. A matrix may be a direct sum of blocks, and then its order is the least common multiple of theirs. A three-fold block costs two dimensions and a five-fold block costs four, so an operation of order fifteen fits in six dimensions, although $\varphi(15)$ is eight. The inequality is therefore not the condition on orders, and the table that illustrated it had been built from it. The six-dimensional column on the page about five-fold symmetry marked fifteen as forbidden, and the matrix that refutes it has six rows.

That leaves a question the correction did not ask. If the totient is not the condition on orders, what is it the condition on? The answer is exact, and it changes what “an $n$-fold rotation” means once a lattice has more than two dimensions. The totient decides which angles a lattice rotation can turn a plane by. A different sum decides which orders the rotations can have. In the plane the two questions are the same question, because a rotation of the plane has only one plane to turn. From four dimensions on they are different. The lattice in four dimensions has not one operation of order twelve but three, and only one of them turns anything by a twelfth.

What names an operation of finite order #

Take any integer matrix $M$ with $M^n = I$ for some $n$. Its eigenvalues are $n$-th roots of unity, and its characteristic polynomial has integer coefficients, so whenever a primitive $m$-th root is an eigenvalue, all $\varphi(m)$ primitive $m$-th roots are eigenvalues with the same multiplicity. The characteristic polynomial is therefore a product of cyclotomic polynomials, $\Phi_{m_1} \Phi_{m_2} \cdots$, and the degrees add up to $d$.

A matrix of finite order is also diagonalisable over the complex numbers, so nothing about it is hidden in a Jordan block. Over the rationals this means it is equivalent to the direct sum of the companion matrices of its factors: one block for $\Phi_{m_1}$, one for $\Phi_{m_2}$, and so on. Two operations are therefore equivalent under a rational change of basis exactly when they are built from the same multiset of cyclotomic blocks. The multiset is the operation’s name.

Everything else can be read off the name. The order is the least common multiple of the $m_i$, because each block returns to the identity after $m_i$ steps and the sum returns when all of them have. The determinant is $+1$ unless an odd number of the blocks are $\Phi_2$, the one-dimensional block $(-1)$. The angles come from each block separately. Over the real numbers the block of $\Phi_m$, for $m \geq 3$, turns $\varphi(m)/2$ perpendicular planes, by $k/m$ of a turn for each $k$ below $m/2$ that shares no factor with $m$. A $\Phi_1$ block keeps a line fixed and a $\Phi_2$ block reverses one.

So a four-dimensional operation built as $\Phi_3 \oplus \Phi_4$ turns one plane by a third and the perpendicular plane by a quarter, and has order twelve. $\Phi_{12}$ on its own is also four-dimensional, since $\varphi(12) = 4$. It turns one plane by a twelfth and the other by five twelfths, and it also has order twelve. The two are not related by any change of basis, since they do not even turn their planes by the same angles.

The picture at the head of this essay is that difference drawn. A point in the first plane of $\Phi_{12}$ runs round twelve positions before it returns, and so does a point in the second, which visits the same twelve in the order of a twelve-pointed star. A point in either plane of $\Phi_3 \oplus \Phi_4$ returns after three steps or after four, and the matrix has order twelve only because no smaller number of steps brings both planes home at once. The third class, $\Phi_4 \oplus \Phi_6$, does the same with four and six. All three are $4 \times 4$ matrices with integer entries, and each carries the lattice $\mathbb{Z}^4$ onto itself. “Twelve-fold” is a true description of the first, and of the other two it is only the arithmetic of their periods.

The angle is the totient’s question #

The argument about the cyclotomic polynomial is correct once it is stated about the right thing. An operation turns some plane by a primitive $n$-th of a turn, $k/n$ with $k$ sharing no factor with $n$, exactly when one of its blocks is $\Phi_n$. That block costs $\varphi(n)$ dimensions. So a $d$-dimensional lattice admits a plane turned by a primitive $n$-th of a turn exactly when $\varphi(n) \leq d$, and the enumeration below confirms it in every dimension to twelve. Nothing in the older essays’ conclusions about quasicrystals depends on the difference, because a quasicrystal’s symmetry is a statement about angles. The dodecagonal phases show twelve-fold symmetry in a plane, and the four-dimensional lattice whose shadow they are is carried by $\Phi_{12}$, not by either of the other two operations of order twelve. The same is true of the octagonal case worked out in eight-fold with no golden ratio, where the plane is turned by an eighth.

