When I try to notate a rank-2 temperament I want a mapping from notes on the staff to pitches in the pitch system that can notate all the pitches (formally a surjective function). By "notes on the staff" I mean nominals (A-G in any octave) with any number of conventional accidentals (
/
) and at most one microtonal accidental (in sagittal these would be single shaft accidentals, and milti-shaft accidentals can be seen as combinations of at most 2 conventional accidentals with at most 1 microtonal one). By "mapping" I mean that every note on the staff has one and only one pitch in the temperament that it notates. Furthermore, in this post I consider only chain-of-fifths notations, meaning that the nominals should relate to each other and to the sharp/flat notes by tempered fifths.Rank-2 temperaments have the nice property that in order to notate them you need only a finite set of microtonal accidentals, which isn't true for temperament of any higher rank. It's also not true for a small collection of rank-2 temperaments, namely those where the 3-limit subgroup is of rank 1, for example limmic (e.g. blackwood), whitewood and compton temperaments. That's because in this case you can (and usually do) choose one generator for the temperament so that both 2 and 3 are represented by multiples of this generator, which means that nominals and conventional accidentals keep you stuck only in multiples of this generator. This necessarily implies that you need a unique microtonal accidental for every value of the second generator's count, which can be arbitrarily large. In essence, in this case the conventional and the microtonal accidentals "switch roles": A finite number of conventional accidentals is enough, but unless you can think of infinite accidetnal shapes, you may need to repeat the microtonal ones an unlimited number of times.
In the other cases, however, we can indeed use only a finite number of microtonal accidentals, and in fact if we choose a generator pair for the 3-limit group \(\textbf{i}_1\) and \(\textbf{i}_2\), for example, an octave and a fifth, or an octave and a twelfth, the number of accidentals needed, including the trivial case (no accidental) and counting complement or negative accidentals as different ones, is exactly what I call the "MOS complexity" of \(\textbf{i}_2\) relative to \(\textbf i_1\) in the temperament (so actually that of \(M\textbf i_2\) relative to \(M\textbf i_1\) where \(M\) is the temperament's mapping matrix), which is also the number of chains of fifths in the temperament (or equivalently the index of the 3-limit subgroup in the temperament).
So what is MOS complexity? The MOS complexity of a tempered interval \(\textbf y_2\) relative to \(\textbf y_1\) is essentially the absolute value of the determinant of the matrix \(\begin{pmatrix}|&|\\\textbf y_1&\textbf y_2\\|&|\end{pmatrix}\). If we choose a unimodular matrix \(U\) so that \(U\textbf y_1=\begin{pmatrix}n\\0\end{pmatrix}\) for some positive integer n, meaning we choose a basis for the temperament so that \(\mathbf y_1\) is a multiple of its first generator, and also \(U\mathbf y_2=\begin{pmatrix}m\\k\end{pmatrix}\) for some non-negative integer \(k\), meaning that it requires \(k\) second generators to get to \(\mathbf y_2\) from an infinite repetition of the first generator, we can see that \[ c=\left|\text{det}\begin{pmatrix}|&|\\\textbf y_1&\textbf y_2\\|&|\end{pmatrix}\right|=\left|\text{det}(U)\right|\left|\text{det}\begin{pmatrix}|&|\\\textbf y_1&\textbf y_2\\|&|\end{pmatrix}\right|=\left|\text{det}\left(U\begin{pmatrix}|&|\\\textbf y_1&\textbf y_2\\|&|\end{pmatrix}\right)\right|=\left|\text{det}\begin{pmatrix}n&m\\0&k\end{pmatrix}\right|=nk.\] This explains why it's called the "MOS complexity": In a MOS of the temperament of size \(s\) where notes \(\mathbf y_1\) apart are regarded as equivalent, there are \(n\) chains of \(\frac sn\) notes separated by second generators, and in each one there are \(k\) notes too high on the chain to have a note that is \(\mathbf y_2\) above them, and so we have \(\frac sn-k\) notes in each chain that do have a note \(\mathbf y_2\) above them, which means that overall there are \(n\left(\frac sn-k\right)=s-nk=s-c\) notes that have a note in the MOS that is \(\mathbf y_2\) above them, or more simply put, \(s-c\) appearances of \(\mathbf y_2\) in the MOS. (Actually you may notice that this calculation may give a negative value for small MOSes which is of course nonsensical, the reason is that if the MOS is too small, i.e. \(\frac sn<k\), there can't be \(k\) notes in a chain of second generators because there are only \(\frac sn\), and so instead of \(k\) we should write \(\min\left(\frac sn, k\right)\) and so we get that there are \(\max(s-c, 0)\) appearances of \(\mathbf y_2\) in the MOS.)
