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== Derivation ==
On some <math>p</math>-limit subgroup with <math>n</math> primes, define the <math>n \times n</math> Tenney weighting matrix <math>W</math>:
On some <math>p</math>-limit subgroup with <math>n</math> primes, define the <math>n \times n</math> Tenney weighting matrix <math>W</math>:


$$
:<math>
W = \begin{bmatrix}
W =
\log_2 2 & 0 & \dots & 0 \\
\begin{bmatrix}
0 & \log_2 3 & \dots &  0\\
\log_2 2 & 0 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & \log_2 3 & \cdots &  0\\
0 & 0 & \dots & \log_2 p
\vdots & \vdots & \ddots & \vdots \\
\end{bmatrix}
0 & 0 & \cdots & \log_2 p
$$
\end{bmatrix}
</math>


And the row vector <math>j</math> containing the log-primes (aka the [[JIP]]): <math>j = \begin{bmatrix}
And the row vector <math>j</math> containing the log-primes (aka the [[JIP]]): <math>j = \begin{bmatrix}
\log_2 2 & \log_2 3 & \dots & \log_2 p \\
\log_2 2 & \log_2 3 & \cdots & \log_2 p \\
\end{bmatrix} </math>
\end{bmatrix} </math>


Then the block matrix <math>X</math> obtained from these:
Then the block matrix <math>X</math> obtained from these:


$$
:<math>
X = \begin{bmatrix}
X =
W \\
\begin{bmatrix}
\hline
W \\
J
\hline
\end{bmatrix}
j
$$
\end{bmatrix}
</math>


Defines an inner product, with positive definite <math>G =  X^{\mathsf T} X </math>:
defines an inner product, with positive definite <math>G =  X^{\mathsf T} X </math>:


$$
:<math>
\left\langle x,y \right\rangle = x^{\mathsf T} X^{\mathsf T} X y = x^{\mathsf T} G y
\left\langle x,y \right\rangle = x^{\mathsf T} X^{\mathsf T} X y = x^{\mathsf T} G y
$$
</math>


and an induced norm <math>||x|| = \sqrt{\left\langle x,x \right\rangle}</math>, which we will call the [[Weil_Norms,_Tenney-Weil_Norms,_and_TWp_Interval_and_Tuning_Space#Weil-Euclidean_Norm|Weil-Euclidean norm]].
and an induced norm <math>||x|| = \sqrt{\left\langle x,x \right\rangle}</math>, which is the [[Weil_Norms,_Tenney-Weil_Norms,_and_TWp_Interval_and_Tuning_Space#Weil-Euclidean_Norm|Weil-Euclidean norm]].


The inner product on the dual space can then be derived by simply inverting <math>G</math>, which then gives the dual norm:
The inner product on the dual space can then be derived by simply inverting <math>G</math>
<ref>Taking the natural map to the dual space <math>\Gamma: V\to V^{\ast}: x \mapsto \left\langle x, \cdot \right\rangle</math>, we require <math>\left\langle \Gamma(x),\Gamma(y) \right\rangle^{\ast} = \left\langle x,y \right\rangle</math>.</ref>
, which gives the dual norm:


$$
:<math>
\left\langle \alpha, \beta \right\rangle = \alpha G^{-1} \beta^{\mathsf T} \\
\left\langle \alpha, \beta \right\rangle^{\ast} = \alpha G^{-1} \beta^{\mathsf T}
</math>


||\alpha|| = \sqrt{\left\langle \alpha,\alpha \right\rangle} = \alpha G^{-1} \alpha^{\mathsf T}
:<math>
$$
||\alpha||^{\ast} = \sqrt{\left\langle \alpha,\alpha \right\rangle^{\ast}} = \sqrt{\alpha G^{-1} \alpha^{\mathsf T}}
</math>


The goal is now to find an expression for <math>G^{-1}</math>.
The goal is now to find an expression for <math>G^{-1}</math>.
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First, note that:
First, note that:


$$
:<math>
G =  X^{\mathsf T} X = W^2 + j^{\mathsf T}j
G =  X^{\mathsf T} X = W^2 + j^{\mathsf T}j
$$
</math>


Since the outer product <math>j^{\mathsf T}j</math> is rank-1 we can use a theorem on the inverse of matrix sums which states:
Since the outer product <math>j^{\mathsf T}j</math> is rank-1 we can use a theorem on the inverse of matrix sums which states:
<ref>Miller, K. S. (1981). On the Inverse of the Sum of Matrices. Mathematics Magazine, 54(2), 67–72. https://doi.org/10.2307/2690437</ref>
<ref>Miller, K. S. (1981). On the Inverse of the Sum of Matrices. Mathematics Magazine, 54(2), 67–72. https://doi.org/10.2307/2690437</ref>


: If <math>A</math> and <math>A+B</math> are invertible, and <math>B</math> has rank 1, then let <math>g = \text{tr}(BA^{-1})</math>. Then <math>g \neq -1</math> and
::If <math>A</math> and <math>A+B</math> are invertible, and <math>B</math> has rank 1, then let <math>g = \text{tr}(BA^{-1})</math>. Then <math>g \neq -1</math> and


