At power \(p=23\), one derivative-moment query has a rank-one adjoint and a translation orbit of dimension
$$ \boxed{(2p)^2=2116.} $$
Both statements are exact. They are not contradictory.
This is the little compression ambush I wanted to understand: a query can remain rank one at every recursive pullback while still visiting every direction in the block state as the preceding prefix changes.
The all-prefix object#
Let
$$ S_N=\sum_{m\le N}a_m,\qquad W_N=S_N^p, $$
and let \(\delta\) be a derivation commuting with conjugation. The three densities I care about are
$$ \sum_N |W_N|^2,\qquad \sum_N (\delta W_N)\overline{W_N},\qquad \sum_N |\delta W_N|^2. \tag{1} $$
They depend on every prefix, not just the completed sum.
For a consecutive block \(B\), write \(x_k,y_k\) for its local value and derivative prefixes. Store the mixed moments
$$ M_B^{\alpha,\beta}(r,s) = \sum_k y_k^\alpha\overline{y_k}^{\,\beta} x_k^r\overline{x_k}^{\,s}, $$
where \(\alpha,\beta\in\{0,1\}\), \(0\le r\le p-\alpha\), and \(0\le s\le p-\beta\).
Concatenating blocks is just translation. Every prefix in the right block is shifted by the value and derivative totals of the left block, so the binomial formula gives an exact associative merge law. The state has
$$ \bigl((p+1)+p\bigr)^2=(2p+1)^2 $$
channels. At \(p=32\), that is \(4225\).
This part is good news: the all-prefix moments in (1) really do have a finite nonlinear state. There is no need to remember every prefix separately.
Why completed frequency is not enough#
If
$$ P_m=W_m-W_{m-1}, $$
then
$$ \sum_N |W_N|^2 = \sum_{m,\ell} \bigl(M-\max(m,\ell)+1\bigr) P_m\overline{P_\ell}. \tag{2} $$
The kernel
$$ H_{m,\ell}=M-\max(m,\ell)+1 $$
is the Brownian max-index Gram matrix. It has determinant one and full rank \(M\).
There is even a rude exact collision at power \(32\):
700000 * 1313000 = 707000 * 1300000.
Pad both sides with thirty copies of \(700000\). The two resulting power-\(32\) monomials have the same completed logarithmic frequency, but different maximum indices. Their difference has exact \(H\)-energy
$$ \boxed{13000}. $$
So a linear summary that merges layers solely by completed frequency has thrown away something the all-prefix norm can still see.
Rank one survives the pullback#
Package the value and derivative monomials into
$$ f(x,y) = (1,x,\ldots,x^p) \oplus (y,yx,\ldots,yx^{p-1}). $$
Translation by block totals \((t,u)\) acts through a triangular matrix:
$$ f(x+t,y+u)=T(t,u)f(x,y). $$
The block state is
$$ M_B=\sum_{k\in B}f(x_k,y_k)f(x_k,y_k)^*. $$
For a selected scalar query, choose covectors \(a,b\) and write
$$ \Phi_B(a,b)=L(aM_Bb^*). $$
If \(AB\) is a concatenation and \(T_A\) is the translation contributed by \(A\), then exactly
$$ \Phi_{AB}(a,b) = \Phi_A(a,b)+\Phi_B(aT_A,bT_A). \tag{3} $$
That is lovely. A rank-one query pulls through the whole block tree as a rank-one query. No dense adjoint matrix appears.
The orbit is still full#
The tempting inference from (3) is that one should now be able to cache a small fixed linear summary for every block.
Nope.
As the preceding totals vary, the translated value row \(X^p\) spans \(p+1\) monomials. The translated derivative row \(YX^{p-1}\) spans the \(2p\) monomials
$$ 1,x,\ldots,x^{p-1}, \qquad y,yx,\ldots,yx^{p-1}. $$
Their outer products are distinct monomials in \(\mathbb C[x,\bar x,y,\bar y]\). Therefore the value, cross, and derivative query orbits have exact dimensions
$$ (p+1)^2,\qquad 2p(p+1),\qquad (2p)^2. \tag{4} $$
At \(p=23\), these are
value 576
cross 1104
derivative 2116
The adjoint is rank one pointwise, but the family of rank-one adjoints seen under all possible translations spans the entire separate-pass state.
That distinction is the whole obstruction.
What remains possible#
Equation (3) still saves matrix materialization for one actual incoming prefix. The finite translation monoid is still exact. A depth-first schedule may still recompute rather than cache, and a source-specific identity may collapse directions that the generic translation algebra keeps independent.
What does not survive is feature-rank-only optimism. A universal linear block cache cannot be made small merely because the selected query is rank one. The next compression has to use the arithmetic source and the final integration kernel together, not just the geometry of the feature vector.
I like this obstruction because it is so cleanly mean: rank one was true. It simply was not the rank that controlled the cost.
Notebook references: R-0549, R-0560