Now, why is it indeed the number of microtonal accidentals? Look at a map of the temperament that has \(n\) first generators equaling tempered \(\mathbf i_1\) and needs \(k\) second generators from that chain to get to tempered \(\mathbf i_2\), and consider the set of all the notes whose intervals above an arbitrary tonic have a representation of \(\begin{pmatrix}i\\j\end{pmatrix}\) in that mapping where \(0\le i<n\) and \(0\le j<k\), which has \(nk=c\) elements.
On one hand, we can show that these notes can't have a 3-limit interval between them: If we choose any two notes of that set and assume there is a 3-limit interval between them, we can represent it as \(\mathbf y-\mathbf y'=aM\mathbf i_1+bM\mathbf i_2\) for integers \(a, b\). Now, on one hand the second entry of that vector is \(j-j'\) which is between \(-k\) and \(k\) exclusive, but on the other hand it is \(b k\) which is divisible by \(k\), and therefore it has to be 0. Similarly, the first entry is between \(-n\) and \(n\) exclusive, but because \(b=0\) it has to be \(a n\) and therefore also 0. This means that the two notes are actually the same, and so no pair of different notes in the set is separated by a 3-limit interval. Therefore each note's accidental has to be different from all of the other notes', and so we need at least \(c\) accidentals.
On the other hand, we can show that every note in the temperament is a 3-limit interval away from a note in the set: Choose any note in the temperament and notate its interval above the tonic as \(\begin{pmatrix}i\\j\end{pmatrix}\) (with no restrictions on \(i\) and \(j\)). We can write \(j=b k+j'\) for intetger \(b\) and \(0\le j'<k\), and note that \(\begin{pmatrix}i\\j\end{pmatrix}=\begin{pmatrix}i'\\j'\end{pmatrix}+bM\mathbf i_2\) for some number \(i'\). Now also write \(i'=a n+i''\) for integer \(a\) and \(0\le i''<n\) and see that \[\begin{pmatrix}i\\j\end{pmatrix}=\begin{pmatrix}i'\\j'\end{pmatrix}+bM\mathbf i_2=\begin{pmatrix}i''\\j'\end{pmatrix}+aM\mathbf i_1+bM\mathbf i_2,\] which means that our original note is some 3-limit interval away from a note whose interval above the tonic is represented as \(\begin{pmatrix}i''\\j'\end{pmatrix}\), with \(0\le i''<n\) and \(0\le j'<k\) which is in the set, and so our original note can use the same accidental as the note in the set.
Therefore we need exactly \(c\) microtonal accidentals to notate the temperament, each one notates exactly one copy of the 3-limit subgroup aka one chain of fifths, and so this is also the number of chains of fifths in the temperament.
For example, let's take hemiaug temperament, which is the 24&27 temperament. One way to calculate the number of accidentals needed is to choose \(\mathbf i_1\) to be an octave and \(\mathbf i_2\) to be a fifth and the map that gives each note the number of steps it takes to get to it in 24edo and 27edo. This gives us \(\left|\det\begin{pmatrix}24&14\\27&16\end{pmatrix}\right|=\left|24\cdot 16-27\cdot14\right|=\left|384-378\right|=6\), which means we need 6 microtonal accidentals to notate this temperament. Another way is to choose \(\mathbf i_1\) to be an octave (prime 2) and \(\mathbf i_2\) to be a twelfth (prime 3) in the standard map of \(\begin{pmatrix}3&1&7&-1&1&13\\0&2&0&5&5&-1\end{pmatrix}\). Because it is already in the normal form we can just take \(n=3\) times \(k=2\) and get again the value of \(6\) for the number of accidentals, and indeed one can notate this temperament with 6 microtonal accidentals. For example, we can use ones for 54/55, 80/81, 1/1, 81/80, 55/54 and 33/32 (sagittal
). However, there is no option to notate it with less.So we found out how many accidentals we need, but it would also be nice if there was some order in them, and luckily there is, but in order to see it we need to talk in the more abstract language of group theory. Let \(G \cong \mathbb Z^2\) be the group of intervals in the temperament, notate by \(H\le G\) the subgroup corresponding to tempered 3-limit intervals, and assume that it is also of rank 2. Now, the elements of the quotient group \(G/H\) are exactly the sets of notes that are separated by 3-limit intervals aka the "chains of fifths" in the temperament, i.e. notes that use the same microtonal accidental. Therefore this is in fact the group of microtonal accidentals, and in particular its size, which is called the index of \(H\) in \(G\) and notated by \([G:H]\), is also the number of mocrotonal accidentals.
