Projective tuning space and moments of symmetry

These notes describe projective tuning space, focusing on its nontrivial relationship to moments of symmetry.

Representing five-limit maps and commas in the plane

Given a five-limit map $[a, b, c]$, where $a$, $b$, and $c$ correspond to primes 2, 3, and 5, draw it at the point $[b/a, c/a]$ in the plane. For example, the map $A = [12, 19, 28]$ is drawn at the point $[19/12, 28/12] = [1.583, 2.333]$, shown in Figure 1.

If a comma can be written as $2^u \cdot 3^v \cdot 5^w$, and so has exponent vector

it is tempered out by the map $A = [a, b, c]$ when $AV = 0$, that is when

and so

So a comma defines a line in $b/a$, $c/a$ space, made up of the points for all the maps that temper it out. For example, the comma $81/80 = 2^{-4} \cdot 3^{4} \cdot 5^{-1}$ has $V = [-4, 4, -1]^{\top}$, and a map $A = [a, b, c]$ will temper it out when

Figure 2 shows this line along with the earlier point $[19/12, 28/12]$ corresponding to the map $A = [12, 19, 28]$. The point lies on the line because the map $[12, 19, 28]$ tempers out 81/80.

Sums of maps and mediants

Two maps $A = [a, b, c]$ and $A' = [a', b', c']$ and their sum $A + A' = [a + a', b + b', c + c']$ are represented as

So addition of maps $[a, b, c]$ corresponds to taking mediants of each coordinate of points $[b/a, c/a]$. The point $A + A'$ lies on the line between $A$ and $A'$. This is because

and likewise with $b$ and $b'$ replaced with $c$ and $c'$, so $A + A'$ lies $a'/(a + a')$ of the way between $A$ and $A'$. For example, taking $A = [12, 19, 28]$ and $A' = [5, 8, 12]$, we have $A + A' = [17, 27, 40]$ and

The point $A + A'$ lies $5/17$ of the way between $A$ and $A'$, since

Figure 3 shows these points and the line they all lie on.

Relation to moments of symmetry

Recall that when stacking a generator $g$ cents within a period $h$ cents, the moments of symmetry are found by the Gral method, a binary search for $g/h$ starting with the interval $[0/1, 1/0]$ and using the mediant to bisect it. Table 1 shows the Gral method applied to a generator of 697 cents and period of 1200 cents. The Left and Right columns show the left and right end-points of the search interval, and the Mediant column shows their mediant (a blank means same value as the row above).

Now say, by analogy, we start with a 'left map' and a 'right map', the left map starting with $1$ and the right map starting with $0$ (corresponding to the denominators of $0/1$ and $1/0$). Taking their sums and replacing the left or right map whenever we replace the left or right search end-point, we obtain a sequence of maps whose first entry always equals the denominator of the mediant in the Gral method. The Left map, Right map, and Sum columns in Table 1 show this process. I chose the other entries of the starting left and right maps to reproduce $[12, 19, 28]$ and $[19, 30, 44]$ lower down the table. The Point column shows the point $[b/a, c/a]$ corresponding to the map $[a, b, c]$ in the Sum column.

All the points in the Point column lie on the same line, corresponding to the comma tempered out by all the maps in the Sum column, and the change in the corresponding mediant is proportional to the distance you move along the line. So you can picture the comma line as a 'number line' of mediant values.

This works because if $L$ and $R$ are the starting left and right maps, for mediant $m/n$ we get sum map $S = nL + mR$. So if $L = [1, b, c]$ and $R = [0, b', c']$, we have

which gives the point

So the coordinates of the point are linear functions of the mediant $m/n$. For example, in Table 1 we have $L = [1, 1, 0]$ and $R = [0, 1, 4]$, so mediant $m/n$ has point $[1 + m/n, 4m/n]$.

Figure 4 shows the points from Table 1, starting at the row with mediant $3/5$, labelled with their corresponding mediants. The cents values are 1200¢ times the mediant value; this gives an idea of the generator size of the corresponding MOS (each mediant $m/n$ corresponds to an $n$-note MOS with the generator at scale degree $m$; the value of $g/h$ can lie anywhere between the left and right end-points of the Gral search interval, but the mediant is always in this interval).

Understanding temperament relationships

To give an example of using these diagrams to understand the relationship between temperaments, Figure 5 shows the lines corresponding to the commas 81/80, 32805/32768, and 15625/15552, tempered out by meantone, schismic, and Hanson temperaments respectively. The points of intersection and their mediants are labelled with the first entry of the corresponding map; this is a shorthand for the maps $[12, 19, 28]$, $[19, 30, 44]$, $[31, 49, 72]$, $[53, 84, 123]$, $[65, 103, 151]$, and $[72, 114, 167]$.

From the intersection labelled $19$, we see that the map $[19, 30, 44]$ is the unique map (along with its multiples) which tempers out $81/80$ and $15625/15552$. From the points labelled $19$ and $53$, we see that 15625/15552 is the unique comma (along with its powers) tempered out by both $[19, 30, 44]$ and $[53, 84, 123]$.

I find the diagram particularly interesting since I play a 19EDO guitar, a Hanson temperament guitar close to 53EDO, and even on occasion a 12EDO guitar.

Relation to A Middle Path

In the above, I've used a projection onto the plane $a = 1, $ $[a, b, c] \mapsto [b/a, c/a]$. In A Middle Path, the axes are first scaled by the $\log_{2}$ of their corresponding prime, giving $a' = a$, $b' = b/\log_{2}3$, $c' = c/\log_{2}5$, then projected onto the plane $a' + b' + c' = 1$, giving the coordinates

So for $[12, 19, 28]$, we get $a' = 12$, $b' = 19/\log_{2}3 = 11.9877$, $c' = 28/\log_{2}5 = 12.0589$, and coordinates $[-0.001806, -0.000296]$, rather than $[19/12, 28/12]$.

In the Middle Path projection, the generator value doesn't vary linearly along a comma line (but it does increase monotonically).

I think the projection $[a, b, c] \mapsto [b/a, c/a]$ makes the connection to moments of symmetry easier to understand.

Naren Ratan, 2026