The order is a different question, with a different cost. An operation of order $n$ needs blocks whose indices have $n$ as their least common multiple, and it is free to choose them to be cheap. For each prime power $p^a$ that divides $n$ exactly, some block’s index must be divisible by $p^a$. The cheapest way to arrange that is one block per prime power, because merging two coprime indices $a$ and $b$ into one block costs $\varphi(a)\varphi(b)$ where keeping them apart costs $\varphi(a) + \varphi(b)$, and a product of two numbers each at least two is at least their sum. The cost comes to

$$\psi(n) = \sum_{p^a \,\|\, n} \varphi(p^a),$$

less one when $n$ is twice an odd number greater than one. The exception holds because the block for the prime power two is $\Phi_2 = (-1)$, and negating a block of odd order $m$ gives order $2m$ at no cost. Hiller published the formula in 1985 as the crystallographic restriction in higher dimensions. It agrees with the totient at twelve by coincidence: $\varphi(12) = \varphi(3)\varphi(4) = 2 \times 2 = 4 = \varphi(3) + \varphi(4)$. That coincidence is why twelve appears on both lists in four dimensions, once as an angle and twice more as an order.

The table sets the two costs side by side for the twenty-four orders that fit in eight dimensions, and the two columns agree through five. The first disagreement is in six, and it comes in a group of four: fifteen as $\Phi_3 \oplus \Phi_5$, twenty as $\Phi_4 \oplus \Phi_5$, twenty-four as $\Phi_3 \oplus \Phi_8$, and thirty as $\Phi_6 \oplus \Phi_5$. Each has order $n$ and turns no plane by a primitive $n$-th of a turn. Eight dimensions add six more orders of that kind, from twenty-one to sixty. The earlier essays’ “exactly when” was right for the first five dimensions and for every statement about angles. As a statement about orders in six dimensions and above it was false, and the figure built from it drew the false version.

The whole list in four dimensions #

The multisets can be listed completely for a small dimension, and four is the first dimension in which the list says something new. There are nine cyclotomic blocks that fit: $\Phi_1$ and $\Phi_2$ of size one, $\Phi_3$, $\Phi_4$ and $\Phi_6$ of size two, and $\Phi_5$, $\Phi_8$, $\Phi_{10}$ and $\Phi_{12}$ of size four. The multisets whose sizes add to exactly four number twenty-four.

Nineteen of the twenty-four have determinant $+1$ and are rotations in the ordinary sense. The other five reverse orientation. The nine orders are the familiar nine: one, two, three, four, five, six, eight, ten and twelve. But the orders spread unevenly across the classes, and the spread is the finding. Order six has seven classes. $\Phi_6 \oplus \Phi_6$ turns both planes by a sixth. $\Phi_3 \oplus \Phi_6$ turns one plane by a third and the other by a sixth. $\Phi_1 \oplus \Phi_1 \oplus \Phi_6$ turns one plane and fixes a whole plane. $\Phi_2 \oplus \Phi_2 \oplus \Phi_3$ turns one plane by a third and the plane perpendicular to it by a half, and so on. Order five has one class, $\Phi_5$, and order eight one, $\Phi_8$. Four-dimensional crystallography does not classify its point groups by order for exactly this reason: the order hides almost everything about what the operation does.

The classification counted here is still coarser than a crystallographer’s. It counts operations up to a rational change of basis, and over the integers some classes split. The plane already shows the split. Its mirror is one rational class, $\Phi_1 \oplus \Phi_2$, and two integral ones: the mirror of a rectangular lattice and the mirror of a centred one, which no change of lattice basis turns into each other. The integral count is the classification into arithmetic classes, and everything on this page lives one level above it. What the rational class does fix, and the integral refinement cannot change, is the order and the set of angles.

Fifteen, and no plane turned by a fifteenth #

The smallest order that exists without its angle is fifteen, and its operation can be written out entry by entry. It is the companion matrix of $\Phi_3 = x^2 + x + 1$, a $2 \times 2$ block, placed beside the companion matrix of $\Phi_5 = x^4 + x^3 + x^2 + x + 1$, a $4 \times 4$ block. The result is a $6 \times 6$ matrix whose entries are all $0$, $1$ or $-1$.

