Manuscript Oai283-A@1 · Oai283-A.tex · proposition · 6fb75895

Proposition: A smooth strictly convex nonquadratic cone

Serial teleportation and first-passage negative controls › Shape regularity does not select a Lorentz cone · lines 1540–1569 · proof included or linked

Oai283-A@1#prop:l4

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The cone

K_{4} = {(t, x, y) : t \geq (x^{4} + y^{4})^{1/4}}

is proper, full-dimensional, strictly convex in the cone sense, and has smooth boundary away from its vertex. It is not linearly isomorphic to a Lorentz cone.

Proof. The $ℓ^{4}$ norm is strictly convex, giving the cone assertions. Its nonzero boundary is a smooth level set of $t^{4} - x^{4} - y^{4}$ .

If a nonzero quadratic form vanished on that boundary, independent sign changes in $x, y$ would force all mixed terms to vanish. It would have the form

a t^{2} + b x^{2} + c y^{2} .

The boundary points with one spatial coordinate zero give $b = c = - a$ . At $(2^{1/4}, 1, 1)$ , this requires $a (2 - 2) = 0$ , hence the quadratic form is zero. A linear image of a Lorentz cone has a nonzero quadratic form vanishing on its boundary, a contradiction.

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unlabeled · line 1543
$K_{4} = {(t, x, y) : t \geq (x^{4} + y^{4})^{1/4}}$
unlabeled · line 1560
$a t^{2} + b x^{2} + c y^{2} .$

Proposition: A smooth strictly convex nonquadratic cone

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