Parallelepipeds of maximal facet area and total edge length in ellipsoids, through prescribed boundary points
Pith reviewed 2026-05-18 06:48 UTC · model grok-4.3
The pith
The total edge length of any inscribed parallelepiped in an ellipsoid reaches exactly 2^n sqrt(tr A), and this maximum is attained even with one prescribed boundary vertex.
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
Core claim
Every inscribed parallelepiped is the A^{1/2}-image of an orthotope on the unit sphere; consequently the maximal total one-skeleton length equals 2^n sqrt(tr A) and is attained for every prescribed vertex on the boundary, while the maximal total facet measure equals 2^n n^{-(n-2)/2} sqrt(det A) sqrt(tr(A^{-1})), with the maximizers for n greater than or equal to 3 being the orthotopes whose edge lengths are all equal and whose diagonals satisfy the Schur-Horn equal-diagonal condition on A inverse.
What carries the argument
The linear map sending each orthotope inscribed in the Euclidean unit sphere to its image under multiplication by A^{1/2}, which produces every centered parallelepiped inscribed in E_A and converts the length and area problems into trace and determinant identities on diagonal matrices.
If this is right
- The edge-length maximum depends only on the trace of A and is therefore the same for every ellipsoid sharing the same trace.
- The prescribed-vertex problem for edge length admits a solution in every dimension and for every boundary point.
- For facet area when n is at least 3 the maximizers on the sphere must have all edges of equal length and must satisfy the Schur-Horn equal-diagonal condition on A inverse.
- In dimension two the facet-area result recovers the known Connes-Zagier property for ellipses.
Where Pith is reading between the lines
- The same reduction technique may apply to other affine-invariant functionals on inscribed polytopes whose vertices lie on a quadratic surface.
- The explicit obstruction found for triaxial ellipsoids at principal-axis vertices suggests that prescribed-vertex problems for facet area become strictly harder in dimension three than in dimension two.
- Because the edge-length maximum is attained for every prescribed vertex, the one-skeleton length functional is constant on the entire boundary in a strong sense.
Load-bearing premise
All parallelepipeds with vertices on the ellipsoid boundary must be centered at the origin and must be precisely the images under A to the one-half of orthotopes on the unit sphere.
What would settle it
An explicit construction of a parallelepiped with vertices on the boundary whose summed edge lengths exceed 2^n sqrt(tr A), or a proof that for some boundary point x0 no inscribed parallelepiped through x0 attains that length.
read the original abstract
Let \[ \mathcal{E}_A=\{x\in\mathbb{R}^n:x^{\top}A^{-1}x\le 1\},\qquad n\ge2, \] where $A$ is real symmetric positive definite. We study full-dimensional parallelepipeds whose $2^n$ vertices lie on $\partial\mathcal{E}_A$. First we show that such parallelepipeds are necessarily centred at the origin and are precisely the images, under $A^{1/2}$, of orthotopes inscribed in the Euclidean unit sphere. This reduces the extremal questions to finite-dimensional linear algebra. For the total length $L$ of the one-skeleton we prove \[ L_{\max}(\mathcal{E}_A)=2^n\sqrt{\operatorname{tr} A}. \] Moreover, the prescribed-vertex problem for $L$ has the same answer in every dimension: for every $x_0\in\partial\mathcal{E}_A$ there is an inscribed parallelepiped with vertex $x_0$ and total edge length $2^n\sqrt{\operatorname{tr} A}$. The proof uses the Schur--Horn theorem applied to the trace-zero matrix $A-\operatorname{tr}(A)y_0y_0^{\top}$, where $y_0=A^{-1/2}x_0$. For the total $(n-1)$-dimensional measure $S$ of the facets we prove \[ S_{\max}(\mathcal{E}_A)=2^n n^{-(n-2)/2}\sqrt{\det A}\,\sqrt{\operatorname{tr}(A^{-1})}. \] For $n\ge3$ the maximisers are more rigid: on the sphere they are orthotopes with all edge lengths equal and with a Schur--Horn equal diagonal condition for $A^{-1}$. The prescribed-vertex facet-area problem is therefore equivalent to a restricted Schur--Horn problem with a prescribed barycentric basis. In dimension two this recovers the Connes--Zagier property for ellipses. In dimension three, however, the direct higher-dimensional analogue fails for triaxial ellipsoids at principal-axis vertices; an exact obstruction is given.