: <math>(A+B)^{−1}=A^{-1} − \frac{1}{1+g}A^{-1}BA^{-1}</math>
::<math>(A+B)^{−1}=A^{-1} − \frac{1}{1+g}A^{-1}BA^{-1}</math>


Now identifying <math>A = W^2</math> and <math>B = j^{\mathsf T}j</math>. We can see that
Now identifying <math>A = W^2</math> and <math>B = j^{\mathsf T}j</math>. We can see that


$$
:<math>
G^{-1} = (W^2 + j^{\mathsf T}j)^{-1} = W^{-2} - \frac{1}{1+g} W^{-2}j^{\mathsf T}jW^{-2}
G^{-1} = (W^2 + j^{\mathsf T}j)^{-1} = W^{-2} - \frac{1}{1+g} W^{-2}j^{\mathsf T}jW^{-2}
$$
</math>
 
Now let <math>
l = \begin{bmatrix}
\frac{1}{\log_2 2} & \frac{1}{\log_2 3} & \cdots & \frac{1}{\log_2 p} \\
\end{bmatrix}
</math>, then


Now let <math>l = \begin{bmatrix}
:<math> l = W^{-2}j </math>
\frac{1}{\log_2 2} & \frac{1}{\log_2 3} & \dots & \frac{1}{\log_2 p} \\
\end{bmatrix} </math>, then


$$
:<math>
l = W^{-2}j \\
G^{-1} = W^{-2} - \frac{1}{1+g} l^{\mathsf T}l
G^{-1} = W^{-2} - \frac{1}{1+g} l^{\mathsf T}l
</math>
$$


Now we only need to find <math>g</math>. The trace of a matrix product is equal to the sum of the elements of their Hadamard (elementwise) product. Since
Now we only need to find <math>g</math>. The trace of a matrix product is equal to the sum of the elements of their Hadamard (elementwise) product. Since


$$
:<math>
j^{\mathsf T}j \circ W^{-2} = I_n \\
j^{\mathsf T}j \circ W^{-2} = I_n
g = \text{tr}(BA^{-1}) = \text{tr}(BA^{-1}) = n
</math>
$$
 
:<math>
g = \text{tr}(BA^{-1}) = \text{tr}(j^{\mathsf T}j W^{-2}) = n
</math>


Which leads to the final expression:
Which leads to the final expression:


$$
:<math>
G^{-1} = W^{-2} - \frac{1}{1+n} l^{\mathsf T}l
G^{-1} = W^{-2} - \frac{1}{n+1} l^{\mathsf T}l
$$
</math>
 
== Relation to other metrics ==
 
[[Graham Breed]] gives the following formula (adapted for the notation introduced here):<ref>See formula (16) in section 3.1 "Cross-Weighted Metrics" In Breed, G. (2008). RMS-Based Error and Complexity Measures Involving Composite Intervals http://x31eq.com/composite.pdf</ref>
 
:<math>
\begin{aligned}
G_a(\lambda) &= W^{-2} - \lambda \frac{W^{-2}j^{\mathsf T}jW^{-2}}{jW^{-2}j^{\mathsf T}} \\
&= W^{-2} - \lambda \frac{l^{\mathsf T}l}{n}
\end{aligned}
</math>
 
So this is equivalent to <math>G^{-1}</math> when we pick <math>\lambda = \frac{n}{n+1}</math>.
 
His [[Cangwu badness|parametric badness]] is given:<ref>Breed, G. (2016). http://x31eq.com/badness.pdf</ref>
 
:<math>
\begin{aligned}
G_b(E) &= \frac{W^{-2}}{jW^{-2}j^{\mathsf T}} (1+E^2) -  \frac{W^{-2}j^{\mathsf T}jW^{-2}}{(jW^{-2}j^{\mathsf T})^2} \\
&= \frac{W^{-2}}{n} (1+E^2) -  \frac{l^{\mathsf T}l}{n^2}
\end{aligned}
</math>
 
Since the metric is equivalent up to scaling, we multiply by <math>\frac{n}{1+E^2}</math> to obtain:
 
:<math>
G^{\prime}_b(E) = W^{-2} - \frac{1}{n(1+E^2)}l^{\mathsf T}l
</math>
 
Again, this is equivalent to <math>G^{-1}</math>, when we pick <math>E = \sqrt{\frac{1}{n}}</math>
 
== Generalized norm ==
For some parameter <math>k > 0</math>, set:
 
:<math>
X_k =
\begin{bmatrix}
W \\
\hline
k\cdot j
\end{bmatrix}
</math>
 
:<math>
G(k) =  X_k^{\mathsf T} X_k = W^2 + k^2j^{\mathsf T}j
</math>
 
Going through the same derivation, we find:
 
:<math>
G^{-1}(k) = W^{-2} - \frac{k^2}{nk^2+1} l^{\mathsf T}l
</math>
 
Which leads to a simple relation to <math>E</math>:
 
:<math>
\begin{aligned}
nk^2E^2 &= 1\\
E &= \sqrt{\frac{1}{nk^2}}\\
k &= \sqrt{\frac{1}{nE^2}}
\end{aligned}
</math>
== References ==
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