Magic permutations for linear forms

At least 20 years old · documented by 2005

Let $f_1,\ldots,f_m$ be positive linear forms on $\mathbb{R}^d$ , and let $s$ be the subspace on which they all vanish. For a permutation $\sigma$ of $[d]$ , define the reverse dominance order on positive linear forms by comparing all partial sums of coefficients in the $\sigma$ -ordering. Magic permutations for linear forms. A permutation $\sigma$ of $[d]$ is realizable by a positive point in $s$ if and only if $\{f_1,\ldots,f_m\}$ is an antichain in the reverse dominance order of forms. This is presented as the corresponding extension of the covering-clutter conjecture from sets to linear forms; its general status is not resolved in the supplied text.

References

Primary source

Matthias Beck and Thomas Zaslavsky, “An Enumerative Geometry for Magic and Magilatin Labellings”, arXiv:math/0506315 (2005).

Posted 1 month ago

Progress summary

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Open

A reader-posted construction claims to refute the conjecture in dimension eight, but no independent verification has been found.

The conjecture asserts that a permutation is realizable by a positive point satisfying the stated linear-form constraints exactly when the forms form an antichain under reverse dominance. Its supplied formulation is presented as a linear-form extension of the covering-clutter conjecture, but no established resolution is recorded.

Posted attempt

An explicit $d=8$ construction with three strictly positive, equal-weight forms claims a complete counterexample: the forms are pairwise incomparable, yet the identity permutation is not realizable, even after replacing common zero by common value. The construction also gives a distinct positive point realizing a different labeling under the corrected interpretation. This claimed counterexample has not been independently verified.

Current status (as of August 2026): The conjecture is claimed false by an explicit $d=8$ example, but that counterexample has not been independently verified, so the mathematical question remains open.

CounterexampleThis solution needs a summarySee full solutionHide full solution

The conjecture fails both as stated and under the natural corrected interpretation that the positive forms should take a common value.

Let $d=8$ , take the identity permutation, and write

J=x_1+x_2+\cdots+x_8.

Define three strictly positive linear forms of equal total weight $12$ :

f_A=J+x_1+x_2+x_7+x_8,

f_B=J+x_1+x_3+x_4+x_7,

f_C=J+x_3+x_4+x_5+x_6.

Every coefficient belongs to $\{1,2\}$ . Their suffix-coefficient vectors in the identity order are

v_A=(12,10,8,7,6,5,4,2),

v_B=(12,10,9,7,5,4,3,1),

v_C=(12,11,10,8,6,4,2,1).

They are pairwise incomparable: $v_A,v_B$ differ in opposite directions at coordinates $3,5$ ; $v_A,v_C$ at coordinates $2,6$ ; and $v_B,v_C$ at coordinates $2,7$ . Thus the forms satisfy the required reverse-dominance antichain condition.

Literally, the conjectured common-zero subspace cannot contain a positive point because each form is strictly positive throughout the positive orthant. More substantially, replacing common zero by common value does not repair the assertion. Any common-value point must satisfy

\begin{aligned} 0&=f_A(x)+f_C(x)-2f_B(x)\\ &=(x_2-x_1)+(x_5-x_3) +(x_6-x_4)+(x_8-x_7). \end{aligned}

The right-hand side is strictly positive whenever

0<x_1<x_2<\cdots<x_8.

Therefore the identity permutation is not realizable.

Nevertheless the common-value subspace contains the pairwise-distinct positive point

x=(1,2,3,6,4,5,8,7).

Here $J=36$ and

f_A(x)=f_B(x)=f_C(x)=54.

Thus the antichain criterion fails even for equal-weight strictly positive integer-coefficient forms whose common-value subspace admits a strongly magic labeling.