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The manuscript studies full-dimensional parallelepipeds with all 2^n vertices on the boundary of the ellipsoid E_A = {x in R^n : x^T A^{-1} x <= 1}, A symmetric positive definite. It first establishes that any such parallelepiped is centered at the origin and equals the image under A^{1/2} of an orthotope inscribed in the unit sphere. This reduces both extremal problems to linear-algebraic questions on the sphere. For the total one-skeleton length L the paper proves L_max(E_A) = 2^n sqrt(tr A) and shows that the same value is attained for every prescribed boundary vertex x_0 via an application of the Schur-Horn theorem to the trace-zero matrix A - tr(A) y_0 y_0^T with y_0 = A^{-1/2} x_0. For the total (n-1)-dimensional facet measure S it proves S_max(E_A) = 2^n n^{-(n-2)/2} sqrt(det A) sqrt(tr(A^{-1})), with rigidity statements for n >= 3 that require equal edge lengths and a Schur-Horn equal-diagonal condition on A^{-1}. The prescribed-vertex facet-area problem is recast as a restricted Schur-Horn problem; the two-dimensional case recovers the Connes-Zagier property for ellipses, while a concrete obstruction is exhibited for triaxial ellipsoids in dimension three at principal-axis vertices.
Significance. If the derivations hold, the work supplies explicit, closed-form maxima for two natural geometric functionals on inscribed parallelepipeds in ellipsoids, together with sharp rigidity and prescribed-vertex results. The reduction to the sphere via A^{1/2} and the subsequent direct use of the Schur-Horn theorem yield parameter-free expressions involving only tr A, det A and tr(A^{-1}). The explicit obstruction in dimension three and the recovery of the known two-dimensional case add concrete value. The manuscript therefore contributes precise extremal geometry that is accessible to linear-algebraic techniques.
major comments (2)
- [§2] §2 (reduction step): the claim that every inscribed parallelepiped is centered at the origin follows from the symmetry of the sign-sum barycenter, but the manuscript should explicitly verify that the barycenter of the 2^n vertices is necessarily the origin when all vertices lie on the ellipsoid boundary; this step is load-bearing for the subsequent linear-algebraic reduction.
- [Theorem 3.1] Proof of Theorem 3.1 (edge-length maximum): the application of Schur-Horn to the matrix A - tr(A) y_0 y_0^T produces the claimed value 2^n sqrt(tr A), yet the manuscript must confirm that the off-diagonal entries of the transformed matrix remain compatible with the orthotope constraint sum ||v_i||^2 = 1 after the linear change of variables; a short calculation showing that the maximum is attained precisely when the v_i are orthogonal would strengthen the argument.
minor comments (3)
- [Abstract and §3] Notation: the symbol L is used both for the total edge length functional and for its maximum value; introducing L_max(E_A) consistently from the first appearance would improve readability.
- [Figure 1] Figure 1 (schematic of orthotope on sphere): the caption should state the dimension n and the explicit relation between the edge vectors v_i and the matrix whose eigenvalues are controlled by Schur-Horn.
- [References] Reference list: the citation to the Schur-Horn theorem should include the original 1954 paper or a standard modern reference (e.g., Horn & Johnson, Matrix Analysis) rather than only a secondary source.
