# Monotonicity of canonical ellipsoids

Let $\mathcal{C}$ be the set of compact convex centrally symmetric sets in $\mathbb{R}^d$, and let $\mathcal{E} \subset \mathcal{C}$ be the set of ellipsoids centered at the origin.

I'm looking for a mapping $\pi:\mathcal{C}\to\mathcal{E}$ that satisfies the following properties:

1. continuity (with respect to the Hausdorff topology, say);
2. equivariance under linear isomorphisms of $\mathbb{R}^d$, i.e., $$C\in \mathcal{C},\ L \in GL(d,\mathbb{R}) \ \Rightarrow \ \pi(L(C))=L(\pi(C));$$
3. mononicity, i.e., $$C \subseteq D \ \Rightarrow \ \pi(C) \subseteq \pi(D).$$

The Löwner ellipsoid (meaning the unique ellipsoid of minimal volume containing the given compact set) is continuous and equivariant, but unfortunately it is not monotone: see this MO question and answer.

Question: Is there another kind of ellipsoid that satisfies the three properties? Or, if no such construction is known explicitly, can it proved abstractly (feel free to use axiom of choice) that such a map $\pi$ exists?

PS: One would expect $\pi$ to be a projection, but I don't need that property.

• Hmm... Considering gauge functions, I can identify $\mathcal{C}$ and $\mathcal{E}$ with ordered cones. Maybe I need a Hanh-Banach kind of argument (to get a positive "abstract" answer). – Jairo Bochi Sep 8 '17 at 22:06
• There are trivial answers, like the map which is constantly a unit sphere (or more interesting, the smallest sphere containing $C$.) Perhaps you want a projection to kill these. – Tim Carson Sep 9 '17 at 0:08
• @TimCarson The smallest sphere is not equivariant under the group action. – Jairo Bochi Sep 9 '17 at 0:54
• I noticed that (2) (continuity) +(3) (monotonicity) implies (1) (continuity). In fact, equivariance w.r.t. homothecies plus monotonicity already implies continuity. – Jairo Bochi Sep 9 '17 at 0:57
• Ah, I had imagined it said $SO(d)$ for some reason. – Tim Carson Sep 9 '17 at 3:01

No, there is no ellipsoid satisfying equivariance and monotonicity.

Suppose we have such an ellipsoid. Consider the square whose corners are $(\pm 1, \pm 1)$. Since this is invariant under a quarter-rotation, so is its ellipsoid, and the ellipsoid must be a circle of radius $r$.

By equivariance, we get the following table of shapes: \begin{matrix} \text{Quadrilateral with Corners at} & \text{Semi-axes of corresponding ellipsoid}\\ (\pm 1, \pm 1) & r,\ r \\ (\pm \cos(t), \pm \sin(t)) & r \cos(t),\ r \sin(t) \\ (\pm \sqrt{2}, 0) \text{ and } (0, \pm \sqrt{2}) & r,\ r \\ (\pm \sec(t), 0) \text{ and } (0, \pm \csc(t)) & r\sec(t)/\sqrt{2},\ r\csc(t)/\sqrt{2} \\ \end{matrix}

But the second quadrilateral is contained in the fourth. So by monotonicity, $r \cos(t) < r \sec(t) / \sqrt{2}$, and $\cos^2(t) < 1/\sqrt{2}$. We can choose $t$ for which this is false, showing the impossibility.

EDIT (by Jairo):

I'll illustrate Matt's solution (or a minor variation of it) with a figure.

As Matt explained, $\pi$(square) = circle. Multiplying $\pi(.)$ be a constant if necessary, we can assume that $\pi$(square) = inscribed circle. By equivariance, $\pi$(rectangle) = inscribed ellipse. But then monotonicity is violated: the red rectangle is contained in the blue square, but the red ellipse is not contained in the blue circle. • Nice argument...! – Igor Rivin Sep 9 '17 at 17:19
• Thanks Matt for the nice solution! I hope it's ok if I edit it to add a figure. – Jairo Bochi Sep 9 '17 at 17:37
• @JairoBochi, please do! – Matt F. Sep 9 '17 at 17:43