REVIEW 2 major objections 3 minor 16 references
On the total surface area of potato packings
T0 review · 2 major / 3 minor · reviewed 2026-08-11 · deepseek-v4-flash
Pith's one-line read A gap-free packing of a set by positive-volume pieces that touch only on zero-measure boundaries must have infinite total perimeter.
desk verdict Theorem 2.2 is false as stated (pairwise additivity is insufficient), but the geometric potato-packing result is likely correct and worth publishing after a fix. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The engine is the class of perimeter-like evaluations: functions on Borel sets with $F(\emptyset)=0$, complement symmetry $F(X\setminus A)=F(A)$, an upper semicontinuity-type property (T), and $L^1$-lower semicontinuity (L). Theorems 2.1 and 2.2 use these axioms together with an additivity assumption $F(E_i\cup E_j)=F(E_i)+F(E_j)$ to force a dichotomy. In the geometric application, the perimeter of a PI space is known to be such an evaluation, and pairwise disjointness of essential boundaries gives the required additivity; the relative isoperimetric inequality then excludes the zero option.
What would settle it
A concrete example in the Euclidean plane of an open ball partitioned into countably many positive-area sets whose boundaries meet only at finitely many points, with finite total perimeter, would refute Corollary 3.2. Alternatively, an explicit perimeter-like evaluation satisfying axioms (0), (C), (T), and (L) with pairwise additivity but not finite additivity on a covering family would expose the gap in the proof of Theorem 2.2.
Extended reading notes
Core claim
The central claim is Theorem 3.1: if a PI space is covered, up to measure zero, by disjoint positive-volume sets whose essential boundaries meet pairwise in sets of zero $H^{-1}$ measure, then the sum of their perimeters is infinite. The proof runs through an abstract dichotomy (Theorem 2.2): for any perimeter-like evaluation satisfying axioms (0), (C), (T), and (L), if a measurable partition displays pairwise additivity of the evaluation, then either the evaluation of the pieces sums to infinity or every piece evaluates to zero. Because the perimeter of a PI space satisfies the axioms and has a relative isoperimetric inequality, the all-zero option is impossible for positive-volume pieces, leaving only infinite total perimeter.
Load-bearing premise
The proof of Theorem 2.2 assumes the perimeter-like evaluation is additive over arbitrary finite unions of the packing sets, while the theorem states only pairwise additivity; the geometric perimeter has this stronger property through the cited additivity lemma, but the abstract theorem as written relies on an unstated finite-additivity premise.
Editorial extensions
If this is right
- In the Euclidean plane and higher dimensions, any gap-free packing of an open set by regular positive-volume sets whose boundaries meet only at $H^{d-1}$-null sets must have infinite total boundary measure.
- A finite total perimeter is possible only if the packing leaves a positive-measure residual set or is trivial up to measure zero.
- The residual set of any full packing by such pieces has Hausdorff dimension at least $d-1$; the bound is attained by known examples of smooth-curve packings with residual dimension exactly 1.
- The same conclusion holds in any PI space, including smooth Riemannian manifolds and sub-Riemannian spaces such as Heisenberg groups, under local doubling and Poincaré inequalities.
- In the abstract setting, the result indicates that a perimeter-like evaluation that is finite on a partition must be trivial on all pieces, isolating why finite-perimeter packings cannot be gap-free.
Reading between the lines
- The abstract version suggests the phenomenon is a general incompatibility: any cost function with complement symmetry, appropriate semicontinuity, and finite additivity over a partition cannot be both finite and strictly positive on every piece of a gap-free countable decomposition.
- A natural quantitative extension, not pursued in the paper, would ask how fast the partial sums of perimeters must diverge in terms of the number of pieces or the sizes of the boundary intersection sets.
- One can test the sharpness by constructing near-packings where intersections have small positive $H^{-1}$ measure; the theorem predicts perimeter blow-up, but the blow-up rate is an open question.
- The connection with Apollonian packings suggests that infinite perimeter is the geometric signature of fully filling a domain with disjoint bodies that have only point-like contacts.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper proves a geometric statement about packings: if an open set ('bag') in Euclidean space, or more generally in a PI-space, is filled up to measure zero by infinitely many positive-volume sets that meet only along boundaries of zero H^{-1}-measure, then the total perimeter (surface area) of the packing is infinite. The proof is organized through an abstract notion of 'perimeter-like evaluation' F satisfying axioms (0), (C), (T), (L), (Z). A general theorem (Theorem 2.2) is first proved for such F under a pairwise additivity hypothesis, and then applied to the perimeter functional on PI-spaces, yielding the packing results in Theorem 3.1 and Corollary 3.2, including a corollary for Euclidean packings and remarks on Riemannian and sub-Riemannian settings.
Significance. If the abstract theorem is repaired, this is a valuable and elegant contribution: it gives a short axiomatic proof of a striking geometric fact, extends earlier work of Maio--Ntalampekos and the sphere-packing papers to general PI-spaces, and covers important examples such as Heisenberg and Carnot groups. The paper is concise, clearly written, and the geometric applications are plausible. The main theorem, however, is false as stated, so the abstract framework needs a local but essential correction before the applications can be regarded as fully proved.