More often than not, this group is also cyclic, which means that there is an element \(g\in G\) (\(g\) from "group element") which can be called a spacer, so that the multiples of \(g\) visit all of the cosets of \(H\) in \(G\). In other words, for every \(g'\in G\) there is some multiple \(ng\) of \(g\) so that \(ng-g'\in H\), which means that \(g\) stacked on itself \(n\) times uses the same microtonal accidental as \(g'\). This can also more conveniently be written as \(ng+H=g'+H\) which literally means "the set of notes that differ from \(ng\) by a 3-limit interval is the same set that differ from \(g'\) by a 3-limit interval" but can be more simply thought of as meaning "the microtonal accidental of \(ng\) is the same as that of \(g'\)." Another way to say it is saying that \(g+H\) is a generator of \(G/H\), which means that when added to itself \(g+H\) can reach all of the elements of the quotient group. This means we can index the microtonal accidentals by the number of spacers it takes to get to them (from a natural note), which is very similar to the modular generator counts used to summarize the notations of temperaments with octave period. This may suggest they are in fact the same thing, and indeed they are. It's easiest to see this in the special case of temperaments with octave period, since in that case the generator has the stronger property that every other interval differs from a multiple of it by a 2-limit interval, but this is also true in the other cases.
In fact, in the general case of a cyclic microtonal accidental (aka quotient) group, even if the 3-limit equivalence interval divides into equal non-3-limit first generators, there is some choice for the second generator that is also a valid spacer. Formally, let's take some generator pair \(g_1, g_2\) of \(G\) and a spacer \(g\in G\). We want to see that there is some other choice \(g_2'=g_2+mg_1\) for the second generator so that \(g_2'\) is also a spacer. First, the generators \(g_1\) and \(g_2\) can be given spacer counts modulo \([G:H]\), let's call them \(k_1\) and \(k_2\) respectively. Now, let's consider every prime that divides the index \(p|[G:H]\), and prove that \(p\) can't divide both \(k_1\) and \(k_2\). Since they generate the group \(G\), we can write \(g=ag_1+bg_2\), and so we'll get that if \(p\) divides both \(k_1\) and \(k_2\) the modular spacer count of \(g\) modulo \(p\) is on one hand \(1\) by definition, and on the other it's \(0\), since \[g+H=ag_1+bg_2+H=ak_1g+bk_2g+H=(ak_1+bk_2)g+H,\] and \(ak_1+bk_2\equiv a\cdot0+b\cdot0 = 0\mod p\) (note that the modular spacer count is defined modulo \(p\) since \(p\) divides the number of microtonal accidentals). This means that either \(k_1\) or \(k_2\) is not divisible by \(p\). Let's take \(m\) to be a number that is \(1\) modulo \(p\) if \(p|k_2\) and otherwise it's \(0\), for each prime \(p|[G:H]\) that divides the index (such \(m\) exists because of the Chinese remainder theorem), and look at the second generator \(g_2'=g_2+mg_1\). For every \(p\) that divides \([G:H]\), the spacer count of this generator modulo \(p\) is not 0: If \(p|k_2\) then \(p\nmid k_1\) and therefore \(k_2'=k_2+mk_1\equiv 0+1\cdot k_1=k_1\not\equiv 0\mod p\) and otherwise \(k_2'=k_2+mk_1\equiv k_2+0k_1=k_2\not\equiv0\mod p\). This means that \(k_2'\), the modular spacer count of \(g_2'\), is coprime with \([G:H]\).
