# Detempering

(Redirected from Epimorphism)

In regular temperament theory, detempering is the process of taking a tempered tuning system and replacing each of its pitches with one or more pitches from its preimage, that is, the just or tempered pitches that the pitch represents. It is the opposite of tempering. Specifically, a detempered system (aka detemperament or detempering) has each pitch of a tempered system (according to a fixed regular temperament) replaced with some set of interpretations of the pitch under the temperament mapping. If exactly one interpretation is used for each degree of a scale, then the detempered scale is called a transversal, epimorphic scale or one-to-one detempering. Ideally the resultant detempered scale will have a compact lattice. A higher rank temperament is also called a detempering of a lower-rank temperament if the lower-rank temperament results from tempering out one or more commas in the higher-rank temperament. For example, meantone is a detempering of 12edo.

Detempering is one way among many to create a neji, or a JI scale approximating a given scale.

## Epimorphic scales

A JI scale S is epimorphic if on the JI subgroup $A \leq \mathbb{Q}_{\gt 0}$ generated by the intervals of S, there exists a linear map v: A → ℤ, called an epimorphism, such that v(S[i]) = i for all i ∈ ℤ. Equivalently, it is a detempering of an equal temperament under some mapping where each note of the equal temperament is matched to exactly one note.

Epimorphicity is strictly stronger than constant structure (CS). When one assumes S is CS but not that it is epimorphic, there is a unique set map $v : \{\text{intervals of S}\} \to \mathbb{Z}$ that witnesses that S is CS and satisfies v(S[i]) = i for all i. Thus a CS scale S is epimorphic if and only if this mapping v extends to a linear map on the entirety of A.

This definition extends naturally to asking whether a higher-dimensional mapping $S:\mathbb{Z}^n \to P$ for an arbitrary codomain $P$ of relative pitches is epimorphic, in the same sense of there existing an abelian group $A$ and a linear map $v : A \to \mathbb{Z}^n$ such that $v(S(x)) = x.$ This can be of practical interest: one might ask whether an isomorphic keyboard mapping $S : \mathbb{Z}^2 \to P$ (for a theoretical infinite 2D isomorphic keyboard) is epimorphic.

Temperament supported by epimorphisms for epimorphic scales have occasionally been considered. Some temperaments (including vals for small edos) can be viewed this way for small epimorphic scales despite their relatively low accuracy:

### Example

Consider the Ptolemaic diatonic scale, {9/8, 5/4, 4/3, 3/2, 5/3, 15/8, 2/1}, which is nicetone with L = 9/8, M = 10/9, and s = 16/15. This scale is epimorphic because we can apply ⟨7 11 16], the 7edo patent val, to map the intervals into the number of scale steps:

$\left(\begin{array} {rrr} 7 & 11 & 16 \end{array} \right) \left(\begin{array}{rrrrrrr} -3 & -2 & 2 & -1 & 0 & -3 & 1 \\ 2 & 0 & -1 & 1 & -1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 & 1 & 0 \end{array}\right) = \left(\begin{array}{rrrrrrr} 1 & 2 & 3 & 4 & 5 & 6 & 7 \end{array}\right)$

where the columns of the 3×7 matrix are the scale intervals written in monzo form. Hence, 7edo (equipped with its patent val) is an epimorphic temperament of the Ptolemaic diatonic scale. Indeed, 7edo supports dicot temperament.

### Facts

#### Definition: constant structure (CS)

Given a periodic scale $S : \mathbb{Z} \to (0,\infty)$ (with codomain written as ratios from S(0) = 1 in the linear frequency domain), let $C_k = \{ S[i+k]/S[i] : i \in \mathbb{Z}\}$ be the set of k-steps of S. Then S is constant structure (CS) if for any $i, j \in \mathbb{Z}, i \neq j,$ we have $C_i \cap C_j = \varnothing.$

#### Epimorphic scales are CS

Proof
Let v: A → ℤ be the epimorphism for s. Let $x \in C_j.$ Then there exists $i \gt 0$ such that $S[i+j]/S[i] = x.$ Suppose by way of contradiction there exist $k \neq j$ and $i \gt 0$ such that $S[i+k]/S[i] = x.$ Then $v(x) = v(S[i+j]/S[i]) = v(S[i+j]) - v(S[i]) = i + j - i = j,$ but also $v(x) = v(S[i^\prime+k]/S[i^\prime]) = v(S[i^\prime+k]) - v(S[i^\prime]) = k,$ a contradiction. $\square$

#### If the steps of a CS scale are linearly independent, then the scale is epimorphic

Theorem: Suppose S is a 2/1-equivalent increasing constant structure JI scale of length n. Let $C_1$ be the set of 1-steps of S, and suppose that $C_1$ is a basis for the JI subgroup A generated by it. Then there exists an epimorphism $v: A \to \mathbb{Z}$ which is a val of n-edo (and a similar statement holds for other equaves).

(The condition of $C_1$ being a basis rather than merely a generating set cannot be omitted, since the scale {5/4, 32/25, 2/1} is CS but not epimorphic. The converse of this conditional also fails, as {9/8, 5/4, 3/2, 25/16, 2/1} is epimorphic under 5edo's patent val.)

Proof
Define the linear map $v:A \to \mathbb{Z}$ by defining $v(\mathbf{s}) = 1$ for any step $\mathbf{s} \in C_1$ and extending uniquely by linearity. Then for any $i \in \mathbb{Z}$ we have $v(S[i]) = v(S[i]/S[i-1]\cdots S[1]) = v(S[i]/S[i-1]) + \cdots + v(S[1]) = i,$ whence v is an epimorphism. That $v(2) = n$ is also automatic. $\square$

## Deregularization

The term deregularization can be used as a JI-agnostic alternative to detempering.

In this sense, diasem (LMLSLMLSL) is a deregularization of semiquartal (LSLSLSLSL) which "detempers" the S step of semiquartal into two steps sizes M and S.