RESEARCH / Focused target
Independent extreme-point tests in polyhedral Banach spaces
RELEASED FOUNDATIONS / PROPOSED CONNECTION
Research connections
All source anchors (1)
- D5/S3/Analytic/BanachGeometry/LinfKNumber.resultAbsent from this graph
Problem
The question
Manna, Mandal, Paul and Sain, An improvement of the Blanco-Koldobsky-Turnšek characterization of isometries, arXiv:2511.20345v1, final Open Question, S2.p45.1:
Open Question: In an n-dimensional polyhedral Banach space, does the preservation of Birkhoff-James orthogonality by a bounded linear operator at any n linearly independent extreme points necessarily imply that the operator is a scalar multiple of an isometry?
This dossier anchors only this question. The standing field is real or complex; local preservation means every direction and every base-field scalar. The test tuple is chosen independently, rather than requiring preservation simultaneously at every independent tuple. The conclusion permits any one common scalar, including zero, and a linear isometry of the same carrier.
Motivation
Our foothold
The distinction between a spanning test set and the geometry of its normal
cones determines whether local tests force a global isometry. The positive
sup-norm theorem LinfKNumber.result shows that independence suffices for
extreme-point bases of cubes. A different polyhedral norm can distinguish
this special case from the universal question.
Gap
Missing bridges
The all-extreme-point and all-k-smooth theorems of arXiv:2504.11161v2 require larger test sets. Theorem 2.33 of arXiv:2501.02714v1 gives negative examples on nonpolyhedral $\ell_p$ spaces. Its Corollary 2.29 only gives bijectivity from the selected tests in a polyhedral plane. The complete bounded comparisons and source locators are in the primary Library note. They supply no applicable whole resolution in the two inspected full texts, and make no global absence or priority claim.
Route
Proposed approach
Use the complete real normed plane with
$$ N(a,b)=\max(|a|,|a+b|,|a/2+b|),\qquad p=(1,0),\quad q=(0,1),\quad r=(-1,3/2),\quad T(a,b)=(a+b,-3b/2). $$
The native carrier is the plane $w_2=w_1-w_0/2$ in Fin 3 → ℝ, with the
inherited metric and norm. Algebraic coordinates give dimension two;
finite-dimensional completeness and continuity give a Banach space and an
actual bounded endomorphism. This coordinate equivalence does not assert
that the usual product norm is preserved.
The ball is exactly the three slab constraints $|w_i|\le1$. An extreme point must have at least two active constraints: a nonzero tangent that fixes a sole active constraint gives a small feasible segment on both sides. The active pairs put every actual extreme point inside the six-point set ${\pm p,\pm q,\pm r}$, proving finiteness. Two saturated independent supporting equations prove actual extremality of $p,q$. They are the images of the standard basis and both have norm one.
For every direction $(a,b)$, input orthogonality at $p$ implies $a(a+b)\le0$, and at $q$ implies $(a+b)(a/2+b)\le0$. The output signed products are
$$ (a+b)(a-b/2)=\tfrac32 a(a+b)-\tfrac12(a+b)^2,\qquad (a+b)(b-a/2)=3(a+b)(a/2+b)-2(a+b)^2. $$
They remain nonpositive. For each real perturbation scalar, an active signed
facet with nonnegative increment proves the output BJ inequality. Necessity
uses the existing LinfKNumber.active_not_BJ finite-coordinate descent proof
at its owning declaration. No direction or scalar is sampled.
If $T=cU$ for a native linear isometry, every norm is multiplied by $|c|$. But $N(p)=N(r)=N(Tp)=1$ and $N(Tr)=2$, a contradiction.
Falsifier
What would falsify this route
The example would fail if the induced carrier or its metric differed from the specified norm, if the ball's actual extreme set were infinite, if the selected vectors were dependent or not extreme unit points, if any real direction violated either full preservation implication, or if some common scalar and native linear isometry represented $T$. Each of these is an explicit obligation in the native proof. The claim must retain both source fields and all dimensions; a norm-ratio statement or a finite-direction calculation alone would not refute it.
Evidence
Evidence to collect
D5.S3.Analytic.BanachGeometry.PolyhedralBJRefutation.result : ¬ claim
instantiates the real branch of a literal closed claim : Prop, containing
the universal real and complex assertions. It proves the entire native
counterexample without assumed geometry, continuity or preservation premises.
The public data definitions are fieldClaim, claim, facetPlane, Plane,
point, coordinates and operator; all other new mathematical lemmas are
private and lie on the resolution's proof path.
The native implementation was prospectively registered in
issue #19309.
The earlier ordinary proof was not prospectively registered there.
The admission basis is open-problem-resolution; the utility is
certified-instance with refutes pointing to this claim and result.
The Scribe attaches OpenProblemResolutionClaim with Refuted to the result.
Triage
Scope assessment
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Proved in the native theorem — the original universal assertion is false. The real two-dimensional instance settles this one question. Neither a classification of polyhedral norms nor a K-number is asserted.
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Explanatory calculation — the omitted facet causes the failure of the proposed test rule. The selected adjacent vertices lie on the facet $[p,q]$, and $T$ maps them to the adjacent vertices $p,-r$. The two required active-slope products are preserved by subtracting nonnegative squares. This does not constrain the remaining vertex $r$. In the stated norm, $r\perp_B q$ since its first facet reading stays $-1$ under every real perturbation by $q$, whereas $N(Tr-\tfrac12 Tq)=N(0,-3/2)=3/2<2=N(Tr)$. These last boundary identities explain the scope of the counterexample; they are not additional named Lean theorems in the resolution module.
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Existing positive results retain their scope. Requiring preservation at every extreme point is the premise of the cited all-extreme theorem; the calculation at $r$ shows that premise fails here. The native positive sup-norm result still proves that every independent extreme-point basis suffices for a cube. Bijectivity alone is too weak: this operator has determinant $-3/2$ and still has unequal norm ratios.
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Unresolved separate questions. The K-number of this norm, the existence of another two-point K-set for it, and a classification of sufficient tuples for other polyhedral norms are not settled by this result. No new companion theorem or classification is included.
ASSUMED-UNVERIFIED
Unverified assumptions
The literature comparison is limited to the authenticated source and the two named versioned full texts. Later unindexed or unpublished work, journal-version equivalence and mathematical priority are unverified. Independent formal review, required CI and final development-branch delivery are separate from the native theorem and are not asserted by this dossier.