Simulated Author's Rebuttal
We thank the referee for the positive evaluation and the recommendation of minor revision. The comments are helpful for clarifying the reduction and strengthening the proof of the edge-length maximum. We address each major comment below.
read point-by-point responses
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Referee: [§2] §2 (reduction step): the claim that every inscribed parallelepiped is centered at the origin follows from the symmetry of the sign-sum barycenter, but the manuscript should explicitly verify that the barycenter of the 2^n vertices is necessarily the origin when all vertices lie on the ellipsoid boundary; this step is load-bearing for the subsequent linear-algebraic reduction.
Authors: We agree that an explicit verification of the centering claim will improve clarity. In the revised manuscript we will add a short direct argument in §2: the vertices of any parallelepiped may be written as c + ∑_{i=1}^n ε_i v_i with ε_i = ±1. Their barycenter is exactly c, because the signed sums over all 2^n choices cancel all v_i terms. When all vertices lie on the centrally symmetric boundary ∂E_A, suppose c ≠ 0. Then the quadratic form values at c + w and c − w (for appropriate sign patterns w) cannot simultaneously equal 1 for every combination unless the linear term 2c^T A^{-1} w vanishes identically, which forces c = 0. This calculation will be inserted verbatim before the reduction to the sphere. revision: yes
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Referee: [Theorem 3.1] Proof of Theorem 3.1 (edge-length maximum): the application of Schur-Horn to the matrix A - tr(A) y_0 y_0^T produces the claimed value 2^n sqrt(tr A), yet the manuscript must confirm that the off-diagonal entries of the transformed matrix remain compatible with the orthotope constraint sum ||v_i||^2 = 1 after the linear change of variables; a short calculation showing that the maximum is attained precisely when the v_i are orthogonal would strengthen the argument.
Authors: We thank the referee for this suggestion. The matrix M = A − tr(A) y_0 y_0^T is trace-zero, so its eigenvalues sum to zero. Schur–Horn supplies an orthogonal matrix U such that diag(U^T M U) equals any vector majorized by the eigenvalues. After the change of variables the orthotope edges v_i on the unit sphere satisfy ∑ ||v_i||^2 = 1, which is precisely the sum of the diagonal entries of the Gram matrix in the eigenbasis; this sum is invariant and equals tr(M) + tr(A) y_0 y_0^T = tr(A) (the extra term cancels the trace-zero condition). The off-diagonal entries of the transformed matrix therefore do not enter the constraint. We will append a short paragraph to the proof of Theorem 3.1 exhibiting the equality case: the total length ∑ 2 ||A^{1/2} v_i|| attains its upper bound 2^n sqrt(tr A) if and only if the v_i are pairwise orthogonal, which is realized precisely when the diagonal of U^T M U consists of the appropriate repeated eigenvalues. revision: yes
Circularity Check
No significant circularity identified
full rationale
The paper's central reduction—that inscribed parallelepipeds in the ellipsoid are centered at the origin and are images under A^{1/2} of orthotopes on the unit sphere—is established by a direct geometric argument: any such parallelepiped pulls back to one in the unit ball, and the equal-norm condition on all 2^n vertices forces the generating vectors to be pairwise orthogonal (via vanishing cross terms in the expanded squared norm over all sign patterns) with sum of squared lengths equal to 1. Centering follows from symmetry of the sign sums. This is an independent fact, not defined circularly in terms of the claimed maxima. Subsequent applications of the external Schur-Horn theorem to the trace-zero matrix A - tr(A) y_0 y_0^T (with y_0 = A^{-1/2} x_0) and the facet-area formula derived from equal-edge orthotopes with Schur-Horn diagonal conditions are likewise independent of the results themselves; no parameters are fitted to the target quantities, no self-citations are load-bearing for the core claims, and the derivations remain self-contained against external mathematical benchmarks.
Axiom & Free-Parameter Ledger
axioms (2)
- standard math Schur-Horn theorem on diagonals of matrices with prescribed eigenvalues
- domain assumption Every full-dimensional parallelepiped with vertices on the ellipsoid boundary is centered at the origin and is the image under A^{1/2} of an orthotope on the unit sphere
discussion (0)
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