# User:Sintel/Dual Weil-Euclidean norm

## Derivation

On some $p$-limit subgroup with $n$ primes, define the $n \times n$ Tenney weighting matrix $W$:

$$W = \begin{bmatrix} \log_2 2 & 0 & \cdots & 0 \\ 0 & \log_2 3 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \log_2 p \end{bmatrix}$$

And the row vector $j$ containing the log-primes (aka the JIP): $j = \begin{bmatrix} \log_2 2 & \log_2 3 & \cdots & \log_2 p \\ \end{bmatrix}$

Then the block matrix $X$ obtained from these:

$$X = \begin{bmatrix} W \\ \hline j \end{bmatrix}$$

defines an inner product, with positive definite $G = X^{\mathsf T} X$:

$$\left\langle x,y \right\rangle = x^{\mathsf T} X^{\mathsf T} X y = x^{\mathsf T} G y$$

and an induced norm $||x|| = \sqrt{\left\langle x,x \right\rangle}$, which is the Weil-Euclidean norm.

The inner product on the dual space can then be derived by simply inverting $G$ [1] , which gives the dual norm:

$$\left\langle \alpha, \beta \right\rangle^{\ast} = \alpha G^{-1} \beta^{\mathsf T} \\ ||\alpha||^{\ast} = \sqrt{\left\langle \alpha,\alpha \right\rangle^{\ast}} = \sqrt{\alpha G^{-1} \alpha^{\mathsf T}}$$

The goal is now to find an expression for $G^{-1}$.

First, note that:

$$G = X^{\mathsf T} X = W^2 + j^{\mathsf T}j$$

Since the outer product $j^{\mathsf T}j$ is rank-1 we can use a theorem on the inverse of matrix sums which states: [2]

If $A$ and $A+B$ are invertible, and $B$ has rank 1, then let $g = \text{tr}(BA^{-1})$. Then $g \neq -1$ and
$(A+B)^{−1}=A^{-1} − \frac{1}{1+g}A^{-1}BA^{-1}$

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

$$G^{-1} = (W^2 + j^{\mathsf T}j)^{-1} = W^{-2} - \frac{1}{1+g} W^{-2}j^{\mathsf T}jW^{-2}$$

Now let $l = \begin{bmatrix} \frac{1}{\log_2 2} & \frac{1}{\log_2 3} & \cdots & \frac{1}{\log_2 p} \\ \end{bmatrix}$, then

$$l = W^{-2}j \\ G^{-1} = W^{-2} - \frac{1}{1+g} l^{\mathsf T}l$$

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

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

Which leads to the final expression:

$$G^{-1} = W^{-2} - \frac{1}{n+1} l^{\mathsf T}l$$

## Relation to other metrics

Graham Breed gives the following formula (adapted for the notation introduced here):[3]

\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}

So this is equivalent to $G^{-1}$ when we pick $\lambda = \frac{n}{n+1}$.

\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}

Since the metric is equivalent up to scaling, we multiply by $\frac{n}{1+E^2_k}$ to obtain:

$$G^{\prime}_b(E) = W^{-2} - \frac{1}{n(1+E^2)}l^{\mathsf T}l$$

Again, this is equivalent to $G^{-1}$, when we pick $E = \sqrt{\frac{1}{n}}$

## Generalized norm

For some parameter $k \gt 0$, set:

$$X_k = \begin{bmatrix} W \\ \hline k\cdot j \end{bmatrix}\\ G(k) = X_k^{\mathsf T} X_k = W^2 + k^2j^{\mathsf T}j$$

Going through the same derivation, we find:

$$G^{-1}(k) = W^{-2} - \frac{k^2}{nk^2+1} l^{\mathsf T}l$$

Which leads to a simple relation to $E$:

$$nk^2E^2 = 1\\ E = \sqrt{\frac{1}{nk^2}}\\ k = \sqrt{\frac{1}{nE^2}}$$

## References

1. Taking the natural map to the dual space $\Gamma: V\to V^{\ast}: x \mapsto \left\langle x, \cdot \right\rangle$, we require $\left\langle \Gamma(x),\Gamma(y) \right\rangle^{\ast} = \left\langle x,y \right\rangle$.
2. Miller, K. S. (1981). On the Inverse of the Sum of Matrices. Mathematics Magazine, 54(2), 67–72. https://doi.org/10.2307/2690437
3. 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