From this the fact that \(g_2'\) is also a spacer follows like so: Because \(k_2'\) is coprime with \([G:H]\), the extended Euclidean algorithm gives us some integers \(a\) and \(b\) satisfying \(ak_2'+b[G:H]=1\). Multiplying this by \(g\) gives us that \(ak_2'g+b[G:H]g=g\), and \([G:H]g\) is also in \(H\) since this is where we get after the cycle of all microtonal accidentals. Therefore \[ag_2'+H=ak_2'g+H=(1-b[G:H])g+H=g-b[G:H]g+H=g+H\] i.e. \(g+H\) is a multiple of \(g_2'+H\) and so every multiple of \(g+H\) is also, but these are all the elements of \(G/H\)! This means that \(g_2'+H\) is a generator of \(G/H\) i.e. \(g_2'\), which was constructed to complete \(g_1\) to a generator pair, is also a valid spacer.
Let's take for example the case of a miracle scale that repeats at the fifth. The first generator would be a secor and a natural choice for the second generator is Archytas's comma. However, it is not a spacer, since miracle needs 6 microtonal accidentals but \[\left(\frac{64}{63}\right)^3=\frac{2^{18}}{3^6\cdot7^3}=\frac{1024}{1029}\frac{256}{243}\sim\frac{256}{243}\] i.e. three Archytas's commas give you a Pythagorean limma which is already 3-limit. If we choose the syntonic comma to be a spacer we see that the secor gets a spacer count of \(k_1=1\) (which makes sense since in the sagittal notation for miracle the secor is notated as a limma with a syntonic comma upwards) and Archytas's comma gets \(k_2=2\). The number of accidentals, 6, is divisible by the primes 2 and 3. 2 divides \(k_2\) but 3 doesn't so we want a number \(m\) that is 1 modulo 2 but 0 modulo 3, and the Chinese remainder theorem gives us \(m=3\).
Finally, 3 secors give you a neutral third of \(\frac{11}9\sim\frac{27}{22}\) and indeed \[\frac{64}{63}\frac{27}{22}=\frac{32}7\frac3{11}=\frac{96}{77}\sim\frac{96}{77}\frac{385}{384}=\frac54\] which is both a valid second generator and a valid spacer, since it orders the accidentals as, for example, 1/1, 80/81, 63/64, 33/32, 64/63, 81/80 (sagittal
). In such a case the modular spacer counts aren't the same as its generator counts, because, for instance, one secor is 1 first generator and 0 second generators, but it's a perfect fourth minus a major third, and so it has a modular spacer count of \(-1\) (alternatively, a secor above C is D, where has a modular spacer count of \(5\equiv-1\)). However, the second generator counts can be corrected by adding the modular spacer count of the first generator times the first generator counts, a step which isn't needed in when the first generator is a 3-limit interval since in this case its modular spacer count is 0. For instance, 7/6, which is the same in miracle as a major third minus a secor, should get a modular spacer count of \(1+(-1)(-1)=2\), and indeed \[\frac76\sim\frac{225}{224}\frac76=\frac{75}{64}=\left(\frac54\right)^2\frac34,\] i.e. it's two major thirds above a 3-limit interval. (Another way to see it is that 7/6 over C is E where is the second generator in the major third's order.)In fact, we can find a relation between generators and spacers in the opposite direction as well: For every spacer that is primitive (i.e. is not a tempered multiple of any other interval) there is some 3-limit interval which completes it to a generator pair of the temperament. This is more easily proved in the vector setting, so let's prove it there. Let \(\mathbf y\) be a primitive spacer, and let's notate by \(c\) the MOS complexity of the fifth relative to the octave (or any other generator pair of the 3-limit intervals). First, note that like before, \(c\mathbf y\) is where we