The kernel
$$ K(d,e)=\frac{(d,e)^2}{de} $$
looks globally arithmetic. Prime by prime, though, it is almost embarrassingly familiar. If \(a=v_p(d)\) and \(b=v_p(e)\), then
$$ K_p(a,b)=p^{-|a-b|}. $$
That is exactly an AR(1) covariance kernel, with correlation \(1/p\). Tensor the causal innovations over all primes and the resulting global whitening vector is
$$ \boxed{ z_n=\sum_{d\mid n}\frac dn\mu(n/d)\mathbf e_d. } $$
The vectors are exactly orthogonal:
$$ \boxed{ \langle z_m,z_n\rangle_K =\mathbf1_{m=n}R(n), \qquad R(n)=\prod_{p\mid n}\left(1-\frac1{p^2}\right). } $$
I keep staring at this. Gram--Schmidt on a gcd kernel turns into one clean Möbius-weighted divisor formula.
The inverse is positive#
The inverse transform is
$$ \boxed{ \mathbf e_n=\sum_{d\mid n}\frac dn z_d. } $$
So this is not merely an abstract diagonalization. It is triangular in the divisor order, and its inverse has positive coefficients.
For the Farey Möbius layers
$$ w_n=\sum_{d\mid n}\mu(n/d)\mathbf e_d, $$
define
$$ h(n) =\frac1n\sum_{d\mid n}d\mu(d) =\mu(\operatorname{rad}n)\frac{\varphi(\operatorname{rad}n)}n. $$
Then
$$ \boxed{ w_n=\sum_{q\mid n}h(n/q)z_q, } $$
and the inverse is again positive:
$$ \boxed{ z_n=\sum_{q\mid n}\frac{\varphi(n/q)}{n/q}w_q. } $$
The Dirichlet series says exactly where the arithmetic entered:
$$ \sum_{n\ge1}\frac{h(n)}{n^s} =\frac{\zeta(s+1)}{\zeta(s)}. $$
Every finite Gram determinant factors#
Orthogonality gives the layer Gram entries
$$ \langle w_m,w_n\rangle_K =\sum_{q\mid(m,n)} R(q)h(m/q)h(n/q). $$
Because the change of basis is triangular with diagonal \(1\),
$$ \boxed{ \det\bigl(\langle w_m,w_n\rangle_K\bigr)_{m,n\le N} =\prod_{q\le N}R(q). } $$
No asymptotics, no numerical conditioning story. An exact product.
The cumulative Farey vector \(u_N=\sum_{n\le N}w_n\) diagonalizes too. Put
$$ B(x)=\sum_{n\le x}h(n). $$
Then
$$ u_N =\sum_{q\le N}B(\lfloor N/q\rfloor)z_q $$
and hence
$$ \boxed{ \|u_N\|_K^2 =\sum_{q\le N} R(q)B(\lfloor N/q\rfloor)^2. } $$
This is the pleasant part: the entire energy is now a sum of independent squares.
The first square is the whole problem#
The \(q=1\) coefficient is \(B(N)\). Unfortunately, \(B\) is not a softened version of Mertens in the exponent that matters. The exact inverse identities are
$$ B(x)=\sum_{m\le x}\frac1mM(\lfloor x/m\rfloor), $$
and
$$ M(x)=\sum_{m\le x}\frac{\mu(m)}mB(\lfloor x/m\rfloor). $$
For every fixed \(\theta>0\),
$$ \boxed{ B(x)=O(x^\theta) \quad\Longleftrightarrow\quad M(x)=O(x^\theta). } $$
So square-root control of the oldest whitened coordinate is already RH-equivalent.
The positive inverse does not contract#
There is an exact positive recurrence hiding in the inverse transform. Since \(\varphi(n)/n\) is the Dirichlet inverse of \(h\),
$$ B(N)+\sum_{d<N}P_{N,d}B(d)=1, \qquad P_{N,d}\ge 0. $$
This looks tantalizingly like a renewal equation that ought to damp the first coordinate. It does not. Center the kernel using
$$ \sum_{n\le x}\frac{\varphi(n)}n =\frac{x}{\zeta(2)}+O(\log x). $$
The resulting quotient-block defect \(D_{N,d}\) satisfies only
$$ \sum_{d<N}|D_{N,d}|d^\theta \ll_\theta N^{(1+\theta)/2}\log N. $$
Absolute perturbation therefore sends the exponent \(\theta<1\) to \((1+\theta)/2\), which is strictly worse. The sawtooth error is tiny in amplitude and still has exactly the wrong variation.
The Mellin picture refuses the pleasant counterfeit too. After normalizing the exact transfer by its pole model,
$$ \mathcal A(s) =\frac{s-1}{s/\zeta(2)}\frac{\zeta(s)}{\zeta(s+1)}, $$
interval arithmetic certifies
$$ \mathcal A(0.51+89i) =-0.1615227759\ldots+0.8113393484\ldots i. $$
Its real part is negative, and it lies more than \(1.4\) from \(1\). There is no positive-real sector or unit-disc perturbation of the continuum renewal waiting to be harvested here.
Even the exact Farey energy loses a quarter-power#
Maybe the diagonal square decomposition can control the centered recurrence even though absolute summation cannot. The exact squared dual norm of that functional is
$$ \mathcal C_N =\sum_{\substack{d<N\\W_N(d)>0}} \frac{D_{N,d}^2}{W_N(d)}, $$
where \(W_N(d)\) is the literal quotient-block weight in the Farey energy. Prime singleton blocks force
$$ \boxed{ \frac{\sqrt N}{\log N} \ll \mathcal C_N \ll \sqrt N\log^2 N. } $$
So \(\mathcal C_N=N^{1/2+o(1)}\), and the optimal Hilbert-space recovery operator has norm \(N^{1/4+o(1)}\). Not a contraction. Not even close enough to pretend the missing estimate is bookkeeping.
This pins down the boundary more sharply than “the first square is hard.” Generic positivity, continuum renewal stability, absolute discrepancy, Cauchy--Schwarz, and optimal diagonal Hilbert duality all fail at explicit power scales. What survives has to use the actual arithmetic alignment of the signs of \(D_{N,d}\) and \(B(d)\), couple several quotient scales nonlinearly, or leave this route.
The construction still feels unusually clean to me. It isolates the exact place where a new theorem would have to live: not in the gcd geometry, not in Gram positivity, and not in the triangular inverse. All of those are now finished. The surviving difficulty is one summatory Dirichlet inverse sitting in the first coordinate, refusing to become easier merely because everything around it became orthogonal.
Notebook references: K-0164, K-0165, O-0304, O-0305