Quadratic ellipsoids in Minkowski geometries

A Minkowski plane is Euclidean if and only if at least one ellipse is a quadric. We discuss the higher dimensional consequences too.


Introduction
Let I be an open, strictly convex, bounded domain in R n , (centrally) symmetric to the origin. Then the function d : R n × R n → R defined by A hypersurface in R n is called a quadric if it is the zero set of an irreducible polynomial of degree two in n variables. We call a hypersurface quadratic if it is part of a quadric. Since every isometric mapping between two Hilbert geometries is a restriction of a projectivity, and every projectivity maps quadrics to quadrics, the quadraticity of a metrically defined hypersurface is a geometric property in each Hilbert geometry. Thus the question arises whether the 568Á. Kurusa AEM metrically defined hypersurfaces are quadrics. In [5] we answered this question for conics.
We prove that (Theorem 4.3) a Minkowski plane is a model of the Euclidean plane if and only if at least one of the ellipses is a quadric, and that (Theorem 4.4) a Minkowski plane is analytic if and only if at least one of the ellipses is analytic.
As for higher dimensions, we prove (Theorem 5.1) that a Minkowski geometry is a model of the Euclidean geometry if and only if every central planar section of at least one quadric is an ellipse.

Notations and preliminaries
Points of R n are labeled as A, B, . . . , vectors are denoted by − − → AB or a, b, . . . , but we use these latter notations also for points if the origin is fixed. The open segment with endpoints A and B is denoted by AB, while AB denotes the open ray starting from A passing through B, and AB = AB ∪ AB.
On It is easy to observe in (D 1 ) that an ellipsoid intersects line F 1 F 2 , the main axis, in exactly two points whose distance is twice the radius. Further notions are the (linear) eccentricity c = d(F 1 , F 2 )/2, the numerical eccentricity ε = c/a. The metric midpoint of the segment F 1 F 2 is called the center.
In the plane we use the notations u ϕ = (cos ϕ, sin ϕ) and u ⊥ ϕ := (cos(ϕ + π/2), sin(ϕ + π/2)). It is worth noting that, by these, we have d dϕ u ϕ = u ⊥ ϕ . A quadric in the plane has the equation of the form Q σ s := (x, y) : in a suitable affine coordinate system s, and we call it elliptic, parabolic, or hyperbolic according to whether σ = 1, σ = 0, or σ = −1, respectively. We usually polar parameterize the boundary ∂D of a non-empty domain D in R 2 star-like with respect to a point P ∈ D so that r : [−π, π) → R 2 is defined by r(ϕ) = r(ϕ)u ϕ , where r is the radial function of D with base point P .
We call a curve in the plane analytic if the coordinates of its points depend on its arc length analytically. Vol. 96 (2022) Quadratic ellipsoids in Minkowski geometries 569

Utilities
In the presented technical lemmas the underlying plane is Euclidean.
hence the function s : ξ → s(ξ) is strictly monotonously increasing, and therefore its inverse function σ : s(ξ) → ξ exists and is strictly monotonously increasing. First, assume the analyticity of r. Then, as r is bounded from below by a positive number, the integrand on the right-hand side of (3.1) is analytic, and therefore s is analytic. Asṡ(ξ) is positive by (3.1), the analyticity of σ follows from the analytic inverse function theorem [3,Theorem 4.2], and this implies the analyticity of p(s) = r(σ(s)) = r(σ(s))u σ(s) .
The lemma is proved.
Let r 1 and r 2 be analytic arc length parametrizations of curves, such that r 1 (0) = r 2 (0) andṙ 1 (0) =ṙ 2 (0). Let be the line through r 1 (0) that is orthogonal toṙ 1 (0), and let F 1 , F 2 , and B be different points on such that B / ∈ F 1 F 2 and r 1 Let e be an analytic arc length parametrization of a curve, such that B = e(0) andė(0) = u π/2 . Every point E = e(s) determines two straight lines 1 := F 1 E and 2 := F 2 E forming small angles α and β with , respectively. Let the straight line¯ j (j = 1, 2) through the midpoint O of the segment F 1 F 2 be parallel to j . See Fig. 2.
Proof. If r Let E ⊥ be the orthogonal projection of E onto . Using L'Hôpital's rule we get δ(β) exists, then L'Hôpital's rule can be used, so we obtain which proves the lemma.

One ellipse in a Minkowski plane
We start by considering the Minkowski plane (R 2 , d I ) with indicatrix I. Since every ellipse is bounded, if an ellipse is a quadric, then it is an elliptic quadric.
Take The symmetry of I entails that t I t J , and it also follows that the affine center of the quadric E a dI ;F1,F2 coincides with its metric center O, hence t A t B too.
As x → 1/x is analytic in a neighborhood of r(0) > 0, r(β) is analytic in a neighborhood of the zero too, so we only need to prove thatr(β) := 1/r(β) is analytic in some neighborhood of the zero. With this in mind (4.1) becomes .

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of that unique analytic solution, so ∂I O is analytic around I, and, by its symmetry, around J too.

Theorem 4.3. A Minkowski plane is a model of the Euclidean plane if and only if at least one ellipse is a quadric.
Proof. Every ellipse is a quadric in the Euclidean plane, so we only have to prove that a Minkowski plane is Euclidean if at least one ellipse is a quadric.
Assume that E a dI ;F1,F2 is a quadric. If F 1 = F 2 , then E a dI ;F1,F2 is a homothetic image of ∂I, hence ∂I is a quadric, and therefore d I is Euclidean by [1, 25.4].
From now on we assume that  Then both E a de;F1,F2 and E a dI ;F1,F2 are quadrics, and they have four common points A, B and C 1 , C 2 , and two common tangents t A and t B , hence they coincide.
Taking the limit of this as ϕ → 0, we obtain F 1 ; B).
On the other hand, (4.4) implies that if δ(β 0 ) = 0 for some β 0 , then δ(β 2i ) never vanishes in the process described in Lemma 3.2, meanwhile β 2i tends to  This kind of implication extends over to analyticity too. Proof. First, assume that the Minkowski plane (R 2 , d I ) is analytic. Then the circles are also analytic, because they are homothetic to the boundary of the indicatrix, so we only need to prove the analyticity of ellipses E a dI ;F1,F2 with different focuses.
This shows that ifr is analytic in an interval (−ε, ε), then it is also analytic in the interval (−β(ε), β(ε)). According to Lemma 3.2, this means that the boundary of the indicatrix is analytic.

Quadrics in a Minkowski geometry
Note that in the planar case Theorem 5.1 states that if one ellipse is a quadric, then the Minkowski plane is a model of the Euclidean geometry. Proof. As every central planar section of each elliptic quadric is an ellipse in the Euclidean geometry, we only have to prove that a Minkowski geometry is Euclidean if every central planar section of at least one quadric is an ellipse. Let the quadric Q be such that its every central planar section is an ellipse. Then Theorem 4.3 implies, that every central planar section of the indicatrix is an ellipse, hence the statement of the theorem follows immediately from [2, II.16.12] which states for any integers 1 < k < n that the border ∂K of a convex body K ⊂ R n is an ellipsoid if and only if every k-plane through an inner point of K intersects ∂K in a k-dimensional ellipsoid.
For the other deduction we only have the following result that we put here without its easy proof.

Theorem 5.2. A Minkowski geometry is a model of the Euclidean geometry if and only if there is a hyperplane and a point in that hyperplane such that every line in the hyperplane through the point is parallel to the main axis of some ellipsoid that is a quadric.
It is worth noting that the question whether finding an ellipsoid that is a quadric would imply that a Minkowski geometry is Euclidean remains open for the non-planar cases.