go after cycling through all the microtonal accidentals, and so it has the trivial microtonal accidental, i.e. it's a tempered 3-limit interval \(c\mathbf y=M\mathbf i\) for some \(\mathbf i=\begin{pmatrix}\text i_1\\\text i_2\end{pmatrix}\). Now let's prove that \(\text i_1\) and \(\text i_2\) have no common divisor greater than 1: If there was one, let's call it \(d\), this would mean that \(\frac1d\mathbf i\) is also an integer vector. Therefore, because \(\mathbf y\) is primitive, \(M\frac1d\mathbf i=\frac cd\mathbf y\) would necessarily be an integer multiple of \(\mathbf y\), i.e. \(\frac cd\) would be an integer. However, it would also be less than \(c\), which means that \(\frac cd\mathbf y\) would have a microtonal accidental, but this is in contradiction to it being a tempered 3-limit interval. Therefore they have no common divisor, so we can use the extended Euclidean algorithm again to find some numbers \(a\) and \(b\) satisfying \(a\text i_1+b\text i_2=1\), which means that we can choose \(\mathbf i'=\begin{pmatrix}b\\-a\end{pmatrix}\) and get that \(\left|\det\begin{pmatrix}|&|\\\mathbf i&\mathbf i'\\|&|\end{pmatrix}\right|=\left|\det\begin{pmatrix}\text i_1&b\\\text i_2&-a\end{pmatrix}\right|=|-a\text i_1-b\text i_2|=1\), which means that \(\mathbf i\) and \(\mathbf i'\) generate the 3-limit group. Now, the MOS complexity of \(\mathbf y\) relative to \(\mathbf i'\) is \[\left|\det\begin{pmatrix}|&|\\M\mathbf i'&\mathbf y\\|&|\end{pmatrix}\right|=\frac 1c\left|\det\begin{pmatrix}|&|\\M\mathbf i'&c\mathbf y\\|&|\end{pmatrix}\right|=\frac 1c\left|\det\begin{pmatrix}|&|\\M\mathbf i'&M\mathbf i\\|&|\end{pmatrix}\right|,\] but this absolute value determinant is the MOS complexity of \(M\mathbf i\) relative to \(M\mathbf i'\) which by definition equals \(c\) (and we showed this definition of \(c\) doesn't depend on the choice of the basis for the 3-limit intervals because it always equals the number of necessary microtonal accidentals) and therefore the MOS complexity of \(\mathbf y\) relative to \(M\mathbf i'\) is in fact 1, which means that a MOS of any size \(s\) that repeats at \(\mathbf i'\) contains exactly \(s-1\) appearances of \(\mathbf y\). They can't form closed loops because then there would be a vector that both \(\mathbf y\) and \(M\mathbf i'\) are multiples of which would mean that their determinant is 0 and so the MOS complexity would also be 0. Therefore the whole MOS has to be one chain of \(\mathbf y\)s which means that \(\mathbf y\) indeed completes \(M\mathbf i'\) to a generator pair.
For example let's take ennealimmal with the nimor third of \(\frac65\) as a spacer. The common mapping matrix of \(\begin{pmatrix}9&1&1&12\\0&2&3&2\end{pmatrix}\) shows that the MOS complexity of the fifth (or actually the twelfth) relative to the octave is \(9\cdot 2=18\), and multiplying the spacer by that gives you \[18\begin{pmatrix}9&1&1&12\\0&2&3&2\end{pmatrix}\begin{pmatrix}1\\1\\-1\\0\end{pmatrix}=18\begin{pmatrix}9\\-1\end{pmatrix}=\begin{pmatrix}162\\-18\end{pmatrix}.\] This is indeed a tempered primitive 3-limit interval: \[\begin{pmatrix}9&1&1&12\\0&2&3&2\end{pmatrix}\begin{pmatrix}19\\-9\\0\\0\end{pmatrix}=\begin{pmatrix}162\\-18\end{pmatrix},\] and the extended Euclidean algorithm for \(19\) and \(-9\) gives us \(a=1, b=2\). Therefore \(\mathbf i'=\begin{pmatrix}2\\-1\end{pmatrix}\) i.e. a fourth, and indeed a fourth and a minor third are a valid generator pair for ennealimmal, which gives you the same modular generator counts for the microtonal accidentals as using a minor third as a spacer in any other setting.
