---
title: "A pure-tensor counterexample to a literal analytic-rank tensorization inequality"
date: 2026-08-30
version: "0.2.0-candidate"
doi: 10.5281/zenodo.22181327
pdf: https://github.com/ipitchford/pure-tensor-analytic-rank-counterexample/releases/download/v0.2.0-candidate/paper.pdf
repository: https://github.com/ipitchford/pure-tensor-analytic-rank-counterexample
archive: https://zenodo.org/records/22181327
license: CC0-1.0
status: unrefereed (internally replayed; not peer reviewed, not independently reproduced, not formally verified)
---
# A pure-tensor counterexample to a literal analytic-rank tensorization inequality
## Summary
An archived AIM problem-list page displays a tensorization inequality for
analytic rank or partition rank. Under one literal reading of its analytic-rank
branch, the inequality fails for the simplest possible kind of tensor: a pure
tensor.
The result depends on the wording. It refutes the **literal, unnormalised
analytic-rank instantiation** under standard base-$p$ analytic rank and
fixed-order Kronecker powers. It does not refute a normalized or
powered-identity formulation, establish what the proposer intended, or settle
partition rank.
## The exact statement: the literal comparator
The archived page prints a bound of the form
$$
\mathbf r(A^{\otimes m})<\mathbf r(I)^m c^m,
$$
where $I$ is the identity tensor on the same space, $c<1$, and $\mathbf r$ is
analytic rank or partition rank. For the analytic-rank branch, the candidate
uses
$$
\operatorname{arank}_p(T)=-\log_p\operatorname{bias}(T)
$$
and groups corresponding tensor legs in each Kronecker power. The archived
display does not specify that logarithm base, grouping, normalization or the
relationship between its two rank branches. Those conventions are assumptions
of the conditional theorem, not recovered historical facts.
## The technical mechanism: the pure-tensor witness
Work over $\mathbb F_2$, at order three and side length two. In fixed coordinate
bases, take
$$
A=e_1^*\otimes e_1^*\otimes e_1^*,\qquad
I=e_1^*\otimes e_1^*\otimes e_1^*+
e_2^*\otimes e_2^*\otimes e_2^*.
$$
For a nonzero pure order-$d$ tensor over $\mathbb F_p$, character orthogonality
gives
$$
\operatorname{bias}(A)=1-(1-1/p)^{d-1}.
$$
Here that bias is $3/4$. Writing $a=\log_2(4/3)$,
$$
\operatorname{arank}(A)=a,
\qquad
\operatorname{arank}(I)=2a.
$$
The hypothesis $\operatorname{arank}(A)<\operatorname{arank}(I)$ therefore
holds strictly.
## Why every positive power eventually fails
Under the fixed-order Kronecker product, a pure tensor remains pure. Its active
diagonal support has size one at every power, so
$$
\operatorname{arank}(A^{\boxtimes m})=a
$$
for every positive integer $m$. The proposed scalar right-hand side is
$(2ac)^m$. The exact rational comparison
$$
(4/3)^2=16/9<2
$$
implies $2a<1$. Thus $2ac<1$ for every fixed $0
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