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→Finding the transversal generators: work through example of computing transversal generators
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We can find transveral generators for V by the following procedure:
We can find transveral generators for V by the following procedure:


<ul><li>Take the transpose of the [[Tenney-Euclidean_Tuning#The pseudoinverse|pseudoinverse]] of V, call that U</li><li>Find a basis for the commas of V</li><li>For each row U[i] of U, clear denominators and append the monzos of the comma basis for V</li><li>[[Saturation|Saturate]] the result to a list of monzos, call that S</li><li>Apply the ith val V[i] (dot product) to each element of S</li><li>Insert V[i].S[j] in front of the elements of S[j] as the first element, obtaining the jth element T[j] of a modified list T</li><li>Hermite reduce the modified list T, take the first row, and remove the first element (which should be a 1.)</li><li>Consider the rest to be a monzo and convert it to a rational number</li><li>This is a corresponding transveral generator to the ith val V[i] of V; it may be reduced to an equivalent generator of minimal [[Tenney_Height|Tenney height]] by multiplying by the commas of V</li></ul>      [[Category:generator]]
<ul><li>Take the transpose of the [[Tenney-Euclidean_Tuning#The pseudoinverse|pseudoinverse]] of V, call that U</li><li>Find a basis for the commas of V</li><li>For each row U[i] of U, clear denominators and append the monzos of the comma basis for V</li><li>[[Saturation|Saturate]] the result to a list of monzos, call that S</li><li>Apply the ith val V[i] (dot product) to each element of S</li><li>Insert V[i].S[j] in front of the elements of S[j] as the first element, obtaining the jth element T[j] of a modified list T</li><li>Hermite reduce the modified list T, take the first row, and remove the first element (which should be a 1.)</li><li>Consider the rest to be a monzo and convert it to a rational number</li><li>This is a corresponding transveral generator to the ith val V[i] of V; it may be reduced to an equivalent generator of minimal [[Tenney_Height|Tenney height]] by multiplying by the commas of V</li></ul>       
 
== Example ==
 
Note: I've followed the algorithm as described above. But clearly I am transposing things here way more often than is necessary than if the algorithm was superficially revised.
 
For 5-limit meantone, <span><math>V</math></span> is
 
<math>
\begin{bmatrix}
1 & 2 & 4 \\
0 & -1 & -4 \\
\end{bmatrix}
</math>
 
The pseudoinverse <span><math>V⁺</math></span> is
 
<math>
\frac{1}{33}
\begin{bmatrix}
17 & 18 \\
16 & 15 \\
-4 & -12 \\
\end{bmatrix}
</math>
 
And that transposed is <span><math>(V⁺)ᵀ</math></span>, which we'll call <span><math>U</math></span>:
 
<math>
\frac{1}{33}
\begin{bmatrix}
17 & 16 & -4 \\
18 & 15 & -12 \\
\end{bmatrix}
</math>
 
And here's the comma basis for V:
 
<math>
\begin{bmatrix}
-4 \\
4 \\
-1 \\
\end{bmatrix}
</math>
 
Beginning with <span><math>i = 1</math></span>, we'll create a matrix out of <span><math>U[i]</math></span> with denominators cleared and the comma basis appended (transposed so each comma is a row):
 
<math>
\begin{bmatrix}
17 & 16 & -4 \\
-4 & 4 & -1 \\
\end{bmatrix}
</math>
 
Saturate (eliminate common factors), and call this <span><math>S</math></span>:
 
<math>
\begin{bmatrix}
1 & 32 & -8 \\
0 & 4 & -1 \\
\end{bmatrix}
</math>
 
Now take the dot product of <span><math>V[1]</math></span> to each element of <span><math>S</math></span>, or in other words, left multiply <span><math>Sᵀ</math></span> by <span><math>V[1]</math></span>:
 
<math>
\begin{bmatrix}
1 & 2 & 4 \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
32 & 4 \\
-8 & -1 \\
\end{bmatrix} =
\begin{bmatrix}
33 & 4 \\
\end{bmatrix}
</math>
 
Now prepend that result, transposed, to <span><math>S</math></span>:
 
<math>
\begin{bmatrix}
33 & 1 & 32 & -8 \\
4 & 0 & 4 & -1 \\
\end{bmatrix}
</math>
 
Take the Hermite Normal Form:
 
<math>
\begin{bmatrix}
1 & 1 & 0 & 0 \\
0 & 4 & -4 & 1 \\
\end{bmatrix}
</math>
 
Take the first row:
 
<math>
\begin{bmatrix}
1 & 1 & 0 & 0 \\
\end{bmatrix}
</math>
 
Remove the first element (which should and indeed is a 1):
 
<math>
\begin{bmatrix}
1 & 0 & 0 \\
\end{bmatrix}
</math>
 
So indeed that gives the period for meantone, 2/1.
 
Let's repeat the latter steps but now with <span><math>i = 2</math></span>. Here's <span><math>U[2]</math></span> with denominators cleared and the comma basis appended:
 
<math>
\begin{bmatrix}
18 & 15 & -12 \\
-4 & 4 & -1 \\
\end{bmatrix}
</math>
 
Saturate, to get our new <span><math>S</math></span>:
 
<math>
\begin{bmatrix}
2 & 31 & -16 \\
0 & 2 & -1 \\
\end{bmatrix}
</math>
 
Now take the dot product of <span><math>V[2]</math></span> to each element of <span><math>S</math></span>, or in other words, left multiply <span><math>Sᵀ</math></span> by <span><math>V[2]</math></span>:
 
<math>
\begin{bmatrix}
0 & -1 & -4 \\
\end{bmatrix}
\begin{bmatrix}
2 & 0 \\
31 & 2 \\
-16 & -1 \\
\end{bmatrix} =
\begin{bmatrix}
33 & 2 \\
\end{bmatrix}
</math>
 
Now prepend that result, transposed, to <span><math>S</math></span>:
 
<math>
\begin{bmatrix}
33 & 2 & 31 & -16 \\
2 & 0 & 2 & -1 \\
\end{bmatrix}
</math>
 
Take the Hermite Normal Form:
 
<math>
\begin{bmatrix}
1 & 2 & -1 & 0 \\
0 & 4 & -4 & 1 \\
\end{bmatrix}
</math>
 
Take the first row:
 
<math>
\begin{bmatrix}
1 & 2 & -1 & 0 \\
\end{bmatrix}
</math>
 
Remove the first element (which should and indeed is a 1):
 
<math>
\begin{bmatrix}
2 & -1 & 0 \\
\end{bmatrix}
</math>
 
So indeed that gives the generator for meantone, 4/3. We're done!
 
[[Category:generator]]
[[Category:theory]]
[[Category:theory]]
[[Category:todo:reduce_mathslang]]
[[Category:todo:reduce_mathslang]]
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