Its order is fifteen because of the arithmetic of its two periods and for no other reason. Its cube is the identity on the first block and a five-fold operation on the second. Its fifth power is the identity on the second block and a three-fold operation on the first. Only its fifteenth power returns everything. Its three invariant planes turn by a third, a fifth and two fifths of a turn. No plane turns by a fifteenth, and no plane could: a plane turned by a primitive fifteenth would need the block $\Phi_{15}$, which has size eight.

This is where the older essays’ phrase “an $n$-fold rotation” stops being unambiguous. A reader who hears “a lattice in six dimensions has a fifteen-fold rotation” pictures a plane with fifteen positions in it, the way a hexagonal lattice has a plane with six. That picture needs eight dimensions. What six dimensions actually permit is a three-fold turn and a five-fold turn performed at once in perpendicular planes, which together have period fifteen. Both statements are true. They are true of different objects, and they are the objects the two inequalities count.

Space already does this #

The separation between order and angle can look like a curiosity of high dimensions. It is not one, and every crystallographer already uses an example of it. Among the ten rational classes of operation in three dimensions is $\Phi_2 \oplus \Phi_3$: a turn by a third in one plane, together with a reversal of the line perpendicular to it. Its order is six, the least common multiple of three and two. Its only plane turns by a third. Its crystallographic name is $\bar{6}$, the six-fold rotoinversion, and the Tables have long noted that it is the same operation as a three-fold rotation followed by a mirror across the axis, written $3/m$.

The symbol says six, and the operation turns nothing by a sixth. That is exactly the order-fifteen situation, one dimension up from the plane and with a reversal standing in for the second turn. Its companion, $\bar{3}$, is $\Phi_2 \oplus \Phi_6$: a turn by a sixth together with a reversal, so its order is six and its angle is a sixth, and the symbol $\bar{3}$ says three. Eleven, eleven and ten builds the improper classes from these operations, and the rational classification explains why the symbols read the way they do. A rotoinversion is named by the rotation it is composed from, not by the order it has, and the order is the least common multiple of that rotation’s order and two.

The enumeration behind this figure knows nothing about blocks. It forms every $3 \times 3$ matrix whose entries are $-1$, $0$ or $1$, multiplies each by itself up to twelve times, and keeps the 1,584 that return to the identity. It then factors each one’s characteristic polynomial by trial division against the cyclotomic polynomials. Every one of the 1,584 factors completely into cyclotomics, and they fall into exactly the ten classes the block count predicts for three dimensions: the identity, the inversion, the mirror, the two-fold rotation, the proper three-, four- and six-fold rotations, and the three rotoinversions. The orders found are one, two, three, four and six, which is the restriction in space arrived at by brute force on small matrices.

The counts are symmetric, and the symmetry has a simple cause. Negating a matrix exchanges $\Phi_1$ with $\Phi_2$ and $\Phi_3$ with $\Phi_6$, so negation carries each proper rotation to its rotoinversion and keeps the counts equal. There are 308 matrices of class $3$ and 308 of class $\bar{3}$, 228 of class $6$ and 228 of class $\bar{6}$. Four and $\bar{4}$ are exchanged by negation too, because the negative of a quarter-turn is a quarter-turn the other way, and they come out at 174 each. The identity and the inversion occur once each, as they must: they are the only matrices in their classes whose entries lie in $\{-1, 0, 1\}$.

Counting both ways, dimension by dimension #

The two conditions give two counts for every dimension, and the enumeration supplies both from the same list of classes. For each $d$ it lists every multiset of cyclotomic blocks whose sizes sum to $d$. It builds each multiset as an integer matrix, measures the order by multiplication, and factors the characteristic polynomial back into blocks. Then it counts the orders that occur and the primitive angles that occur. To eight dimensions that is 500 classes built and measured. Past eight the lists are counted from the multisets alone, and the two rules are checked against them to twelve.

Three things are visible, and all three are exact.

Odd dimensions add nothing, to either list, in every dimension checked. Every block past $\Phi_2$ has even size, since $\varphi(m)$ is even for every $m$ above two, so an odd dimension can only hold what the even dimension below it holds plus one more line kept or reversed, and a line changes neither the order past two nor any angle. The argument the restriction in three dimensions made for space, that a third dimension adds no rotation order to the plane’s, is the case $d = 3$ of a statement true for every odd $d$. The number of classes does grow at odd dimensions, from twenty-four to thirty-eight between four and five for instance. The extra classes are the even dimension’s operations with one more line, kept or reversed.