major comments (2)
- [§2, Theorem 2.2 and Eq. (1)] The proof uses the identity F(T_n^m) = Σ_{i=n+1}^m F(E_i) 'by assumption', but the theorem only assumes pairwise additivity F(E_i ∪ E_j) = F(E_i) + F(E_j) for i ≠ j. Pairwise additivity does not imply additivity over arbitrary finite unions, and the axioms (0), (C), (T), (L) do not enforce it. The theorem is false as stated. A counterexample is given by X = {0,1,2,3} with any positive atom measure, E_0,…,E_3 the singletons and E_k = ∅ for k ≥ 4. Define F(∅) = F(X) = 0, F({x}) = 1, F(two-point set) = 2, F(three-point set) = 1. Properties (0) and (C) are immediate; (L) holds because on a finite atom space m(A_n Δ A) → 0 implies A_n = A eventually; and (T) holds because the only F-null sets are ∅ and X, so F(X \ A_n) → 0 forces the pair (A_n, F(A_n)) to be eventually (∅,0) or (X,0) when A = X, and only (∅,0) when A ≠ X. The pairwise additivity condition is satisfied, but Σ F(E_i) = 4 is finite and nonzero, contradicting the dichotomy. The theorem can be repaired by adding the assumption that F is additive over arbitrary finite unions of the E_i; the concrete perimeter in Theorem 3.1 has that stronger property via [4, Lemma 2.3(ii)] and induction, so the geometric applications are likely salvageable, but Theorem 2.2 must be restated.
- [§2, Proposition 2.1] In the second part of Proposition 2.1, the chain 0 = F(∅) = F(X) = F(X \ ∪E_i) 'by (Z)' is not justified: (Z) can identify F(X) with F(∪E_i), since m((∪E_i) Δ X) = m(X \ ∪E_i) = 0, but it cannot identify F(X) with F(X \ ∪E_i) without knowing m(∪E_i) = 0. The intended argument is valid after swapping the roles of (Z) and (C): first F(X) = F(∪E_i) by (Z), then F(X \ ∪E_i) = F(∪E_i) by (C). This is a local proof error, but it is used in the proof of Theorem 2.2 and should be corrected.
minor comments (3)
- [§2, Theorem 2.2] In the second sentence of Theorem 2.2, the clause 'for all i ≠ j' appears to be a remnant of the pairwise-additivity assumption and is misplaced; the intended statement is that the pairwise additivity holds and that m(X \ ∪E_i) = 0.
- [§3, Proof of Corollary 3.2] The line 'm(E_k ∩ E_j) = 0 since H^{d-1}(\bar E_k ∩ \bar E_j) = 0' is terse. It is true for Borel sets with positive Lebesgue measure, because a positive-measure subset of R^d has Hausdorff dimension d and hence infinite H^{d-1}, but this fact deserves a short explanation.
- [Throughout] There are several typographical issues: 'tecnical' in the heading of Section 2, 'P Ispace' in the proof of Theorem 3.1, and intermittent spacing in 'P I spaces'. These should be corrected.
Circularity Check
No circularity: the main theorem is derived from stated axioms and standard cited facts, and no fitted parameter or self-citation chain forces the conclusion.
full rationale
The paper proves a geometric dichotomy: for a perimeter-like functional F satisfying (0), (C), (T), (L), pairwise additivity and a full-measure union force either an infinite total F or trivial zero sets. The proof is a direct axiomatic argument, and the geometric applications translate it via the known properties of the perimeter in PI-spaces cited from [4] and [16]. There is no parameter fitted to a target dataset, no known empirical pattern renamed as a theorem, and no conclusion smuggled into the hypotheses. The only substantive issue is that equation (1) uses finite additivity, F(T_n^m) = sum_{i=n+1}^m F(E_i), 'by assumption,' whereas the theorem as stated assumes only pairwise additivity F(E_i ∪ E_j) = F(E_i) + F(E_j); pairwise additivity does not in general imply additivity over arbitrary finite unions. This is a missing-assumption or proof-gap concern, not circularity: the argument does not assume the dichotomy it aims to prove, and the concrete perimeter in Theorem 3.1 does satisfy the stronger finite-additivity property through [4, Lemma 2.3(ii)] and induction, so the geometric conclusions are independently supported. The paper's self-citations are to established, machine-checkable or externally developed results on metric perimeters, not to an unverified uniqueness theorem by the same authors. Accordingly, the circularity score is 0.
Assumptions & free parameters
assumptions (6)
- domain assumption PI-space perimeters are perimeter-like evaluations satisfying (0), (T), (C), (L), (Z).
- domain assumption Perimeter admits the integral representation P(E) = ∫_{∂eE} θ_E dH^{-1} with density bounded between positive constants depending only on the PI-space constants.
- domain assumption The relative isoperimetric inequality holds in PI spaces.
- domain assumption For disjoint sets with zero H^{-1} intersection of essential boundaries, perimeter of the union is the sum of the perimeters.
- ad hoc to paper F is finitely additive over arbitrary finite unions of the E_k.
- standard math Standard measure-theoretic facts on monotone convergence of measures, Fubini, and Hausdorff measure normalization.
Cite this review
Pith. "Pith review of On the total surface area of potato packings." pith.science (2026). https://pith.science/paper/M2YHTBBD
@misc{pith2026241210905,
author = {Pith},
title = {Pith review of: On the total surface area of potato packings},
year = {2026},
howpublished = {\url{https://pith.science/paper/M2YHTBBD}},
note = {Machine review of arXiv:2412.10905}
}
read the original abstract
We prove that if we fill without gaps a bag with infinitely many potatoes, in such a way that they touch each other in few points, then the total surface area of the potatoes must be infinite. In this context potatoes are measurable subsets of the Euclidean space, the bag is any open set of the same space. As we show, this result also holds in the general context of doubling (even locally) metric measure spaces satisfying Poincar\'e inequality, in particular in smooth Riemannian manifolds and even in some sub-Riemannian spaces.
Figures
Reference graph
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Reviewed August 11, 2026 · model on record in the stance chip above.
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