This theorem is in fact stronger than it may appear, as it says something about the temperaments that have spacers themselves: In each one of them there has to be a 3-limit interval that tempers to a primitive vector, i.e. does not split into a stack of smaller equal tempered intervals. (Actually we proved it only for primitive spacers, but this is not a problem because if a spacer is a multiple of another interval then every multiple of the original spacer is also a multiple of that interval, and so by definition the new interval is also a spacer.) This is not true, for instance, for hemipyth temperaments, like decimal, sruti, harry, semimiracle and quadritikleismic, for ones that divide primes 2 and 3 into 3 equal parts like tritikleismic and also theoretically for ones that divide them both into 5, 7, etc. parts, although I don't think any of them were ever thought of. In fact, it turns out that these are the only limitations on the existence of a spacer, as one exists in any temperament where no prime divides all the elements of the 3-limit subgroup: For any prime \(p|[G:H]\) notate by \(h_p\) an element of \(H\) that does not divide in \(G\) into \(p\) equal parts, and look at \[h=\sum_{p|[G:H]}\left(\prod_{p\ne q|[G:H]}q\right)h_p.\] Let's prove that this \(h\) does not split into any prime \(p\) that divides \([G:H]\): For each such prime, \(h\) can be expressed as \[h=\left(\prod_{p\ne q|[G:H]}\right)h_p+p\sum_{p\ne q|[G:H]}\left(\prod_{p,q\ne r|[G:H]}r\right)h_q,\] so if it splits into \(p\) parts then \(\left(\prod_{p\ne q|[G:H]}\right)h_p\) also splits into \(p\) parts. However, the product only contains primes different from \(p\) and so this would mean that \(h_p\) splits into \(p\) parts, but we chose it so that it doesn't. That being said, \(h\) may split into a number of equal parts \(n\) that is coprime with \([G:H]\), i.e. \(h=ng\) for some \(g\in G\). However, in this case \(n(g+H)=ng+H=h+H=H\) which means that \(n\) is divisible by the order of \(g+H\) in the quotient group \(G/H\). On the other hand, by a consequence of Lagrange's theorem, \([G:H]\) has to also be divisible by its order. Therefore its order is 1 which means it's trivial i.e. \(g+H=H\) and hence \(g=h'\in H\). We take such \(h'\) that does not split further and complete it to a generator pair of \(G\) using some element \(g\). Because every element of \(G\) can be expressed as \(ah'+bg\) for integer \(a, b\) this is an expression of it as a multiple of \(g\) plus a 3-limit interval which means that \(g\) is indeed a spacer.
This theorem shows that no matter how complex and weird a temperament can be, unless there is a number that all 3-limit intervals split into, it has a spacer. Therefore it can only be demonstrated fully in really horribly complex temperaments, and so as an example I crafted a probably impractical temperament on the 13-limit no-11 subgroup, which can be described by the matrix \(\begin{pmatrix}7&4&34&22&33\\0&6&-15&-2&-6\end{pmatrix}\). In this temperament, where the fifth has octave-relative MOS complexity of \(7\cdot6=42\), both the perfect fourth and the twelfth divide into 2 parts, the fifth divides into 3 and the octave into 7, which means none of them can be used as a first generator. However, each of them divides into a different prime, which means we can choose the fifth as the one that doesn't divide into 2 parts, the octave as the one that doesn't divide into 3 parts and the fourth as the one that doesn't divide into 7 parts, and this is enough since \(42=2\cdot3\cdot7\). Now, the proof tells us to use 6 fourths, 21 fifths and 14 octaves, i.e. \[\left(\frac43\right)^6\left(\frac32\right)^{21}2^{14}=2^{12-21+14}3^{-6+21}=2^53^{15}\] which does split but into 5 3-limit parts of 54/1. (a pythagorean major forty-first?) However it does not split any further in this temperament:
\[\begin{pmatrix}7&4&34&22&33\\0&6&-15&-2&-6\end{pmatrix}\begin{pmatrix}1\\3\\0\\0\\0\end{pmatrix}=\begin{pmatrix}19\\18\end{pmatrix},\] and therefore it can be completed to a basis of the temperament by another vector which would necessarily also be a spacer, for example, \(\begin{pmatrix}1\\1\end{pmatrix}\). This spacer is, among other interpretations, a tempered 56/45, and so any interval that differs from it by a 3-limit interval, for example 7/5, is a spacer as well.
Finally, it goes without saying that group theory is a very wide mathematical subject that can be quite complicated sometimes, and while I tried to make this post as short as possible to convey my ideas but as long as necessary to convey them clearly I may have missed some key parts that are clear to me but may not be clear to everyone who reads this. So If you don't understand something in this post it's probably because I glossed over some important detail, and in that case please feel free to comment and tell me what part you didn't understand so I can fill the gaps.