The lists part at six and diverge from there. At six there are seventeen orders against thirteen angles, at eight twenty-four against eighteen, at ten thirty-five against twenty, and at twelve forty-eight against twenty-six. The share of orders that are also angles falls from all of them to about half. In high dimension most operations of a given order turn no plane by a primitive fraction of it.

The largest order runs away from the largest angle. The largest angle in twelve dimensions is a forty-second of a turn, from $\Phi_{42}$, whose size is twelve. The largest order is 210, which is $2 \cdot 3 \cdot 5 \cdot 7$, from $\Phi_3 \oplus \Phi_5 \oplus \Phi_7$ negated: two plus four plus six dimensions, with the factor of two free. The pattern is the one number theory knows from Landau’s function, which gives the largest order of a permutation of $d$ objects. The best way to get a long period from a fixed budget is to spend it on small coprime pieces, and one large cycle is the worst way. A permutation matrix is itself an integer matrix of finite order, so every order a permutation of $d$ objects can have is also an order here. The lattice does better than the permutations by paying $\varphi(p^a)$ rather than $p^a$ for each prime-power piece.

What the enumeration has to refuse #

The first refused claim is the one this essay corrects. An operation of order $n$ is said to need $\varphi(n)$ dimensions, and the $6 \times 6$ matrix $\Phi_3 \oplus \Phi_5$ of order fifteen refutes it. The second is the natural repair: that an operation of order fifteen in six dimensions is a genuine fifteen-fold rotation of some plane. Its planes turn by a fifth, a third and two fifths, and a check that accepted any operation of the right order as “fifteen-fold” would have accepted it. The third is the assumption the older essays made without stating it: that an order names an operation, so that “the twelve-fold rotation of a four-dimensional lattice” picks out one thing. It picks out three.

The tests that pass are the ones that could have gone wrong silently. The orders read from the enumerated classes are compared with the orders $\psi(n) \leq d$ admits, number by number, in every dimension to twelve and for every order to four hundred. The angles read from the classes are compared with the totient rule in the same range. Both comparisons run between computations that do not share code. One builds matrices from a list of blocks. The other is a formula on the prime factorisation of $n$. An error in the list of blocks would show up as a disagreement in some dimension.

What this settles, and what it leaves alone #

The older statement survives as a statement about angles, and the correction is small in the dimensions that matter to real crystals. Nothing in the plane or in space changes. Every quasicrystal symmetry that has been observed is a symmetry of a plane, and so an angle, and so on the totient’s list, as the cut-and-project construction uses it. What changes is the meaning of the words. “A six-dimensional lattice admits order fifteen” is true, “it admits a fifteen-fold rotation” is ambiguous, and the reading in which a plane has fifteen positions is false.

The picture has a limit of its own, and the dials hide it. Each dial shows one invariant plane on its own, and the lattice is not drawn at all. The dials look the same for any lattice carried by the operation. They cannot show that $\Phi_3 \oplus \Phi_4$ and $\Phi_{12}$ preserve different lattices. They cannot show how an operation’s planes sit relative to the lattice points, which is the information the second plane of a four-dimensional rotation carries in the cut-and-project construction. They also cannot show the integral splitting of a class, since two integrally different mirrors draw the same dial.

Nor does anything here count the finite groups. A point group is a set of operations closed under composition, and which multisets can sit together in one group is a far harder question than which multisets exist. That is the question behind the 710 arithmetic classes in four dimensions that why there is a list at all quotes and does not derive. The rational classes of single operations are the letters that list is spelled in, and there are twenty-four of them in four dimensions against its 710 words.

Still open: which groups the three twelves live in #

The three operations of order twelve raise a question the enumeration of single operations cannot answer: in which four-dimensional point groups does each appear? $\Phi_{12}$ generates a cyclic group whose projection onto its first plane is a twelve-fold rotation group, and it sits in the symmetry groups of the dodecagonal lattices. $\Phi_3 \oplus \Phi_4$ generates a cyclic group that is the direct product of a three-fold and a four-fold group, and it appears in any four-dimensional lattice that is the sum of a hexagonal plane and a square one. $\Phi_4 \oplus \Phi_6$ appears in the same lattices, as the product of the square lattice’s quarter-turn with the hexagonal lattice’s sixth-turn.

So in the direct sum of a square and a hexagonal lattice, two of the three twelves appear together, and the third does not. Whether some four-dimensional lattice admits all three, and what the smallest point group containing all three looks like, is a question about groups of $4 \times 4$ integer matrices. It is the first place where the letters would be assembled into words, and nothing here has computed it.